THE MAGNETIC VECTOR POTENTIAL, KLEIN-GORDON EQUATION AND KLEIN’S
PARADOX IN RELATIVISTIC QUANTUM MECHANICS
A Thesis by
David L. Schaffer
Bachelor’s Degree, University of Utah, 1982
Submitted to the Department of Physics
and the faculty of the Graduate School of
Wichita State University
in partial fulfillment of
the requirements for the degree of
Master of Physics
May 2007
 copyright by David L. Schaffer
All Rights Reserved
THE MAGNETIC VECTOR POTENTIAL, THE KLEIN-GORDON EQUATION AND
KLEIN’S PARADOX IN RELATIVISTIC QUANTUM MECHANICS
I have examined the final copy of this thesis for form and content and recommend that it
be accepted in partial fulfillment of the requirement for the degree of Master of Physics.
_________________________________
Dr. Elizabeth Behrman, Committee Chair
We have read this thesis
and recommend its acceptance:
___________________________________
Dr. Waldemar Axmann, Committee Member
___________________________________
Dr. Tom DeLillo, Committee Member
___________________________________
Dr. Jason Ferguson, Committee Member
iii
DEDICATION
To my family, from whom I took much time,
and
to Dr. Ashton Pyrefitte,
who showed me the way
iv
It’s always easiest to find the hard way first.
Corollary:
The hardest thing to find is that which isn’t there.
Unrelated Corollary
Never sell your books.
v
ACKNOWLEDGMENTS
I would like to thank my advisor, Elizabeth Behrman, for all her encouragement,
starting with a long conversation years ago about the feasibility of a graduate degree in
Physics, and continuing to the present. I am also grateful to my committee members,
Tom DeLillo, Jason Ferguson and Wally Axmann, who also taught some of the classes I
took on the way to this degree, for their help and time in evaluating this thesis. Finally, I
want to thank Wichita State University for the flexibility that allowed me, while working
and raising a family, to finish this degree.
vi
ABSTRACT
This Thesis evaluates solutions to the Klein-Gordon equation with scalar and vector
potentials in the Symmetric and Landau gauges. The solution for the Klein-Gordon
equation in the Symmetric gauge does reproduce elements of the two dimensional
quantum harmonic oscillator. The Landau gauge solution is used in the velocity selector
situation where particles can have acceleration free motion for a selected velocity.
The Klein Paradox for spinless particles is reviewed and its application to the KleinGordon equation shows that the vector potential can have an effect on particle pair
production without the violation of conservation of momentum or energy, that this effect
is to suppress the particle pair production associated with Klein’s paradox and that the
suppressive effect can, under certain conditions, become strong enough to prevent
particle pair production in Klein’s paradox when the scalar potential is otherwise many
times what would be necessary to create Klein’s paradox particle pairs.
vii
PREFACE
This thesis grew out of a problem encountered in graduate quantum
mechanics. This was not an assigned problem but it involved Klein’s Paradox
and not with the usual scalar potential but with the vector potential instead. That
provoked a lot of curiosity.
A major development during research was the fact that by itself the vector
potential generally cannot generate particle pairs in Klein’s Paradox in the same
way in as does the scalar potential without violating the conservation laws.
Superficially, that seems like the end of the issue entirely. However, there are still
at least two major questions to deal with: “Is there any way in which the vector
potential can be involved in particle pair creation in any Klein’s Paradox situation?” and
“Why does the Klein-Gordon Equation, based partly on the conservation of energy, so
explicitly show particle pair creating involving the vector potential”.
This thesis is an attempt to answer these questions as thoroughly as possible in
the available time. In a practical sense, most of the time that went into this thesis was
absorbed by deriving, researching or verifying solutions to the Klein-Gordon equation
with scalar and vector potentials mainly because this seemed a necessary first step to
going on to investigate Klein’s Paradox. What really happened was that it is possible to
get some information on the role of the vector potential in Klein’s Paradox that is
independent of the details of the wave functions but that the wave functions became
more interesting than just a means to an end.
viii
TABLE OF CONTENTS
Chapter
Page
1.
INTRODUCTION……………………………………………………………..……………1
2.
LITERATURE REVIEW………………………………………………………………..….3
3.
METHODOLOGY…………………………………………………………………….…….6
4.
THE KLEIN-GORDON EQUATION AND KLEIN’S PARADOX…………………….…7
5.
THE NATURE OF THE VECTOR POTENTIAL……………………………………….13
6.
CYCLOTRON MOTION AND THE SYMMETRIC GAUGE…………………...……..16
7.
A WAVE FUNCTION FOR CYCLOTRON MOTION………………………………….19
8.
THE EFFECT OF THE VECTOR POTENTIAL ON KLEIN’S PARADOX
IN CYCLOTRON MOTION………………………………………………………………26
9.
THE VELOCITY SELECTOR…………………………………………………………..29
10. THE VELOCITY SELECTOR AND THE SCALAR AND VECTOR
POTENTIALS…………………………………………………………………………….34
11. RESULTS AND ANALYSIS…….…………………………………………………...….40
11.1
11.2
11.3
11.4
11.5
11.6
11.7
Klein’s Paradox and Cyclotron Motion……………………………………..…..40
The Effect of the Vector Potential……………………………………………....41
Vector Potential Suppression with Gauge and Wave Function…………..…43
The Magnitude of the Suppression………………………………………….….44
Wave Function Similarities Between Solutions…………………….…………67
Behavior of the Functions in the Direct Solution………………………………68
Results for the Velocity Selector and Cyclotron Motion……………………...75
11.7.1 The Velocity Selector Wave Function………………………………...75
11.7.2 Similarities Between the Cyclotron and Velocity Selector
Wave Functions…………………………………………………………77
11.7.3 Derivation of Cyclotron Energy Quantization………………………..78
11.7.4 Reduction to Non-Relativistic Quantization………………………….79
11.7.5 Quantization in the Direct Solution for the Klein-Gordon
Equation for Relativistic Cyclotron Motion……………………………79
11.7.6 Klein’s Paradox and the Velocity Selector…………………………...80
LIST OF REFERENCES………………………………………………………………….…..84
ix
APPENDICES………………………………………………………………………………….87
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
Derivation of the “Classic” Klein’s Paradox…………………….….……..…….....88
Conservation Laws and the “Classic” Klein’s Paradox……….……….……..…108
The Scalar and Vector Potentials……………………………….………………..112
Short Notes on the Aharanov-Bohm Effect……………………………….……..114
The Non-divergence of the Vector Potential in the Symmetric Gauge……….116
Commutation of A and P in the Symmetric Gauge……………..………………123
Angular Dependence of the Cyclotron Wave Function……………….………..125
Search for a Complex Radial Wave Function for Cyclotron Motion…….…….133
The Relevance of Bessel Functions………………………………………..…....142
Derivation of the Harmonic Oscillator Wave Function by Analogy………......158
Direct Solution of the Klein-Gordon Equation with Symmetric Gauge……….168
Testing the Harmonic Oscillator Wave Function in the Particle
Density Equation………………………..………………………………….……...187
The Vector Potential, Klein’s Paradox and Cyclotron Motion………….……..192
The Klein-Gordon Equation and the Compton Wavelength…………….…….201
The Velocity Selector, the Landau Gauge, a Wave Function and
Klein’s Paradox……………………………………………………………………202
Sign Conventions for Key Quantities…………………………………………….223
General Form for Boundary Conditions for a Step Potential………………….229
Derivation of the Quantization of Cyclotron Energy……………………………237
Quantitative Analysis of Terms…………………………………………………...240
Gaussian Versus MKS Units……………………………………………………..283
Maple Graphics…………………………………………………………………….285
x
LIST OF TABLES
Table
Page
A7.1
Lower Degree 2-d Hermite Polynomials………………………………………….126
A7.2
Lower Degree Hermite Polynomials in Rectangular and Polar
Coordinates…………………………………………………………………………...127
A7.3
2-d Harmonic Oscillator Wave Functions………………………………………....128
A7.4
Lower Order Hydrogen Atom Wave Functions……………….…………………..129
A16.1 Elements and Sign Conventions…………………………………………………...225
xi
LIST OF FIGURES
Figure
Page
2.1
The Aharanov-Bohm Experimental Setup……………………………………………14
6.1
Cyclotron Motion for a Positive Particle…………………………………….………...16
6.2
Vector Potential in the Symmetric Gauge…………………………………………….17
6.3
The Symmetric Gauge by the Right Hand Rule……………………………………..18
9.1
Vector Relationships in the Velocity Selector for a Positive Particle………………30
9.2
Vector Relationships in the Velocity Selector for a Negative Particle…….……….30
9.3
Motion in the Velocity Selector……………………………………………………...…31
9.4
Velocity Selector Step Potential……...………………………………………………..32
9.5
The “Classic” Klein’s Paradox Step Potential…………………………………..……33
10.1
The Scalar Potential and Electric Field in the Velocity Selector…..…..…………35
10.2
The Vector Potential in the Landau Gauge…………………………………………37
10.3
The Vector Potential and the Symmetric Gauge…………………………………...38
11.1
Relative Strength of Terms: mL to 104, B to 104 T………………………………….48
11.2
Radii of Cyclotron Motion: mL to 104, B to 104 T……………………………………49
11.3
Velocities Associated with: mL to 104, B to 104 T…………………………………..50
11.4
Relative Strength of Terms: mL to 1015, B to 1 T…………………………………...51
11.5
Radii of Cyclotron Motion: mL to 1015, B to 1 T…..…………………………………52
11.6
Velocities Associated with: mL to 1015, B to 1 T..…………………………………..53
11.7
Relative Strength of Terms: mL to 1019, B to 10-4 T………………………………..54
11.8
Radii of Cyclotron Motion: mL to 1019, B to 10-4 T………………………………….55
11.9
Velocities Associated with: mL to 1019, B to 10-4 T…………………………………56
xii
11.10 Relative Strength of Terms: mL to 1014, B to 104 T………………………………...57
11.11 Radii of Cyclotron Motion: mL to 1014, B to 104 T…….…………………………….58
11.12 Velocities Associated with: mL to 1014, B to 104 T………………………………….59
11.13 Relative Strength of Terms: mL to 1019, B to 102 T………………………………...60
11.14 Radii of Cyclotron Motion: mL to 1019, B to 102 T…………………………………..61
11.15 Velocities Associated with: mL to 1019, B to 102 T………………………………….62
11.16 Relative Strength of Terms: mL to 1023, B to 109 T………………………………...64
11.17 Radii of Cyclotron Motion: mL to 1023, B to 109 T…………………………………..65
11.18 Velocities Associated with: mL to 1032, B to 109 T………………………………….66
11.19 Gaussian Exponential at 10-2 T………………………………………………………69
11.20 The Gaussian Exponential Versus the Hypergeometric Function………………..70
11.21 The WhittakerM Function for 10-2 T………………………………………………….71
11.22 Gaussian Exponential Compared to WhittakerM Function………………………..72
11.23 The First 3 Terms of the KummerU Function………………………………………73
11.24 The Gaussian Exponential Versus the WhittakerW Function…………………….74
A1.1
Particle Incident on Potential Step…………………………………………………..89
A4.1
Schematic for the Experimental Setup for the Aharanov-Bohm Effect………...114
A5.1
Vector Relationships in Cyclotron Motion in Cylindrical Coordinates………….117
A5.2
Vector Relationships Between A, v, p, B and r………………………………….118
A5.3
Vector Potential in the Symmetric Gauge…………………………………………119
A5.4
The Vector Potential in the r,φ Plane……….……………………………………...120
A10.1 Vector Relationships for a Positive Particle Between B, Lz and FB…………….160
A11.1 Gaussian Exponential for Positive Radii……………………………………..……175
A11.2 Hypergeometric Function Terms Compared to Gaussian Exponential……..…181
xiii
A11.3 Leading Terms of the WhittakerM Function………………………………………182
A11.4 Leading Terms of the KummerU Function……………………...………………...185
A11.5 Leading Terms of the WhittakerW Function………….…………………………186
A13.1 Cyclotron Motion for a Positively Charged Particle….…………………………295
A15.1 Vector Relationships in the Velocity Selector……..……….……………………203
A15.2 The Vector Potential for A = Byx……..……………………………….………….205
A15.3 The Vector Potential for A = -Bxy……..………………………………….………206
A15.4 Vector Potential for Combined Landau Gauges…..……………………….……207
A15.5 The Vector Potential in a Symmetric Gauge…..………………………………...208
A19.1 Relative Strength of Terms: mL to 104, B to 104 T………..……………………..252
A19.2 Radii of Cyclotron Motion: mL to 104, B to 104 T………..……………………….253
A19.3 Velocities Associated With: mL to 104, B to 104 T………...……………………..254
A19.4 Relative Strength of Terms: mL to 107, B to 107 T………..……………………..255
A19.5 Radii of Cyclotron Motion: mL to 107, B to 107 T………..……………………….256
A19.6 Velocities Associated With: mL to 107, B to 107 T……………………...………..257
A19.7 Relative Strength of Terms: mL to 108, B to 107 T………..……………………..258
A19.8 Radii of Cyclotron Motion: mL to 108, B to 107 T………..……………………….259
A19.9 Velocities Associated With: mL to 108, B to 107 T………...……………………..260
A19.10 Relative Strength of Terms: mL to 1015, B to 1 T………………………………...261
A19.11 Radii of Cyclotron Motion: mL to 1015, B to 1 T…………………………………..262
A19.12 Velocities Associated With: mL to 1015, B to 1 T…………………………………263
A19.13 Relative Strength of Terms: mL to 1019, B to 10-4 T……………………………..264
A19.14 Radii of Cyclotron Motion: mL to 1019, B to 10-4 T……………………………….265
A19.15 Velocities Associated With: mL to 1019, B to 10-4 T……………………………...266
xiv
A19.16 Relative Strength of Terms: mL to 1014, B to 104 T………………………………267
A19.17 Radii of Cyclotron Motion: mL to 1014, B to 104 T………………………………..268
A19.18 Velocities Associated With: mL to 1014, B to 104 T……………………………...269
A19.19 Relative Strength of Terms: mL to 1017, B to 10 T………………………………270
A19.20 Radii of Cyclotron Motion: mL to 1017, B to 10 T…………………………………271
A19.21 Velocities Associated With: mL to 1017, B to 10 T……………………………….272
A19.22 Relative Strength of Terms: mL to 1019, B to 102 T……………………………...273
A19.23 Radii of Cyclotron Motion: mL to 1019, B to 102 T………………………………..274
A19.24 Velocities Associated With: mL to 1019, B to 102 T………………………………275
A19.25 Relative Strength of Terms: mL to 1015, B to 109 T………………………………276
A19.26 Radii of Cyclotron Motion: mL to 1015, B to 109 T………………………………..277
A19.27 Velocities Associated With: mL to 1015, B to 109 T………………………………278
A19.28 Relative Strength of Terms: mL to 1023, B to 109 T………………………………279
A19.29 Radii of Cyclotron Motion: mL to 1023, B to 109 T………………………………..280
A19.30 Velocities Associated With: mL to 1023, B to 109 T………………………………281
xv
LIST OF ABBREVIATIONS/NOMENCLATURE
cos
Cosine Function
cyl
Pertains to Cylindrical Coordinates
dsolve
Differential Equation Solver, Maple 9 Command
gamma
Unit-less Relativistic Factor
hypergeom
the Hypergeometric Series Function
hypergeometric
the Hypergeometric Series Function
inc
Pertains to Incident Particle
ln
Natural Logarithm Function
KGE
Klein-Gordon Equaition
KummerU
the KummerU Series Function
ode
Ordinary Differential Equation, Maple 9 Designation
Pochammer
Pochammer Factorial Function
Psi
Psi Mathematical Function
refl
Pertains to Reflected Particle
sin
Sine Function
trans
Pertains to Transmitted Particle
WhittakerM
the WhittakerM Series Function
WhittakerW
the WhittakerW Series Function
xvi
LIST OF SYMBOLS
Lz
Angular Momentum in z Direction
mL = m(L)
Angular Momentum Quantum Number
Ω
Angular Velocity
ωe
Angular Frequency of Electron in Magnetic Field
Q
Bessel Function Component
R
Bessel Function Component
∆
Change in or Width of Variable
λC
Compton Wavelength
b
Coefficient for r for Harmonic Oscillator Wave Function
b’
Coefficient for r for Harmonic Oscillator Wave Function
C
Constant
C1
Constant
C2
Constant
Ci
Constant
CΘ
Constant of Integration, Scalar Potential
z
Cylindrical Coordinate
Φ
Cylindrical and Polar Azimuthal Coordinate
r
Cylindrical Radial Coordinate
∇
Del Operator
∇r
Del Operator in r
∇z
Del Operator in z
δ
Delta Function
xvii
f’
Derivative (first) of the Function f
f”
Derivative (second) of the Function f
EL
Electric Field
E
Energy, Total Relativistic
n
Energy Quantum Number, Principle
nx
Energy Quantum Number x Direction
ny
Energy Quantum Number y Direction
Fi
Forces
f
Frequency
∞
Infinity
∫
Integral
є
Harmonic Oscillator Energy
Hn
Hermite Polynomial
F( , , )
Hypergeometric Function
ζ
Gaussian Wavepacket Width Parameter
Cb
KGE Substitute Variable
D
KGE Substitute Variable
F
KGE Substitute Variable
G
KGE Substitute Variable
η1
KGE Substitute Variable
η’
KGE Substitute Variable
T’
KGE Substitute Variable
∇2
Laplacian Operator
∇φ2
LaPlacian Operator in φ
xviii
∇r2
Laplacian Operator in r
∇z2
Laplacian Operator in z
p
Linear Momentum
p1 = p(2)
Linear Momentum of Incident Particle
p2 = p(2)
Linear Momentum of Reflected Particle
p3 = p(3)
Linear Momentum of Transmitted Particle
px
Linear Momentum along x Axis
pz
Linear Momentum along z Axis
B
Magnetic Field
V= eΦ = V0 Magnitude of Potential Step
●
Multiplication or Dot Product
ex
Natural Exponential Function
a
Normalization Constant Complex Wave Function (Incident Particle)
d
Normalization Constant for Complex Wave Function (Reflected Particle)
f
Normalization Constant for Complex Wave Function (Transmitted Particle)
Cn
Normalization Constant for Harmonic Oscillator Wave Function
j
Particle Current
ρ
Particle Density
v
Particle Linear Velocity
v1
Particle Linear Velocity (Incident Particle)
v2
Particle Linear Velocity (Reflected Particle)
dL
Path Length Differential
π
Pi
h
Planck’s Constant/2π
xix
n-
Quantum Number, Rotational
n+
Quantum Number, Rotational
r
Radial Variable, Cylindrical and Polar Coordinates
x
Rectangular Coordinate
y
Rectangular Coordinate
z
Rectangular Coordinate; Series Variable
ω
Relativistic Cyclotron Frequency; Cyclotron Frequency
γ
Relativistic “Gamma” Factor
γ1
Relativistic “Gamma” Factor (Incident Particle)
γ2
Relativistic “Gamma” Factor (Reflected Particle)
M = γmv
Relativistic Mass
m
Rest Mass (Non-Relativistic)
Θ
Scalar Potential
A0
Scalar Potential
Ξ
Separation Constant
µ
Separation Constant
O( )
Series Higher Order Terms
i
Square Root of Negative One
c
Speed of Light in a Vacuum
θ
Spherical Coordinate Angular Variable
Σ
Summation
t
Time
E= E(p)
Total Relativistic Energy
E1 = E(1)
Total Relativistic Energy (Incident Particle)
xx
E2 = E(2)
Total Relativistic Energy (Reflected Particle)
E3 = E(3)
Total Relativistic Energy (Transmitted Particle)
e=q
Unit Electric Charge
x
Unit Vector x Direction
y
Unit Vector y Direction
z
Unit Vector z Direction
r
Unit Vector, Radial
φ
Unit Vector, Azimuthal
h
Planck’s Constant/2π
V
Potential Energy
V0
Potential Energy Step Magnitude
A
Vector Potential
Aφ
Vector Potential, Angular Component
Ar
Vector Potential, Radial Component
Az
Vector Potential, z Component
v
Velocity
vG
Velocity, Group
Vol
Volume
Ψ
Wave Function
S(φ)
Wave Function, Angular Component
Ψ*
Wave Function, Complex Conjugate
Ψ T*
Wave Function, Complex Conjugate Transpose
ξ
Wave Function Exponential Factor
Ψ1
Wave Function, Incident Particle
xxi
Ψ1*
Wave function, Incident Particle, Complex Conjugate
f(r)
Wave Function, Proposed for Complex Wave Function
R(r)
Wave Function, Radial Component
Ψ2
Wave Function, Reflected Particle
Ψ2*
Wave Function, Reflected Particle, Complex Conjugate
T(t)
Wave Function, Time Component
ψ3
Wave Function, Transmitted Particle
ψ3*
Wave Function, Transmitted Particle Complex Conjugate
Z(z)
Wave Function, z Component
k
Wave Number
xxii
CHAPTER 1
INTRODUCTION
The Klein-Gordon equation was one of the first results of the attempts to
marry special relativity to quantum mechanics. Due in part to its negative energy
solutions, it fell out of favor for a while, but these solutions turn out to be a
general feature of equations derived for relativistic quantum mechanics, their
source being the squared energy term in the relativistic energy/momentum
relation, not just a characteristic of the Klein-Gordon equation. The physical
counterparts to these negative energy solutions, antimatter particles, have turned
out to be real.
Nonetheless, the Klein-Gordon equation does have some serious
limitations. It is basically a single particle equation [2]. Only one mass is explicitly
included in the equation so while the Klein-Gordon can handle particles with
different masses it is not constructed to deal with more than one kind of particle
at a time. This restriction actually precludes more than one particle of any kind at
a time (one of the interesting things about Klein’s Paradox is that boundary
conditions clearly indicate that more than one particle is present).
Today, quantum field theory is used to deal with multiparticle situations.
However, the Klein-Gordon equation contains valuable physics
and is still the subject of occasional research. The solutions of the Klein-Gordon
equation are in some cases (Landau levels and the scalar potential) [14] well
known but solutions for combinations of scalar and vector potentials that in
the same place and time are not so common but do exist [15].
1
This thesis examines some of those solutions and their characteristics, along
with any application to Klein’s paradox.
2
CHAPTER 2
LITERATURE REVIEW
Literature on the Klein-Gordon equation, spinless particles, Landau levels,
the two dimensional quantum harmonic oscillator and on some of the solutions of
the Klein-Gordon equation is common. However, one aspect of the literature
search turned out to be rather difficult and that had to do with an aspect of Klein’s
paradox.
The “classic” Klein’s Paradox (that with no vector potential), in general, is
literally a textbook example in graduate [2] or relativistic quantum mechanics [25]
and is usually used to illustrate one of the ways in which negative energy
solutions (antimatter particles) tend to sprout up when using the equations of
relativistic quantum mechanics like the Klein-Gordon and Dirac Equations.
Although there are some other views, such as the single particle interpretation
[9], the “classic” Klein’s Paradox usually invokes particle pair creation to explain
oddities like charge densities of opposite charge than the incident particle,
transmission or reflection coefficients greater than 1 or the presence of particles
where none should be. In this form, the Klein’s Paradox involves an incident
particle interacting with a potential barrier (the origin of which is the scalar
potential) [2]. Information about this is common and easy to come by in textbooks
and on the internet. The behavior of particles in electric fields, magnetic fields or
both is also common. Information about a Klein’s Paradox involving the vector
potential instead of the scalar potential (essentially involving a magnetic field
instead of or with an electric field) is quite scarce.
There appear to be several reasons for this.
3
The first has to do with the conservation laws. Particle pair creation with a
scalar potential does not violate the conservation of energy or momentum
because particles can be created with no net expenditure of energy providing the
potential is of great enough strength. A vector potential generated magnetic field
has no known corollary.
This is probably the main reason for the dearth of information in the
journals, internet and textbooks. When a mention is made of vector potential
involvement in a Klein’s Paradox type situation, it is usually in passing along with
a note of the violation of conservation laws [12,17]. Such a basic obstacle likely
severely reduces research interest.
There is research that shows that some of the particle pair production
characteristics associated with Klein’s paradox come as a result of the
assumption that the exact location of the step potential can be known [6]. A step
potential with a finite step (V0 <∞) but an infinite gradient at the step involves
knowing the precise location of the step and of the particle as the interaction
takes place while still retaining some knowledge of the particle momentum. This
violates the Heisenberg Uncertainty Principle and particle pair production has
been established as a consequence of, i.e., basically attempting to confine a
particle beyond the Heisenberg limits (sometimes approximated by the Compton
wavelength) [2,9].
In any case, it may be possible to use a vector potential Klein’s Paradox to
illustrate some of the characteristics of equations like the Klein-Gordon equation
but scalar entities are usually easier to deal with so with the scalar potential
providing such a nice example there is not much motivation to look further to
4
satisfy that requirement.
Another reason is likely my level of understanding. Much of quantum
electrodynamics and quantum field theory are still beyond my present grasp so
the literature at those levels may furnish information on the issue not used here,
such as Schwinger’s Instanton Method [21].
Finally, the possibility that the vector potential can have a role in Klein’s
Paradox when accompanied by a strong scalar potential, without violating
conservation laws, does not appear to have received much attention.
There is some literature that bears at least peripherally on the issue of
the vector potential and particle pair production [11], such as the behavior of
intense magnetic fields around pulsars [17,11] or the stimulation of particle
pairs involving photons interacting with magnetic fields [4,7]. For the most part,
though, when particle pair formation is mentioned in connection with the
potentials this is usually with A = A0 = Θ only, of the 4 component [A0,A1,A2,A3] (in
cylindrical coordinates [A0,Ar,Aφ,Az}); in other words just the scalar potential [16]
There is enough literature on general solutions to the Klein-Gordon
equation in scalar and vector potentials, in one or two component form [8,2],
that is directly relevant to be quite useful [15].
5
CHAPTER 3
METHODOLOGY
The original methodology was simple: derive wave functions for the KleinGordon equation with a vector potential, examine the solutions and find any
relevance to Klein’s paradox by using the charge density and current equations
derived from the Klein-Gordon equation. However, these equations are not
applicable to some of the actual wave functions. Other wave functions in
seemingly simple physical situations have unexpectedly complex solutions.
The expectation that deriving the wave functions would be a relatively
simple preliminary step followed by the straightforward application of the charge
density and current equations had to yield to the experience of spending more
time deriving wave functions than on any other part of this research, none of
which were compatible with the application of the the charge
density or current equations and of, in the case of cyclotron motion, a clear effect
of the vector potential on Klein’s paradox that is independent of some of the
details of the wave function and is the result of a relatively simple calculation.
6
CHAPTER 4
THE KLEIN-GORDON EQUATION AND KLEIN’S PARADOX
The Klein-Gordon equation of relativistic quantum mechanics has its
origin in the energy-momentum formulation in Special Relativity:
E2 = p2c2 + m2c4
(4.1)
Solve this equation for E:
(E2 = p2c2 + m2c4)1/2
(4.2)
E = ±(p2c2 + m2c4)1/2
(4.3)
and the logical consequence of the squared energy in the equation shows up:
both positive and negative energies are possible.
Reworking the energy equation:
E2 = p2c2 + m2c4
(4.4)
E2/c2 = p2 + m2c2
(4.5)
a little bit:
and inserting the quantum mechanical operators [2]:
E → ih∂/∂t
(4.6)
p → (h/i)∇
∇
(4.7)
and
results in the basic Klein-Gordon equation:
{(ih∂/∂t)2/c2 = ((h/i)∇
∇)2 + m2c2}ψ(r,t)
where ψ(r,t) is an appropriate wave function.
Considering the origin of the Klein-Gordon equation it’s not surprising that
the issue of positive and negative energies surfaces with it as well. One of the
most dramatic examples of this is Klein’s Paradox.
7
(4.8)
The scalar potential Θ and the vector potential A are associated with the
Klein-Gordon equation via the usual minimal coupling [2]:
ih∂/∂t → ih∂/∂t - eΘ
(4.9)
(h/i)∇
∇ → (h/i)∇
∇ - eA/c
(4.10)
∇ - eA/c)2 + m2c2}ψ(r,t)
{(ih∂/∂t - eΘ)2/c2 = ((h/i)∇
(4.11)
and:
to get:
The version of Klein’s paradox usually encountered in textbooks on
relativistic quantum mechanics, referred to in this thesis as the “classic” Klein’s
paradox, involves the scalar potential only, so the Klein-Gordon equation reduces
to:
{(ih∂/∂t - eΘ)2/c2 = ((h/i)∇
∇)2 + m2c2}ψ(x,t)
The wave function is listed as only a function of x in the space variables since the
classic Klein’s Paradox can be laid out along or parallel to the x axis:
Figure 1.1. The Step Potential of Klein’s Paradox with V = eΘ
Θ
8
(4.12)
This is just a version of the non-relativistic step potential common to most
undergraduate books on quantum mechanics and if it is employed in that fashion
the same non-relativistic results ensue. However, applying the Klein-Gordon
equation results in, for those unfamiliar with it and the other equations of
relativistic quantum mechanics, some very odd behavior.
To see why, here’s the Klein-Gordon equation again:
∇)2 + m2c2}ψ(x,t)
{(ih∂/∂t - V)2/c2 = ((h/i)∇
(4.13)
for the area under the step potential to the right of x = 0. Keeping in mind the one
dimensional nature of the classic Klein’s paradox, a workable wave function is
[2]:
ψ(x,t) = ei(p•x – Et)/h
(4.14)
where h = planck’s constant divided by 2π. Expanding the squares:
{(-h2∂2/∂t2 – 2ihV∂/∂t + V2)/c2 = (-h2∇2) + m2c2}ψ(x,t)
(4.15)
and applying the energy and momentum operators to the wave function yields:
(E – V)2ψ = p2ψ + m2c2ψ
(4.16)
At this point, for purposes of analysis, the wave function is superfluous
and solving for p yields:
p = ±[(E – V)2 - m2c2]1/2
If the barrier is made strong enough so that V is a little greater in magnitude than
E, the momentum is imaginary, indicating that the incident particle cannot
significantly penetrate the barrier. This is a familiar result encountered with strong
barriers in non-relativistic quantum mechanics.
One would think that increasing the strength of the barrier would merely
allow it to continue to reflect the incident particle. However, increasing the
9
(4.17)
strength of the barrier so that:
V > E + mc2
(4.18)
actually allows real momenta again. Particles are found, seemingly inexplicably,
to the right of the barrier. The paradox appears to deepen when the equation
traditionally used to calculate probability density [24]:
ρ = (1/2mc2)[ψ*(ih∂/∂t – V)ψ + ψ(ih∂/∂t – V)ψ*]
(4.19)
where ψ* is the complex conjugate of ψ, gives results negative in magnitude, an
apparent impossibility since probabilities are not negative.
The similar equation for the probability current [25]:
j = (1/2m)[*((h/i)∇
∇)ψ + ψ((h/i)∇
∇)ψ*]
also gives seemingly impossible negative results. Furthermore, the reflection
coefficient indicates values greater than one, meaning more particles are being
reflected than the original incident particle (derivation of the classic Klein’s
paradox and consequences is in Appendix 1).
The resolution of this weird behavior, which gives Klein’s paradox its
name, is now known to be rooted in the production of particle pairs at the barrier,
one of the pair matter and the other antimatter [25]. The probability density
equation is interpreted as a particle density equation with negative
values indicating antiparticles and positive values indicating particles (the
probability current equation is interpreted as particle current [2].
Multiplication of either of these equations by e with the proper sign yields the
charge density and electric current equations). The negative charge density and
negative current found to the right of the barrier is then seen as the antiparticle
members of the particle pairs being created at the barrier finding the barrier,
10
(4.20)
repulsive to particles, attractive to antiparticles, accounting for the particles to the
right of the barrier. The incident particle really is reflected and it and the matter
members of the created particle pairs linger to the left of the barrier, creating a
reflection coefficient greater than one [2].
A very important feature of the classic Klein’s paradox is that this particle
creation can occur without violating any conservation laws (Appendix 2),[2]. If the
members of the pair have electric charges, the charges are opposites, allowing
particle pair creation without violating the conservation of charge. When the
potential (V > E + mc2) allows the creation of these pairs, one has
positive energy (matter) and the other an equal amount of negative energy
(antimatter). The total energy sum used in creating the pair is zero and the law
of conservation of relativistic energy remains unbroken [2].
Does the other potential, the vector potential, have any role in Klein’s
paradox?
Adding this potential to the Klein-Gordon equation results in:
{(ih∂/∂t - V)2/c2 = ((h/i)∇
∇ - eA/c)2 + m2c2}ψ(x,t)
Unlike the scalar potential, the vector potential is coupled to the momentum, not
the energy. There is no way to use the vector potential to induce the negative
energies necessary to allow particle creation to occur with no net energy
expenditure and no violation of the law of conservation of relativistic energy [12].
This appears to shut the door quite decisively against any possibility of the vector
potential as having any role in any Klein’s paradox situation.
The literature and textbooks on relativistic quantum mechanics frequently
make use of Klein’s paradox to illustrate some of the features of relativistic
11
(4.21)
quantum mechanics but invariably use the scalar potential for these
demonstrations [2,9,25]. In the infrequent cases where the vector potential is
mentioned at all it is always with a quick dismissal citing the violation of
conservation laws [12].
There is really no way around this problem in any Klein’s paradox that
involves only the vector potential.
However, suppose a scalar potential is also present of
sufficient strength that particle pair creation can occur. Will the addition of the
vector potential to this classic Klein’s paradox change the results? If so, the
vector potential then can have a role in Klein’s paradox without conservation law
violation since the scalar potential is now present to create particles.
Whether or not such a role exists and what it might be is one the subject of this
thesis.
The other subject of this thesis developed on the way to investigating
Klein’s paradox. This has to do with solutions of the Klein-Gordon equation with
scalar and vector potentials.
12
CHAPTER 5
THE NATURE OF THE VECTOR POTENTIAL
The vector potential itself can at first appear to a student of physics as a
mere mathematical aberration. Gauss’s Law of Magnetism:
∇•B = 0
(5.1)
one of Maxwell’s equations, states that the divergence of a magnetic field B is
zero. Since the divergence of the curl of some vector field A:
∇•(∇
∇ x A) = 0
(5.2)
is valid for all vector fields there is then no mathematical contradiction in writing
Gauss’s law as (Appendix 3) [13]:
∇•(∇
∇ x A) = 0
(5.3)
B=∇xA
(5.4)
where:
But what IS A? A mere mathematical oddity? A somewhat interesting but
meaningless detail? Are magnetic fields, measurable and real, the actual
manifestation of some deeper vector field, a ghost potential that is itself not
directly measurable? If so, how can anyone ever tell?
Enter the Aharonov-Bohm effect [2]:
13
Figure 2.1. The Aharonov-Bohm Experimental Setup
In the Aharonov-Bohm experiment, particles are emitted by a particle
source. These particles then travel through a double slit setup and result in an
interference pattern on the detecting screen. The key to the experiment is the
cylindrical solenoid that occurs in the barrier in between the slits. This solenoid
contains a magnetic field, pointing into or out of the paper but which is confined
to the solenoid. The particles passing through the slits never experience a nonzero magnetic field but the associated vector potential is not so restricted and is
free to have an effect, if it exists.
The crux of the matter concerns what happens to the interference effects
detected on the screen when a non-zero magnetic field appears in the solenoid.
The answer is that the observed interference effects do change when the
magnetic field comes on (Appendix 4),[2]. This cannot be the work of the
magnetic field since the particles are shielded from it. It is, instead, the work of
the vector potential associated with that magnetic field.
14
This experimental evidence indicates that the vector potential is real and is
actually the fundamental entity, not the magnetic field.
In a further interesting development, the vector potential comes with a
seemingly dangerously unspecified degree of freedom in that an exact form is
not generally required. Different forms of the vector potential can apply to
different situations or even to the same situation as long as they satisfy:
B=∇xA
to generate the necessary magnetic field. Any of these specific forms of the
vector potential is known as a gauge.
It is against this interesting backdrop that the rest of this thesis takes
place.
15
(5.5)
CHAPTER 6
CYCLOTRON MOTION AND THE SYMMETRIC GAUGE
One of the more well known situations involving a vector potential and
magnetic field is the circulation of a charged particle in a uniform, unchanging
magnetic field resulting in particle motion known as cyclotron motion:
Figure 6.1. The Constant Angular Velocity of Cyclotron Motion for a
Positively Charged Particle. Lz is the Angular Momentum.
There is more than one gauge that will produce a magnetic field sufficient
for cyclotron motion. However, noting that the physical situation is one of
cylindrical symmetry, a good gauge to use would be one that also reflects this
16
symmetry. This gauge is the symmetric gauge [24]:
φ
A = (B x r)/2 = (1/2)Bzrφ
Figure 6.2. The Vector Potential in the Symmetric Gauge. This diagram
occurs in the same Plane (r,φ
φ) as the Circulation of the Particle. The vectors
are in the negative φ direction.
17
(6.1)
Figure 6.2 reveals an important detail of the symmetric gauge. The vector
field composing A is clearly one of pure rotation. As such, it can have no
divergence. In the symmetric gauge, this is qualitative proof that the vector
potential is nondivergent. This non-divergence can also be confirmed by working
through the vector mechanics of A•∇
∇ in cylindrical coordinates (Appendix 5).
Hints of another important characteristic of the vector potential in the
symmetric gauge show up in figure 6.3.
Figure 6.3. The Symmetric Gauge by the Right Hand Rule. The Vectors of A, using
cylindrical coordinates are in the -φ
φ direction with B in the –z direction.
The vector potential A is antiparallel with the momentum p of the particle.
This is true of cyclotron motion at any r. The direction and magnitude of the
vector potential, relative to the momentum, in the symmetric gauge is a constant
of the motion, implying that p, in the Hamiltonian, and A commute. This
qualitative observation can also be confirmed by direct calculation (Appendix 6).
18
CHAPTER 7
A WAVE FUNCTION FOR CYCLOTRON MOTION
In the usual demonstration of Klein’s paradox, the wave function often
used is a plane wave of the form [2]:
ψ(x,t) = ei(p•x – Et)/h
(7.1)
Due to the inclusion of the imaginary number i this is an complex wave function
and as such it has a distinct complex conjugate:
ψ(x,t)* = e-i(p•x – Et)/h
(7.2)
In the Klein-Gordon equation the wave function and its complex conjugate
are a reflection of the existence of particles and antiparticles. Indeed, the KleinGordon equation for particles:
{(ih∂/∂t - V)2/c2 = ((h/i)∇
∇ - eA/c)2 + m2c2}ψ(x,t)
(7.3)
itself has a complex conjugate:
{(-ih∂/∂t - V)2/c2 = ((-h/i)∇
∇ - eA/c)2 + m2c2}ψ(x,t)*
(7.4)
of which the complex conjugate of the wave function is a solution.
The existence of these two forms of the Klein-Gordon Equation are the basis for
the derivation of the particle density equation [25]:
ρ = (1/2mc2)[ψ*(ih∂/∂t – V)ψ + ψ(ih∂/∂t – V)ψ*]
(7.6)
and particle current equation:
j = (1/2m)[*((h/i)∇
∇)ψ + ψ((h/i)∇
∇)ψ*]
These two equations are a relatively straightforward way to evaluate
particle/antiparticle production with Klein-Gordon (spinless) particles like pions,
particularly the charge density.
With the charge density equation, negative charge density means
19
(7.7)
antiparticles and a positive charge density means particles [2]. Since particles
occur, like the electron, with negative charges, and antiparticles exist, like the
positron, with positive charges the label of charge density equation might seem a
little misleading. However, if there is a negative charge density and one is aware
that the particles in question are positrons the quantization of positive charge
with positrons allows easy computation of the actual electric charge involved, the
negative charge density notwithstanding. Similarly, if one knows that the particles
involved with a positive charge density are electrons, the quantization of negative
charge with electrons allows an easy computation of the negative charges
actually involved, the positive charge density notwithstanding.
The charge density equation does allow the calculation of the charge density but
a more accurate name might involve the actual role of particle/antiparticle
classification and counting.
The current equation is a little less straightforward because a particle
traveling with velocity v will appear like an antiparticle moving with velocity –v
and vice versa [2]. However, with the sign of the charge density available this is
not difficult to sort out and actually yields some velocity information as well.
These relatively minor issues do not prevent the charge density and current
equations from representing a very straightforward way to investigate particle
and antiparticle pair production in Klein’s paradox situations, which is a
goal of this thesis. However, a wave function with a distinct complex conjugate
(not a real wave function) seems to be a fundamental requirement for their use.
Does such a complex wave function for cyclotron really exist?
One way to find out is to propose such a wave function in cylindrical
20
coordinates:
ψ(r,φ,t) = R(r)S(φ)T(t)Z(z) = eif(r)eim(L)φe-iEt/h
(7.8)
where the time component:
T(t) = e-iEt/h
(7.9)
is of standard form, the angular component [5]:
S(φ) = eim(L)φ
(7.10)
is the standard form for allowing standing waves, the boundary conditions of
which (see Appendix 7 for symmetry and the necessity for an angular wave
function) require the existence in the usual way of an angular momentum
quantum number mL that is in integer values. The z component is:
Z(z) =1
(7.11)
Most of this wave function is already complex; the issue now is whether or
not a radial component exists which is also complex, allowing the unfettered use
of the charge density and current equations. The method for determining this is
to apply the Klein-Gordon equation:
(ih∂/∂t - V)2/c2 = ((h/i)∇
∇ - eA/c)2 + m2c2
(7.12)
with cylindrical coordinate radial operators of:
∇2 = (1/r)∂/∂r(r∂/∂r)
(7.13)
and:
∇ = r∂/∂r
(7.14)
to the radial part of the proposed wave function:
R(r) = eif(r)
to see if a viable f(r) exists and, if so, what it is. The isolation of the radial
equation is possible through the usual separation of variables and which is then
21
(7.15)
refined using the commutability of a nondivergent vector potential A in the
symmetric gauge with p.
Executing this strategy results in an imaginary differential equation in r:
(1/c2)(E - V)2R(r) = {[(-hif’/r) + (-ihf”) + (f’)2 + h2mL2/r2]}R(r)
(7.16)
+ (heBmL/c)R(r) + (e2B2r2/4c2)R(r) + m2c2R(r)
This equation has two imaginary terms so it must be separated into complex and
real equations:
f’2 + h2mL2/r2 + heBmL/c + e2B2r2/4c2 + m2c2 + (1/c2)(E - V)2 = 0
(7.17)
which is entirely real and:
-ihf’/r - ihf” = 0
(7.18)
f’/r + f” = 0
(7.19)
which is complex but reduces to:
These two equations do not have compatible solutions [3].
This inconsistency means that the basic premise of
the calculation, that a complex radial wave function exists, is flawed and that, at
least in the symmetric gauge, such a radial wave function does not exist
(Appendix 8).
The direct implication is that the radial wave function is real and the
problem now is determining what it is and if that can still be effectively used to
make calculations using the charge density and current equations.
One might expect, with the cylindrical symmetry of cyclotron motion, that
the natural result would be a radial wave function involving Bessel functions.
Indeed, this is the result with the assumption of a non-divergent, commuting
vector potential as long as one does not specify the symmetric gauge (Appendix
22
9) [13]. If, however, the symmetric gauge, which is non-divergent and does
commute with p, is fully applied a Hamiltonian results that is of the form of the
harmonic oscillator, not too surprising since cyclotron motion is just such an
oscillation. Through analogy with the two dimensional quantum harmonic
oscillator this Hamiltonian can be used to find a harmonic oscillator radial wave
function, which is a real not complex wave function [24]:
R(r) = e-br^2
(7.20)
(Appendix 10).
A way to confirm this solution is to make the same use of the KleinGordon equation but without using the analogy of the two dimensional quantum
harmonic oscillator. Using the symmetric gauge in this manner results in a radial
equation:
d2R(r)/dr2 + (1/r)dR(r)/dr - (mL2/r2)R(r) - eBmLR(r)/hc
(7.23)
- e2B2r2R(r)/4h2c2 – (m2c4/h2c2)R(r) + (1/h2c2)(E – V)2R(r) = 0
which is not a Bessel’s equation. Simplifying this a little bit results in:
R(r)” + (1/r)R(r)’ – (D/r2)R(r) - (Fr2)R(r) + GR(r) = 0
(7.24)
where:
R(r)” = d2R(r)/dr2
(7.25)
R(r)’ = dR(r)/dr
(7.26)
D = mL2
(7.27)
F = e2B2/4h2c2
(7.28)
and:
G = (1/h2c2)(E – V)2 - eBmL/hc - m2c4/h2c2
The solutions of this differential equation are [18]:
23
(7.29)
R(r) = (1/r){C1WhittakerM[G/(4F1/2), D1/2, F1/2r2]
(7.30)
+ C2WhittakerW[G/(4F1/2), D1/2, F1/2r2]}
The Whittaker functions can be decomposed to [18]:
R(r) = C1F1/2rD^1/2e-(F^1/2r^2)/2hypergeometric[1/2 + D1/2 – G/(4F1/2), (7.31)
1 + 2D1/2, F1/2r2]
+ C2F1/2rD^1/2e-(F^1/2r^2)/2KummerU[1/2 + D1/2
– G/(4F1/2), 1 + 2D1/2, F1/2r2]
where:
C1, C2 = constants
(7.32)
(Appendix 11)
The significance of this is that a Gaussian exponential has emerged as
part of the solutions, which is the same form seen in the radial wave function
derived by analogy to the two dimensional quantum harmonic oscillator [24].
Upon investigation, the exponential coefficient of this Gaussian, associated with
the hypergeometric and Kummer functions, -F1/2r2/2, is exactly the same as the
Gaussian coefficient evolved using the harmonic oscillator analogy, -b:
e-(F^1/2r^2)/2 → F1/2/2 = (e2B2/4h2c2)1/2/2 = (eB/2hc)/2 = eB/4hc = b → e-br^2 (7.33)
When using the symmetric gauge with cyclotron motion all roads seem to lead to
Rome, in this case a Gaussian radial function instead of Bessel functions.
The question now is which approach will generate the correct wave
function: the analogy with the two dimensional harmonic oscillator or the exact
solution of the relevant Klein-Gordon equation? The former is easier to evaluate
but one might suspect that a direct solution is more accurate than an analogy. In
this case, the reasons are that when using the analogy, the operator –h2∇2 stays
24
in p2 form and does not directly operate on the wave and that the two
dimensional harmonic oscillator used in many books is non-relativistic (energy is
not squared). This makes the exact determination of the energy quantization
problematical using the analogy.
In any case, using either approach there is one component of the wave
function, R(r), that is real and not complex, and it remains to be seen if this will
work with the charge density and current equations. To see if the wave function
is viable in this way, it is possible to devise a simple test: use the wave function
in a non-relativistic way with one particle and no barrier. In this case, the particle
density should be one and the current should be that expected for the passage of
a single particle, relativistically or otherwise.
What happens when the wave function is evaluated in this manner is that
the charge density and current results fail and the failure is a direct result of the
real nature of the radial part of the wave function (Appendix 12). The evaluation
uses the simpler form of radial wave function derived by the analogy to the two
dimensional quantum harmonic oscillator but given the degree of greater
complexity of the polynomials associated with the wave function derived by the
direct solution and the fact that failure involved the real nature of the radial
function indicate that failure will occur using the exact solution as well.
For cyclotron motion, the charge density and current equations remain
unavailable. Other methods might work, such as using equations involving
scattering from a potential involving Fermi’s golden rule [10] or Schwinger’s
instanton method [21] but these are beyond the present scope of this thesis.
25
CHAPTER 8
THE EFFECT OF THE VECTOR POTENTIAL ON KLEIN’S
PARADOX IN CYCLOTRON MOTION
Fortunately, there is a way around this difficulty with wave functions that
actually turns out to be quite simple by comparison. To see this it is necessary to
go back to an “early” form of the Klein-Gordon equation, equation (A10.32):
(p2 + e2B2r2/4c2)Ψ(r,φ,t) = ((E – V)2/c2 - eLzB/c - m2c2)Ψ(r,φ,t)
(8.1)
At this point in the development of the Klein-Gordon equation, the symmetric
gauge has been used along with the angular momentum substitution but the
operator –h2∇2 is still in p2 form [24]. For an exact solution and wave function p2
must be applied to the general wave function, Ψ(r,φ,t), and the derivatives carried
out. However, for the purposes of investigating the effect of the vector potential
on Klein’s paradox under cyclotron motion this more general but valid form of the
Klein-Gordon equation is actually more useful.
Dividing out the wave function leaves:
p2 + e2B2r2/4c2 = (E – V)2/c2 - eLzB/c - m2c2
(8.2)
a statement of the relativistic energy/momentum relationship with added
potentials. Solving for the general momentum p:
p = ±{(E – V)2/c2 - eLzB/c - e2B2r2/4c2 - m2c2}1/2
(8.3)
The result of applying the ∇ operator to ψ(φ) instead of using vector rules
to substitute the angular momentum doesn’t change anything:
p = ±{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
(8.4)
This just says that in magnitude:
Lz = mLh
26
(8.5)
and mL, since Lz is known independently to be positive, must be positive as well.
(Appendix 13)
What are the implications of these equations?
p = ±{(E – V)2/c2 - eLzB/c - e2B2r2/4c2 - m2c2}1/2
(8.6)
p = ±{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
(8.7)
and:
An interesting comparison can now be made with the result of the same
derivation concerning the Klein-Gordon equation but without the vector potential.
p = ±{(1/c2)(E – V)2 - m2c2}1/2
(8.8)
This is the “classic” Klein’s paradox that is in the books on relativistic quantum
mechanics [2],[9],[25]. The most interesting aspect concerns what
happens when V > E. At first, after increasing the strength of V from a level of V
= E, this results in imaginary values of p, no surprise considering the strength of
the energy barrier. However, if:
V ≥ E + mc2
(8.9)
real values of momentum are again somehow possible. This is generally
attributed to the ability of a very strong scalar potential to allow the creation of
particle pairs at the barrier without the violation of conservation of relativistic
energy [2, 25].
The vector potential cannot, by itself, create particle pairs in this manner.
However, rather than outright dismissing any role in Klein’s paradox, now go
back to the result that includes the vector potential:
p = ±{(1/c2)(E – V)2 - eBLz/c - e2B2r2/4c2 - m2c2}1/2
The threshold for real values of the momentum now is:
27
(8.10)
V ≥ E + mc2 + ecBLz + e2B2r2/4
(8.11)
It is possible to put this in a somewhat more quantum mechanical form by
allowing the momentum operator to interact with the angular part of the wave
function:
(p2 + e2B2r2/4c2)Ψ(r,φ,t) = ((E – V)2/c2 – (ehB/ic)∂/∂φ - m2c2)Ψ(r,φ,t)
(8.12)
(p2 + e2B2r2/4c2)Ψ(r,φ,t) = Ψ(r,φ,t)((E – V)2/c2
(8.13)
– (ehB/ic)∂[Ψ(r,φ,t)]/∂φ - m2c2Ψ(r,φ,t)
and:
(p2 + e2B2r2/4c2)Ψ(r,φ,t) = Ψ(r,φ,t)((E – V)2/c2
(8.14)
– Ψ(r,t)(ehB/ic)∂[eim(L)φ]/∂φ - m2c2Ψ(r,φ,t)
(p2 + e2B2r2/4c2)Ψ(r,φ,t) = ((E – V)2/c2 – (ehBmL/c) - m2c2)Ψ(r,φ,t)
(8.15)
Solving for p now yields:
p = ±{(1/c2)(E – V)2 - eBhmL/c - e2B2r2/4c2 - m2c2}1/2
The presence of the vector potential has caused a change. For real
momenta, the scalar potential has to grow stronger. The vector potential has a
suppressing effect on the particle pair formation of Klein’s paradox and there is
no conservation of energy violation because a strong scalar potential is still
present to induce particle pair formation at no net energy.
28
(8.16)
CHAPTER 9
THE VELOCITY SELECTOR
In deriving wave functions for cyclotron motion using the Klein-Gordon
equation, cylindrical coordinates are by far the best choice [13]. However,
rectangular coordinates has the advantage of usually being easier to use.
The technical problem is how to arrange a situation that can exploit these
advantages. The solution of this technical problem is called the velocity selector
[26]. The velocity selector allows a charged particle to move through properly
crossed electric and magnetic fields in a straight line at a constant speed. This
allows the use of rectangular coordinates.
The calculation of how this is done uses the well known rule [26]:
ΣF = Felectric + Fmagnetic = qEL + qv x B
(9.1)
where ΣF is the vector sum of the forces on the particle due to the electric and
magnetic fields and the charge (q = e) and movement (v = velocity) of the particle
itself.
The lack of acceleration means that:
ΣF = Felectric + Fmagnetic = qEL + qv x B = 0
The electric and magnetic forces must be equal in magnitude and opposite in
direction.
29
(9.2)
Figure 9.1. Vector Relationships for a Positively Charged Particle in the Velocity
Selector. The Magnetic Field is in the Negative z Direction.
Interestingly, a change in the sign of the charge of the particle doesn’t really
change the setup:
Figure 9.2. Vector Relationships for a Negatively Charged Particle in the Velocity
Selector. The Magnetic Field is in the Negative z Direction.
This is actually rather important because if particles and antiparticles are
30
both involved and have the same velocity the track of either does not change.
Regarding just the magnitudes:
qEL = qvB
(9.3)
v = = qEL/qB = EL/B
(9.4)
Solving this for the velocity v:
For a particular ratio of EL/B there will only be one non-accelerated velocity so
there is an interesting limitation but there are an infinite number of EL and B fields
that can do this so there is a lot of flexibility as well.
31
Figure 9.3. Motion in the velocity selector in the positive and negative x
directions. The Magnetic Field must Reverse with Velocity Since Magnetic Forces
are Velocity Dependent
The actual form of the step potential otherwise remains as close to that of
the “classic” Klein’s paradox as well. In cross section:
Figure 9.4. The Velocity Selector Step Potential in Klein’s Paradox. Motion is
along or parallel to the x axis. The y axis, which partly defines the potentials, is
into and out of the page.
32
Figure 9.5. The “Classic” Klein’s Paradox Step Potential. Motion is along or
parallel to the x axis.
33
CHAPTER 10
THE VELOCITY SELECTOR AND SCALAR AND VECTOR POTENTIALS
Athough rectangular coordinates are generally easier to use than
cylindrical or polar coordinates there is one point that is actually more
complicated. In the “classic” Klein’s Paradox and that involving cyclotron motion
the scalar potential is very simple: just a constant scalar potential with no
gradient at all. The reason this works is that no electric field had to be present.
The electric field is a product of the scalar and vector potentials, namely [10]:
∇Θ - (1/c)∂A/∂t
EL = -∇
(10.1)
where the scalar potential is Θ and the vector potential is A.
Through this entire thesis the vector potential doesn’t change in time:
∂A/∂t = 0
(10.2)
∇Θ
EL = -∇
(10.3)
so the electric field equation reduces to:
and if the electric field is zero so is the gradient of the scalar potential.
In the velocity selector, an electric field must be present so this gradient has to
be non-zero.
What form exactly is the scalar potential going to take? In the vector
situation set up with the velocity selector in this thesis the electric field direction
must be in the negative y direction.
34
Figure 10.1. The Scalar Potential Θ and the Electric Field Vector in the Velocity
Selector. Since the Electric Field must be Constant the Gradient of the Scalar
Potential must be Linear.
This reduces the electric field equation from:
EL = -∇Θ
(10.4)
to:
EL = -ydΘ/dy
(10.5)
EL = -dΘ/dy
(10.6)
and in scalar form:
keeping in mind that the time rate of change of the vector potential is still zero. To
get the actual electric field it is now necessary to integrate:
∫dΘ = - EL∫dy
(10.7)
Θ(y) = -ELy + CΘ
(10.8)
and the result is:
Is it necessary now to do the same with the vector potential?
35
The fact that rectangular coordinates can be used in the velocity selector
doesn’t necessarily mean that a new gauge in A has to be found. The symmetric
gauge will still produce a constant, uniform magnetic field and in the symmetric
gauge A will also still be nondivergent; changing coordinate systems won’t alter
those facts. What it will do is make describing and using the vector potential in
the symmetric gauge more complicated mathematically.
Is there a gauge suited to rectangular coordinates that will produce the
required magnetic field? Keeping in mind that any gauge in A has to satisfy:
B=∇xA
(10.9)
A = Byx
(10.10)
it is possible to find a suitable new gauge:
which is a Landau gauge (Appendix 15). This Landau gauge is structured
differently than the symmetric gauge:
36
Figure 10.2. The Vector Potential in the Landau Gauge A = Byx, where x is the unit
vector in the x direction.
37
Figure 10.3. The Vector Potential in the Symmetric Gauge. This diagram occurs in
the same Plane (r,φ
φ) as the Circulation of the Particle. The vectors are in the
negative φ direction.
…but both gauges produce the same magnetic field.
In the symmetric gauge A was non-divergent, a quality that helps lead to the
commutation of the momentum with A. Since in the Landau gauge the vector
potential and the momentum are both constants of the motion, the implication is
that the attributes of commutability and non-divergence also exist. A direct,
quantitative calculation to make sure reveals that in the Landau gauge:
∇•A = 0
38
(10.11)
(Appendix 15) and:
[A,p] = 0
so that the non-divergence of A and its commutability with p do indeed hold
(Appendix 15).
39
(10.12)
CHAPTER 11
RESULTS AND ANALYSIS
11.1
Klein’s Paradox and Cyclotron Motion
In cyclotron motion the particle is under the influence of magnetic forces
but the vector potential and magnetic field are not capable of causing the
production of particle pairs without violating the conservation of energy and/or
momentum [17,12]. However, a sufficiently strong scalar potential can do so.
The methodology of this thesis concerning Klein’s paradox has been to use the
scalar potential to get around the limitation of conservation of energy and then
see if the vector potential has any role to play in what follows.
For cyclotron motion, the simplest way to do this is to use a scalar
potential that has no gradient so that there is no electric field and no
corresponding electrical forces on the circulating charged particles to make
motion more complicated and “change” the wave function.
One way to do this is to form a potential energy barrier in the path of the
circulating particle strong enough not only to reflect the particle but to allow the
creation of particle pairs in the vicinity of the barrier. This would be the cyclotron
analog of the “classic” Klein’s paradox. Interestingly, since the radius of
circulation of the incident particle is not infinite, the potential barrier is not infinite
in spacial extent as usually depicted in the textbook illustrations of the “classic”
Klein’s paradox. In effect, it would resemble a sort of two dimension version of
the one dimensional finite step potential barrier found in undergraduate textbooks
on quantum mechanics.
40
11.2
The Effect of the Vector Potential on Klein’s Paradox in Cyclotron
Motion
The use of the symmetric gauge:
A = (B x r)/2
(11.1)
is the natural gauge to use for cyclotron motion. It is not the only gauge to
produce a constant, uniform magnetic field but it does fit very well with the
coordinate system natural to use in cyclotron motion, the cylindrical coordinates.
This gauge is non-divergent and does commute with the momentum operator,
leading to a Klein-Gordon equation (Appendix 13):
p2 + e2B2r2/4c2 = (E – V)2/c2 - eLzB/c - m2c2
(11.2)
Solving this for p:
p = ±{(E – V)2/c2 - eLzB/c - e2B2r2/4c2 - m2c2}1/2
(11.3)
This can also be done with quantized angular momentum:
p = ±{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
The sign of the terms on the right is very clear in the cases of (1/c2)(E – V)2,
e2B2r2/4c2, and m2c2 because all the elements squared. The remaining term,
ehBmL/c, has the same sign as e2B2r2/4c2 and m2c2 due to the relative
orientations of the magnetic field and angular momentum vectors. This can be
expressed in terms of an angular momentum quantum number, mL, but that
shouldn’t change the sign of the term. mL in general can be either sign (or 0).
In this case, the sign of mL must be such that the sign of - ehBmL/c in equation
(11.4) is negative (positive mL). The reason is that the sign of the term is already
known from the vector relationships in (e/c)Lz•B. This classical term in equation
(11.3) doesn’t give the magnitude of the quantum number mL but it does tell the
41
(11.4)
sign of mL.
The net effect is that the terms derived from the vector potential act in the
same way as the m2c2 term: for real momenta in high V situations the magnitude
of V must be sufficiently larger than the magnitude of E for:
(1/c2)(E – V)2 = ehBmL/c + e2B2r2/4c2 + m2c2
(11.5)
which is threshold for Klein’s paradox with a vector potential in cyclotron motion.
The effect of the vector potential is clearer with comparison to the
threshold of Klein’s paradox with a strong but gradient free scalar potential and
no vector potential:
(1/c2)(E – V)2 = m2c2
(11.6)
[(1/c2)(E – V)2 = m2c2]1/2
(11.7)
[(1/c2)(E – V)2]1/2 = [m2c2]1/2
(11.8)
-(1/c)(E – V) = mc
(11.9)
(-E – V) = mc2
(11.10)
V - E = mc2
(11.11)
Vthreshold = mc2 + E
(11.12)
Solving this for V:
Taking the appropriate roots:
although it will be more convenient to leave the result in this form:
Vthreshold = c(m2c2)1/2 + E
(11.13)
For cyclotron motion the same procedure produces:
(1/c2)(E – V)2 = ehBmL/c + e2B2r2/4c2 + m2c2
(11.14)
[(1/c2)(E – V)2 = ehBmL/c + e2B2r2/4c2 + m2c2]1/2
(11.15)
[(1/c2)(E – V)2]1/2 = [ehBmL/c + e2B2r2/4c2 + m2c2]1/2
(11.16)
42
-(1/c)(E – V) = [ehBmL/c + e2B2r2/4c2 + m2c2]1/2
(11.17)
-(E – V) = c[ehBmL/c + e2B2r2/4c2 + m2c2]1/2
(11.18)
V - E = c[ehBmL/c + e2B2r2/4c2 + m2c2]1/2
(11.19)
Vthreshold = c[ehBmL/c + e2B2r2/4c2 + m2c2]1/2 + E
(11.20)
Comparing this result with that from the “classic” Klein’s paradox [2]:
Vthreshold = c(m2c2)1/2 + E
(11.21)
Vthreshold cyclotron > Vthreshold “classic” klein
(11.22)
it’s clear that:
The difference is the magnetic terms;
Vthreshold cyclotron - Vthreshold “classic” klein
(11.23)
= c[ehBmL/c + e2B2r2/4c2 + m2c2]1/2 + E – (c(m2c2)1/2) + E
(11.24)
= c[ehBmL/c + e2B2r2/4c2 + m2c2]1/2 + E – (c(m2c2)1/2) – E
(11.25)
= c[ehBmL/c + e2B2r2/4c2 + m2c2]1/2 – c[m2c2]1/2
(11.36)
both of which originate from the vector potential. The vector potential then, in this
sense, acts to suppress particle pair formation in Klein’s paradox when cyclotron
motion is occurring.
11.3
Relation of the Vector Potential Suppression of Klein’s Paradox in
Cyclotron Motion to the Gauge and Wave Function:
What information was critical to finding this result? The use of the
symmetric gauge is necessary to commute the vector potential with the
momentum operator and also to make clear the sign of the term ehBmL/c.
The results of the previous section are, however, partially independent of
the wave function. Using the vector relationships between r, p, A and B it is
possible to conclude that in cyclotron motion the vector potential does have a
43
suppressive effect on Klein Paradox production of particle pairs without having
complete knowledge of the wave function.
Quantum mechanically, the form using the angular momentum quantum
number mL is a more consistent equation. This means assuming an angular
component to the wave function:
S(φ) = eim(L)φ
(11.37)
but details about the radial component are still unknown (although it’s not hard to
surmise what, in general, those might be).
This is a hint that the vector potential might exhibit a suppressing effect on
Klein’s Paradox particle pair production with a wide variety of wave functions
and, by implication, a wide variety of physical situations, i.e., including those
other than cyclotron motion. This is also a hint that using a gauge other than the
symmetric gauge in some different situation may complicate the clear nature of
the aforesaid hint.
11.4
The Magnitude of the Suppression
A scale analysis will go a long way toward quantifying the suppressive
effect.
e = 1.602 x 10-19 C
(11.38)
h = (Planck’s constant)/2π = 1.055 x 10-34 J-s
(11.39)
mpion = 2.489 x 10-28 kg
(11.40)
c = 2.998 x 108 meters/sec
(11.41)
To help with thinking of the elements in terms of energy instead of momentum it
is possible to convert:
p = ±{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
44
(11.42)
to:
pc = ±c{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
(11.43)
pc = ±{(E – V)2 - ehBcmL - e2B2r2/4 - m2c4}1/2
(11.44)
and:
To quantify the suppressive effect of the vector potential upon particle pair
production in a Klein’s paradox situation where:
V > E + mc2
(11.45)
it is necessary to quantify the two terms in equation (11.44) that have their
source in the vector potential, the ehBcmL and e2B2r2/4 terms, against a
reference, the most convenient of which is in the same equation: the m2c4 term or
what can be called the rest mass term.
The two vector potential terms represent the suppressive effect of the
vector potential, then, and the rest mass term represents the “suppression” that
is present even in the “classic” Klein’s paradox. The effect of the vector potential
terms can then be labeled “significant” if their combined effect approaches or
exceeds the suppressive effect of the rest mass term.
The evaluation of the vector potential terms is not entirely straightforward
since they need to be converted from the Gaussian system of units to the mks
system (Appendix 20), which involves transformations of B/c → B and A/c → A
and also because in cyclotron motion B, r and mL are all related. This last feature
is actually a hidden advantage since it allows both vector potential terms to be
written in the same variables (Appendix 19) and then combined:
ehBcmL + e2B2r2/4 = (5/4)(hec2)mLB
= (5/4)(1.055 x 10-34 J-sec)(1.602 x 10-19 Coul)(2.998 x 108 m/sec)2mLB
45
(11.46)
= (1.899 x 10-36 J-coul-m2/sec)mLB
The quantization of the m2c4 term is, for a positively charged pion,
m = 2.489 x 10-28 kg
(11.47)
For this mass the rest energy term has a magnitude:
m2c4 = (2.489 x 10-28 kg)2(2.998 x 108 meters/sec)4
(11.48)
= 5.005 x 10-22 Joule2
For relatively minimal mL and B, the rest energy term m2c4 is many orders
of magnitude larger than the vector potential term. For an mL = 100 and B = 10-2
Tesla (about 100 times the average surface magnitude of the Earth’s magnetic
field):
m2c4/[(5/4)(hec2)mLB]
(11.49)
= 5.005 x 10-22 Joule2/[(5/4)(1.899 x 10-36)(100)(10-2)
Joule2] = 2.109 x 1014
For relatively low values of mL and B the suppressive effect of the vector
potential is very weak, many orders of magnitude less than the magnitude of the
rest energy term.
However, there is no set limit on the size of mL and B. This is something of
an idealization since energy loss can occur due to synchrotron radiation, etc.,
and there are technical limits in how powerfully current engineering can build
cyclotrons and synchrotrons, although in the latter case it is possible to move
from human laboratories to those natural laboratories where relatively, high B
and mL can actually exist (such as the vicinity of neutron stars with exceptionally
strong magnetic fields). In this thesis, these limiting issues are assumed
overcome or compensated for in some fashion. This allows a full exploration of
46
the implications of being able to greatly increase B and, especially, mL.
The results are that if B and mL increase sufficiently, the suppressive
effect of the vector potential on Klein’s paradox does become significant, even
overpowering (Appendix 19). This can be graphically shown by plots of
(5/4)(hec2)mLB versus m2c4.
It is possible to write the particle velocity v and the radius of particle
rotation r in terms of B and mL (Appendix 19):
r = (mLh/eB)1/2
(11.50)
and:
v = [(hmLeB/m2)/(1 + hmLeB/m2c2)]1/2
so each plot of (5/4)(hec2)mLB versus m2c4 is accompanied by two associated
plots, one of r(B,mL) and one of v(B,mL).
The first set of these (all the figures of 11.4 are in Appendix 19) shows the
weakness of the vector potential suppressive effect, for relatively low mL and B,
compared to that of the rest energy term:
47
(11.51)
Figure 11.1. The (5/4)(hec2)mLB Term, with mL to 104 and B to 104 Tesla, versus
the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 11.1 shows that even with a magnetic field millions of times the
average strength of the Earth’s magnetic field that the vector potential term
((5/4)(hec2)mLB) is still much lower in magnitude than the rest energy term
(m2c4).
48
Figure 11.2. Radii of Cyclotron Motion (in meters) for mL to 104 and B to = 104
Tesla.
Figure 11.2 shows that rotation is occurring over scales << 1 meter. For
cyclotron motion, mL (and the angular momentum) and B are inversely
proportional to each other.
49
Figure 11.3. Velocities Associated with mL to 104 and B to 104. The Vertical Axis is
in Meters/Second.
Velocities in figure 11.3 are still non-relativistic.
The next three sets of figures (9 in all) show combinations of B and mL
where the suppressive effect of the vector potential on particle pair production in
Klein’s paradox becomes significant, rivaling or exceeding the effect of the rest
energy term:
50
Figure 11.4. The (5/4)(hec2)mLB Term, with mL to 1015 and B to 1 Tesla, versus the
m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 11.4 shows some additional combinations of mL and B for which
the magnitude of the vector potential term ((5/4)(hec2)mLB) exceeds the
magnitude of the rest energy term.
51
Figure 11.5. Radii of Cyclotron Motion (in meters) for mL to 1015 and B to= 101
Tesla.
Figure 11.5 shows that radii of cyclotron motion is on the room size scale.
52
Figure 11.6. Velocities Associated with mL to 1015 and B to 101 Tesla. The Red
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
53
Figure A11.7. The (5/4)(hec2)mLB term, with mL to 1019 and B to 10-4 Tesla, versus
the m2c4 term (blue). The vertical axis is in units of Joule2.
Figure 11.7 shows values of mL for which the magnitude of the vector
potential term ((5/4)(hec2)mLB) exceeds that of the rest energy term for magnetic
fields in strength up to about the average surface strength of the Earth’s
magnetic field.
54
Figure 11.8. Radii of Cyclotron Motion for mL = 1019 and B = 10-4 Tesla.
Figure 11.8 shows that rotation is now taking place with radii on the order
of several tens of kilometers.
55
Figure 11.9. Velocities Associated with mL to 1019 and B to 10-4 Tesla. The Blue
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
56
Figure 11.10. The (5/4)(hec2)mLB Term, with mL to 1014 and B to 104 Tesla, versus
1000 times the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 11.10 shows that for some values of mL and B the magnitude of the
vector potential term ((5/4)(hec2)mLB) can exceed 1000 times the value of the
rest energy term.
57
Figure 11.11. Radii of Cyclotron Motion (in meters) for mL = 1014 and B = 104 Tesla.
Figure 11.11 shows that the radii of circulation are again below 1 meter.
58
Figure 11.12. Velocities Associated with mL to 1014 and B to 104 Tesla. The Purple
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
The last set of figures is an example of conditions where the
suppressive effect of the vector potential versus that of the rest energy term
becomes very strong:
59
Figure 11.13. The (5/4)(hec2)mLB Term, with mL to 1019 and B to 102 Tesla, versus
2,000,000 times the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 11.13 shows that the strength of the vector potential term
((5/4)(hec2)mLB) can exceed 2 million times the magnitude of the rest energy
term.
60
Figure 11.14. Radii of Cyclotron Motion (in meters) for mL = 1019 and B = 102 Tesla.
The Vertical Axis is in Meters.
Figure 11.14 shows radii of rotation from room scale to somewhat larger
than room scale.
61
Figure 11.15. Velocities Associated with mL to 1019 and B to 102 Tesla. The Blue
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
As evaluated against the magnitude of the rest energy term the
62
suppressive effect of the vector potential term over the particle pair production
effects of Klein’s paradox, while very weak under magnetic conditions naturally
encountered on Earth , can become very strong under conditions of higher
angular momentum and stronger magnetic field, to the point of suppressing
particle pair formation even when V is millions of times greater than E. This
generally occurs under strongly relativistic conditions and is possible even
though the value of the total relativistic energy E increases, and the difference
between E and a set V then then decreases, under relativistic conditions.
Some of these conditions, where the vector potential can act powerfully in
suppression, can occur naturally in the universe around neutron stars that have
particularly strong magnetic fields (even for neutron stars) [17].
The threshold of surface magnetic field strength for a magnetar, those neutron
stars with the strongest magnetic fields, is 4.4 x 109 Tesla [17], at which point the
energy corresponding to the cyclotron frequency equals that of the electron rest
energy:
hωe = mec2
The following set of graphics is meant to evaluate the parameters under
magnetar conditions:
63
(11.52)
Figure 11.16. The (5/4)(hec2)mLB Term, with mL to 1023 and B to 109 Tesla, versus
1017 times the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
64
Figure 11.17. Radii of Cyclotron Motion (in meters) for mL = 1023 and B = 109 Tesla.
The Vertical Axis is in Meters.
65
Figure 11.18. Velocities Associated with mL to 1023 and B to 109 Tesla. The Blue
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
66
11.5
Wave Function Similarities Between Solutions
This is basically concerning the similarities between the exact solution
(Appendix 11) of the relativistic Klein-Gordon equation [18] with those derived by
partial analogy (Appendix 10) of the Schröedinger equation (non-relativistic) two
dimensional harmonic oscillator [24]. Both solutions are identical in terms of their
angular components:
S(φ) = e im(L)φ
(11.53)
Though there are some differences in the radial equations the similarities are
striking:
R(r)non-relativistic = e-br^2H0 (ground state)
(11.54)
R(r)relativistic = e-br^2Gn
(11.55)
where Hn are the Hermite polynomials and the Gn are the generalized
hypergeometric polynomials.
The exponent coefficient b is exactly the same in both solutions and the
polynomial series are similar. The quantization of energy is similar in that the
term (2n + 1) with n as the energy quantum number appears in both the nonrelativistic version and in the general solution for the Klein-Gordon equation with
scalar vector potentials although this comes with the significant caveat that in
energy quantization the (2n + 1) is proportional to E and E2, respectively. In the
relativistic case [15]:
(E – ELpx/B)2 = ((B2 - EL2)/B2)[(2n + 1)e(B2 – EL2) + px2] + m2
This is for the c = h = 1 system with rectangular coordinates. The quantization of
energy will not change with a switch to cylindrical coordinates. As the electric
field EL → 0, for cyclotron motion, this simplifies to:
67
(11.56)
E2 = [(2n + 1)eB2 + px2] + m2
(11.57)
E2 ∝ 2n + 1
(11.58)
and it is easy to see that
as E is in the two dimensional quantum harmonic oscillator [24].
With this difference in mind, the analogy between the non-relativistic and
relativistic cases obviously only goes so far.
11.6
Behavior of the Functions in the Direct Solution:
Another difference in the wave functions derived by direct solution
of the Klein-Gordon equation and those derived by analogy to the two
dimensional quantum harmonic oscillator mainly comes down to the polynomial
series associated with each. In the 2 dimensional non-relativistic version the
polynomials are the Hermite polynomials. In the direct solution of the KleinGordon equation, the polynomials are those of the general hypergeometric series
and the KummerU series. These functions can be related to each other [15] as
seen later in this chapter.
How do the functions of the direct solution behave?
First, the exponential term e-br^2:
68
Figure 11.19. The Gaussian Exponential at 10-2 Tesla
This is well behaved (figure 11.5) at all r, with a finite value at r = 0 and the
function approaches zero as r becomes very large. The terms of the polynomial
generated by the hypergeometric function are quite different (Basic Graphics in
Appendix 19). Figure 11.6 is the plot for the exponential term versus the first
terms of the polynomial series:
69
Figure 11.20. The Gaussian Exponential at 10-2 Tesla Versus the Associated
Terms of the Hypergeometric Function.
The polynomial terms are not well behaved at all at even modest r. What
happens to the entire function (figure 11.7)?
70
Figure 11.21. The WhittakerM Function for B = 10-2 tesla.
The Gaussian dominates the result and the WhittakerM function does not
diverge if the hypergeometric series is truncated properly. What the series terms
do is to modulate the radius interval over which the Gaussian in significantly nonzero (figure 11.8):
71
Figure 11.22. The Gaussian Exponential Compared to the entire WhittakerM
Function.
What about the WhittakerW function, which includes both the exponential
e-br^2 and the polynomials generated by the KummerU function?
First a series of factors from the KummerU function.
72
Figure 11.23. The First Three Terms of the KummerU Function with mL = 3 and B =
10-2 Tesla.
The KummerU terms are convergent for all but low r. Here, because the
KummerU series starts out with negative exponents of br2 (instead of positive
exponents as with the hypergeometric function) [18]:
series(KummerU(1,5,z),z);
6 z -4 + 6 z-3 + 3 z-2 + z-1
73
(11.59)
The terms are not well behaved for low r, individually or collectively, since there
is a vertical asymptote for each term at r = 0. The WhittakerW function, a
combination formed by the multiplication of the Gaussian exponential by the
KummerU function also diverges at low r because the Gaussian, with a value of 1
at r = 0, cannot overcome the KummerU terms that explode at r = 0:
Figure 11.24. The Gaussian Exponential Versus the WhittakerW Function.
The KummerU series must be terminated to get a manageable
WhittakerW function. This done, the KummerU function serves to modulate the
Gaussian exponential at low r.
The Hypergeometric and KummerU functions serve to modulate the
74
Gaussian exponential at mainly different r although a range of r over which they
overlap does occur. In terms of relativistic (Klein-Gordon) and non-relativistic
(Schroedinger) solutions the Hypergeometric and KummerU functions serve
make the difference, although, as seen in the next section, the differences are
not as great as they might seem.
11.7
General Results for Velocity Selector and Cyclotron Motion
The velocity selector features straight line, non-accelerated motion and a
Landau gauge;
A = Byx
(11.60)
instead of circular motion and the symmetric gauge:
A = (B x r)/2
(11.61)
Despite these apparent simplifications, the derivation of a wave function for the
velocity selector is actually more difficult and more complicated [15] than was the
direct solution of the Klein-Gordon equation for cyclotron motion. Most of the
reason for this lies in the dependence of the vector and scalar potentials upon y.
11.7.1 The Velocity Selector Wave Function
The velocity selector wave function is a special case of the general rule
for wave functions for Klein-Gordon particles [15] in perpendicular electric and
magnetic fields.
As such its wave function is of the form [15] in the c = h = 1 mode:
ψ(x,y,z,t) ≈ ei(p(x)x – Et) – (ξ^2)/2Hn(ξ)
(11.62)
where:
ξ = (|e|(B2 – EL2)1/4[(y – (EEL – Bpx)/[e(B2 – EL2)]]
Following the general structure of Gaussian wave packets:
75
(11.63)
Gaussian = e-ζ^2/(∆ζ)^2
(11.64)
width = ∆ζ
(11.65)
where the:
Squaring ξ yields:
ξ2 = {(|e|(B2 – EL2)1/4[(y – (EEL – Bpx)/[e(B2 – EL2)]]}2
(11.66)
ξ2 = (|e|2(B2 – EL2)1/2[(y – (EEL – Bpx)]2/[e2(B2 – EL2)2]
(11.67)
ξ2 ∝ y2/[e(B2 – EL2)]2
(11.68)
width = e(B2 – EL2)
(11.69)
So:
with:
This width is not dependent on time so it doesn’t spread with time. Instead, it is
dependent on e, B and EL, which, for a given velocity selector situation, are
constant.
The width can also be expressed relativistically. Using the velocity selector
condition:
v = E/B
(11.70)
the width becomes:
width = eB2(1 – EL2/B2)
(11.71)
and:
width = eB2(1 – v2)
(11.72)
This is still in c = h = 1 mode so re-inserting the c results in:
width = eB2(1 – v2/c2)
(11.73)
(1 – v2/c2) = 1/γ2
76
(11.74)
where:
where gamma is the relativistic factor:
γ = 1/(1 – v2/c2)1/2
(11.75)
Continuing:
if:
y → y0
(11.76)
then:
ξ2 = (|e|2(B2 – EL2)1/2[(y0 – (EEL – Bpx)]2/[e2(B2 – EL2)2]
(11.77)
the Gaussian is centered on y0, which is constant in time since it depends only
on other constants: px, B, E, EL and e.
This is the quantum analog to the classical situation wherein one might take the
variable y as being constant. Instead, the width of the distribution of wave packet
y around a constant y0 is constant.
11.7.2 Similarities Between Cyclotron and Velocity Selector Wave
Functions
The velocity selector wave function is [15]:
ψ(x,y,z,t) ≈ ei(p(x)x – Et) – (ξ^2)/2Hn(ξ)
(11.78)
where:
ξ = (|e|(B2 – EL2)1/4[(y – (EEL – Bpx)/[e(B2 – EL2)]]
(11.79)
and the Hn(ξ) are the Hermite polynomials in ξ.
The cyclotron wave function derived by direct solution of the Klein-Gordon
equation is, from equations (7.29-7.33):
ψ(r,φ) = e-br^2(C1[Hypergeometric Series] + C2[KummerU Series])
Many of the details of the two solutions are different but it is interesting, given
77
(11.80)
some of the physical differences in the settings under which they occur, that the
large scale features of the two wave functions are similar. Both wave functions
contain a Gaussian exponential that is real and in both cases the Gaussian has
as a coefficient a polynomial series.
11.7.3 Derivation of Relativistic Cyclotron Energy Quantization
The general solution for the energy for the Klein-Gordon equation for
electric and magnetic fields perpendicular to each other (h = c = 1) [15]:
(E – ELpx/B)2 = [(B2 – EL2)1/2/B]2[(2n + 1)|e|(B2 – EL2)1/2 + pz2 + m2] (11.81)
This includes the velocity selector if:
pz = 0
(11.82)
(E – ELpx/B)2 = [(B2 – EL2)1/2/B]2[(2n + 1)|e|(B2 – EL2)1/2 + m2]
(11.83)
so:
Taking the electric field EL to zero (to get cyclotron conditions):
E2 = [(B2)1/2/B]2[(2n + 1)|e|B + m2]
(11.84)
E2 = [(2n + 1)|e|B + m2
(11.85)
and:
Now it’s time to go back to the c = c and h = h mode. With the rest energy term:
E2 = [(2n + 1)|e|B + m2c4
Including the other term (Appendix 18 for details):
E2 = (2n + 1)|e|Bhc2 + m2c4
(A11.86)
for relativistic cyclotron motion. Since eBh units are momentum units squared
this resembles:
E2 = (2n + 1)p2c2 + m2c4
which is just a quantized version of the relativistic energy-momentum relation.
78
(11.87)
11.7.4 Reduction to Non-Relativistic Quantization
For non-relativistic motion [15]:
E = [(2n + 1)|e|B]/2m + ELpx/B + m
(11.88)
To find the non-relativistic limit for a particle in a uniform, constant
magnetic field (details in Appendix 18):
EL → 0
(11.89)
E = [(2n + 1)|e|B]/2m + ELpx/B + m
(11.90)
which turns:
into:
E = [(2n + 1)|e|B]/2m + m
(11.91)
This is still in the h = c = 1 mode. Restoring this:
E = (2n + 1)|e|Bh/2m + mc2 = (|e|Bh/2m)(2n + 1) + mc2
(11.92)
ω = (|e|B/2m)
(11.93)
Since:
(the non-relativistic cyclotron frequency) the result is:
E = (2n + 1)ωh + mc2 = ωh(2n + 1) + mc2
(11.94)
which is the energy quantization of the two dimensional, non-relativistic harmonic
oscillator [23].
11.7.5 Quantization in the Direct Solution for the Klein-Gordon Equation
for Relativistic Cyclotron Motion
The derivation of the wave function by direct solution of the Klein-Gordon
resulted in (from equations (7.30) and (7.31):
R(r) = (1/r){C1WhittakerM[G/(4F1/2), D1/2, F1/2r2]
+ C2WhittakerW[G/(4F1/2), D1/2, F1/2r2]}
79
(11.95)
The Whittaker functions can be decomposed to [18]:
R(r) = C1F1/2rD^1/2e-(F^1/2r^2)/2hypergeometric[1/2 + D1/2 – G/(4F1/2), (11.96)
1 + 2D1/2, F1/2r2]
+ C2F1/2rD^1/2e-(F^1/2r^2)/2KummerU[1/2 + D1/2
– G/(4F1/2), 1 + 2D1/2, F1/2r2]
There is some quantization contained already since:
D = mL2
(11.97)
where mL is the angular momentum quantum number.
In addition, the Hypergeometric function is related to the Hermite
polynomials in such a way that the (2n + 1) energy quantization does emerge
[15]:
H2n+1 = [(-1)n(2n + 1)!/n!F(-n,3/2,τ2)2τ
and KummerU functions are related to the hypergeometrics so the same factors
in quantization will emerge with those as well.
11.7.6
Klein’s Paradox and the Velocity Selector
The similarities between the wave functions associated with cyclotron
motion and the velocity selector extend to the difficulty in using either to
determine the effects of the vector potential on Klein’s paradox. One of the less
complicated ways to determine what is happening with particle pair production in
a Klein’s paradox situation is to use the particle density and particle current
equations. Once the wavefunctions are derived the application of the particle
density and particle current equations is relatively simple and works IF the wave
functions have no major components that are real.
Unfortunately, the Gaussian exponentials that occur in both wave
80
(11.98)
functions are real.
With cyclotron motion it is possible to get around this problem and
determine the general effect of the vector potential by using the radial equation
with angular momentum eigenvalues and then solving the resulting equation for
p3.
With the velocity selector wave function [15] of equation (A15.55):
ψ = P(x)P(y) = ei(p(3)x – E(3)t)/he-(ξ’^2)/2Hn(ξ’)
(11.99)
it is possible to derive, from the Klein-Gordon equation in the Landau gauge,
an equation for p3, the momentum in the velocity selector area to the right of the
step potential equation (A15.125):
p32 - p3(2eA/c) + e2A2/c2 - η1 = 0
(11.100)
η1 = h2{|e|η’} – h2{ξ’2(|e|η’)} + m2c2 - (E – V)2/c2
(11.101)
where equation (A15.126):
Using the quadratic formula to solve for p3 (equation A15.133):
p3 = eA/c ± (η1)1/2
(11.102)
Solving for the boundary conditions (Appendix 15), the results are equation
(A15.183):
f/a = [(p1 + p2)/(eA/c ± (h2{|e|η’} – h2{ξ’2(|e|η’)}
(11.103)
+ m2c2 - (E – V)2/c2)1/2 + p2)]e(ξ’^2)/2
and equation (A15.184):
d/a = (p1 – p3)/(p2 + eA/c ± (h2{|e|η’}
– h2{ξ’2(|e|η’)} + m2c2 – (E – V)2/c2)1/2)
where p3 is included with all its detail.
The only purely vector potential term, eA/c, occurs only in the denominators of
81
(11.104)
each coefficient. To find out the effect of this term is it is necessary to know
if it is positive or negative. From Appendix 16, for a positively charged particle
encountering a repulsive barrier, e and Θ are both positive.
With Θ = -ELy, EL is negative so y has to be positive for Θ to be positive.
This makes A positive since:
A = By
(11.105)
and y is positive (Appendix 16) in the case of a positively charged particle
encountering a repulsive barrier and B is positive as well.
The effect of the single term eA/c in general acts to decrease both the
reflection and transmission coefficients, implying that with the velocity selector
the vector potential acts to inhibit or suppress particle pair production associated
with Klein’s paradox, as in the case of cyclotron motion.
However, this situation is also much more complex than that. Both the
factors ξ’ and η’ contain the magnetic field magnitude B in squared form.
Expressing the magnitude of B in terms of the magnitude of A using:
B = A/y
(11.106)
A now occurs in several different spots multiplied against different factors in
multiple terms of differing sign. The net effect of these terms and whether or not
this net effect could overwhelm the easier to see effects of eA/c is not at all clear.
It seems unlikely that the effect of all these terms including A would be to cancel
each other out so the vector potential very likely does have an effect on Klein’s
paradox in the velocity selector but without further analysis and time, perhaps
quite a bit of both, what the effect is cannot be determined.
A note for further research concerns p3. From equation (A15.179):
82
p3 = eA/c ± (η1)1/2
(11.107)
there is the choice of two roots. The obvious choice might seem to be the
positive root, making p3 positive entirely. However, there are several compelling
reasons to choose the negative root.
The first is circumstantial. The term eA/c seems to inhibit particle pair
production. The radical, which contains the term (E – V)2, the generating term for
Klein’s paradox, would then have the opposite effect, as it should.
The other reason is more concrete. In the “classic” Klein’s paradox:
p3 = ±[(E – V)2/c2 - m2c4/c2]1/2
(11.108)
the proper choice turned out to be the negative root. This makes sense since
what is really moving around to the right of the barrier are not particles but
antiparticles. Particle motion on the right should be in the positive x direction
since the source has to be the barrier. Antiparticles look like particles with the
opposite sign of momenta. In this case that’s negative. That makes the negative
root the right choice.
Since p3 in total must be negative, then with:
p3 = eA/c - (η1)1/2
(11.109)
(η1)1/2 > eA/c
(11.110)
either:
or the sign of the term eA/c should be re-evaluated since this actually deals with
antiparticles instead of particles. In either case, this does not clear any of the
mud in the water regarding how the vector potential in the velocity selector
influences the particle pair production of Klein’s paradox other than that such an
influence very likely exists.
83
REFERENCES
84
LIST OF REFERENCES
[1]
Dr. Waldemar Axmann, Physics Department, Wichita State University
[2]
Baym, Gordon. 1990. “Lectures on Quantum Mechanics”. Third Edition.
New York. Westview Press.
[3]
Dr. Elizabeth Behrman, Physics Department, Wichita State University
[4]
M.L. Burns and A.K. Harding. “Comparison of Photon-Photon and PhotonMagnetic Field Pair Production Rates”, Conference Proceedings,
American Institute of Physics, Volume 101, Issue 1, pp. 416-420.
[5]
Cohen-Tannouji, et. al. 1977.”Quantum Mechanics”. Second edition.
John Wiley and Sons, Paris, France.
[6]
J. Drell and S. Bjorken. “Relativistic Quantum Mechanics”. McGraw-Hill.
1964.
[7]
Di Piazza, Antonino. “Pair Production in a Slowly Varying Magnetic Field:
the Effect of a Background Gravitational Field”, International Journal of
Modern Physics, A21 (2006) 251
[8]
H. Feshback and F. Villars. “Elementary Relativistic Wave Mechanics for
Spin 0 and Spin ½ Particles”. Review of Modern Physics, 1958.
[9]
Greiner, Walter. 2000. “Relativistic Quantum Mechanics
(Wave Equations)”. Third Edition. New York. Springer Books.
[10]
Griffiths, David. 1987. “Introduction to Elementary Particles”. New York.
John Wiley and Sons.
[11]
Harding, A.K. and Baring, M.G. “Photon Splitting/Pair Cascades in
Ultra-Strong Magnetic Fields”. American Association of Science Meeting
#193. 1999. Austin, Texas.
[12]
Haro, J. “The Semi-classical Theory of Quantized Fields in
Electromagnetic Backgrounds”. Mexicana de Fisica 50. 2004.
[13]
Jackson, John David. 1998. “Classical Electrodynamics”. Third Edition.
New York. John Wiley and Sons.
[14]
Landau, L.D. and Lifschitz, E.M. “Quantum Mechanics: Non-Relativistic
Theory”. 2002. Third Revised Edition. Butterworth-Heinemann. Oxford.
85
[15]
Lam, Liu. ”Motion in Electric and Magnetic Fields I: Klein-Gordon
Particles”. Journal of Mathematical Physics. Volume 12, number 2.
February 1971.
[16]
LaPoint, Michael R. “Antimatter Production at a Potential Boundary”.
Nasa, Ohio Aerospace Institute, Brooke Park, Ohio.
[17]
Manchester, Dick and Melrose, Don. “Observations and Theory of
Anomalous X-ray Pulsars”. University of Sydney workshop: Supernova
Remants, Pulsars and the Interstellar Medium, 1999.
[18]
Maple 9 Mathematics Software. Maplesoft, Waterloo Maple.
Waterloo, Ontario, Canada.
[19]
Messiah, Albert. 1958. “Quantum Mechanics”. Second Edition. New
York. Dover Publications.
[20]
Ozana, M. and Shelankov, A. L. “Squeezed States of a Particle in
Magnetic Field”. Physics of the Solid State, Volume 40, #8. 1998.
[21]
Page, Don N. and Sang, Pyo Kim. “Schwinger Pair Production in
Electric and Magnetic Fields”. Internet source, University of Alberta,
Kunsan National University, Korea.
[22]
Semionova, L. and Leahy, D. “Remarks Concerning Pair Creation in
Strong Magnetic Fields”. Astronomy and Astrophysics 373, 272-280.
2001.
[23]
Peleg, Yoav, et. al. “Theory and Problems of Quantum Mechanics”.
New York. McGraw-Hill. 1998.
[24]
Schwinger, Julian. “Quantum Mechanics”. New York. Springer. 1965.
[25]
Strange, Paul. “Relativistic Quantum Mechanics”. Cambridge. Cambridge
University Press. 1998.
[26]
Young, Hugh and Freedman, Roger. “University Physics”. 9th Edition.
Addison-Wesley Publishing Co. 1996.
86
APPENDICES
87
APPENDIX 1
DERIVATION OF THE “CLASSIC” KLEIN’S PARADOX
Figure A1.1. Particle Incident on Potential Step
In the “classic” Klein’s paradox [2], a particle is incident on a potential step
of:
V = eΘ
(A1.1)
where V is the potential energy and Θ is the scalar potential. This is a relativistic
version of the non-relativistic step potential problem invariably included in
undergraduate textbooks on quantum mechanics.
In the relativistic version the solution of this one dimensional problem
proceeds in basically the same way except that the relativistic equation for
spinless particles, the Klein-Gordon equation, is used instead of the nonrelativistic Schroedinger equation:
{(1/c2)(ih∂/∂t)2 = ((h/i)∇)2 + m2c2}ψ(x,t)
(A1.2)
on the left with the solutions:
ψincident = aψ1 = aei(p(1)x – E(1)t)/h
88
(A1.3)
and:
ψreflected = dψ2 = de-i(p(2)x + E(2)t)/h
(A1.4)
and on the right:
{(1/c2)(ih∂/∂t – eΘ)2 = ((h/i)∇)2 + m2c2}ψ(x,t)
(A1.5)
ψtransmitted = fψ3 = fei(p(3)x - E(3)t)/h
(A1.6)
with the solution:
Running this solution through the Klein-Gordon equation on the right:
{(1/c2)(ih∂/∂t – eΘ)2 = ((h/i)∇)2 + m2c2}fψ3
(A1.7)
{(1/c2)(ih∂/∂t – V)2 = ((h/i)∇)2 + m2c2}fei(p(3)x - E(3)t)/h
(A1.8)
{(1/c2)(ih∂/∂t – V)2fei(p(3)x - E(3)t)/h = {-h2∂2/∂x2 + m2c2}fei(p(3)x - E(3)t)/h
(A1.9)
{(1/c2)(-h2∂2/∂t2 - 2ihV∂/∂t + V2)}fei(p(3)x - E(3)t)/h
= {-h2∂2/∂x2 + m2c2}fei(p(3)x - E(3)t)/h
{(1/c2)(-h2∂2/∂t2 - 2ihV∂/∂t + V2)}fei(p(3)x - E(3)t)/h
(A1.10)
= {-h2∂2/∂x2 + m2c2}fei(p(3)x - E(3)t)/h
{(1/c2)(-h2∂2[fei(p(3)x - E(3)t)/h]/∂t2 - 2ihV∂[fei(p(3)x - E(3)t)/h]/∂t
(A1.11)
+ V2fψ3)fei(p(3)x - E(3)t)/h) = {-h2∂2[fei(p(3)x - E(3)t)/h]/∂x2 + m2c2fψ3}
Taking the derivatives:
(1/c2)(-h2(-iE3/h)2 - 2ihV(-iE3/h) + V2fψ3)fei(p(3)x - E(3)t)/h)
(A1.12)
= {-h2∂2[fei(p(3)x - E(3)t)/h]/∂x2 + m2c2fψ3}
(1/c2)(E32 - 2VE3 + V2)fψ3 = {-h2(ip3/h)2 + m2c2}fψ3
(A1.13)
(1/c2)(E3 - V)2fψ3 = {p32 + m2c2}fψ3
(A1.14)
Dividing out the wave function now that all the operators have operated:
(1/c2)(E3 - V)2 = p32 + m2c2
Solving for p3:
89
(A1.15)
p32 = (1/c2)(E3 - V)2 - m2c2
(A1.16)
p32 = (1/c2)(E3 - V)2 - m2c4/c2
(A1.17)
{p32 = (1/c2)(E3 - V)2 - m2c4/c2}1/2
(A1.18)
{p32}1/2 = {(1/c2)(E3 - V)2 - m2c4/c2}1/2
(A1.19)
p3 = ±(1/c){(E3 - V)2 - m2c4}1/2
(A1.20)
The next equations needed for analysis are the reflection and
transmission coefficients [2], [25]. These derive from the boundary conditions that
the wave functions and their first derivatives must be continuous across the
barrier at x = 0. The wave functions:
aψ1 + dψ2 = fψ3
(A1.21)
aei(p(1)x – E(1)t)/h + de-i(p(2)x + E(2)t)/h = fei(p(3)x - E(3)t)/h
At:
x=0
(A1.22)
t=0
(A1.23)
a+d=f
(A1.24)
and:
this reduces to:
For the second boundary conditions equation the first derivatives of all the
wave functions are necessary:
∂[aψ1]/∂x + ∂[dψ2]/∂x = ∂[fψ3]/∂x
(A1.25)
∂[aei(p(1)x – E(1)t)/h]/∂x + ∂[de-i(p(2)x + E(2)t)/h]/∂x = ∂[fei(p(3)x - E(3)t)/h]/∂x
(A1.26)
[aip1/h]ψ1 + [-dip2/h]ψ2 = [ifp3/h]ψ3
(A1.27)
[ap1]ψ1 + [-dp2]ψ2 = [fp3]ψ3
90
(A1.28)
The (i/h) factor divides out:
and at:
x=t=0
(A1.29)
ap1 - dp2 = fp3
(A1.30)
this simplifies to:
…so the two boundary condition equations are:
a+d=f
(A1.31)
and:
ap1 - dp2 = fp3
(A1.32)
To get the reflection coefficient solve both equations for f, equate the two
equations using f, which will write f out of the combined equation and solve the
equation for d/a:
The first equation is already solved for f:
a+d=f
(A1.33)
ap1 - dp2 = fp3
(A1.34)
fp3 = ap1 - dp2
(A1.35)
(fp3 = ap1 - dp2)/p3
(A1.36)
(fp3)/p3 = (ap1 - dp2)/p3
(A1.37)
f = (ap1 - dp2)/p3
(A1.38)
(ap1 - dp2)/p3 = a + d
(A1.39)
ap1/p3 - dp2/p3 = a + d
(A1.40)
but the second is not:
Equating:
Now solving for d/a:
Grouping like terms:
91
ap1/p3 – a = dp2/p3 + d
(A1.41)
a(p1/p3 – 1) = d(p2/p3 + 1)
(A1.42)
a(p1/p3 – 1)(p3) = d(p2/p3 + 1)(p3)
(A1.43)
a(p1 – p3) = d(p2 + p3)
(A1.44)
(p1 – p3)/(p2 + p3) = d/a
(A1.45)
d/a = (p1 – p3)/(p2 + p3)
(A1.46)
and now solving for d/a:
Much the same procedure will produce a transmission coefficient: solve
the two source equations for d, equate the two resulting equations to each other
using d, which will write d out of that equation, and then solve for f/a.
Solving the equations for d:
a+d=f
(A1.47)
d=a–f
(A1.48)
ap1 - dp2 = fp3
(A1.49)
- dp2 = fp3 - ap1
(A1.50)
dp2 = ap1 - fp3
(A1.51)
(dp2)/p2 = (ap1 - fp3)/p2
(A1.52)
d = (ap1 - fp3)/p2
(A1.53)
becomes:
and the second equation:
Now equating to write d out of the resulting equation:
a – f = (ap1 - fp3)/p2
(A1.54)
(a – f = (ap1 - fp3)/p2)p2
(A1.55)
(a – f)p2 = ap1 - fp3
(A1.56)
92
ap2 – fp2 = ap1 - fp3
(A1.57)
ap2 - ap1 = fp2 - fp3
(A1.58)
a(p2 - p1) = f(p2 - p3)
(A1.59)
(p2 - p1)/(p2 - p3) = f/a
(A1.60)
f/a = (p2 - p1)/(p2 - p3)
(A1.61)
Grouping like terms:
and solving for f/a:
The final equation necessary for analysis is the group velocity vG on the left [2]:
x∂p3/∂E3 = x∂[±(1/c){(E3 - V)2 - m2c4}1/2]/∂E3
(A1.62)
x∂p3/∂E3 = ± x(1/c)∂[{(E3 - V)2 - m2c4}1/2]/∂E3
(A1.63)
x∂p3/∂E3 = ±x(1/c)(1/2){(E3 - V)2 - m2c4}-1/2∂{(E3 - V)2 - m2c4}]/∂E3 (A1.64)
x∂p3/∂E3 = ±x(1/c)(1/2){(E3 - V)2
(A1.65)
- m2c4}-1/2{∂[(E3 - V)2]/∂E3 - ∂[m2c4]/∂E3}
x∂p3/∂E3 = ±x(1/c)(1/2){(E3 - V)2
(A1.66)
- m2c4}-1/2{(2)(E3 - V)∂(E3 - V)/∂E3 - 0}
x∂p3/∂E3 = ±x(1/c)(2/2){(E3 - V)2
(A1.67)
- m2c4}-1/2(E3 - V){∂E3/∂E3 - ∂V/∂E3}
x∂p3/∂E3 = ±x(1/c){(E3 - V)2 - m2c4}-1/2(E3 - V){1 - 0}
(A1.68)
x∂p3/∂E3 = ±x(1/c){(E3 - V)2 - m2c4}-1/2(E3 - V)
(A1.69)
vG = x∂E3/∂p3 = ±xc{(E3 - V)2 - m2c4}1/2/(E3 - V)
(A1.70)
and since:
p3 = ±(1/c){(E3 - V)2 - m2c4}1/2
vG = x∂E3/∂p3 = ±x(c2/c){(E3 - V)2 - m2c4}1/2/(E3 - V)
93
(A1.71)
(A1.72)
vG = x∂E3/∂p3 = ±x(c2)(1/c){(E3 - V)2 - m2c4}1/2/(E3 - V)
(A1.73)
vG = x∂E3/∂p3 = xc2p3/(E3 - V)
(A1.74)
The last set of equations to develop here are those for the charge density and
current. Listing the wave functions again, since that will be a starting point:
The wave functions are [2]:
Ψincident = aΨ1 = aei(p(1)x – E(1)t)/h
(A1.75)
Ψincident* = aΨ1* = ae-i(p(1)x – E(1)t)/h
(A1.76)
Ψreflected = dΨ2 = de-i(p(2)x + E(2)t)/h
(A1.77)
Ψreflected* = dΨ2* = dei(p(2)x + E(2)t)/h
(A1.78)
Ψtransmitted = fΨ3 = fei(p(3)x – E(3)t)/h
(A1.79)
Ψtransmitted* = fΨ3* = fe-i(p(3)x – E(3)t)/h
(A1.80)
The charge density equation is [2], [25]:
ρ(x,t) = (1/2mc2)(Ψ*(ih∂/∂t - V)Ψ + Ψ(-ih∂/∂t - V)Ψ*)
(A1.81)
Ψleft = aΨ1 + dΨ2
(A1.82)
V=0
(A1.83)
On the left,
and:
so:
ρ = (1/2mc2) ((aΨ1* + dΨ2*)(ih∂/∂t)(aΨ1 + dΨ2)
(A1.84)
+ (aΨ1 + dΨ2)(-ih∂/∂t)( aΨ1* + dΨ2*))
Working in the wave functions:
ρ = (1/2mc2)((aΨ1* + dΨ2*)((iha)∂Ψ1/∂t + (ihd)∂Ψ2/∂t)
(A1.85)
+ (aΨ1 + dΨ2)((-iha)∂Ψ1*/∂t + (-ihd)∂Ψ2*/∂t))
ρ = (1/2mc2)((aΨ1* + dΨ2*)((iha)∂ei(p(1)x – E(1)t)/h/∂t
94
(A1.86)
+ (ihd)∂e-i(p(2)x + E(2)t)/h/∂t)
+ (aΨ1 + dΨ2)((-iha)∂e-i(p(1)x – E(1)t)/h/∂t + (-ihd)∂ei(p(2)x + E(2)t)/h/∂t))
Now taking the time derivatives:
ρ = (1/2mc2)((aΨ1* + dΨ2*)((iha)(-iE1/h)Ψ1 + (ihd)(-iE2/h)Ψ2)
(A1.87)
+ (aΨ1 + dΨ2)((-iha)(iE1/h)Ψ1* + (-ihd)(iE2/h)Ψ2*)
ρ = (1/2mc2)((aΨ1* + dΨ2*)((aE1)Ψ1
(A1.88)
+ (dE2)Ψ2) + (aΨ1 + dΨ2)((aE1)Ψ1* + (dE2)Ψ2*)
Consolidating terms:
ρ = (1/2mc2)(a2E1Ψ1Ψ1* + adE2Ψ1*Ψ2
(A1.89)
+ adE1Ψ1Ψ2* + d2E2Ψ2Ψ2*
+ a2E1Ψ1Ψ1* + adE2Ψ1Ψ2* + adE1Ψ1*Ψ2 + d2E2Ψ2Ψ2*)
ρ = (1/2mc2)(a2E1 + adE2Ψ1*Ψ2 + adE1Ψ1Ψ2* + d2E2
(A1.90)
+ a2E1 + adE2Ψ1Ψ2* + adE1Ψ1*Ψ2 + d2E2)
ρ = (1/2mc2)(2a2E1 + + 2d2E2 + adE2Ψ1*Ψ2 + adE1Ψ1Ψ2*
(A1.91)
+ adE2Ψ1Ψ2* + adE1Ψ1*Ψ2)
ρ = (1/2mc2)(2a2E1 + 2d2E2 + adE1Ψ1Ψ2* + adE1Ψ1*Ψ2
(A1.92)
+ adE2Ψ1Ψ2*+ adE2Ψ1*Ψ2)
ρ = (1/2mc2)(2a2E1 + 2d2E2 + adE1 (Ψ1Ψ2* + Ψ1*Ψ2)
(A1.93)
+ adE2(Ψ1Ψ2*+ Ψ1*Ψ2))
ρ = (1/2mc2)(2a2E1 + 2d2E2 + ad(E1 + E2)(Ψ1Ψ2* + Ψ1*Ψ2))
(A1.94)
Concerning the wave functions remaining in the equations:
(Ψ1Ψ2* + Ψ1*Ψ2) = ei(p(1)x – E(1)t)/hei(p(2)x + E(2)t)/h
(A1.95)
+ e-i(p(1)x – E(1)t)/he-i(p(2)x + E(2)t)/h
(Ψ1Ψ2* + Ψ1*Ψ2) = ei(p(1)x – E(1)t)/h + i(p(2)x + E(2)t)/h
95
(A1.96)
+ e-i(p(1)x – E(1)t)/h + -i(p(2)x + E(2)t)/h
(Ψ1Ψ2* + Ψ1*Ψ2) = ei([p(1) – p(2)]x + [E(2) – E(1)]t)/h
(A1.97)
+ e-i([p(1) – p(2)]x + [E(2) – E(1)]t)/h
This fits the Euler formula:
2cos(x) = eix + e-ix
(A1.98)
where:
x = ([p1 – p2]x + [E2 – E1]t)/h
(A1.99)
so:
(Ψ1Ψ2* + Ψ1*Ψ2) = ei([p(1) – p(2)]x + [E(2) – E(1)]t)/h + e-i([p(1) – p(2)]x + [E(2) – E(1)]t)/h (A1.100)
becomes:
(Ψ1Ψ2* + Ψ1*Ψ2) = 2cos{([p1 – p2]x + [E2 – E1]t)/h}
(A1.101)
Substituting this back into the charge density equation:
ρ = (1/2mc2)(2a2E1 + 2d2E2
(A1.102)
+ ad(E1 + E2)2cos{([p1 – p2]x + [E2 – E1]t)/h})
ρ = (2/2mc2)(a2E1 + d2E2
(A1.103)
+ ad(E1 + E2)cos{([p1 – p2]x + [E2 – E1]t)/h})
ρ = (1/mc2)(a2E1 + d2E2
(A1.104)
+ ad(E1 + E2)cos{([p1 – p2]x + [E2 – E1]t)/h})
For the current on the left the general current equation is [2]:
j(x,t) = (1/2m)(Ψ*((h/i)∇
∇ - eA/c)Ψ + Ψ((-h/i)∇
∇ - eA/c)Ψ*)
(A1.105)
On the left, there is no vector potential, so:
j(x,t) = (1/2m)(Ψ*((h/i)∇
∇)Ψ + Ψ((-h/i)∇
∇)Ψ*)
(A1.106)
Ψ = aΨ1 + dΨ2
(A1.107)
and:
96
so:
∇)(aΨ1 + dΨ2)
j(x,t) = (1/2m)((aΨ1* + dΨ2*)((h/i)∇
(A1.108)
+ (aΨ1 + dΨ2)((-h/i)∇
∇)(aΨ1* + dΨ2*))
In this situation,
∇ = x∂/∂x
(A1.109)
j(x,t) = (1/2m)((aΨ1* + dΨ2*)((h/i)x∂/∂x)(aΨ1 + dΨ2)
(A1.110)
so:
+ (aΨ1 + dΨ2)((-h/i)x∂/∂x)(aΨ1* + dΨ2*))
Working in the wave functions:
j(x,t) = (1/2m)((aΨ1* + dΨ2*)((ah/i)x∂Ψ1/∂x) + (dh/i)x∂Ψ2/∂x))
(A1.111)
+ (aΨ1 + dΨ2)((-ah/i)x∂Ψ1*/∂x)+ (-dh/i)x∂Ψ2*/∂x))
j(x,t) = (1/2m)((aΨ1* + dΨ2*)((ah/i)x∂ei(p(1)x – E(1)t)/h/∂x)
(A1.112)
+ (dh/i)x∂e-i(p(2)x + E(2)t)/h/∂x))
+ (aΨ1 + dΨ2)((-ah/i)x∂e-i(p(1)x – E(1)t)/h/∂x)
+ (-dh/i)x∂ei(p(2)x + E(2)t)/h/∂x))
Now taking the derivatives:
j(x,t) = (1/2m)((aΨ1* + dΨ2*)((ah/i)(ip1/h)xΨ1 + (dh/i)(-ip2/h)xΨ2) (A1.113)
+ (aΨ1 + dΨ2)((-ah/i)(-ip1/h)xΨ1* + (-dh/i)(ip2/h)xΨ2*)
and consolidating the results:
j = (1/2m)((aΨ1* + dΨ2*)((ap1)xΨ1 + (-dp2)xΨ2)
(A1.114)
+ (aΨ1 + dΨ2)((ap1)xΨ1* + (-dp2)xΨ2*)
j = (1/2m)((a2p1Ψ1Ψ1* - adp2Ψ1*Ψ2 + adp1Ψ1*Ψ2 - d2p2Ψ2Ψ2* (A1.115)
+ a2p1Ψ1Ψ1* - adp2Ψ1Ψ2* + adp1Ψ1Ψ2* - d2p2Ψ2Ψ2*)x
j = (1/2m)((a2p1 - adp2Ψ1*Ψ2 + adp1Ψ1*Ψ2 - d2p2
97
(A1.116)
+ a2p1 - adp2Ψ1Ψ2* + adp1Ψ1Ψ2* - d2p2)x
j = (1/2m)((2a2p1 - 2d2p2 - adp2Ψ1*Ψ2 + adp1Ψ1*Ψ2
(A1.117)
- adp2Ψ1Ψ2* + adp1Ψ1Ψ2*)x
j = (1/2m)((2a2p1 - 2d2p2 - adp2Ψ1*Ψ2 - adp2Ψ1Ψ2* + adp1Ψ1*Ψ2 (A1.118)
+ adp1Ψ1Ψ2*)x
Grouping like terms:
j = (1/2m)((2a2p1 - 2d2p2 - adp2(Ψ1*Ψ2 + Ψ1Ψ2*)
(A1.119)
+ adp1(Ψ1*Ψ2 + Ψ1Ψ2*))x
j = (1/2m)((2a2p1 - 2d2p2 + ad(p1 – p2)(Ψ1*Ψ2 + Ψ1Ψ2*))x
(A1.120)
Concerning the wave functions remaining in the equations:
(Ψ1Ψ2* + Ψ1*Ψ2) = ei(p(1)x – E(1)t)/hei(p(2)x + E(2)t)/h
(A1.121)
+ e-i(p(1)x – E(1)t)/he-i(p(2)x + E(2)t)/h
(Ψ1Ψ2* + Ψ1*Ψ2) = ei(p(1)x – E(1)t)/h + i(p(2)x + E(2)t)/h
(A1.122)
+ e-i(p(1)x – E(1)t)/h + -i(p(2)x + E(2)t)/h
(Ψ1Ψ2* + Ψ1*Ψ2) = ei([p(1) – p(2)]x + [E(2) – E(1)]t)/h
(A1.123)
+ e-i([p(1) – p(2)]x + [E(2) – E(1)]t)/h
This fits the Euler formula:
2cos(x) = eix + e-ix
(A1.124)
x = ([p1 – p2]x + [E2 – E1]t)/h
(A1.125)
where:
so:
(Ψ1Ψ2* + Ψ1*Ψ2) = ei([p(1) – p(2)]x + [E(2) – E(1)]t)/h
+ e-i([p(1) – p(2)]x + [E(2) – E(1)]t)/h
becomes:
98
(A1.126)
(Ψ1Ψ2* + Ψ1*Ψ2) = 2cos{([p1 – p2]x + [E2 – E1]t)/h}
(A1.127)
Substituting this result back into the current equation:
j = (1/2m)((2a2p1 - 2d2p2 + ad(p1 – p2)(Ψ1*Ψ2 + Ψ1Ψ2*))x
(A1.128)
j = (1/2m)((2a2p1 - 2d2p2 + ad(p1 – p2)2cos{([p1 – p2]x
(A1.129)
yields:
+ [E2 – E1]t)/h})x
j = (2/2m)((a2p1 - d2p2 + ad(p1 – p2)cos{([p1 – p2]x
(A1.130)
+ [E2 – E1]t)/h})x
j = (1/m)((a2p1 - d2p2 + ad(p1 – p2)cos{([p1 – p2]x
(A1.131)
+ [E2 – E1]t)/h})x
The next step is to derive the charge density on the right. The equation for this is:
ρ(x,t) = (1/2mc2)(Ψ*(ih∂/∂t - V)Ψ + Ψ(-ih∂/∂t - V)Ψ*)
(A1.132)
This time the potential term V = eΘ must stay in. The wave function on the right
is:
Ψtransmitted = fΨ3 = fei(p(3)x – E(3)t)/h
(A1.133)
Substituting this into the charge density equation:
ρ(x,t) = (1/2mc2)(fΨ3*(ih∂/∂t - V)fΨ3 + fΨ3(-ih∂/∂t - V)fΨ3*)
(A1.134)
ρ(x,t) = (1/2mc2)(fΨ3*((ihf)∂Ψ3/∂t - VfΨ3)
(A1.135)
+ fΨ3((-ihf)∂Ψ3*/∂t - VfΨ3*))
ρ(x,t) = (1/2mc2)(fΨ3*((ihf)∂ei(p(3)x – E(3)t)/h/∂t - VfΨ3)
(A1.136)
+ fΨ3((-ihf)∂e-i(p(3)x – E(3)t)/h/∂t - VfΨ3*))
Working out the derivatives:
ρ(x,t) = (1/2mc2)(fΨ3*((ihf)(-iE3/h)Ψ3 - VfΨ3)
+ fΨ3((-ihf)(iE3/h)Ψ3* - VfΨ3*))
99
(A1.137)
and consolidating terms:
ρ(x,t) = (1/2mc2)(fΨ3*((fE3)Ψ3 - VfΨ3) + fΨ3((fE3)Ψ3* - VfΨ3*))
(A1.138)
ρ(x,t) = (1/2mc2)(fΨ3*(fE3)Ψ3 - fΨ3*VfΨ3 + fΨ3(fE3)Ψ3* - fΨ3VfΨ3*) (A1.139)
ρ(x,t) = (1/2mc2)(f(fE3)Ψ3*Ψ3 - fVfΨ3*Ψ3 + f (fE3)Ψ3Ψ3* - fVfΨ3Ψ3*)(A1.140)
ρ(x,t) = (1/2mc2)(f(fE3) - fVf + f (fE3) - fVf)
(A1.141)
ρ(x,t) = (1/2mc2)(f2E3 – f2V + f2E3 – f2V)
(A1.142)
ρ(x,t) = (1/2mc2)(2f2E3 – 2f2V)
(A1.143)
ρ(x,t) = (2f2/2mc2)(E3 – V)
(A1.144)
ρ(x,t) = (f2/mc2)(E3 – V)
(A1.145)
j(x,t) = (1/2m)(Ψ*((h/i)∇
∇ - eA/c)Ψ + Ψ((-h/i)∇
∇ - eA/c)Ψ*)
(A1.146)
For the current on the right:
There is no vector potential, so:
j(x,t) = (1/2m)(Ψ*((h/i)∇
∇)Ψ + Ψ((-h/i)∇
∇)Ψ*)
(A1.147)
fΨ3 = fei(p(3)x – E(3)t)/h
(A1.148)
Using:
as the wave function:
j(x,t) = (1/2m)(fΨ3*((h/i)∇
∇)fΨ3 + fΨ3((-h/i)∇
∇)fΨ3*)
(A1.149)
In this one dimensional problem:
∇ = x∂/∂x
(A1.150)
so:
j(x,t) = (1/2m)(fΨ3*((h/i)x∂/∂x)fΨ3 + fΨ3((-h/i)x∂/∂x)fΨ3*)
j = (1/2m)(fΨ3*((hf/i)∂Ψ3/∂x) + fΨ3((-hf/i)∂Ψ3*/∂x)x
j = (1/2m)(fΨ3*((hf/i)∂ei(p(3)x – E(3)t)/h/∂x )
+ fΨ3((-hf/i)∂e-i(p(3)x – E(3)t)/h/∂x ))x
100
(A1.151)
(A1.152)
(A1.153)
Taking the derivatives:
j = (1/2m)(fΨ3*((hf/i)(ip3/h)Ψ3) + fΨ3((-hf/i)(-ip3/h)Ψ3*))x
(A1.154)
and consolidating the terms:
j = (1/2m)(fΨ3*((fp3)Ψ3) + fΨ3((fp3)Ψ3*)x
(A1.155)
j = (1/2m)(fΨ3Ψ3*(fp3) + fΨ3Ψ3*fp3)x
(A1.156)
j = (1/2m)(f2p3 + f2p3)x
(A1.157)
j = (1/2m)(2f2p3)x
(A1.158)
j = (2f2/2m)(p3)x
(A1.159)
j = (f2/m)(p3)x
(A1.160)
The overall results are, for the right and the left:
ρleft = (1/mc2)(a2E1 + d2E2
(A1.161)
+ ad(E1 + E2)cos{([p1 – p2]x + [E2 – E1]t)/h})
jleft = (1/m)((a2p1 - d2p2
(A1.162)
+ ad(p1 – p2)cos{([p1 – p2]x + [E2 – E1]t)/h})x
ρright = (f2/mc2)(E3 – V)
(A1.163)
jright = (f2/m)(p3)x
(A1.164)
The set of equations derived for analysis of the “classic” Klein’s paradox
are, in order:
p3 = ±(1/c){(E3 - V)2 - m2c4}1/2
(A1.165)
d/a = (p1 – p3)/(p2 + p3)
(A1.166)
f/a = (p2 - p1)/(p2 - p3)
(A1.167)
vG = ∂E3/∂p3 = c2p3/(E3 - V)
(A1.168)
ρleft = (1/mc2)(a2E1 + d2E2
(A1.169)
+ ad(E1 + E2)cos{([p1 – p2]x + [E2 – E1]t)/h})
101
jleft = (1/m)((a2p1 - d2p2
(A1.170)
+ ad(p1 – p2)cos{([p1 – p2]x + [E2 – E1]t)/h})x
ρright = (f2/mc2)(E3 – V)
(A1.171)
jright = (f2/m)(p3)x
(A1.172)
There are a variety of interesting effects, including real and imaginary
momenta, that occur as V is increased from 0 in equation (A1.165). This thesis is
concerned mainly with particle pair production that occurs with a “supercritical”
potential step, in which case:
V > E3 + mc2
(A1.173)
V > E3 + mc2
(A1.174)
so the analysis here will be at that level.
First of all, with:
the momenta on the left in:
p3 = ±(1/c){(E3 - V)2 - m2c4}1/2
(A1.175)
are real. This is a strange result since the potential step is so strong that the
incident particle should be easily reflected. An imaginary p3, signaling reflection
and no transmission, would be easier to understand.
The next step is to look at the group velocity:
vG = x∂E3/∂p3 = xc2p3/(E3 - V)
(A1.176)
With:
V >> E3
(A1.177)
the group velocity is negative if p3 is positive but with a transmitted particle,
combined with the fact that there are no particles incident from the right, the
group velocity should be positive. This can be made so if p3 is negative. Though
102
this doesn’t seem like much of an improvement, there actually is a negative p3
available since:
p3 = ±(1/c){(E3 - V)2 - m2c4}1/2
(A1.178)
includes a negative root.
Moving on to the charge density on the right:
ρright = (f2/mc2)(E3 – V)
(A1.179)
this turns out to be negative. That means antiparticles to the right of the barrier,
which is puzzling since the incident particle is definitely not antimatter. However,
it does help explain some of the other anomalies. An antiparticle moving to the
right (with positive velocity) will look like a particle moving to the left (with
negative velocity). This explains the negative group velocity and negative
momentum on the right.
The current on the right:
jright = (f2/m)(p3)x
(A1.180)
is also negative, which it should be if antiparticles are involved (for the same
reason as the group velocity on the right is negative), which is weird but
consistent.
Finally, the reflection coefficient:
d/a = (p1 – p3)/(p2 + p3)
(A1.181)
is greater than 1 because of the negative p3. More particles are being reflected
than are incident!
This is all consistent with a picture of particle pair production at the barrier.
The incident particle is reflected as a particle pair is produced at the barrier. The
particle member of the pair, combined with the reflected incident particle, make
103
the reflection coefficient greater than 1. The antiparticle of the pair finds the
barrier attractive and moves to the right, accounting for the negative group
velocity, negative momentum, negative charge density and negative current, all
on the right.
A final note on the “classic” Klein’s paradox concerns the wave function on
the right:
fψ3 = fei(p(3)x – E(3)t)/h
(A1.182)
How is this wave function reconciled with the negative momentum on the
left and the fact that antiparticles have negative energy instead of positive energy
(in other words, is our initial assumption that the transmission would involve a
particle and not an antiparticle consistent with all else)? To answer this question,
there is a final analysis for this appendix:
The wave function for a particle moving toward greater x is:
Ψ(x,t) = ei(p●x – Et)/h
(A1.183)
The wave function for an antiparticle moving toward greater x is:
Ψ(x,t) = e-i(p●x – Et)/h
(A1.184)
Since taking the complex conjugate of either of these equations is effectively
changing the sign of the momentum it’s not a surprise at this point to see that of
p3 = ±[(E – V)2/c2 - m2c4/c2]1/2x
the negative root is required.
However, changing the sign of i also effectively changes the sign of E.
This is interesting because although wave functions with either positive or
negative momentum will solve the Klein-Gordon equation IF with the time
104
(A1.185)
component is:
T(t) = e-iEt/h
(A1.186)
they will not solve the Klein-Gordon equation if:
T(t) = eiEt/h
(A1.187)
In other words, the antiparticle wave function:
Ψ(x) = e-i(p●x – Et)/h
(A1.188)
will not work in the relevant Klein-Gordon equation even though antiparticles are
involved.
The relevant Klein-Gordon equation is:
(1/c2)(ih∂/∂t – V)2Ψ = ((h/i)∇)2Ψ + m2c2Ψ
(A1.189)
(1/c2)(-h2∂2/∂t2 – 2Vih∂/∂t + V2)Ψ = (-h2∂2/∂x2)Ψ + m2c2Ψ
(A1.190)
(1/c2)(-h2∂2Ψ/∂t2 – 2Vih∂Ψ/∂t + V2Ψ) = -h2∂2Ψ/∂x2 + m2c2Ψ
(A1.191)
Now inserting the antiparticle wave function:
(1/c2)(-h2∂2e-i(p●x – Et)/h/∂t2 – 2Vih∂e-i(p●x – Et)/h/∂t + V2Ψ)
(A1.192)
= -h2∂2e-i(p●x – Et)/h/∂x2 + m2c2Ψ
(1/c2)(-h2(iE/h)2Ψ – 2Vih(iE/h)Ψ + V2Ψ) = -h2(-ip/h)2Ψ + m2c2Ψ (A1.193)
(1/c2)(E2Ψ + 2VEΨ + V2Ψ) = p2Ψ + m2c2Ψ
(1/c2)(E + V)2Ψ = p2Ψ + m2c2Ψ
(A1.194)
(A1.195)
Now dividing out the wave function:
(1/c2)(E + V)2 = p2 + m2c2
(A1.196)
Now, going back to the scalar form of the momentum:
p = ±[(E – V)2/c2 - m2c4/c2]1/2
and substituting the negative root (although either will have exactly the same
105
(A1.197)
effect; not surprising since they are both solutions) into:
(1/c2)(E + V)2 = p2 + m2c2
(A1.198)
(1/c2)(E + V)2 = (-[(E – V)2/c2 - m2c4/c2]1/2)2 + m2c2
(A1.199)
with the result:
Expanding the squares:
(1/c2)(E2 +2EV + V2) = (E – V)2/c2 - m2c4/c2 + m2c2
(1/c2)(E2 +2EV + V2) = (E2 – 2EV + V2)/c2 - m2c4/c2 + m2c2
(A1.200)
(A1.201)
The rest mass terms cancel out:
(1/c2)(E2 +2EV + V2) = (E2 – 2EV + V2)/c2 - m2c2 + m2c2
(A1.202)
(1/c2)(E2 +2EV + V2) = (E2 – 2EV + V2)/c2
(A1.203)
E2 +2EV + V2 = E2 – 2EV + V2
(A1.204)
Multiplying by c2 yields:
Moving all the terms on the right hand side to the left hand side leaves:
E2 +2EV + V2 - E2 + 2EV - V2 = 0
(A1.205)
(E2 - E2) + (2EV + 2EV) + (V2 - V2) = 0
(A1.206)
Grouping like terms:
with the result:
0 + 4EV + 0 = 0
(A1.207)
4EV = 0
(A1.208)
and:
and the wave function for antiparticles fails with:
T(t) = eiEt/h
(A1.209)
which is usually required for antiparticles and:
T(t) = e-iEt/h
106
(A1.210)
which is used to describe particles, works.
What this boils down to is that the wave function is behaving in a familiar
way describing antiparticles concerning the momentum but not the energy.
Inconsistent behavior with the wave function energy in precisely the
circumstances when there is unusual behavior in the potential is not a
coincidence. It is the very strong potential, V > E, capable of producing
antiparticles with negative energies, that allows the inconsistency to occur.
107
APPENDIX 2
CONSERVATION LAWS AND THE “CLASSIC” KLEIN’S PARADOX
The main issue with the “classic” Klein’s paradox concerns the production
of particle pairs. Because particle formation occurs in pairs there is no violation
conservation of charge but there is no obvious source for the energy necessary
to create a new particle pair, even if this only involves the rest energy.
What can account for this apparent massive and basic violation?
Well, the energy of a particle on the left is Eparticle while the energy of an
antiparticle on the right is the solution for Eantiparticle can be derived from
the results of the Klein-Gordon equation on the right (in the area of strong
potential), equation (A1.15):
For the creation of a particle on the right:
(1/c2)(Eparticle - V)2 = p2particle + m2c2
(A2.1)
c2[(1/c2)(Eparticle - V)2 = p2particle + m2c2]
(A2.2)
(Eparticle - V)2 = p2particlec2 + m2c4
(A2.3)
[(Eparticle - V)2 = p2particlec2 + m2c4]1/2
(A2.4)
[(Eparticle - V)2]1/2 = [p2particlec2 + m2c4]1/2
(A2.5)
(Eparticle – V) = [p2particlec2 + m2c4]1/2
(A2.6)
Eparticle – V = [p2particlec2 + m2c4]1/2
(A2.7)
Eparticle = [p2particlec2 + m2c4]1/2 + V
(A2.8)
Eparticle = [p2particlec2 + m2c4]1/2 + V
(A2.9)
(1/c2)(Eantiparticle - V)2 = p2antiparticle + m2c2
(A2.10)
Solving this for Eparticle:
For an antiparticle:
108
Solving this for Eantiparticle:
c2[(1/c2)(Eantiparticle - V)2 = p2antiparticle + m2c2]
(A2.11)
(Eantiparticle - V)2 = p2antiparticlec2 + m2c4
(A2.12)
[(Eantiparticle - V)2 = p2antiparticlec2 + m2c4]1/2
(A2.13)
[(Eantiparticle - V)2]1/2 = -[p2antiparticlec2 + m2c4]1/2
(A2.14)
where the negative root is the choice due to the negative energy of an
antiparticle.
(Eantiparticle – V) = -[p2antiparticlec2 + m2c4]1/2
(A2.15)
Eantiparticle – V = -[p2antiparticlec2 + m2c4]1/2
(A2.16)
Eantiparticle = -[p2antiparticlec2 + m2c4]1/2 + V
(A2.17)
It is necessary to change the sign of V here since the potential has the opposite
sign for an antiparticle. This gives:
Eantiparticle = -[p2antiparticlec2 + m2c4]1/2 – V
(A2.18)
Now adding the energies of creation of a particle and an antiparticle:
Eparticle + Eantiparticle = [p2antiparticlec2 + m2c4]1/2 + V
(A2.19)
- [p2antiparticlec2 + m2c4]1/2 – V
Grouping like terms:
Eparticle + Eantiparticle = {[p2antiparticlec2 + m2c4]1/2 - [p2antiparticlec2 + m2c4]1/2} (A2.20)
+ {V – V}
The negative signs necessary for expressing the differences between particle
and antiparticle are expressed in the sign of the roots of the term
[p2antiparticlec2 + m2c4]1/2 so:
[p2antiparticlec2 + m2c4]1/2 = [p2antiparticlec2 + m2c4]1/2
and:
109
(A2.20)
Eparticle + Eantiparticle = {0} + {0} = 0
(A2.21)
and the net relativistic energy used to create a particle pair is 0. There is no
violation of the conservation of relativistic energy.
A last detail concerns the conservation of momentum. The negative
current On the left indicates that the antiparticles are moving. Where did this
momentum come from?
The momentum of the antiparticles on the left results from the attractive
nature of the barrier to antiparticles. The momentum that the antiparticles gain
can be balanced by the introduction of the same amount of momentum in the
opposite direction in the mechanism producing the barrier.
Lastly, the current on the right, equation (A1.162):
jleft = (1/m)((a2p1 - d2p2
(A2.22)
+ ad(p1 – p2)cos{([p1 – p2]x + [E2 – E1]t)/h})x
is zero since:
p1 = p2
The momenta of the incident and reflected particle are of the same
magnitude but opposite sign due to the opposing directions of motion. There is
no “left-over” momentum on the left to account for, meaning that the particle of
the created particle pair, created at the barrier interface with its antiparticle
partner, is not required to move. It can add to the reflection coefficient without
being involved in any momentum calculations.
Thus, despite all its peculiarities, the “classic” Klein’s paradox does not
violate charge, energy or momentum conservation thus avoiding the sight of an
equation built directly upon the fundamental energy/momentum relation of
110
(A2.23)
special relativity:
E2 = p2c2 + mc4
violating its founding rule!
111
(A2.24)
APPENDIX 3
THE SCALAR AND VECTOR POTENTIALS
Where do these potentials come from?
The homogeneous pair of Maxwell’s equations are:
∇•B = 0
(A3.1)
∇ x EL + (1/c)∂B/∂t = 0
(A3.2)
and:
Starting with the nondivergence of the magnetic field, equation (A3.1), it is
possible to take advantage of the law of vector mechanics that says that the
divergence of any curl (or field of pure rotation) is zero [10]:
∇•(∇
∇ x A) = 0
(A3.3)
In general this is true of any vector field A but here A is called by a specific
name: the vector potential. This means that:
∇•(∇
∇ x A) = 0
(A3.4)
using equations (A3.1) and (A3.3). This effectively means that:
B=∇xA
(A3.5)
it is consistent mathematically to define the magnetic field B as the curl of the
vector potential A.
Starting now with equation (A3.2):
∇ x EL + (1/c)∂B/∂t = 0
(A3.6)
Making the substitution for B results in:
∇ x EL + (1/c)∂[∇
∇ x A]/∂t = 0
(A3.7)
Because the curl operator is distributive:
∇ x [EL + (1/c)∂A/∂t] = 0
112
(A3.8)
Now it’s time to invoke another law of vector mechanics (or vector
identity), the curl or rotation of the gradient of a scalar vector field (gradients are
irrotational):
∇ x (∇
∇Θ) = 0
(A3.9)
In general, this is true of the gradient of any scalar field Θ. In the case of the
present development this is named the scalar potential. Using equations (A3.8)
and (A3.9):
EL + (1/c)∂A/∂t = ∇Θ
(A3.10)
EL = ∇Θ - (1/c)∂A/∂t
(A3.11)
and:
The gradient can also be negative so this is also true of:
∇Θ - (1/c)∂A/∂t
EL = -∇
which is the preferred version.
So, as the result of mathematical rules about vectors, the electric and
magnetic fields can be described in terms of these two potentials, the scalar and
vector potentials. Interestingly, these potentials are not mere mathematical
constructs but actually are real [2].
113
(A3.12)
APPENDIX 4
SHORT NOTES ON THE AHARONOV-BOHM EFFECT
Figure A4.1. Schematic for the Experimental Setup for the Aharonov-Bohm effect.
In this experimental setup there are two paths, one through each slit in the
barrier, for emitted particles to move from the emitter to the detection screen. In
keeping with the idea that particles travel all paths from source to destinations,
the path integral formulation of quantum mechanics, the phase differences along
the different paths result in an interference pattern seen on the detector. In this
experiment the phase difference between the two sets of paths (one set for each
slit) depends on the vector potential A [2]:
∫1A•dL - ∫2A•dL = ∆ϕphase
Changing the vector potential A changes the difference in phase, which in turn
changes the interference effects seen on the detection screen. Since the
magnetic field is confined to the solenoid in the center section of the barrier, none
of the particles passing through the experimental apparatus “feel” any magnetic
114
(A4.1)
field. Thus the physical changes in interference have to be attributed to the
vector potential A, proving that it is a real physical entity and not merely a
mathematical abstraction or convenience and allowing one to make the argument
that the vector potential, not the magnetic field, is fundamental.
115
APPENDIX 5
THE NON-DIVERGENCE OF THE VECTOR POTENTIAL IN THE SYMMETRIC
GAUGE
The symmetric gauge, which is in terms of the magnetic field vector B and
the cylindrical radial coordinate vector r (giving the direction of the radius of
particle circulation) [2, 24]:
A = (B x r)/2
(A5.1)
is listed in the literature as one of a set of gauges belonging to the general
category of the Coulomb gauge:
∇●A = 0
where the divergence of A explicitly = 0.
The symmetric gauge results in a uniform, constant
magnetic field in the z direction. The path of a charged particle in such a
field, barring any initial motion in the z direction, is circular at a constant angular
velocity.
116
(A5.2)
Figure A5.1. The vector relationship between the cylindrical coordinates angular
unit vector Φ (phi), velocity v, momentum p, and the magnetic field B with a
constant, uniform B field in the positive or negative z direction during cyclotron
motion r in the symmetric gauge for a positively charged particle.
117
Figure A5.2. The vector relationship between the potential A, velocity v,
momentum p, the magnetic field B and the radius of cyclotron motion r in the
symmetric gauge for a positively charged particle.
In the symmetric gauge the angle between r and B is always (θ =) 90
degrees and A is always in the positive or negative Φ direction. Using the
magnitude formula for the vector cross product, the symmetric gauge becomes:
A = (B x r)/2 = ±Φ(Brsinθ)/2 = ±Φ(Brsin90)/2 = ±Φ(Br/2)
With B both constant in magnitude and direction, the magnitude of A must
118
(A5.3)
increase linearly as the magnitude of r increases and vice versa. Thus the
symmetric mode yields the vector field for A [18]:
Figure A5.3. The Vector Field of the Vector Potential in the Symmetric Gauge. The
vectors are all in the Negative φ Direction due to the Magnetic Field being in the
negative z direction.
119
Although the magnitude of A varies as r changes in magnitude, in cyclotron
motion r does not change in magnitude so A does not change in magnitude
either along the path of motion:
Figure A5.4. Vector Potential A in the r, Φ plane viewed from the positive and
negative z axis.
Whatever the magnitude of the radius of the cyclotron motion, A vector is
always tangent to the motion of the circulating particle. Qualitatively, then, A
must be non-divergent.
It is true that if the center of the A field is not at the center of the path of
motion of the particle that A is then divergent along the path but this centering is
implicit in the symmetric gauge. In any case, if the center of the path of motion is
not at the center or axis of the A field it can be moved there with no change in the
physical problem since the energies and angular momenta are infinitely
degenerate in that fashion.
Actually calculating the divergence of A should yield the same nondivergent result.
In cylindrical coordinates:
120
∇ = r∂/∂r + Φ(1/r)∂/∂Φ + z∂/∂z
(A5.4)
A = rAr + ΦAΦ + zAz = ΦAΦ
(A5.5)
and:
So:
∇●A = (r∂/∂r + Φ(1/r)∂/∂Φ + z∂/∂z)●(ΦAΦ)
(A5.6)
∇●A = r∂(ΦAΦ)/∂r + Φ(1/r)∂(ΦAΦ)/∂Φ + z∂(ΦAΦ)/∂z
(A5.7)
∇●A = (r●Φ)∂(AΦ)/∂r + rAΦ∂(Φ)/∂r + (Φ●Φ)(1/r)∂(AΦ)/∂Φ
(A5.8)
+ Φ(AΦ/r)∂(Φ)/∂Φ + (z●Φ)∂(AΦ)/∂z + zAΦ∂(Φ)/∂z
The dot products:
(r●Φ) = (z●Φ) = 0
(A5.9)
(Φ●Φ) = 1
(A5.10)
and:
so:
∇●A = rAΦ∂(Φ)/∂r + (1/r)∂(AΦ)/∂Φ + Φ(AΦ/r)∂(Φ)/∂Φ + zAΦ∂(Φ)/∂z (A5.11)
Evaluating the derivatives of the unit vectors:
Since:
dr/dΦ = Φ, dΦ/dΦ = -r and dΦ/dz = 0
(A5.12)
∇●A = (r●Φ)AΦ + (1/r)∂(AΦ)/∂Φ + (Φ●[-r])(AΦ/r)
(A5.13)
Also:
(r●Φ) = (Φ●[-r]) = 0
(A5.14)
∇●A = (1/r)∂(AΦ)/∂Φ
(A5.15)
so:
The magnitude of the A vector doesn’t change with Φ so:
∂(AΦ)/∂Φ = 0
121
(A5.16)
so:
∇●A = 0
(A5.17)
and A is indeed non-divergent.
A seeming loose end is that in figure 2 the vectors p and A are antiparallel.
As such, the vector dot product of the two:
p●A ≠ 0
(A5.18)
apparently contradicting the result in (A5.17). The resolution of this little paradox
is that divergence is measured over a field and that (A5.17) uses the quantum
mechanical operator for p, in (A5.18) p is treated as only a classical vector. Since
this is a quantum mechanical calculation, (A5.17) and not (A5.18), is the correct
result.
122
APPENDIX 6
COMMUTATION OF A AND P IN THE SYMMETRIC GAUGE
The quantum mechanical momentum operator is:
p = (h/i)∇
∇
(A6.1)
The operator ∇, in cylindrical coordinates, is:
∇ = r∂/∂r + Φ(1/r)∂/∂Φ + z∂/∂z
(A6.2)
while A, in the symmetric gauge and in cylindrical coordinates, is:
A = -ΦAΦ
(A6.3)
The simplest way of calculating the commutator is:
∇,A]Ψ = ((h/i)∇
∇●A - A●(h/i)∇
∇)Ψ
[p,A]Ψ = [(h/i)∇
(A6.4)
The next step is a little tricky as the wave function is applied:
((h/i)∇
∇●A - A●(h/i)∇
∇)Ψ = (h/i)∇
∇●(AΨ) - (h/i)A●∇
∇Ψ
(A6.3)
as the del operator in the first term on the right must by applied to both
A and Ψ using the product rule of derivatives:
∇Ψ = (h/i)Ψ∇
∇●A + (h/i)A●∇
∇Ψ - (h/i)A●∇
∇Ψ
(h/i)∇
∇●(AΨ) - (h/i)A●∇
(A6.4)
In the symmetric gauge:
∇●A = 0
(A6.5)
so:
[p,A]Ψ = [(h/i)∇
∇,A]Ψ = 0 + (h/i)A●∇
∇Ψ - (h/i)A●∇
∇Ψ
(A6.6)
= (h/i)A●∇
∇Ψ - (h/i)A●∇
∇Ψ
and:
[p,A]Ψ = [(h/i)∇
∇,A]Ψ = 0
and:
123
(A6.7)
[p,A] = [(h/i)∇
∇,A] = 0
and the two operators do commute.
One interesting thing about this result is that although a wave function is
needed to carry out the commutation calculation, the result of that calculation is
independent of the exact nature of the wave function; there is no information
about which of cylindrical coordinate variables r, Φ and/or z are included in Ψ.
124
(A6.8)
APPENDIX 7
ANGULAR DEPENDENCE OF THE CYCLOTRON WAVE FUNCTION
Angular motion is obviously occurring when particles undergo cyclotron
motion so it is not immediately clear why any associated wave function should
not have a dependence on the angular variables.
It is possible to model cyclotron motion, or any periodic circular motion, by
the use of the harmonic oscillator. Two one dimensional harmonic oscillators that
are of the same energy but out of phase by the proper amount will reproduce
circular periodic motion in a given plane. The wave function of one of these
harmonic oscillators is [23]:
Ψ1-d h.o(x). = C1-d h.o.Hn(x/C2)e-x^2/(2C(2)^2)
(A7.1)
where:
C1-d h.o. = an n-based constant for the 1-d harmonic oscillator
(A7.2)
and:
C2 = C(2) = a constant based on the harmonic oscillator frequency. (A7.3)
and:
Hn(x/C2) = Hermite Polynomials
(A7.4)
H = -(h2/2m)d2/dx2 + mω2x2/2
(A7.5)
with a Hamiltonian of:
With the addition of another 1-d harmonic oscillator, of the same energy
and correct phase difference, in the y dimension the combination becomes a 2-d
harmonic oscillator. The 2-d wave function is simply the product Ψ(x)Ψ(y), so:
Ψ2-d h.o. = X(x)Y(y) = C2-d h.o.Hn(x,y)e-[(x^2 + y^2)/C(2)^2]
Since:
125
(A7.6)
x = rcosΦ, y = rsinΦ,
(A7.7)
then:
x2 + y2 = (rcosΦ)2 + (rsinΦ)2 = r2cos2Φ + r2sin2Φ = r2(cos2Φ + sin2Φ) = r2 (A7.8)
any dependence on Φ drops out of the exponential part of the wave function [23]:
Ψ2-d h.o. = C2-d h.o.Hn(x,y)e-r^2/(C(2)^2)
However, the same effect does not generally occur with the associated Hermite
polynomials. A list of the lower degree Hermite polynomials for the 1-d harmonic
oscillator is [23]:
TABLE A7.1
LOWER DEGREE 2-D HERMITE POLYNOMIALS
Degree (n)
Polynomial Hn(x)
Polynomial Hn(y)
0
H0(x) = 1
H0(y) = 1
1
H1(x) = 2x
H1(y) = 2y
2
H2(x) = 4x2 – 2
H2(y) = 4y2 – 2
3
H3(x) = 8x3 – 12x
H3(y) = 8y3 – 12y
126
(A7.9)
TABLE A7.2
LOWER DEGREE 2-D HERMITE POLYNOMIALS IN RECTANGULAR AND POLAR
COORDINATES
(n)
Hn(x)Hn(y) = Hn(x,y)
Hn(rcosΦ)Hn(rsinΦ) = Hn(r,Φ)
0
H0(x,y) = 1
H0(r,Φ) = 1
1
H1(x,y) = 4xy
H1(r,Φ) = 4r2cosΦsinΦ
2
H2(x,y) = 16x2y2 – 8x2 – 8y2 + 4
H2(r,Φ) = 16r4cos2Φsin2Φ – 8r2 +4
3
H3(x,y) = 64x3y3 – 96xy3
H3(r,Φ) = 64r6cos3Φsin3Φ
– 96x3y + 144xy
- 96r4cosΦsin3Φ
- 96r4cosΦsin3Φ
+ 144r2cosΦsinΦ
It’s clear from the trend here and from the recursion formula for the higher
order Hermite polynomials:
Hn+1(x) = 2xHn(x) – 2nHn-1(x)
(A7.10)
that, unlike what occurred with the exponential term, the Hermite polynomials in
general will show a dependence on the angular variable.
It’s also possible to calculate the wave functions for the 2-d harmonic
oscillator from the relevant differential equations, the Hamiltonian of which is:
H(r,Φ) = (1/r)∂/∂r(r[∂/∂r]) + (1/r2)(∂2/∂Φ2)
The application of this Hamiltonian to the wave function Ψ will generally
necessitate a separation of the variables r and Φ into two distinct differential
equations. This results in a set of quantum numbers. From the radial equation
comes the energy quantum number n. The angular component to the wave
function, S(Φ), is generally presumed to have the form [1]:
127
(A7.11)
Ψ(Φ) = eim(L)Φ
partly because it will generate eigenvalues, and contains the quantum number
mL, associated with angular momentum [5]:
TABLE A7.3
2-D QUANTUM HARMONIC OSCILLATOR WAVE FUNCTIONS
n
mL
Wave Function Ψ(r,Φ)
(C = constant; no coefficients)
0
0
e-C^2r^2
1
1
re-C^2r^2eiΦ
1
-1
re-C^2r^2e-iΦ
2
2
((Cr)2 – 1)e-C^2r^2e2iΦ
2
0
((Cr)2 – 1)e-C^2r^2
2
-2
((Cr)2 – 1)e-C^2r^2e-2iΦ
There are eigenfunctions with no angular dependence (m = 0) but these
are the states where the difference in phase between the two 1-d harmonic
oscillators that make up the 2-d harmonic oscillator is such that the composite
motion is along a straight line and there is no angular momentum at all.
The situation is much the same with the 3-d case of the hydrogen atom
and the spherical harmonics in the relevant wave functions [19]:
128
(A7.12)
TABLE A7.4
LOWER ORDER HYDROGEN ATOM WAVE FUNCTIONS
Shell Quantum Numbers
General Wave Function
Ψ1S
Ψ100
e-r/c
Ψ2S
Ψ200
(r/c)e-r/2c
Ψ2P
Ψ210
(r/c)e-r/2ccosθ
Ψ2P
Ψ211
(r/c)e-r/2csinθeiΦ
Ψ2P
Ψ21(-1)
(r/c)e-r/2csinθe-iΦ
The S states show no angular dependence but that, like the case of the
2-d harmonic oscillator wave functions, is because these are precisely the states
with no angular momentum. States with some angular momentum are invariably
accompanied by angular variable(s). Furthermore, all these 2-d harmonic
oscillator and hydrogen wave functions do or can involve isotropic potentials
(invariant under rotation) so this symmetry does nothing to eliminate angular
dependence in any of the relevant wave functions.
Since cyclotron motion has to involve angular momentum and symmetry in
general does not at all preclude angular variables in the wave function it is logical
to conclude that the case has been made that any wave function developed
regarding cyclotron motion and this thesis must contain S(Φ). This is absolutely
necessary if one wants to develop the angular momentum quantum numbers mL
from the boundary conditions.
It is possible in the case of the 2-d harmonic oscillator to use a somewhat
classical method to derive some angular momentum information without the
explicit involvement of the angular component of the wave function. Going back
129
to the Klein-Gordon equation [2, 24]:
(1/c2)(E – V)2ψ = (-h2∇2 – (2e/c)A•p + e2A2/c2 + m2c2)ψ
(A7.13)
Using the symmetric gauge:
A = (B x r)/2
(A7.14)
(1/c2)(E – V)2ψ = (-h2∇2 – (2e/c)[(B x r)/2]•p
(A7.15)
the Klein-Gordon equation becomes:
+ e2(Br/2)2/c2 + m2c2)ψ
(1/c2)(E – V)2ψ = (-h2∇2 – (e/c)(B x r)•p
(A7.16)
+ e2(Br/2)2/c2 + m2c2)ψ
(1/c2)(E – V)2ψ = (-h2∇2 – (e/c)(B x r)•p
(A7.17)
+ e2B2r2/4c2 + m2c2)ψ
Using a vector identity:
(B x r)•p = B•(r x p)
(A7.18)
so the Klein-Gordon equation becomes:
(1/c2)(E – V)2ψ = (-h2∇2 – (e/c)B•(r x p) + e2B2r2/4c2 + m2c2)ψ
(A7.19)
rxp=L
(A7.20)
Since:
where L is the angular momentum, the Klein-Gordon equation now becomes:
(1/c2)(E – V)2ψ = (-h2∇2 – (e/c)B••L + e2B2r2/4c2 + m2c2)ψ
(A7.21)
and since the angular momentum in this thesis is in the z direction:
L → LZ
(A7.22)
(1/c2)(E – V)2ψ = (-h2∇2 – (e/c)B••Lz + e2B2r2/4c2 + m2c2)ψ
(A7.23)
so:
130
In the case of this thesis where a positively charged particle is undergoing
cyclotron motion, B and L are in the opposite direction so the dot product is
negative and:
(1/c2)(E – V)2ψ = (-h2∇2 + (e/c)BLz + e2B2r2/4c2 + m2c2)ψ
(A7.24)
What has happened here is that the momentum operator has been “used”
with the symmetric gauge and there is some explicit angular momentum
information in the equation. In this case this has not involved the wave function
directly.
The Klein-Gordon equation can be left in this form:
(1/c2)(E – V)2ψ = (p2 + (e/c)BLz + e2B2r2/4c2 + m2c2)ψ
(A7.25)
which can render some significant information about the role of the vector
potential in Klein’s paradox, as is discussed in the main body of this thesis. What
is lost are the angular momentum quantum numbers. And any attempt to further
apply the derivatives in the remaining operator, p2 → -h2∇2, will involve error if
there is no angular part of the wave function. The reason is that the LaPlacian
operator in polar coordinates contains angular derivatives:
∇2 = (1/r)∂/∂r(r∂/∂r) + (1/r2)∂2/∂φ2
(A7.26)
Finally, if the angular part of the wave function reflects boundary
conditions and has the form:
S(φ) = eim(L)φ
then the relationship between the using the symmetric gauge to interact with the
momentum operator to generate angular momentum and just using the
momentum operator in operator form on the wave function becomes clear. Using
the term in question:
131
(A7.27)
(2e/c)A••pψ = (2e/c)A••∇ψ
(A7.28)
∇ = r∂/∂r + (1/r)φ
φ∂/∂φ
(A7.29)
In this case:
Since the vector potential A has only a component in the negative φ
direction, the radial portion of the operator drops away in the calculation of this
term and:
∇ = (1/r)φ
φ∂/∂φ
(A7.30)
This means that the only part of the wave function that will interact with this
operator is:
S(φ) = eim(L)φ
(A7.31)
(2e/c)A••pψ = (2he/ic)A••∇ψ = (2eh/ic)A(1/r)∂ψ/∂φ
(A7.32)
so:
(2eh/ic)A(1/r)∂[eim(L)φ]/∂φ = (mLe/rc)Aψ(φ)
(A7.33)
Now submitting the symmetric gauge in magnitude form:
(2mLe/rc)(Br/2) = (mLeB/c)
(A7.34)
Comparing this to the result for this term using the angular momentum
substitution:
(e/c)BLz = (mLehB/c)
(A7.35)
LZ = hmL
(A7.36)
which is consistent if:
132
APPENDIX 8
SEARCH FOR A COMPLEX RADIAL WAVE FUNCTION FOR CYCLOTRON
MOTION
The search starts with a proposed wave function of:
ψ = ψ(r,φ,z,t) = R(r)S(φ)Z(z)ψ(t)
(A8.1)
Because the motion of the particle is constrained to the r,φ plane the wave
function in z is minimal:
Z(z) = 1
(A8.2)
ψ(t) = e-iEt/h
(A8.3)
The wave function in t is of standard form:
and the wave function in φ, due to the boundary conditions of standing waves, is
the familiar:
S(φ) = eim(L)φ
(A8.4)
where mL is the angular momentum quantum number.
The exact form of R(r) is unknown so the proposed form is general:
R(r) = eif(r)/h
(A8.5)
with the strategy being to put R(r) through the Klein-Gordon equation to find the
exact form of f(r):
{(1/c2)(ih∂/∂t - eΘ)2 = ((h/i)∇
∇ - eA/c)2 + m2c2}ψ
(A8.6)
Expanding the square on the left:
{(1/c2)(-h2∂2/∂t2 – 2iheΘ∂/∂t + e2Θ2) = ((h/i)∇
∇ - eA/c)2 + m2c2}ψ
(A8.7)
Applying the wave function to the left hand side:
{(1/c2)(-h2∂2/∂t2 – 2iheΘ∂/∂t + e2Θ2)}ψ = {((h/i)∇
∇ - eA/c)2 + m2c2}ψ
133
(A8.8)
(1/c2)(-h2∂2ψ/∂t2 – 2iheΘ∂ψ/∂t + e2Θ2ψ) = {((h/i)∇
∇ - eA/c)2 + m2c2}ψ (A8.9)
(1/c2)(-h2∂2[R(r)ψ(φ)ψ(t)]/∂t2 – 2iheΘ∂[R(r)ψ(φ)ψ(t)]/∂t + e2Θ2ψ)
(A8.10)
= {((h/i)∇
∇ - eA/c)2 + m2c2}ψ
R(r)ψ(φ)(1/c2)(-h2∂2[ψ(t)]/∂t2 – 2iheΘ∂[ψ(t)]/∂t + e2Θ2ψ(t))
(A8.11)
= {((h/i)∇
∇ - eA/c)2 + m2c2}ψ
The time derivatives are:
∂ψ(t)/∂t = ∂e-iEt/h/∂t = (-iE/h)ψ(t)
(A8.12)
∂2ψ(t)/∂t2 = ∂[(-iE/h)ψ(t)]/∂t = (-iE/h)∂[ψ(t)]/∂t = (-iE/h)2ψ(t) = (-E2/h2)ψ(t)
Inserting these derivatives back into the Klein-Gordon equation:
R(r)ψ(φ)(1/c2)(-h2(-E2/h2)ψ(t) – 2iheΘ(-iE/h)ψ(t) + e2Θ2ψ(t))
(A8.13)
= {((h/i)∇
∇ - eA/c)2 + m2c2}ψ
Consolidating the terms on the left:
R(r)S(φ)ψ(t)(1/c2)(-h2(-E2/h2) – 2iheΘ(-iE/h) + e2Θ2)
(A8.14)
= {((h/i)∇
∇ - eA/c)2 + m2c2}R(r)S(φ)ψ(t)
R(r)S(φ)ψ(t)(1/c2)(E2 – 2eΘE + e2Θ2)
(A8.15)
= {((h/i)∇
∇ - eA/c)2 + m2c2}R(r)S(φ)ψ(t)
Now dividing ψ(t) out of the equation:
R(r)S(φ)(1/c2)(E2 – 2eΘE + e2Θ2)
(A8.16)
= {((h/i)∇
∇ - eA/c)2 + m2c2}R(r)S(φ)
and setting:
eΘ = V
(A8.17)
then:
(1/c2)(E2 – 2EV + V2)R(r)S(φ) = {((h/i)∇
∇ - eA/c)2 + m2c2}R(r)S(φ) (A8.18)
(1/c2)(E - V)2R(r)S(φ) = {((h/i)∇
∇ - eA/c)2 + m2c2}R(r)S(φ)
134
(A8.19)
Now expanding the square on the right hand side:
∇•A – (he/ic)A••∇
(1/c2)(E - V)2R(r)S(φ) = {-h2∇2 – (he/ic)∇
(A8.20)
+ e2A2/c2 + m2c2}R(r)S(φ)
Since in the symmetric gauge:
A = (B x r)/2
(A8.21)
(1/c2)(E - V)2R(r)S(φ) = {-h2∇2 – (2he/ic)∇
∇•A + e2A2/c2
(A8.22)
the vector potential A commutes with ∇:
+ m2c2}R(r)S(φ)
Bringing in the wave function on the right:
∇•AR(r)S(φ)
(1/c2)(E - V)2R(r)S(φ) = -h2∇2R(r)S(φ) – (2he/ic)∇
(A8.23)
+ e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
(1/c2)(E - V)2R(r)S(φ) = -h2∇2R(r)S(φ) – (2he/ic)[A••∇(R(r)S(φ))
(A8.24)
+ R(r)S(φ)∇
∇•A] + e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
Since the vector potential A is non-divergent in the symmetric gauge:
∇•A = 0
(A8.25)
(1/c2)(E - V)2R(r)S(φ) = -h2∇2R(r)S(φ) – (2he/ic)A••∇(R(r)S(φ))
(A8.26)
and:
+ e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
Now it is time to work out the derivatives on the right hand side. In
cylindrical coordinates:
∇2 = (1/r)∂/∂r(r∂/∂r) + (1/r2)∂2/∂φ2
(A8.27)
Applying the wave function to the LaPlacian:
∇2R(r)S(φ) = (1/r)∂/∂r(r∂[R(r)S(φ)]/∂r) + (1/r2)∂2[R(r)S(φ)]/∂φ2
135
(A8.28)
∇2R(r)S(φ) = (S(φ)/r)∂/∂r(r∂[R(r)]/∂r) + (R(r)/r2)∂2[S(φ)]/∂φ2
(A8.29)
Working out the radial part of the LaPlacian first:
(S(φ)/r)∂/∂r(r∂[R(r)]/∂r) = (S(φ)/r)∂/∂r(r∂[eif(r)/h]/∂r)
(A8.30)
f” = ∂f(r)/∂r
(A8.31)
(S(φ)/r)∂/∂r(r∂[R(r)]/∂r) = (S(φ)/r)∂/∂r(rif’/h)R(r)
(A8.32)
where:
= (S(φ)/r)(i/h)∂/∂r(rf’R(r))
(S(φ)/r)(i/h)∂/∂r(rf’R(r)) = (S(φ)/r)(i/h)[f’R(r)∂r/∂r
(A8.33)
+ rR(r)∂f’/∂r + rf’∂R(r)/∂r]
(S(φ)/r)(i/h)∂/∂r(rf’R(r)) = (S(φ)/r)(i/h)[f’R(r)
(A8.34)
+ rf”R(r) + rf’(if’/h)R(r)]
where:
f’ = ∂2f(r)/∂r2
(A8.35)
(S(φ)/r)(i/h)∂/∂r(rf’R(r)) = (S(φ)/r)(i/h)[f’R(r)
(A8.36)
Continuing:
+ rf”R(r) + (f’)2(ir/h)R(r)]
(S(φ)/r)(i/h)∂/∂r(rf’R(r)) = (S(φ)/r)[(if’/h)R(r)
(A8.37)
+ (irf”/h)R(r) - (f’)2(r/h2)R(r)]
(S(φ)/r)(i/h)∂/∂r(rf’R(r)) = S(φ)[(if’/rh)R(r)
(A8.38)
+ (if”/h)R(r) - (f’)2(1/h2)R(r)]
(S(φ)/r)(i/h)∂/∂r(rf’R(r)) = {(if’/rh)R(r)
(A8.39)
+ (if”/h)R(r) - (f’)2(1/h2)}R(r)S(φ)
Now for the angular part of the LaPlacian:
(1/r2)∂2S/∂φ2 = (1/r2)∂2[R(r)S(φ)]/∂φ2
136
(A8.40)
(1/r2)∂2S/∂φ2 = (R(r)/r2)∂2[S(φ)]/∂φ2
(A8.41)
(1/r2)∂2S/∂φ2 = (R(r)/r2)∂2[eim(L)φ]/∂φ2
(A8.42)
(1/r2)∂2S/∂φ2 = (R(r)/r2)[imL]2S(φ)
(A8.43)
(1/r2)∂2S/∂φ2 = (R(r)/r2)[-mL2]S(φ)
(A8.44)
(1/r2)∂2S/∂φ2 = (1/r2)[-mL2]R(r)S(φ)
(A8.45)
Inserting the results for the radial and angular parts of the LaPlacian back into
the Klein-Gordon equation:
(1/c2)(E - V)2R(r)S(φ) = {-h2[(if’/rh) + (if”/h)
(A8.46)
- (f’)2(1/h2)R(r)S(φ) + (1/r2)[-mL2]}R(r)S(φ)]
– (2he/ic)A••∇(R(r)S(φ)) + e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
Working out some of the algebra:
(1/c2)(E - V)2R(r)S(φ) = [(-hif’/r)R(r)S(φ) + (-ihf”)R(r)S(φ)
(A8.47)
+ (f’)2R(r)S(φ) + (-h2/r2)[-mL2]R(r)S(φ)
– (2he/ic)A••∇(R(r)S(φ)) + e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.48)
+ (-h2/r2)[-mL2]}R(r)S(φ)
– (2he/ic)A••∇(R(r)S(φ)) + e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
The only operator left to apply is the ∇ in the dot product term. The del operator
in cylindrical coordinates is:
∇ = r∂/∂r + φ(1/r)∂/∂φ
(A8.49)
Applying the wave function to this operator:
∇R(r)S(φ) = r∂[R(r)S(φ)]/∂r + φ(1/r)∂[R(r)S(φ)]/∂φ
(A8.50)
∇R(r)S(φ) = rS(φ)∂[R(r)]/∂r + φ(R(r)/r)∂[S(φ)]/∂φ
(A8.51)
137
and working out the derivatives:
∇R(r)S(φ) = rS(φ)∂[eif(r)/h]/∂r + φ(R(r)/r)∂[eim(L)φ]/∂φ
(A8.52)
∇R(r)S(φ) = rR(r)S(φ)(if’/h) + φ(1/r)(imL)R(r)S(φ)
(A8.53)
∇R(r)S(φ) = rR(r)S(φ)(if’/h) + φ(imL/r)R(r)S(φ)
(A8.54)
Inserting this back into the Klein-Gordon equation yields:
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.55)
+ (-h2/r2)[-mL2]}R(r)S(φ)
– (2he/ic)A•[rR(r)S(φ)(if’/h) + φ(imL/r)R(r)S(φ)]
–
2 2
– + e A R(r)S(φ)/c2 + m2c2R(r)S(φ)
Since the only component of the vector potential in the symmetric gauge is in the
negative φ direction:
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.56)
+ (-h2/r2)[-mL2]}R(r)S(φ)
+ (2he/ic)Aφ
φ•[rR(r)S(φ)(if’/h) + φ(imL/r)R(r)S(φ)]
+ e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.57)
+ (-h2/r2)[-mL2]}R(r)S(φ)
+ (2he/ic)A[(φ
φ•r)R(r)S(φ)(if’/h) + (φ
φ•φ)(imL/r)R(r)S(φ)]
+ e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
Working out the dot products between the unit vectors:
φ•r = 0
(A8.58)
φ•φ = 1
(A8.59)
and:
results in:
138
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.60)
+ (-h2/r2)[-mL2]}R(r)S(φ)
+ (2he/ic)A(imL/r)R(r)S(φ) + e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
Consolidating terms again:
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.61)
+ h2mL2/r2]}R(r)S(φ)
+ (2heAmL/cr)R(r)S(φ) + e2A2R(r)S(φ)/c2 + m2c2R(r)S(φ)
and invoking the symmetric gauge, in scalar form now:
A = Br/2
(A8.62)
(1/c2)(E - V)2R(r)S(φ) = {[(-hif’/r) + (-ihf”) + (f’)2
(A8.63)
+ h2mL2/r2]}S(r)ψ(φ)
+ (heBmL/c)R(r)S(φ) + (e2B2r2/4c2)R(r)S(φ) + m2c2R(r)S(φ)
Dividing out R(r)S(φ) leaves:
(1/c2)(E - V)2 = {[(-hif’/r) + (-ihf”) + (f’)2 + h2mL2/r2]}
(A8.64)
+ (heBmL/c) + (e2B2r2/4c2) + m2c2
where:
f” = d2f(r)/dr2
(A8.65)
f’ = df(r)/dr
(A8.66)
and:
Equating the imaginary parts of equation (A8.64):
(-hif’/r) + (-ihf”) = 0
(A8.67)
f” + f’/r = 0
(A8.68)
The solution to this differential equation is:
f(r) = ln r + Cr
139
(A8.69)
where Cr is a constant of integration.
The real equation, from the real terms of equation (A8.64), is:
(1/c2)(E - V)2 = {(f’)2 + h2mL2/r2}
(A8.70)
+ (heBmL/c) + (e2B2r2/4c2) + m2c2
Re-arranging the terms:
(f’)2 + h2mL2/r2
(A8.71)
+ (heBmL/c) + (e2B2r2/4c2)
+ m2c2 - (1/c2)(E - V)2 = 0
Simplifying:
(f’)2 + D/r2 + Fr2 + G = 0
(A8.72)
D = h2mL2
(A8.73)
F = e2B2/4c2
(A8.74)
G = heBmL/c + m2c2 - (1/c2)(E - V)2
(A8.75)
where:
and:
Integrating this latest differential equation yields [18]:
ode1 := {(diff(f(r),r))^2
+ D/r^2 + F*r^2 + G = 0};
2
d
D


ode1 := {  f( r )  + 2 + F r 2 + G = 0 }
d
r
r


dsolve(ode1,{f(r)});

⌠ −D − F r 4 − G r 2
 
⌠


 

 f( r ) = 
,  f( r ) = −
d
r
+
_C1


 

r


 


⌡
 
⌡
140
(A8.76)
(A8.77)
(A8.78)
(A8.79)

−D − F r 4 − G r 2

dr + _C1 
r


Comparing this rather complex result to the product of the imaginary
equation:
f(r) = ln r + Cr
it is clear that the two solutions are not consistent with each other [3].
This must mean that the premise underlying this calculation, that a
complex radial wave function exists, is in error and no such function exists, at
least when using the symmetric gauge.
141
(A8.80)
APPENDIX 9
THE RELEVANCE OF BESSEL FUNCTIONS
The physical situation for this derivation is that of a positively charged,
spinless particle undergoing cyclotron motion in a constant, uniform magnetic
field.
This is a physical situation that has cylindrical symmetry and is made for
cyclindrical coordinates [13]. Often when using cylindrical coordinates, the result
for the radial equation are Bessel functions. However, the radial wave function
derived in this thesis is a Gaussian wave function connected with harmonic
oscillation. What is it that causes this shift?
The conventional development of situations of cylindrical symmetry with
wave functions that have radial and angular components does result in a radial
equation made of Bessel functions if a vector potential is not included [13]. This
is even true of the Klein-Gordon equation.
Suppose the wave function involved is:
ψ = R(r)S(φ)Z(z)
(A9.1)
and, with time derivatives already applied to the time component ψ(t) there is
now a Klein-Gordon equation of:
{(E – V)2/c2) = {[(h/i)∇]2 + m2c2}Ψ
(A9.2)
{(E – V)2/c2) = {-h2∇2 + m2c2}Ψ
(A9.3)
((E – V)2/c2)Ψ = -h2∇2Ψ + m2c2Ψ
(A9.4)
Applying the wave function:
((E – V)2/c2)Ψ = -h2∇2R(r)S(φ)Z(z) + m2c2Ψ
The LaPlacian in cylindrical coordinates is:
142
(A9.5)
∇2 = (1/r)∂/∂r(r∂/∂r) + (1/r2)∂2/∂Φ2 + ∂2/∂z2
(A9.6)
Applying the wave function:
∇2R(r)S(φ)Z(z) = (1/r)∂/∂r(r∂[R(r)S(φ)Z(z)]/∂r)
(A9.7)
+ (1/r2)∂2[R(r)S(φ)Z(z)]/∂Φ2 + ∂2[R(r)S(φ)Z(z)]/∂z2
∇2R(r)S(φ)Z(z) = (S(φ)Z(z)/r)∂/∂r(r∂[R(r)]/∂r)
(A9.8)
+ (R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + R(r)S(φ)∂2[Z(z)]/∂z2
Working out the radial component:
∇2R(r)S(φ)Z(z) = (S(φ)Z(z)/r)[∂r/∂r∂R(r)/∂r + r∂/∂r∂ψ/∂r)]
(A9.9)
+ (R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + R(r)S(φ)∂2[Z(z)]/∂z2
∇2R(r)S(φ)Z(z) = (S(φ)Z(z)/r)[∂R(r)/∂r + r∂2ψ/∂r2)]
(A9.10)
+ (R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + R(r)S(φ)∂2[Z(z)]/∂z2
∇2R(r)S(φ)Z(z) = S(φ)Z(z)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.11)
+ (R(r)S(z)/r2)∂2[S(φ)]/∂Φ2 + R(r)S(φ)∂2[Z(z)]/∂z2
Putting these results back into the Klein-Gordon equation:
((E – V)2/c2)Ψ = -h2∇2R(r)S(φ)Z(z) + m2c2Ψ
(A9.12)
((E – V)2/c2)Ψ = -h2[S(φ)Z(z)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.13)
yields:
+ (R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + R(r)S(φ)∂2[Z(z)]/∂z2] + m2c2Ψ
((E – V)2/c2)Ψ = [-h2S(φ)Z(z)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.14)
+ -h2(R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + -h2R(r)S(φ)∂2[Z(z)]/∂z2] + m2c2Ψ
Dividing by:
ψ = R(r)S(φ)Z(z)
(A9.15)
{((E – V)2/c2)Ψ = -h2ψ(φ)Z(z)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.16)
results in:
143
+ -h2(R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + -h2R(r)S(φ)∂2[Z(z)]/∂z2]
+ m2c2Ψ}/R(r)S(φ)Z(z)
((E – V)2/c2) = -h2S(φ)Z(z)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.17)
+ -h2(R(r)Z(z)/r2)∂2[S(φ)]/∂Φ2 + -h2R(r)S(φ)∂2[Z(z)]/∂z2]/R(r)S(φ)Z(z)
+ m2c2
((E – V)2/c2) = -h2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.18)
+ -h2(1/S(φ)r2)∂2[S(φ)]/∂Φ2 + (-h2/Z(z))∂2[Z(z)]/∂z2 + m2c2
Re-arranging the terms to prepare for separation of the z variables:
-h2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.19)
+ -h2(1/S(φ)r2)∂2[S(φ)]/∂Φ2 + m2c2 - (E – V)2/c2 = (h2/Z(z))∂2[Z(z)]/∂z2
A separation constant is now necessary between the r, φ equation and the z
equation:
-h2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.20)
+ -h2(1/S(φ)r2)∂2[S(φ)]/∂Φ2 + m2c2 - (E – V)2/c2 = Q = (h2/Z(z))∂2[Z(z)]/∂z2
For the purposes of seeing what happens to the radial equation the z
equation can be neglected, leaving:
-h2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.21)
+ -h2(1/S(φ)r2)∂2[S(φ)]/∂Φ2 + m2c2 - (E – V)2/c2 = Q
Multiplying by r2:
{-h2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
(A9.22)
+ -h2(1/S(φ)r2)∂2[S(φ)]/∂Φ2 + m2c2 - (E – V)2/c2 = Q}r2
yields:
-h2r2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)]
+ -h2(1/S(φ))∂2[S(φ)]/∂Φ2 + m2c2r2 – r2(E – V)2/c2 = Qr2
144
(A9.23)
Now a little re-arrangement separates r and φ:
-h2r2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)] + m2c2r2 – r2(E – V)2/c2 - Qr2
(A9.24)
= h2(1/S(φ))∂2[S(φ)]/∂Φ2
Another separation constant is now necessary:
-h2r2/R(r)[(1/r)∂R(r)/∂r + ∂2ψ/∂r2)] + m2c2r2 – r2(E – V)2/c2 - Qr2
(A9.25)
= R = h2(1/ψ(φ))∂2[S(φ)]/∂Φ2
Again, for purposes seeing whether or not Bessel functions occur as solutions to
the radial equation the angular portion can be discarded, leaving an r equation
with derivatives that are now total:
-h2r2/R(r)[(1/r)dR(r)/dr + d2ψ/dr2)] + m2c2r2 – r2(E – V)2/c2 - Qr2 = R (A9.26)
Dividing by r2:
{-h2r2/R(r)[(1/r)dR(r)/dr + d2ψ/dr2)] + m2c2r2
(A9.27)
– r2(E – V)2/c2 - Qr2 = R}/r2
-h2/R(r)[(1/r)dR(r)/dr + d2ψ/dr2)] + m2c2 – (E – V)2/c2 - Q = R/r2
(A9.28)
-h2/R(r)[(1/r)dR(r)/dr + d2ψ/dr2)] + m2c2 – (E – V)2/c2 - Q - R/r2 = 0 (A9.29)
Now multiplying by R(r):
{-h2/R(r)[(1/r)dR(r)/dr + d2ψ/dr2)] + m2c2
(A9.30)
– (E – V)2/c2 - Q - R/r2 = 0}R(r)
{-h2/R(r)[(1/r)dR(r)/dr + d2ψ/dr2)] + m2c2
(A9.31)
- (E – V)2/c2 - Q - R/r2}R(r) = 0
-h2[(1/r)dR(r)/dr + d2R(r)/dr2)] + {m2c2 – (E – V)2/c2
(A9.32)
- Q - R/r2}R(r) = 0
Now dividing by –h2:
{-h2[(1/r)dR(r)/dr + d2R(r)/dr2)] + {m2c2
145
(A9.33)
– (E – V)2/c2 - Q - R/r2}R(r) = 0}/-h2
[(1/r)dR(r)/dr + d2R(r)/dr2)] + {m2c2
(A9.34)
– (E – V)2/c2 - Q - R/r2}R(r)/-h2 = 0
Doing a little more re-arrangement:
d2R(r)/dr2 + (1/r)dR(r)/dr + {-m2c2/h2
(A9.35)
+ (E – V)2/h2c2 + Q/h2 + R/h2r2}R(r) = 0
and:
d2R(r)/dr2 + (1/r)dR(r)/dr + {(E – V)2/h2c2 + Q/h2
(A9.36)
+ R/h2r2 - m2c2/h2}R(r) = 0
This radial equation becomes:
d2R(r)/dr2 + (1/r)dR(r)/dr + {C1 + C2/r2}R(r) = 0
(A9.37)
This is a form of Bessel’s equation [13], here a natural outgrowth of using
cylindrical coordinates in a general sense without a vector potential.
Suppose in developing a radial equation there is a vector potential.
Suppose further that this vector potential, although the gauge is not completely
specified, commutes with the momentum operator and is non-divergent. What
does the resulting radial equation look like then?
The Klein-Gordon Equation with a minimally coupled vector potential is:
{(1/c2)(ih∂/∂t)2 = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.38)
{(1/c2)([ih]2∂2/∂t2) = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.39)
{(1/c2)(-h2∂2/∂t2) = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.40)
In this case the point can be made without including a specific dependence on φ.
The wave function, then, is:
Ψ = Ψ(r,t) = R(r)T(t)
146
(A9.41)
where:
T(t) = e-iEt/h
(A9.42)
With this the time derivatives are, in the familiar way:
∂T(t)/∂t = ∂(e-iEt/h)/∂t = (e-iEt/h)∂(-iEt/h)/∂t = (-iE/h)e-iEt/h
∂2T(t)/∂t2 = ∂[(-iE/h)e-iEt/h]/∂t = (-iE/h)2e-iEt/h
(A9.43)
(A9.44)
= (-E2/h2)e-iEt/h = (-E2/h2)T(t)
Substituting these results back into the Klein-Gordon Equation:
{(1/c2)(-h2∂2/∂t2) = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.45)
{(1/c2)(-h2∂2/∂t2) = [(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.46)
{(1/c2)(-h2∂2/∂t2)}R(r)T(t) = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.47)
R(r){(1/c2)(-h2∂2/∂t2)}T(t) = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.48)
R(r){(1/c2)(-h2∂2[T(t)]/∂t2)} = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.49)
R(r){(-h2/c2)(∂2[T(t)]/∂t2)} = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.50)
R(r)(-h2/c2)(-E2/h2)T(t) = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.51)
R(r)(-h2/c2)(-E2/h2)T(t) = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.52)
R(r)(E2/c2)T(t) = {[(h/i)∇ - eA/c]2 + m2c2}R(r)T(t)
(A9.53)
Dividing by T(t):
R(r)(E2/c2) = {[(h/i)∇ - eA/c]2 + m2c2}R(r)
(A9.54)
Now working out the squared term on the right::
R(r)(E2/c2) = {(-h2)∇2 – (eh/ic)∇●A – (eh/ic)A●∇ +
(A9.55)
e2A2/c2 + m2c2}R(r)
R(r)(E2/c2) = (-h2)∇2R(r) – (eh/ic)∇●(AR(r))
– (eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r)
147
(A9.56)
R(r)(E2/c2) = (-h2)∇2R(r) – (eh/ic)R(r)(∇●A) – (eh/ic)A●∇R(r)
(A9.57)
– (eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r)
R(r)(E2/c2) = (-h2)∇2R(r) – (eh/ic)R(r)(∇●A)
(A9.58)
– (2eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r)
Since the divergence of A is zero:
R(r)(E2/c2) = (-h2)∇2R(r) - (2eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r) (A9.59)
The next step is applying the LaPlacian, which in cylindrical coordinates is:
∇2 = (1/r)∂/∂r(r∂R(r)/∂r) + (1/r2)∂2R(r)/∂Φ + ∂2R(r)/∂z2
(A9.60)
Because of dependence only on r, the middle and right hand terms vanish to
leave:
∇2 = (1/r)∂/∂r(r∂/∂r)
∇2R(r) = (1/r)∂/∂r(r∂R(r)/∂r)
(A9.61)
(A9.62)
and since there is only one variable to deal with now:
∇2R(r) = (1/r)d/dr(rdR(r)/dr)
(A9.63)
Working this out:
∇2R(r) = (1/r)d/dr(rR(r)’)
(A9.64)
where:
R(r)’ = dR(r)/dr
(A9.65)
∇2R(r) = (1/r)d(r R(r)’)/dr
(A9.66)
∇2R(r) = (1/r)[(dr/dr) R(r)’ + r(d R(r)’/dr)]
(A9.67)
and:
∇2 R(r) = (1/r)[ R(r)’ + r(d R(r)’/dr)]
(A9.68)
∇2 R(r) = (1/r)[ R(r)’ + r R(r)’’]
(A9.69)
where:
148
R(r)” = d2R(r)/dr2
(A9.70)
Inserting this back into the Klein-Gordon Equation:
R(r)(E2/c2) = (-h2)∇2R(r) – (2eh/ic)A●∇+ e2A2/c2R(r) + m2c2R(r)
(A9.71)
yields:
R(r)(E2/c2) = (-h2)[(1/r)R(r)’ + R(r)’’] – (2eh/ic)A●∇R(r)
(A9.72)
+ e2A2/c2R(r) + m2c2R(r)
R(r)E2/c2 = (-h2)[(1/r)dR(r)/dr + d2R(r)/dr2]
(A9.73)
– (2eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r)
R(r)E2/c2 + (h2)[(1/r)dR(r)/dr + d2R(r)/dr2] =
(A9.74)
– (2eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r)
(h2/r)dR(r)/dr + h2d2R(r)/dr2 + E2/c2R(r) =
(A9.75)
– (2eh/ic)A●∇R(r) + e2A2/c2R(r) + m2c2R(r)
h2d2R(r)/dr2 + (h2/r)dR(r)/dr + (E2/c2 + (2eh/ic)A●∇
(A9.76)
- e2A2/c2 - m2c2)R(r) = 0
Dividing by h2:
(h2d2Ψ/dr2 + (h2/r)dΨ/dr + (E2/c2 + (2eh/ic)A●∇
(A9.77)
- e2A2/c2 - m2c2)R(r) = 0)/h2
d2Ψ/dr2 + (1/r)dΨ/dr + (E2/h2c2 + (2eh/h2ic)A●∇
(A9.78)
- e2A2/h2c2 - m2c2/h2)R(r) = 0
Consolidating the constants into one:
Cb = E2/h2c2 + (2eh/h2ic)A●∇ - e2A2/h2c2 - m2c4/h2c2
(A9.79)
yields:
d2R(r)/dr2 + (1/r)dR(r)/dr + CbR(r) = 0
149
(A9.80)
which can be transformed into a Bessel’s Equation.
The Bessel Equation is [13]:
d2R(r)/dr2 + (1/r)dR(r)/dr + (s2 – µ2/r2)R(r) = 0
(A9.81)
where µ is the separation constant relating the Φ equation to the rest of the
Klein-Gordon equation. Since there is no Φ dependence the separation
constant (assumed for simplification):
µ=0
(A9.82)
The zero value here is because the separation equation in Φ is:
d2S(Φ)/dΦ2 + µ2S(Φ) = 0
(A9.83)
The generalized wave function for the space variables is:
Ψ(r,Φ,z) = R(r)S(Φ)Ψ(z)
(A9.84)
In this case:
Ψ(r,Φ,z) = R(r)
(A9.85)
S(Φ) = Ψ(z) = 1
(A9.86)
then:
The separation equation in Φ then becomes:
d2(1)/dΦ2 + µ2S(Φ) = 0
(A9.87)
and:
0 + µ2S(Φ) = 0
(A9.88)
µ2S(Φ) = 0
(A9.89)
S(Φ) ≠ 0
(A9.90)
µ2 = µ = 0
(A9.91)
and since:
then:
150
Since there is no z dependence the separation constant:
s=0
(A9.92)
The reason is that the case of the separation equation in z is much the
same as was the separation equation in Φ:
d2Ψ(z)/dΦ2 - s2Ψ(z) = 0
(A9.93)
Since:
Ψ(z) = 1
(A9.94)
d2(1)/dΦ2 - s2Ψ(z) = 0
(A9.95)
0 - s2Ψ(z) = 0
(A9.96)
s2Ψ(z) = 0
(A9.97)
s2 = s = 0
(A9.98)
d2R(r)/dr2 + (1/r)dR(r)/dr + (s2 – µ2/r2)R(r) = 0
(A9.99)
and:
becomes:
d2R(r)/dr2 + (1/r)dR(r)/dr = 0
(A9.100)
However, there is a constant, Cb, as defined in equation (45):
Cb = E2/h2c2 + (2eh/h2ic)A●∇ - e2A2/h2c2 - m2c4/h2c2
(A9.101)
such that:
d2R(r)/dr2 + (1/r)dR(r)/dr + (Cb)R(r) = 0
This restores the form of the equation derived in this chapter to a form of
Bessel’s equation. The solutions then to the r dependent part of the wave
function in this derivation seem to be solutions of the Bessel Equation, i.e.,
Bessel functions.
If a uniform, unchanging scalar potential Θ is included with the vector
151
(A9.102)
potential A the Klein-Gordon Equation becomes:
{(1/c2)(ih∂/∂t - eΘ)2 = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.103)
This changes only the left hand side and so adding the scalar potential initially
involves that side only:
{(1/c2)(-h2∂2/∂t2 - 2iheΘ∂/∂t + e2Θ2) = [(h/i)∇ - eA/c]2 + m2c2}Ψ (A9.104)
and now applying the wave function Ψ on the left hand side:
(1/c2)(-h2∂2Ψ/∂t2 - 2iheΘ∂Ψ/∂t + e2Θ2Ψ) = {[(h/i)∇ - eA/c]2 + m2c2}Ψ (A9.105)
The same results apply for the time derivatives as in as before:
(1/c2)(-h2(-E2/h2)Ψ - 2iheΘ(-Ei/h)Ψ + e2Θ2Ψ)
(A9.106)
= {[(h/i)∇ - eA/c]2 + m2c2}Ψ
(1/c2)(E2Ψ - 2eΘEΨ + e2Θ2Ψ) = {[(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.110)
Reconfiguring:
{(1/c2)(E2 - 2eΘE + e2Θ2) = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.111)
{(1/c2)(E - eΘ)2 = [(h/i)∇ - eA/c]2 + m2c2}Ψ
(A9.112)
Through the rest of the calculations this means that E2 → (E – eΘ)2, if a
scalar potential is present, but that otherwise the equations are unchanged.
Also:
E2/c2)R(r)S(Φ) = -h2∇2(R(r)S(Φ)) – (2eh/ic)∇●AR(r)S(Φ)
(A9.113)
+ e2A2/c2R(r)S(Φ) + m2c2R(r)S(Φ)
E2/c2)R(r)S(Φ) = -h2∇2(R(r)S(Φ)) – (2eh/ic)R(r)S(Φ)∇●A – (A9.114)
(2eh/ic)A●∇R(r)S(Φ) + e2A2/c2R(r)S(Φ) + m2c2R(r)S(Φ)
Since the divergence of A is zero:
E2/c2)R(r)S(Φ) = -h2∇2(R(r)S(Φ)) – (2eh/ic)A●∇R(r)S(Φ)
152
(A9.115)
+ e2A2/c2R(r)S(Φ) + m2c2R(r)S(Φ)
Now the results for the LaPlacian:
∇2R(r)S(Φ) = (S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2 (A9.116)
and substituting this result into equation (A1.31):
E2/c2)R(r)S(Φ) = -h2(S(Φ)[∂2[R(r)]/∂r2
(A9.107)
+ (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
– (2eh/ic)A●∇R(r)S(Φ) + e2A2/c2R(r)S(Φ) + m2c2R(r)S(Φ)
Re-arranging:
-h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2) (A9.108)
– (2eh/ic)A●∇R(r)S(Φ) + e2A2/c2R(r)S(Φ)
+ m2c2R(r)S(Φ) - E2/c2)R(r)S(Φ) = 0
Multiplying by negative 1:
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2) (A9.109)
+ (2eh/ic)A●∇R(r)S(Φ) - e2A2/c2R(r)S(Φ)
- m2c2R(r)S(Φ) + E2/c2)R(r)S(Φ) = 0
Setting:
- e2A2/c2 - m2c2 + E2/c2 = T’
(A9.110)
yields:
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
(A9.111)
+ (2eh/ic)A●∇R(r)S(Φ) + T’R(r)S(Φ) = 0
Unlike in the previous derivation, using the symmetric gauge:
A = (B x r)/2
(A9.112)
on the term with the dot product:
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
153
(A9.113)
+ (2eh/ic)[(B x r)/2]●∇R(r)S(Φ) + T’R(r)S(Φ) = 0
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
(A9.114)
+ (2eh/2ic)[(B x r)]●∇R(r)S(Φ) + T’R(r)S(Φ) = 0
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
(A9.115)
+ (eh/ic)[(B x r)]●∇R(r)S(Φ) + T’R(r)S(Φ) = 0
and:
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
(A9.116)
+ (e/c)[(B x r)●p]R(r)S(Φ) + T’R(r)S(Φ) = 0
Since:
(B x r)●p = B●(r x p)
(A9.117)
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
(A9.118)
then:
+ (e/c)[B●(r x p)]R(r)S(Φ) + T’R(r)S(Φ) = 0
Since the angular momentum:
Lz = r x p
(A9.119)
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2)
(A9.120)
then:
+ (e/c)[B●Lz]R(r)S(Φ) + T’R(r)S(Φ) = 0
and carrying out the dot product:
h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2) (A9.121)
+ (eBLz/c)R(r)S(Φ) + T’R(r)S(Φ) = 0
Now dividing by R(r)S(Φ):
{h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2) (A9.122)
+ (eBLz/c)R(r)S(Φ) + T’R(r)S(Φ) = 0}/R(r)S(Φ)
154
{h2(S(Φ)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (R(r)/r2)∂2[S(Φ)]/∂Φ2) (A9.123)
+ (eBLz/c)R(r)S(Φ) + T’R(r)S(Φ)}/R(r)S(Φ) = 0
h2(1/R(r)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (1/S(Φ)r2)∂2[S(Φ)]/∂Φ2) (A9.124)
+ (eBLz/c) + T’ = 0
Now, in preparation for separation of variables, multiply by r2:
{h2(1/R(r)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (1/S(Φ)r2)∂2[S(Φ)]/∂Φ2) (A9.125)
+ (eBLz/c) + T’ = 0}r2
{h2(1/R(r)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (1/S(Φ)r2)∂2[S(Φ)]/∂Φ2) (A9.126)
+ (eBLz/c) + T’}r2 = 0
r2h2(1/R(r)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r] + (1/S(Φ))∂2[S(Φ)]/∂Φ2) (A9.127)
+ (eBLz/c)r2 + T’r2 = 0
Now separating the variables r and Φ:
r2h2(1/R(r)[∂2[R(r)]/∂r2 + (1/r)∂R(r)/∂r]
(A9..128)
+ (eBLz/c)r2 + T’r2 = -(1/S(Φ))∂2[S(Φ)]/∂Φ2)
Both sides of the equation must now be equal to a separation constant Ξ such
that:
r2h2(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr]
(A9.129)
+ (eBLz/c)r2 + T’r2 = Ξ
and:
-(1/S(Φ))d2[S(Φ)]/dΦ2) = Ξ
(A9.130)
Equation (A1.56) can be solved by the usual assumption of solutions of the form
e-im(L)Φ where mL will be an angular momentum quantum number; somewhat
“redundant” since the angular momentum is already included explicitly in the
equation. For the radial variable:
155
r2h2(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr] + (eBLz/c + T’)r2 = Ξ
(A9.131)
Dividing this by r2:
{r2h2(1/R(r)[d2[R(r)]/∂d2 + (1/r)dR(r)/dr] + (eBLz/c + T’)r2 = Ξ}/r2 (A9.132)
yields:
(r2/r2)h2(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr] + (eBLz/c + T’)(r2/r2) = Ξ/r2 (A9.133)
and:
h2(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr)] + (eBLz/c + T’) = Ξ/r2
(A9.134)
h2(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr)] + (eBLz/c + T’) - Ξ/r2 = 0 (A9.135)
Dividing now by h2:
{h2(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr)]
(A9.136)
+ (eBLz/c + T’) - Ξ/r2 = 0}/h2
(1/R(r)[d2[R(r)]/dr2 + (1/r)dR(r)/dr)]
(A9.137)
+ (eBLz/c + T’)/h2 - Ξ/h2r2 = 0
Now multiplying by R(r):
d2[R(r)]/dr2 + (1/r)dR(r)/dr
(A9.138)
+ (eBLz/h2c + T’/h2)R(r) - ΞR(r)/h2r2 = 0
and:
d2[R(r)]/dr2 + (1/r)dR(r)/dr
(A9.139)
+ (eBLz/h2c + T’/h2)R(r) – (Ξ/h2r2)R(r) = 0
This is Bessel’s equation:
d2[R(r)]/dr2 + (1/r)dR(r)/dr + (C12 – C22/r2)R(r) = 0
(A9.140)
C12 = eBLz/h2c + T’/h2
(A9.141)
where:
and:
156
C22 = Ξ/h2
(A9.142)
This is the general result for the Klein-Gordon equation in cylindrical
coordinates for the r equation. Bessel functions naturally occur with in cylindrical
coordinates and with the Klein-Gordon equation in cylindrical coordinates. In this
appendix they show up in the generalized cylindrical coordinate situation when a
vector potential is not present and also show up when a vector potential is
present but the gauge is not specified.
What will become clear is that when the symmetric gauge is applied
consistently a radial equation with Bessel function solutions does not occur and
that it is possible to conclude that the reason that Bessel functions don’t show up
as solutions to the radial equation in the cyclotron motion in this thesis is
because of the use of the symmetric gauge.
157
APPENDIX 10
DERIVATION OF THE HARMONIC OSCILLATOR WAVE FUNCTION BY
ANALOGY
Starting with the Klein-Gordon Equation with scalar and
vector potentials:
(1/c2)(ih∂/∂t – eΘ)2Ψ(r,t) = (((h/i)∇ – (e/c)A)2 + m2c2) Ψ(r,t)
(A10.1)
where Θ = scalar potential
A = vector potential
(A10.2)
h = (planck’s constant)/2π
(A10.3)
With an added scalar potential the Klein-Gordon Equation becomes:
(1/c2)(ih∂/∂t– eΘ)2Ψ(r,t) = (((h/i)∇ – (e/c)A)2 + m2c2)Ψ(r,t)
(A10.4)
(1/c2)(ih∂/∂t– V)2Ψ = (((h/i)∇ – (e/c)A)2 + m2c2)Ψ
(A10.5)
and
The next step is to work out the squared term containing the magnetic
vector potential:
(1/c2)(ih∂/∂t– V)2Ψ = ((p –(e/c)A)2 + m2c2)Ψ, where p = (h/i)∇
(A10.6)
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (e/c)p●A – (e/c)A●p + (e2/c2)A2 + m2c2)Ψ (A10.7)
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – ((e/c)p●A + (e/c)A●p) + (e2/c2)A2 + m2c2)Ψ (A10.8)
Since A and p commute in the symmetric gauge (which will appear shortly):
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (2e/c)p●A + (e2/c2)A2 + m2c2)Ψ
(A10.9)
Working out the dot product term involves the wave function:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (2e/c)p●(AΨ) + (e2/c2)A2Ψ + m2c2Ψ) (A10.10)
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (2e/c)Ψ(p●A) - (2e/c)A●p(Ψ)
+ (e2/c2)A2Ψ + m2c2Ψ)
158
(A10.11)
With p in operator form in the dot product terms:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (2ehi/c)Ψ(∇●A) - (2he/ic)A●∇(Ψ)
(A10.12)
+ (e2/c2)A2Ψ + m2c2Ψ)
and since in the symmetric gauge A is non-divergent:
(2ehi/c)Ψ(∇●A) = 0
(A10.13)
and:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 - (2he/ic)A●∇(Ψ)
(A10.14)
+ (e2/c2)A2Ψ + m2c2Ψ)
and back to:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 - (2e/c)A●p(Ψ)
(A10.15)
+ (e2/c2)A2Ψ + m2c2Ψ)
With the symmetric gauge:
A = (B x r)/2
(A10.16)
equation (A10.15) becomes:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (2e/c)[(B x r)/2]•p
(A10.17)
+ (e2/c2)A2 + m2c2) Ψ
and:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (e/c)(B x r)•p + (e2/c2)A2 + m2c2) Ψ (A10.18)
Now using the vector identity [13]:
(a1 x a2)•a3 = a1•(a2 x a3)
(A10.19)
to say:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (e/c)B•(r x p) + (e2/c2)A2 + m2c2)Ψ (A10.20)
The cross product is of the familiar form:
Lz = (r x p)
159
(A10.21)
where Lz is the angular momentum [26].
Using cylindrical coordinates with the rotation occurring in the r,Φ plane,
the angular momentum vector is along the z axis. Substituting the angular
momentum into the Klein-Gordon equation (A10.20):
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (e/c)B●Lz + (e2/c2)A2 + m2c2)Ψ
(A10.22)
B and Lz should be in opposite directions:
Figure A10.1. Vector Relationships Between Magnetic Force F on a Positively
Charged Particle, Angular Momentum Lz of the Particle and Magnetic Field B.
The angle between the angular momentum vector Lz and the magnetic field B is
180 degrees. Using the dot product formula for the magnitude:
160
Lz●B = LzBcosθ
(A10.23)
where theta is the angle between the two vectors,
Lz●B = LzBcosθ = LzBcos(180) = -LzB
(A10.24)
the dot product is negative, so:
(1/c2)(ih∂/∂t– V)2Ψ = (p2 – (e/c)B●Lz + (e2/c2)A2 + m2c2) Ψ
(A10.25)
(1/c2)(ih∂/∂t– V)2Ψ = (p2 + (e/c)BLz + (e2/c2)A2 + m2c2)Ψ
(A10.26)
becomes:
The time component T(t) can be the familiar complex form:
T(t) = e-iEt/h
(1/c2)R(r)[-h2∂2(e-iEt/h)/∂t2 - 2ihV∂(e-iEt/h)/∂t + V2Ψ]
(A10.27)
(A10.28)
= (p2 + (eLzB/c) + e2B2r2/4c2 + m2c2)R(r)T(t)
Working out the time derivatives yields:
Ψ(r,t) (1/c2)R(r)[-h2(-iE/h)2 - 2ihV(-iE/h) + V2]
(A10.29)
= (p2 + (eLzB/c) + e2B2r2/4c2 + m2c2)Ψ(r,t)
Ψ(r,t) (1/c2)R(r)[E2 - 2VE + V2]
(A10.30)
= (p2 + (eLzB/c) + e2B2r2/4c2 + m2c2)Ψ(r,t)
A little rearranging yields:
(p2 + e2B2r2/4c2)Ψ(r,t) = ((E – V)2/c2 - eLzB/c - m2c2)Ψ(r,t)
(A10.31)
and dividing by 2MT(t), where:
M = m/(1 – v2/c2)1/2 = relativistic mass
(A10.32)
m = rest mass
(A10.33)
with:
yields:
(p2/2M + e2B2r2/8Mc2)R(r) = ((E – V)2/2Mc2
161
(A10.34)
- eLzB/2Mc – m2c2/2M)R(r)
where each term is in units of energy.
This equation (A10.35) now has the form:
(p2/2M + Mω2r2/2)R(r) = ЄR(r),
(A10.35)
Є = harmonic oscillator energy
(A10.36)
where:
This is an equation for a 2 dimensional harmonic oscillator [23],[24].
This isn’t too surprising since uniform circular motion is essentially two
simultaneous, coplanar, orthogonal (and out of phase) harmonic oscillators, i.e.:
([px2 + py2]/2M + Mω2(x2 + y2) /2)Ψ(x,y) = ЄΨ(x,y)
(A10.37)
In any case, the frequency of the oscillator ω in
(p2/2M + Mω2r2/2)R(r) = ЄR(r)
(A10.38)
is:
Mω2/2 = e2B2/8Mc2,
(A10.39)
and
ω2 = e2B2/8Mc2(2/M) = e2B2/4M2c2,
(A10.40)
and:
ω = (e2B2/4M2c2)1/2 = ±eB/2Mc
(A10.41)
is the relativistic cyclotron frequency, which in this case is the angular frequency
of the velocity vector in the r,Φ plane [24], with the plus and minus referring to the
possibility of rotation in two “directions”, one with Lz, the other with -Lz.
The energy Є of the two dimensional harmonic oscillator is:
Є = hωx(nx + ½) + hωy(ny + ½)
Since ωx = ωy = ω with uniform circular motion:
162
(A10.42)
Є = hω(nx + ny + ½ + ½) = hω(nx + ny + 1) → hω(n+ + n- + 1)
(A10.43)
with the right hand expression a conversion to Schwinger’s two dimensional
rotational quantum numbers, which will make adding the angular momentum a
little easier. Now, returning to the harmonic oscillator equation (A10.35) in detail:
((p2/2M) + e2B2r2/8Mc2)R(r) = ((E – V)2/2Mc2
(A10.44)
- eLzB/2Mc – m2c2/2M)R(r)
Є = hω(n+ + n- + 1) = ((E – V)2/2Mc2 - eLzB/2Mc - m2c2/2M)
(A10.45)
hω(n+ + n- + 1) = ((E – V)2/2Mc2 – (eB/2Mc)Lz - m2c2/2M)
(A10.46)
and since:
ω = eB/2Mc
(A10.47)
then:
hω(n+ + n- + 1) = (E – V)2/2Mc2 – ωLz - m2c2/2M
(A10.48)
and since the angular momentum is also quantized [24]:
ωLz = ωh(n+ – n-)
(A10.49)
then:
hω(n+ + n- + 1) = (E – V)2/2Mc2 – ωh(n+ – n-) - m2c2/2M
hω(n+ + n- + 1) + ωh(n+ – n-) = (E – V)2/2Mc2 - m2c2/2M
hω(n+ + n+ + n- – n- + 1) = (E – V)2/2Mc2 - m2c2/2M
hω(2n+ + 1) = (E – V)2/2Mc2 - m2c2/2M
and with only one quantum number remaining the subscript can be dropped:
(A10.50)
(A10.51)
(A10.52)
(A10.53)
hω(2n + 1) = (E – V)2/2Mc2 - m2c2/2M
(A10.54)
(E – V)2/2Mc2 - m2c2/2M = hω(2n + 1)
(A10.55)
(E – V)2/2Mc2 = hω(2n + 1) + m2c2/2M
(A10.56)
2Mc2[(E – V)2/2Mc2 = hω(2n + 1) + m2c2/2M]
(A10.57)
Solving now for (E – V)2:
163
(E – V)2 = 2Mc2[hω(2n + 1) + m2c2/2M]
(A10.58)
and now solving for E - V:
{(E – V) = 2Mc2[hω(2n + 1) + m2c2/2M]}1/2
(A10.59)
(E – V) = {2Mc2[hω(2n + 1) + m2c2/2M]}1/2
(A10.60)
This is a useful result but it can be used only with the realization that the
quantization in this analogy to the 2 dimensional quantum harmonic oscillator is
proportional to the energy E. In the relativistic case, as will be shown, the
quantization, which also includes the term (2n + 1), is proportional to E2.
The general form of a harmonic oscillator wave function is [23]:
R(r) = CnHn(r)e-br●r
(A10.61)
where Cn is a constant partially dependent on the energy level of the oscillator,
e-br●r is the familiar Gaussian exponential and Hn(r) is a hermite polynomial,
also dependent on the energy level (n).
What is b? This portion of the exponent takes on the form [23]:
e-br●r→ e-br^2 = e-r^2(1/2)(mω/h)
(A10.62)
-br2 = (-r2)(1/2)(Mω/h)
(A10.63)
Dealing only with the exponents:
Dividing by –r2, which can readily be done since:
r=0
(A10.64)
isn’t part of this physical problem,
b = (1/2)(Mω/h)
(A10.65)
b = Mω/2h
(A10.66)
ω = eB/2Mc
164
(A10.66)
Since:
b = (M/2h)(eB/2Mc)
(A10.67)
b = eB/4hc
(A10.68)
Note here that b is derived without any reference to energy quantization.
Indeed, b can be derived solely from the Klein-Gordon equation once one
knows that the radial function is a Gaussian.
The proposed radial wave function is:
R(r) = e-br●r
(A10.69)
If this is a solution of the harmonic oscillator form of the Klein-Gordon Equation,
what is b? Going back to the radial wave function:
(p2/2M + e2B2r2/8Mc2)R(r) = (E2/2Mc2 + eLzB/2Mc – m2c2/2M)R(r) (A10.70)
Converting p2 back to operator form:
[(h/i)∇]2 = -h2∇2
(A10.71)
yields a Klein-Gordon equation:
(-h2/2M∇2 + e2B2r2/8Mc2)R(r) = (E2/2Mc2
(A10.72)
- eLzB/2Mc - m2c2/2M)R(r))
Multiplying by 2M yields:
(-h2∇2 + e2B2r2/4c2)R(r) = (E2/c2 - eLzB/c - m2c2)R(r)
(A10.73)
-h2∇2R(r) + e2B2r2/4c2R(r) = (E2/c2 - eLzB/c - m2c2)R(r)
(A10.74)
-h2∇2e-br●r + e2B2r2/4c2R(r) = (E2/c2 - eLzB/c - m2c2)R(r)
(A10.75)
Using the radial equation on R(r):
∇2Ψ = (1/r)∂/∂r[r∂Ψ/∂r] + (1/r2)∂2Ψ/∂Φ2 + ∂2Ψ/∂z2
(A10.76)
∇2Ψ = (1/r)∂/∂r[r∂Ψ/∂r]
(A10.77)
reduces to:
Working out the derivatives:
165
∂Ψ/∂r = ∂e-br●r/∂r = -2bre-br●r
(A10.78)
(1/r)∂Ψ/∂r = (-2br/r)e-br●r = -2be-br●r
(A10.79)
∂2Ψ/∂r2 = ∂(-2bre-br●r)/∂r = -2b∂(re-br●r)/∂r
(A10.80)
∂2Ψ/∂r2 = -2b(r∂e-br●r/∂r + e-br●r∂r/∂r)
∂2Ψ/∂r2 = -2b(r(-2br)e-br●r + e-br●r) = 4b2r2e-br●r – 2be-br●r
(A10.81)
Putting all these derivatives together builds the LaPlacian:
∇2Ψ = 4b2r2e-br●r – 2be-br●r - 2be-br●r
(A10.82)
∇2Ψ = 4b2r2e-br●r – 4be-br●r
(A10.83)
-h2∇2Ψ = -h2(4b2r2e-br●r – 4be-br●r)
(A10.84)
-h2∇2Ψ = -4h2b2r2e-br●r + 4h2be-br●r = (-4h2b2r2 + 4h2b)Ψ
(A10.85)
Substituting this into the Klein-Gordon equation (plus the other non-LaPlacian
terms):
(-4h2b2r2 + 4h2b)R(r) + e2B2r2/4c2R(r) = (E2/c2 - eLzB/c - m2c2)R(r) (A10.86)
it is possible to match like terms; specifically the r2 term from the LaPlacian must
“match” the other r2 term in the equation and the r0 term from the LaPlacian must
“match” the other r0 terms in the equation. This yields two equations:
1) 4h2b2r2 = e2B2r2/4c2
(A10.87)
2) 4h2b = E2/c2 - eLzB/c - m2c2
(A10.88)
and:
For consistency, it is necessary to divide these equations by 2M.
This yields:
1) 4h2b2r2/2M = e2B2r2/8Mc2
(A10.89)
and:
2) 2h2b/M = E2/2Mc2 - eLzB/2Mc – m2c2/2M
166
(A10.90)
Starting with the first equation (88):
2h2b2/M = e2B2/8Mc2
(A10.91)
Solving for b:
b2 = (e2B2/8Mc2)/(2h2/M) = e2B2/16h2c2
(b2 = e2B2/16h2c2)1/2
(b2)1/2 = (e2B2/16h2c2)1/2 = ±eB/4hc
b = ±eB/4hc
(A10.92)
(A10.93)
(A10.94)
(A10.95)
The negative root leads to an exponential that is not normalizable so:
b = eB/4hc
167
(A10.96)
APPENDIX 11
DIRECT SOLUTION OF THE RADIAL KLEIN-GORDON EQUATION WITH THE
SYMMETRIC GAUGE
What happens for a solution of the Klein-Gordon equation with scalar and
vector potentials if the symmetric gauge is used but the short cut of the analogy
to the two dimensional quantum harmonic oscillator is NOT taken? Instead, the
Klein-Gordon equation must be solved directly?
The Klein-Gordon equation for scalar and vector potentials is:
(1/c2)(ih∂/∂t – V)2ψ = {[(h/i)∇ - eA/c]2 + m2c2}ψ
(A11.1)
If the time component of the wave function is so:
ψ(t) = e-iEt/h
(A11.2)
then the left hand side of the equation develops as usual as the time derivatives
are applied to the wave function, with the result:
(1/c2)(E – V)2ψ = {[(h/i)∇ - eA/c]2 + m2c2}ψ
(A11.3)
Now expanding the square on the right hand side:
(1/c2)(E – V)2ψ = {-h2∇2 – (eh/ic)A•∇ - (eh/ic)∇•A + e2A2/c2 + m2c2}ψ (A11.4)
Using the symmetric gauge here:
A = (B x r)/2
(A11.5)
which commutes with the momentum, so:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)∇•A + e2A2/c2 + m2c2}ψ
(A11.6)
To evaluate the dot product properly, the wave function must be brought in:
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)∇•(Aψ) + e2A2ψ/c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)A•∇ψ
- (2eh/ic)ψ∇
∇•A + e2A2ψ/c2 + m2c2ψ
168
(A11.7)
(A11.8)
The vector potential in the symmetric gauge is also non-divergent, so:
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)A•∇ψ - 0 + e2A2ψ/c2 + m2c2ψ
(A11.9)
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)A•∇ψ + e2A2ψ/c2 + m2c2ψ
(A11.10)
At this point taking the wave function on the right hand side of the
equation back off to the right yields:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)A•∇ + e2A2/c2 + m2c2}ψ
(A11.11)
Applying the symmetric gauge, labeled with the relevant unit vectors:
-φA = (-zB x r)/2
(A11.12)
Keeping track of the signs of the unit vectors is a little challenging but necessary.
This results in:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
∇
(A11.13)
+ e2[(-zB x r)/2]2/c2 + m2c2}ψ
The term with the symmetric gauge squared loses its vector nature
because of the squaring of the cross product of the unit vectors and it can be
reduced to a magnitude only form:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
∇
(A11.14)
+ e2B2r2/4c2 + m2c2}ψ
Continuing:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
∇
(A11.15)
+ e2B2r2/4c2 + m2c2}ψ
Suppose that in the space variables the wave function has the form [1]:
ψ = R(r)S(φ) = R(r)eim(L)φ
where mL is the traditional angular momentum quantum number.
169
(A11.16)
Continuing:
∇ψ
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[(-zB x r)/2]•∇
(A11.17)
+ e2B2r2ψ/4c2 + m2c2ψ
∇R(r)eim(L)φ
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[(-zB x r)/2]•∇
(A11.18)
+ e2B2r2ψ/4c2 + m2c2ψ
∇R(r)eim(L)φ
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[(-zB x r)/2]•∇
(A11.19)
+ e2B2r2ψ/4c2 + m2c2ψ
Before actually operating with del, taking care of the unit vector cross product:
-φ = -z x r
(A11.20)
so now:
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[-φ(B r)/2]•∇
∇R(r)eim(L)φ
(A11.21)
+ e2B2r2ψ/4c2 + m2c2ψ
The del operator in cylindrical coordinates is, minus the z component:
∇ = r∂/∂r + φ(1/r)∂/∂φ
(A11.22)
so:
(1/c2)(E – V)2ψ = -h2∇2ψ – R(r)(2eh/ic)[Br/2](-φ•φ
φ)(1/r)∂/∂φeim(L)φ (A11.23)
– eim(L)φ(2eh/ic)[Br/2](-φ•r)∂/∂rR(r) + e2B2r2ψ/4c2 + m2c2ψ
Since:
φ•r = 0
(A11.24)
the Klein-Gordon equation is now:
(1/c2)(E – V)2ψ = -h2∇2ψ – R(r)(2eh/ic)[Br/2](-φ•φ
φ)(1/r)∂/∂φeim(L)φ (A11.25)
+ e2B2r2ψ/4c2 + m2c2ψ
The other unit vector dot product is:
φ = -1
-φ•φ
170
(A11.26)
so:
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)(eh/ic)Br(1/r)∂/∂φeim(L)φ
(A11.27)
+ e2B2r2ψ/4c2 + m2c2ψ
and again this term turns positive. Taking the remaining derivative:
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)(eh/ic)B∂[eim(L)φ]/∂φ
(A11.28)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)S(φ)(eh/ic)BimL
(A11.29)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)S(φ)(ehBmL/c)
(A11.30)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ + ψ(ehBmL/c)
(A11.31)
+ e2B2r2ψ/4c2 + m2c2ψ
The wave function now commutes with the angular momentum term, so:
(1/c2)(E – V)2ψ = -h2∇2ψ + (ehBmL/c)ψ
(A11.32)
+ e2B2r2ψ/4c2 + m2c2ψ
{(1/c2)(E – V)2 = -h2∇2 + ehBmL/c + e2B2r2/4c2 + m2c2}ψ
(A11.33)
The LaPlacian is the polar coordinates form with radial and angular
sections is:
∇2 = (1/r)∂/∂r(r∂/∂r) + (1/r2)∂2/∂φ2
(A11.34)
∇2ψ = (1/r)∂/∂r(r∂ψ/∂r) + (1/r2)∂2ψ/∂φ2
(A11.35)
∇2ψ = (1/r)∂/∂r(r∂[R(r)S(φ)]/∂r) + (1/r2)∂2[R(r)S(φ)]/∂φ2
(A11.36)
∇2ψ = (S(φ)/r)∂/∂r(r∂[R(r)]/∂r) + (R(r)/r2)∂2[S(φ)]/∂φ2
(A11.37)
Applying the wave function:
171
∇2ψ = (S(φ)/r)∂/∂r(r∂[R(r)]/∂r) + (R(r)/r2)∂2[eim(L)φ]/∂φ2
(A11.38)
Applying the angular derivatives:
∇2ψ = (S(φ)/r)∂/∂r(r∂[R(r)]/∂r) + (R(r)/r2)(imL)2ψ(φ)
(A11.39)
∇2ψ = (S(φ)/r)∂/∂r(r∂[R(r)]/∂r) - (R(r)/r2)(mL2)ψ(φ)
(A11.40)
∇2ψ = (S(φ)/r)∂/∂r(r∂[R(r)]/∂r) - (mL2/r2)R(r)ψ(φ)
(A11.41)
It is not possible now to assume a Gaussian form for R(r) since this time
the Klein-Gordon equation is not in a recognizable harmonic oscillator form.
Nonetheless, it is possible to simplify the radial component somewhat:
∇2ψ = (S(φ)/r)[(∂r/∂r)(∂R(r)/∂r) + (r∂/∂r)∂R(r)/∂r] - (mL2/r2)R(r)S(φ) A11.42)
∇2ψ = (S(φ)/r)[(∂r/∂r)(∂R(r)/∂r) + r∂2R(r)/∂r2] - (mL2/r2)R(r)S(φ) (A11.43)
∇2ψ = (S(φ)/r)[∂R(r)/∂r + r∂2R(r)/∂r2] - (mL2/r2)R(r)S(φ)
(A11.44)
∇2ψ = S(φ)[∂2R(r)/∂r2 + (1/r)∂R(r)/∂r] - (mL2/r2)R(r)S(φ)
(A11.45)
∇2ψ = [∂2R(r)/∂r2 + (1/r)∂R(r)/∂r]S(φ) - (mL2/r2)R(r)S(φ)
(A11.46)
The results for the LaPlacian are now ready to go back into the KleinGordon equation:
{(1/c2)(E – V)2 = -h2∇2 + ehBmL/c + e2B2r2/4c2 + m2c2}ψ
(A11.47)
The Klein-Gordon equation then becomes:
(1/c2)(E – V)2ψ = -h2[[∂2R(r)/∂r2 + (1/r)∂R(r)/∂r]S(φ)
(A11.48)
- (mL2/r2)R(r)S(φ)] + ehBmψ/c + e2B2r2ψ/4c2 + m2c2ψ
Now dividing out the angular part of the wave function and the equation
then becomes a radial equation with total derivatives only:
(1/c2)(E – V)2R(r) = -h2[[d2R(r)/dr2 + (1/r)dR(r)/dr]
- (mL2/r2)R(r)] + ehBmLR(r)/c + e2B2r2R(r)/4c2 + m2c2R(r)
Multiplying by –(1/h2):
172
(A11.49)
(-1/h2c2)(E – V)2R(r) = [[d2R(r)/dr2 + (1/r)dR(r)/dr]
(A11.50)
- (mL2/r2)R(r)] - eBmLR(r)/hc - e2B2r2R(r)/4h2c2 – (m2c4/h2c2)R(r)
and re-arranging the terms:
d2R(r)/dr2 + (1/r)dR(r)/dr - (mL2/r2)R(r) - eBmLR(r)/hc
(A11.51)
- e2B2r2R(r)/4h2c2 – (m2c4/h2c2)R(r) + (1/h2c2)(E – V)2R(r) = 0
The structure of this radial equation will become a little clearer with some
simplification:
R(r)” + (1/r)R(r)’ – (D/r2)R(r) - (Fr2)R(r) + GR(r) = 0
(A11.52)
where:
R(r)” = d2R(r)/dr2
(A11.53)
R(r)’ = dR(r)/dr
(A11.54)
D = mL2
(A11.55)
F = e2B2/4h2c2
(A11.56)
G = (1/h2c2)(E – V)2 - eBmL/hc - m2c4/h2c2
(A11.57)
and:
With so many functions of r present; 1/r, r2 and 1/r2, it’s not at all clear
what the nature of the radial wave function is going to be. The extended power
series method or a numerical solution beckon but the Maple 9 software was
faster. Here’s the result [18]:
(A11.57)
ode2 := {diff(R(r),r,r) + (1/r)*diff(R(r),r) –
(D/r^2)*R(r) - (F*r^2)*R(r) + G*R(r) = 0};
(A11.58)
d


R( r )
 2

  d

d
r
D
R
(
r
)

2

ode2 :=   2 R( r )  +
−
− F r R( r ) + G R( r ) = 0 
2
 dr


r
r



173
dsolve(ode2,{R(r)});
(A11.59)



 R( r ) =


D
D
 G

 G

_C1 WhittakerM 
,
, F r 2  _C2 WhittakerW 
,
, F r 2  
 
4 F 2
+
4 F 2


r
r

The Whittaker functions have turned up before. These yield Gaussian
exponentials multiplied by hypergeometric and Kummer series which in turn
appear to yield Gaussian exponentials multiplied by something that looks
like somewhat like Hermite polynomials.
All roads lead to Rome it seems and the Gaussian will apparently not be
denied.
What is different here compared with the radial result from the analogy
with the two dimensional quantum radial wave function is that the polynomials in
this direct solution are more complex (mainly in the form of
hypergeometric and KummerU functions, which are series of polynomials)
(A11.60)
R(r) = C1F1/2rD^1/2e-(F^1/2r^2)/2hypergeometric[1/2 + D1/2 – G/(4F1/2), 1 + 2D1/2, F1/2r2]
+ C2F1/2rD^1/2e-(F^1/2r^2)/2KummerU[1/2 + D1/2 – G/(4F1/2), 1 + 2D1/2, F1/2r2]
although the Gaussian exponential is exactly the same.
From equation (A10.112):
b = eB/4hc
(A11.61)
F = e2B2/4h2c2
(A11.62)
From equation (A11.56):
and (A11.60):
174
e-(F^1/2r^2)/2 = e-(eB/4hc)r^2 = e-br^2
(A11.63)
b = eB/4hc
(A11.64)
where:
So the exact solution is returning the exponential of the 2
dimensional quantum harmonic oscillator. This is what it looks like:
Figure A11.1. Gaussian exponential on Positive Radii
What about the hypergeometric and KummerU polynomials? First of all,
some analysis about how they behave in general.
The hypergeometric function is a polynomial series. With very simple
parameters [18]:
series(hypergeom([1],[1],z),z);
1+z+
1 2 1 3 1 4
1 5
z + z +
z +
z + O( z6 )
2
6
24
120
175
(A11.65)
(A11.66)
For small r, this series converges to er. For:
z = -r2
(A11.67)
series = e-r^2
(A11.68)
the series is a Gaussian:
To see the effect more complicated parameters, here’s a variation in the
first parameter of the hypergeometric [18]:
series(hypergeom([1-a],[1],z),z);
(A11.69)
(1 − a) (2 − a) 2 (1 − a) (2 − a) (3 − a) 3
z +
z +
1 + (1 − a) z +
4
36
(1 − a) (2 − a) (3 − a) (4 − a) 4 (1 − a) (2 − a) (3 − a) (4 − a) (5 − a) 5
z +
z +
576
14400
O( z 6 )
The first parameter controls series of multiplicative factors in the
numerators that is similar to a factorial. The second parameter does the same
thing in the denominator [18]:
series(hypergeom([1-a],[1-b],z),z);
(A11.70)
1−a
(1 − a) (2 − a) 2
(1 − a) (2 − a) (3 − a) 3
1+
z+
z +
z +
1−b
2 (1 − b) (2 − b)
6 (1 − b) (2 − b) (3 − b)
(1 − a) (2 − a) (3 − a) (4 − a) 4
(1 − a) (2 − a) (3 − a) (4 − a) (5 − a)
z +
24 ( 1 − b ) ( 2 − b ) ( 3 − b ) ( 4 − b )
120 ( 1 − b ) ( 2 − b ) ( 3 − b ) ( 4 − b ) ( 5 − b )
z 5 + O( z 6 )
What are the parameters in the case of cyclotron motion? Going back to
equations:
D = mL2
(A11.71)
F = e2B2/4h2c2
(A11.72)
G = (1/h2c2)(E – V)2 – eBmL/hc – m2c4/h2c2
(A11.73)
and:
These can mostly be expressed in terms of the Gaussian exponential coefficient
176
b in e-br^2 and the angular momentum quantum number mL:
D = mL2
(A11.74)
F = 4b2
(A11.75)
D1/2 = ±mL
(A11.76)
F1/2 = ±2b
(A11.77)
G = (1/h2c2)(E – V)2 – eBmL/hc – m2c4/h2c2
(A11.78)
with:
and:
For:
a little more surgery is helpful. The conditions of most interest in this thesis are
those of Klein’s paradox where the scalar potential term is of sufficient magnitude
to cause particle pair formation. This threshold occurs once V is strong enough
that:
(1/h2c2)(E – V)2 = eBmL/hc + m2c4/h2c2
(A11.79)
The left hand side is the “classic” Klein’s paradox term and when V is
large enough to offset the rest energy term
(1/h2c2)(E – V)2 = m2c4/h2c2
(A11.80)
The effect of the vector potential is to add the term eBmL/hc to raise this
threshold.
The establishment of:
(1/h2c2)(E – V)2 = eBmL/hc + m2c4/h2c2
(A11.81)
has the simplifying effect of, at the threshold of Klein’s Paradox, setting:
G=0
which now accompanies:
177
(A11.82)
D = mL2
(A11.83)
F = 4b2
(A11.84)
The general situation is made clearer by the magnitude of b, for a
magnetic field of 10-2 Tesla, about 100 times the strength of the magnetic field of
the earth:
b = eB/4hc
(A11.85)
= (1.602x10-19C)(10-2 Tesla)/[(4)(1.055x10-34Jsec)(2.998x108m/sec)]
b = 1.266x104
(A11.86)
The factor z in the hypergeometric series [18]:
series(hypergeom([1],[1],z),z);
1+z+
(A11.87)
1 2 1 3 1 4
1 5
z + z +
z +
z + O( z6 )
2
6
24
120
is:
z = br2
(A11.88)
For the series to converge z has to be within roughly an order of
magnitude of 1.
For:
r=1
(A11.89)
this requires a magnetic field B of a reciprocal value to e/4hc:
B = 4hc/e = [(4)(1.055x10-34)(2.998x108)]/(1.602x10-19) = 7.897 x 10-7 (A11.90)
or a magnetic field about 100 times weaker than the Earth’s magnetic field [26]
at most.
If the field is not weak, another way the series can converge is for r to be
rather small. In the case of:
178
B = 10-2
(A11.91)
about 100 times the strength of the Earth’s magnetic field, r must be on the order
of:
r ≈ 10-3
(A11.92)
for the series to converge.
The third way for the hypergeometric series to converge is for b and r to
“share the pain”: both variables reduced in magnitude.
In any event, it’s clear that the hypergeometric function converges for relatively
weak magnetic fields, small r or a combination of both. The series will diverge
strongly even for relatively moderate B and r.
The hypergeometric series can be expressed rather conveniently in terms
of the angular momentum quantum number mL, which, due to limitations of font
will appear in the series parameters as m, and b:
series = hypergeom([(1/2)*(1+m+m/b)],[1+m],br2)
Except for extremely high angular momentum quantum numbers, b >> mL so for
situations not involving extremely weak magnetic fields, the m/b term is much
smaller than m or 1. This means that the following series:
series = hypergeom([(1/2)*(1+m)],[1+m],br2)
will be a very good approximation.
To go further, it is possible to pick an angular momentum quantum
number of:
mL = 1
A(11.93)
which makes:
series = hypergeom([(1/2)*(2)],[1+1],br2)
179
(A11.94)
or:
series = hypergeom([1],[2],br2)
(A11.95)
or, in formal Maple syntax [18]:
series(hypergeom([1],[2],z),z);
1+
1
1
1 3
1 4
1 5
z + z2 +
z +
z +
z + O( z6 )
2
6
24
120
720
How does this interact with the Gaussian exponential function?
Here is its plot:
180
(A11.96)
Figure A11.2. Leading Factors of the Hypergeometric Function Compared to the
Gaussian Exponential
The Gaussian function converges but all of the hypergeometric terms for
this strength B are strongly divergent at radii well below 1. An interesting detail is
that succeeding terms are convergent or nearly so for somewhat higher r but
diverge more rapidly once the divergence really starts.
181
(All.95)
exp(1.266*10^4*r^2)*(1+1.266*10^4*r^2+(1/2)*(1.266*10^4*r^2)^2+
(1/6)*(1.266*10^4*r^2)^3+(1/24)*(1.266*10^4*r^2)^4+(1/120)*
(1.266*10^4*r^2)^5);
e
2
( −12660.000 r )
( 1 + 12660.000 r 2 + 0.8013780000 10 8 r 4 + 0.3381815160 10 12 r 6
+ 0.1070344498 10 16 r 8 + 0.2710112269 10 19 r 10 )
Figure A11.3. The WhittakerM Function for its Leading Terms.
The KummerU series can get quite complicated [18]:
182
series(KummerU(1-a,2,z),z);
(A11.96)
1
ln( z ) − 1 + Ψ ( 1 − a ) + 2 γ
z -1 +
+
Γ( 1 − a )
Γ( −a )
1
5
1 a
( 1 − a )  Ψ ( 2 − a ) − + 2 γ  +  −  ln( z )
2
2

 2 2 
z+
Γ( −a )
1 1 a 
1
10
1
pochhammer ( 1 − a, 2 )  Ψ ( 3 − a ) −
+ 2 γ 
pochhammer
 −  ( 2 − a ) ln( z ) +
6 2 2 
12
3

 z 2 + 144
Γ( −a )
1
131
z 3 + 
pochhammer ( 1 − a, 4 )  Ψ ( 5 − a ) −
+ 2 γ 
30
 2880


1 1 a 
4


+
 −  ( 2 − a ) ( 3 − a ) ( 4 − a ) ln( z ) /Γ( −a ) z + 
1440  2 2 


1
71
pochhammer ( 1 − a, 5 )  Ψ ( 6 − a ) −
+ 2 γ 
86400
15


1 1 a 
5
6

+
 −  ( 2 − a ) ( 3 − a ) ( 4 − a ) ( 5 − a ) ln( z ) /Γ( −a ) z + O( z )
43200  2 2 

…and slightly harder to interprete than the hypergeometric, with the gamma,
pochammer and psi subfunctions. With simpler parameters the essential nature
of the series becomes easier to see [18]:
series(KummerU(1,9,z),z);
(A11.97)
5040 z-8 + 5040 z-7 + 2520 z -6 + 840 z-5 + 210 z -4 + 42 z-3 + O( z-2 )
This series starts with negative powers of z. This part of the series will
converge for larger b and r (stronger magnetic field and larger radii). This is the
opposite in behavior as with the hypergeometric series. In that sense, these
functions compensate for each other. However, this symmetry is not complete.
The higher order terms of the KummerU function have positive exponents of
increasing degree. If carried far enough, then, the KummerU series will suffer
from the same divergence unless it is truncated.
In terms of Klein’s paradox, at the threshold of particle pair production,
(1/h2c2)(E – V)2 = eBmL/hc + m2c4/h2c2
183
(A11.98)
where [18]:
G=0
(A11.99)
series(KummerU(1,2,z),z);
z -1
(A11.100)
For an angular quantum number of:
mL = 1
(A11.101)
KummerU = (1/2br2)
(A11.102)
mL = 3
(A11.103)
For an angular quantum number of:
series(KummerU(2,4,z),z);series(1*z^(-1),z);
-3
2 z + z -2
(A11.104)
z -1
As the angular momentum quantum number rises so does the number of terms
in the series. Separately plotted, these three terms:
184
Figure A11.4. The Leading Terms of the KummerU Series.
185
Figure A11.5. The WhittakerW Function for its Leading Terms
The terms of the KummerU series are strongly divergent for low r [18]
because the terms approach a vertical asymptote at r = 0 and are the dominant
influence in the WhittakerW series over that short range but the Gaussian
exponential is dominant everywhere else in r.
186
APPENDIX 12
TESTING THE HARMONIC OSCILLATOR WAVE FUNCTION IN THE
CHARGE DENSITY EQUATION
A straightforward way to calculate the charge density due to
the passage of a spinless particle is to use the charge density equation
connected with the Klein-Gordon equation.
The charge density equation is:
eρ(r,t) = (1/2mc2)(Ψ*(ih∂/∂t)Ψ + Ψ(-ih∂/∂t)Ψ*)
(A12.1)
This equation is made using the assumption that the wave
function has a distinct complex conjugate. The ground state two dimensional
quantum harmonic oscillator wave function is [5]:
Ψ = ae-br^2 – iEt/heim(L)φ
(A12.3)
This function does have a distinct complex conjugate but only in the time
and angular components. The radial component of the wave function is real and
is its own complex conjugate.
For the test the wave function used will be ground state wave function ;
Ψ = aeim(L)φe-br^2 – Et/h
(A12.4)
Hn = 1
(A12.5)
where:
and the angular component, which, since there are no. This will be sufficient to
see if a wave function with one real component can be subjected to the charge
density and current equations. If a function with a simple real component fails
and the failure is because of that real component, more complicated real
components of the wave function are quite unlikely to succeed.
187
Here is the test that can be administered: apply the
charge density and current equations to Ψ by itself in non-relativistic conditions.
For a positively charged particle the result should be equal to 1.
For that, it is necessary to first normalize the function.
The normalization formula is:
1 = <Ψ|Ψ> = ∫VΨΨdV
(A12.6)
V = volume
(A12.7)
Ψ = aeim(L)φe-br^2 – iEt/h
(A12.8)
Ψ* = aeim(L)φe-br^2 + iEt/h
(A12.9)
dV = rdrdΦdz
(A12.10)
∫∞0 ∫2π0 ∫∞0 raΨaΨ*drdΦdz
(A12.11)
where:
For the normalization:
and:
In cylindrical coordinates:
and the normalization integral is:
The integral in z diverges so an arbitrary upper limit must be set:
∫∞0 ∫2π0 ∫10 raΨaΨ*drdΦdz
(A12.12)
1 = ∫∞0 ∫2π0 ∫10 raΨaΨ*drdΦdz
(A12.13)
1 = a2∫∞0 ∫2π0 ∫10 rΨΨ*drdΦdz
(A12.14)
1 = a2∫∞0 ∫2π0 ∫10re-br^2 – iEt/he-br^2 + iEt/hdrdΦdz
(A12.15)
1 = a2∫∞0 ∫2π0 ∫10re-br^2 – iEt/h – br^2 + iEt/hdrdΦdz
(A12.16)
The normalization equation is:
1 = a2∫∞0 ∫2π0 ∫10re-br^2 – br^2 + iEt/h – iEt/hdrdΦdz
188
(A12.17)
1 = a2∫∞0 ∫2π0 ∫10re-2br^2drdΦdz
(A12.18)
1 = a2∫∞0 ∫2π0re-2br^2 (z|10) drdΦ
(A12.19)
1 = a2∫∞0 ∫2π0re-2br^2drdΦ
(A12.20)
1 = a2∫∞0re-2br^2 (Φ|2π0) dr
(A12.21)
1 = 2πa2∫∞0re-2br^2dr
(A12.22)
Integrating through z:
Integrating through Φ:
The remaining integral in r is a Gaussian. Integrating through r:
1 = 2πa2(1/2b)
(A12.23)
1 = πa2/b
(A12.24)
a2 = b/π
(A12.25)
(a2 = b/π)1/2
(A12.26)
a = (b/π)1/2
(A12.27)
Solving for a:
The complete wave function is:
Ψ = (b/π)1/2 eim(L)φe-br^2 – iEt/h
(A12.28)
If the charge density equation is applied to this wave function under nonrelativistic conditions and works correctly the resulting change density should be
equal to 1 since the assumed charge of one particle is 1. If the result is some
value other than one then the charge density equation cannot be applied to this
wave function.
The charge density equation is [2]:
eρ(r,t) = (e/2mc2)(Ψ*(ih∂/∂t)Ψ + Ψ(-ih∂/∂t)Ψ*)
Inserting the wave function:
189
(A12.29)
eρ(r,t) = (e/2mc2)(ae-im(L)φe-br^2 + iEt/h(ih∂/∂t)aeim(L)φe-br^2 – iEt/h
(A12.30)
+ aeim(L)φe-br^2 – iEt/h(-ih∂/∂t)ae-im(L)φe-br^2 + iEt/h)
The angular part of the wave function, eim(L)φ, is not involved in any of the
derivatives and with the complex conjugation cancels itself out of the equation.
The radial portion, since it is real, does not disappear and this turns out to be
the source of the problem, as we’ll see. Continuing and for the moment retaining
the normalization constant in the form of a for convenience in notation:
eρ(r,t) = (e/2mc2)(a2e-br^2 + iEt/h(ih∂/∂t)e-br^2 – iEt/h
(A12.31)
+ a2e-br^2 – iEt/h(-ih∂/∂t)e-br^2 + iEt/h)
eρ(r,t) = (ea2/2mc2)(e-br^2 + iEt/h(ih∂/∂t)e-br^2 – iEt/h
(A12.32)
+ e-br^2 – iEt/h(-ih∂/∂t)e-br^2 + iEt/h)
Now using the specific normalization constant:
eρ(r,t) = (eb/2πmc2)(e-br^2 + iEt/h(ih∂/∂t)e-br^2 – iEt/h
(A12.33)
+ e-br^2 – iEt/h(-ih∂/∂t)e-br^2 + iEt/h)
Proceeding now with the derivatives:
eρ(r,t) = (eb/2πmc2)(e-br^2 + iEt/h(ih∂[e-br^2 – iEt/h]/∂t)
(A12.34)
+ e-br^2 – iEt/h(-ih∂[e-br^2 + iEt/h]/∂t))
eρ(r,t) = (eb/2πmc2)(e-br^2 + iEt/h(ih)(-iE/h)e-br^2 – iEt/h
(A12.35)
+ e-br^2 – iEt/h(-ih)(iE/h)e-br^2 + iEt/h)
eρ(r,t) = (eb/2πmc2)(e-br^2 + iEt/h(E)e-br^2 – iEt/h + e-br^2 – iEt/h(E)e-br^2 + iEt/h) (A12.36)
eρ(r,t) = (ebE/2πmc2)(e-br^2 + iEt/he-br^2 – iEt/h + e-br^2 – iEt/he-br^2 + iEt/h)
(A12.37)
eρ(r,t) = (bE/2πmc2)(e-2br^2 + e-2br^2)
(A12.38)
eρ(r,t) = (ebE/2πmc2)(2e-2br^2) = (2bE/2πmc2)(e-2br^2)
(A12.39)
= (ebE/πmc2)(e-2br^2)
190
Under non-relativistic conditions:
E ≈ mc2
(A12.40)
so:
eρ(r,t) = (ebmc2/πmc2)(e-2br^2) = (b/π)(mc2/mc2)(e-2br^2)
(A12.41)
= (eb/π)(e-2br^2)
The charge density will be 1 in this case only if:
(eb/π)(e-2br^2) = 1
(A12.42)
This can happen in particular circumstances, such as if b = π and r = 0.
Aside from the fact that this value for r destroys the physical problem, the general
case is much more problematical. It is true only when (b/π) is the reciprocal of
(e-2br^2), a difficulty that relativistic conditions will not ease.
While it is possible to manipulate the value of r for any given b to make
equation (A12.41) true, this is not a satisfactory result. A charge density of 1
should be the general not the particular result.
191
APPENDIX 13
THE VECTOR POTENTIAL, KLEIN’S PARADOX AND CYCLOTRON MOTION
There are two ways to do this. Both involve some simplification of the
Klein-Gordon equation and, since cyclotron motion is occurring, require the
symmetric gauge. However the results in both cases do not require full
knowledge of the details of the structure of the wave function and the two cases
turn out to be equivalent.
No Direct Interaction of the Momentum Operator With the Wave Function:
The first method involves using vector mechanics with the interaction of
the symmetric gauge and the p (but not p2) operator to insert angular momentum
into the equation without allowing either p or p2 to operate directly on the wave
function.
The Klein-Gordon equation for scalar and vector potentials is:
(1/c2)(ih∂/∂t – V)2ψ = {[(h/i)∇ - eA/c]2 + m2c2}ψ
(A13.1)
If the time component of the wave function is so:
ψ(t) = e-iEt/h
(A13.2)
then the left hand side of the equation develops as usual as the time derivatives
are applied to the wave function, with the result:
(1/c2)(E – V)2ψ = {[(h/i)∇ - eA/c]2 + m2c2}ψ
(A13.3)
Now expanding the square on the right hand side:
(1/c2)(E – V)2ψ = {-h2∇2 – (eh/ic)A•∇
(A13.4)
- (eh/ic)∇•A + e2A2/c2 + m2c2}ψ
Using the symmetric gauge here:
A = (B x r)/2
192
(A13.5)
which commutes with the momentum, so:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)∇•A + e2A2/c2 + m2c2}ψ
(A13.6)
To evaluate the dot product properly, the wave function must be brought in:
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)∇•(Aψ) + e2A2ψ/c2 + m2c2ψ
(A13.7)
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)A•∇ψ
(A13.8)
- (2eh/ic)ψ∇
∇•A + e2A2ψ/c2 + m2c2ψ
The vector potential in the symmetric gauge is also non-divergent, so:
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)A•∇ψ - 0 + e2A2ψ/c2 + m2c2ψ (A13.9)
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)A•∇ψ + e2A2ψ/c2 + m2c2ψ
(A13.10)
At this point taking the wave function on the right hand side of the equation
back off to the right yields:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)A•∇ + e2A2/c2 + m2c2}ψ
(A13.11)
Applying the symmetric gauge, labeled with the relevant unit vectors:
-φA = (-zB x r)/2
(A13.12)
Keeping track of the signs of the unit vectors is a little challenging but necessary.
This results in:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
∇
(A13.13)
+ e2[(-zB x r)/2]2/c2 + m2c2}ψ
The term with the symmetric gauge squared loses its vector nature because of
the squaring of the cross product of the unit vectors; reducing it to a
magnitude only form:
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
∇
+ e2B2r2/4c2 + m2c2}ψ
193
(A13.14)
Continuing:
∇
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
(A13.15)
+ e2B2r2/4c2 + m2c2}ψ
Here, converting p from operator form:
(1/c2)(E – V)2ψ = {p2 – (2e/c)[(-zB x r)/2]•p
(A13.16)
+ e2B2r2/4c2 + m2c2}ψ
(1/c2)(E – V)2ψ = {p2 – (e/c)[(-zB x r)]•p
(A13.17)
+ e2B2r2/4c2 + m2c2}ψ
(1/c2)(E – V)2ψ = {p2 – (e/c)(-zB x r•p) + e2B2r2/4c2 + m2c2}ψ (A13.18)
Using a vector identity on the mixed product:
(1/c2)(E – V)2ψ = {p2 – (e/c)(-zB•r x p) + e2B2r2/4c2 + m2c2}ψ (A13.19)
(1/c2)(E – V)2ψ = {p2 – (e/c)(-zB•[r x p])
(A13.20)
+ e2B2r2/4c2 + m2c2}ψ
Now needed is:
L=rxp
(A13.21)
where L is the angular momentum. The unit vectors with r and p are r and φ and
the result of the unit vector cross product:
rxφ=z
(A13.22)
means that the angular momentum is in the positive z direction. Now:
(1/c2)(E – V)2ψ = {p2 – (e/c)(-zB•Lz) + e2B2r2/4c2 + m2c2}ψ
(A13.23)
Sorting out the magnitudes from the dot product:
(1/c2)(E – V)2ψ = {p2 – (e/c)(-z•z)BLz + e2B2r2/4c2 + m2c2}ψ
with a z subscript on the angular momentum variable for consistency with the
194
(A13.24)
rest of the thesis. Carrying out the dot product:
(1/c2)(E – V)2ψ = {p2 + eBLz/c + e2B2r2/4c2 + m2c2}ψ
(A13.25)
has the very important effect of turning that term positive. This is a direct result of
the fact that the angular momentum is in the opposite direction as the magnetic
field. A refresher of the vector setup is a good idea here:
Figure A13.1. Cyclotron motion for a positively charged particle. B is in the
negative z direction.
Now back to the equation:
(1/c2)(E – V)2ψ = {p2 + eBLz/c + e2B2r2/4c2 + m2c2}ψ
(A13.26)
Another important fact is that this result is completely independent of the
details of the wave function since the derivatives in the momentum operator have
not yet been applied to the wave function.
It’s very interesting that the angular momentum appears here with no
apparent application of the derivatives with respect to φ to the wave function,
195
which is how angular momentum information is usually generated.
This occurs at the same time that the operator p
“disappears”. What has happened is that the angular momentum operator, even
though not here in operator form, has indeed “produced” the angular momentum,
although not exactly in quantum form, via the agency of the symmetric gauge.
The results so far are independent of the exact form of the space variables
of the wave function. To find out why that is interesting, divide out the wave
function:
(1/c2)(E – V)2 = p2 + eBLz/c + e2B2r2/4c2 + m2c2
(A13.27)
and re-arrange the terms by solving for p2:
p2 + eBLz/c + e2B2r2/4c2 + m2c2 - (1/c2)(E – V)2 = 0
(A13.28)
p2 = (1/c2)(E – V)2 - eBLz/c - e2B2r2/4c2 - m2c2
(A13.29)
and:
At this point there is no ambiguity at all about the sign of any of the terms
in the equation. Solving now for the momentum:
p2 = (1/c2)(E – V)2 - eBLz/c - e2B2r2/4c2 - m2c2
(A13.30)
by taking the square root:
{p2 = (1/c2)(E – V)2 - eBLz/c - e2B2r2/4c2 - m2c2}1/2
(A13.31)
p = ±{(1/c2)(E – V)2 - eBLz/c - e2B2r2/4c2 - m2c2}1/2
(A13.32)
Interaction of the ∇ Operator and the Wave Function
This time there will be no use of vector mechanics to insert Lz into the
equation. Instead, the operator (h/i)∇ (but not –h2∇2) will operate directly on the
wave function. Because of the way the vector components interact (the r
component of ∇ drops out since the symmetric gauge has only a φ component), it
196
is possible to do this without exact knowledge of the form of the radial component
of the wave function:
Starting with:
∇
(1/c2)(E – V)2ψ = {-h2∇2 – (2eh/ic)[(-zB x r)/2]•∇
(A13.33)
+ e2B2r2/4c2 + m2c2}ψ
with the wave function has the form:
ψ = R(r)S(φ) = R(r)eim(L)φ
(A13.34)
where mL is the traditional angular momentum quantum number.
Anyway, this time allowing the momentum operator in the remaining
vector term of the equation to operate on the wave function:
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[(-zB x r)/2]•∇
∇ψ
(A13.35)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[(-zB x r)/2]•∇
∇R(r)eim(L)φ
(A13.36)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[(-zB x r)/2]•∇
∇R(r)eim(L)φ
(A13.37)
+ e2B2r2ψ/4c2 + m2c2ψ
Before actually operating with del taking care of the unit vector cross product:
-φ = -z x r
(A13.38)
(1/c2)(E – V)2ψ = -h2∇2ψ – (2eh/ic)[-φ(B r)/2]•∇
∇R(r)eim(L)φ
(A13.39)
so now:
+ e2B2r2ψ/4c2 + m2c2ψ
The del operator in cylindrical coordinates is, minus the z component:
∇ = r∂/∂r + φ(1/r)∂/∂φ
so:
197
(A13.40)
φ)(1/r)∂/∂φeim(L)φ (A13.41)
(1/c2)(E – V)2ψ = -h2∇2ψ – R(r)(2eh/ic)[Br/2](-φ•φ
– eim(L)φ(2eh/ic)[Br/2](-φ•r)∂/∂rR(r) + e2B2r2ψ/4c2 + m2c2ψ
Since:
φ•r = 0
(A13.42)
we now have:
(1/c2)(E – V)2ψ = -h2∇2ψ – R(r)(2eh/ic)[Br/2](-φ•φ
φ)(1/r)∂/∂φeim(L)φ (A13.43)
+ e2B2r2ψ/4c2 + m2c2ψ
The other unit vector dot product:
-φ•φ
φ = -1
(A13.44)
so:
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)(eh/ic)Br(1/r)∂/∂φeim(L)φ
(A13.45)
+ e2B2r2ψ/4c2 + m2c2ψ
and again this term turns positive. Taking the remaining derivative:
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)(eh/ic)B∂[eim(L)φ]/∂φ
(A13.46)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)ψ(φ)(eh/ic)BimL
(A13.47)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ + R(r)S(φ)(ehBmL/c)
(A13.48)
+ e2B2r2ψ/4c2 + m2c2ψ
(1/c2)(E – V)2ψ = -h2∇2ψ + ψ(ehBmL/c)
(A13.49)
+ e2B2r2ψ/4c2 + m2c2ψ
The wave function now commutes with the angular momentum term, so:
(1/c2)(E – V)2ψ = -h2∇2ψ + (2πehBmL/c)ψ
+ e2B2r2ψ/4c2 + m2c2ψ
198
(A13.50)
{(1/c2)(E – V)2 = -h2∇2 + ehBmL/c + e2B2r2/4c2 + m2c2}ψ
(A13.51)
Now converting the form of the momentum operator:
{(1/c2)(E – V)2 = p2 + ehBmL/c + e2B2r2/4c2 + m2c2}ψ
(A13.52)
and dividing out the wave function:
(1/c2)(E – V)2 = p2 + ehBmL/c + e2B2r2/4c2 + m2c2
(A13.53)
Now there is an angular momentum term complete with the angular
momentum quantum number.
There is now the very interesting question about the sign of mL, which
can in general be positive, negative or zero. Here:
(1/c2)(E – V)2 = p2 + ehBmL/c + e2B2r2/4c2 + m2c2
(A13.54)
mL has to be positive since it is clear that the angular momentum is positive from
the vector relationships in figure 1.
Solving for p2:
p2 + 2πehBLz/c + e2B2r2/4c2 + m2c2 - (1/c2)(E – V)2 = 0
(A13.55)
p2 = (1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2
(A13.56)
Taking the square root and solving for p:
{p2 = (1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
(A13.57)
p = ±{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
(A13.58)
Here we find the same suppressive role played by the vector potential.
This time the angular momentum term is the explicit result of operator action on a
φ dependent component of the wave function instead of a direct vector
substitution independent of the exact form of the wave function. However, the
effect of the vector potential on Klein’s paradox appears unchanged, as can be
seen by the juxtaposition of equations:
199
p = ±{(1/c2)(E – V)2 - ehBmL/c - e2B2r2/4c2 - m2c2}1/2
(A13.59)
p = ±{(1/c2)(E – V)2 - eBLz/c - e2B2r2/4c2 - m2c2}1/2
(A13.60)
provided:
Lz = hmL
200
(A13.61)
APPENDIX 14
THE KLEIN-GORDON EQUATION AND THE COMPTON WAVELENGTH
The Compton Wavelength, h/mc, is often mentioned as a limit in particle
Physics beyond which something like a particle cannot be forced without
triggering the formation of particle pairs [9]. Here is its’ relationship with the KleinGordon Equation. Starting with the equation:
(1/c2)(ih∂/∂t – eΘ)2Ψ(x,t) = {[(h/i)∇]2 + m2c2}Ψ(x,t)
(A14.1)
(1/c2)(ih∂/∂t – V)2Ψ(x,t) = {-h2∇2 + m2c2}Ψ(x,t)
(A14.2)
(1/c2)(-h2∂2/∂t2 – 2ihV∂/∂t + V2)Ψ(x,t) = {-h2∇2 + m2c2}Ψ(x,t)
(A14.3)
(1/c2)(-h2∂2[Ψ(x,t)]/∂t2 – 2ihV∂[Ψ(x,t)]/∂t + V2Ψ(x,t))
(A14.4)
(1/c2)(E – V)2Ψ = -h2∇2Ψ + m2c2Ψ
(A14.5)
Dividing out the wave function:
(1/c2)(E – V)2 = -h2∇2 + m2c2
(A14.6)
(1/h2c2)(E – V)2 = -∇2 + m2c2/h2
(A14.7)
Now dividing by h2:
If the Compton Wavelength is [9]:
λC = h/mc
(A14.8)
(1/h2c2)(E – V)2 = -∇2 + λC-2
(A14.9)
then the Klein-Gordon Equation:
…so the Compton Wavelength is an integral part of the Klein-Gordon Equation.
201
APPENDIX 15
THE VELOCITY SELECTOR, THE LANDAU GAUGE, A WAVE FUNCTION
AND KLEIN’S PARADOX
The velocity selector is a set of crossed electric and magnetic fields such
that a charged particle will moving through them at a selected velocity will
maintain non-accelerated motion [26]:
ΣF = ma = 0
(A15.1)
ΣF = qE + v x qB = Felectric + Fmagnetic = 0
(A15.2)
From the Lorenz relation:
If these two forces are to sum to zero they must be directionally opposite each
other [26]:
-qE + v x qB = 0
(A15.3)
In the case of a balance of these forces, v will be at a right angle (θ) to B so:
-qE + vqBsinθ = 0 = -qE + vqB
(A15.4)
qE = vqB
(A15.5)
v = E/B = constant
(A15.6)
and that leads to:
and:
202
Figure A15.1. Vector Relationships for a Positively Charged Particle in the
Velocity Selector. The Magnetic Field is in the Negative z Direction.
To use the Klein-Gordon Equation it is necessary to calculate the source
field, the scalar potential. The relationship between the electric field and the
generating potentials is [10]:
EL = -∇Θ – (1/c)∂A/∂t
(A15.7)
where EL is the electric field, Θ the scalar potential and A the vector potential.
In this case, there will be no change in the vector potential with time, so:
∂A/∂t = 0
(A15.8)
EL = -∇Θ
(A15.9)
and :
Since the electric field vector is in the –y direction, the gradient of the
scalar potential needs to be in the positive y direction. The electric field
magnitude is a constant so the magnitude of the gradient of the scalar potential
will also be a constant, making it a linear function of y:
203
EL = -ydΘ/dy
(A15.10)
-dΘ = ELdy
(A15.11)
-∫dΘ = ∫ELdy
(A15.12)
Θ(y) = -ELy + CΘ
(A15.13)
This is easily separable and integrable:
where CΘ is a constant.
The other preliminary task that remains is to calculate a suitable gauge for
the vector potential A. This gauge must result in a uniform, constant magnetic
field B in the – z direction and must conform to the relation:
B=∇xA
(A15.14)
Since particle motion will be in a straight line in the case of the velocity
selector, rectangular coordinates will be a good choice and the coordinate field
can be arranged so that motion is at a constant y (along or parallel to the x axis).
In rectangular coordinates, A is:
A = xAx + yAy + zAz
(A15.15)
and the curl of A is:
∇ x A = (∂Az/∂y - ∂Ay/∂z)x - (∂Az/∂x - ∂Ax/∂z)y + (∂Ay/∂x - ∂Ax/∂y)z (A15.16)
For a constant magnetic field B in the –z direction:
∂Az/∂y = ∂Ay/∂z = ∂Az/∂x = ∂Ax/∂z = 0
(A15.17)
or:
∂Az/∂y - ∂Ay/∂z = ∂Az/∂x - ∂Ax/∂z = 0
(A15.18)
The remaining derivatives, ∂Ay/∂x and ∂Ax/∂y, must be linear in x and/or y for B to
be constant in magnitude.
The symmetric gauge:
204
A = (B x r)/2
(A15.19)
is not at all suited for rectangular coordinates. This gauge does produce the
required magnetic field but the conversion to rectangular coordinates is very
messy and doesn’t relate well to the physical aspects (the result resembles
straight line oscillation) of the problem and it’s easier just to start with a
“rectangular” gauge in the first place.
The required gauge will feature [18]:
∂Ax/∂y = constant,
(A15.20)
Figure A15.2. The Vector Potential with A = Byx, where x is the unit vector in the x
direction.
205
∂Ay/∂x = constant
(A15.21)
Figure A15.3. The Vector Potential with A = -Bxy, where y is the unit vector in the
y direction.
206
or:
∂Ay/∂x - ∂Ax/∂y = constant
(A15.22)
Figure A15.4. Vector Potential as a Combination of the Potentials in Figures A15.2
and A15.3.
The vector potential in figure A15.4 is figure A15.5 from along the z axis on
opposite sides of the origin [18]:
207
Figure A15.5. The Vector Potential in the Symmetric Gauge.
…which is not suited to the velocity selector situation. In any case,
∂Ay/∂x - ∂Ax/∂y = constant
(A15.23)
is more complicated than:
∂Ax/∂y = constant
or:
208
(A15.24)
∂Ay/∂x = constant
(A15.25)
∂Ax/∂y = constant
(A15.26)
A = Byx
(A15.27)
Choosing:
and since Ay must be linear in y:
and:
∇ x A = [-∂Ax/∂y]z = [-∂(By)/∂y]z = -Bz = B
(A15.28)
Prior to using this gauge in the Klein-Gordon equation, it is necessary to
see if A is non-divergent in this gauge because that is important in determining
whether or not A commutes with ∇:
The Divergence of A is:
∇●A = (x∂/∂x + y∂/∂y + z∂/∂z)Byx
(A15.29)
∇●A = x∂[Byx]/∂x + y∂[Byx]/∂y + z∂[Byx]/∂z
(A15.30)
∇●A = B(x∂[yx]/∂x + y∂[yx]/∂y + z∂[yx]/∂z)
(A15.31)
∇●A = B(x●x)∂[y]/∂x + xy∂[x]/∂x + (y●x)∂[y]/∂y
(A15.32)
+ yy∂[x]/∂y + (z●x)∂[y]/∂z + zy∂[x]/∂z)
∇●A = B((1)(0) + y(x●0) + (0)(1) + y(y●0) + (0)(0) + y(z●0)) = 0 (A15.33)
and A is non-divergent in this Landau gauge.
The next question is whether or not A in this gauge will commute with ∇:
[A,∇]Ψ = (A●∇ - ∇●A)Ψ = A●∇Ψ - ∇●(AΨ)
(A15.34)
= A●∇Ψ - A●∇Ψ - Ψ∇●A
[A,∇]Ψ = - Ψ∇●A
(A15.35)
and since A is non-divergent:
[A,∇]Ψ = [A,∇] = 0
209
(A15.36)
and A does commute with ∇ in the gauge:
A = Byx
(A15.37)
as will any other non-divergent A.
The general time independent Klein-Gordon equation, with both potentials,
is:
{(E – eΘ)2/c2 = ((h/i)∇ - eA/c)2 + m2c2}Ψ
(A15.38)
Working out the squared term on the right:
{(E – eΘ)2/c2 = -h2∇2 – (he/ic)A●∇
(A15.39)
- (he/ic)∇●A + e2A2/c2 + m2c2}Ψ
Since A and ∇ commute:
{(E – eΘ)2/c2 = -h2∇2 - (2he/ic)∇●A + e2A2/c2 + m2c2}Ψ
(A15.40)
and, applying the wave function:
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2he/ic)(Ψ∇●A + A●∇Ψ)
(A15.41)
+ e2A2Ψ/c2 + m2c2Ψ
Since the divergence of A is zero:
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2he/ic)(A●∇Ψ) + e2A2Ψ/c2 + m2c2Ψ (A15.42)
For the factor A●∇Ψ:
A●∇Ψ = (Byx)●(x∂Ψ/∂x + y∂Ψ/∂y + z∂Ψ/∂z)
(A15.43)
A●∇Ψ = (x●x)By∂Ψ/∂x + (x●y)By∂Ψ/∂y + (x●z)By∂Ψ/∂z
(A15.44)
A●∇Ψ = By∂Ψ/∂x + 0 + 0 = By∂Ψ/∂x
(A15.45)
Plugging this back into the Klein-Gordon equation:
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2he/ic)(A●∇Ψ) + e2A2Ψ/c2 + m2c2Ψ (A15.46)
yields:
210
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2he/ic)(By∂Ψ/∂x)
(A15.47)
+ e2A2Ψ/c2 + m2c2Ψ
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2Byhe/ic)(∂Ψ/∂x)
(A15.48)
+ e2A2Ψ/c2 + m2c2Ψ
Expanding the ∇2 term and ignoring z:
{(E – eΘ)2/c2}Ψ = -h2(∂2Ψ/∂x2 + ∂2Ψ/∂y2)
(A15.49)
- (2Byhe/ic)(∂Ψ/∂x) + e2A2Ψ/c2 + m2c2Ψ
and adding the y dependence of the scalar potential:
Θ(y) = -ELy + CΘ
(A15.50)
{(E – e(-ELy + CΘ))2/c2}Ψ = -h2(∂2Ψ/∂x2 + ∂2Ψ/∂y2)
(A15.51)
the result is:
- (2Byhe/ic)(∂Ψ/∂x) + e2A2Ψ/c2 + m2c2Ψ
Now taking several steps to reorganize the terms:
-h2(∂2Ψ/∂x2 + ∂2Ψ/∂y2) - (2Byhe/ic)(∂Ψ/∂x)
(A15.52)
+ e2A2Ψ/c2 + m2c2Ψ - (E – e(-ELy + CΘ))2/cΨ = 0
Making a last substitution using:
A = Byx
(A15.53)
the result is:
-h2(∂2Ψ/∂x2 + ∂2Ψ/∂y2) - (2Byhe/ic)(∂Ψ/∂x)
(A15.54)
+ e2B2y2Ψ/c2 + m2c2Ψ - (E – e(-ELy + CΘ))2/c2Ψ = 0
This equation is not separable in this form.
This logjam is breakable because the form of the wave function is already
known [15]:
ψ = X(x)P(y) = ei(p(3)x – E(3)t)/he-(ξ’^2)/2Hn(ξ’)
211
(A15.55)
Where:
Hn(ξ’) → H0(ξ’) → 1
(A15.56)
for the ground state wave function for simplicity. In addition:
pz = 0
(A15.57)
ξ’ = (|e|η’)1/2y – [(|e|η’)1/2(E3EL – Bp3)]/((eη’2)
(A15.58)
η’ = (B2 – E2)1/2
(A15.59)
h=c=1
(A15.60)
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2he/ic)(A●∇Ψ)
(A15.61)
and:
and:
in the:
system.
Going back to equation (A16.46):
+ e2A2Ψ/c2 + m2c2Ψ
In rectangular coordinates:
∇2 = ∂2/∂x2 + ∂2/∂y2
(A15.62)
∇ = x∂/∂x + y∂/∂y
(A15.63)
and:
In this gauge, A has only an x component ( A = yBx) so the dot product term:
- (2he/ic)(A●∇Ψ) = - (2he/ic)(xyB)•(x∂ψ/∂x + y∂ψ/∂y)
(A15.64)
- (2he/ic)(A●∇Ψ) = - (2he/ic)(By)[(x•x)∂ψ/∂x + (x•y)(∂ψ/∂y)]
(A15.65)
- (2he/ic)(A●∇Ψ) = - (2he/ic)(By)[(x•x)∂ψ/∂x + 0(∂ψ/∂y)]
(A15.66)
- (2he/ic)(A●∇Ψ) = - (2he/ic)(By)[(x•x)∂ψ/∂x + 0]
(A15.67)
-
(2he/ic)(A●∇Ψ) = - (2he/ic)(By)∂ψ/∂x
212
(A15.68)
- (2he/ic)(A●∇Ψ) = - (2he/ic)(A)∂ψ/∂x = - (2heA/ic)∂ψ/∂x
(A15.69)
and only the derivative with respect to x survives in that term. This leaves
a Klein-Gordon equation (for simplicity of notation –ELy + CΘ → eΘ):
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2heA/ic)∂ψ/∂x
(A15.70)
+ e2A2Ψ/c2 + m2c2Ψ
To finish with this term:
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - (2heA/ic)∂[X(x)P(y)]/∂x
(A15.71)
+ e2A2Ψ/c2 + m2c2Ψ
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - P(y)(2heA/ic)∂[X(x)]/∂x
(A15.72)
+ e2A2Ψ/c2 + m2c2Ψ
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - P(y)(2heA/ic)∂[ei(p(3)x – E(3)t)/h]/∂x
(A15.73)
+ e2A2Ψ/c2 + m2c2Ψ
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - P(y)X(x)(2heA/ic)(ip3/h)
(A15.74)
+ e2A2Ψ/c2 + m2c2Ψ
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - Ψ(2heA/ic)(ip3/h)
(A15.75)
+ e2A2Ψ/c2 + m2c2Ψ
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - Ψ(2eAp3/c)
(A15.76)
+ e2A2Ψ/c2 + m2c2Ψ
Now the LaPlacian:
∇2 = ∂2/∂x2 + ∂2/∂y2
(A15.77)
∇2ψ = ∂2ψ/∂x2 + ∂2ψ/∂y2 = ∂2[X(x)P(y)]/∂x2 + ∂2[X(x)P(y)]/∂y2
(A15.78)
∇2ψ = P(y)∂2[X(x)]/∂x2 + X(x)∂2[P(y)]/∂y2
(A15.79)
For the x dependent part of the LaPlacian:
213
P(y)∂2[X(x)]/∂x2 = P(y)∂2[ei(p(3)x – E(3)t)/h]/∂x2 = X(x)P(y)(ip3/h)2
(A15.80)
= -X(x)P(y)p32/h2 = -ψ p32/h2
(A15.81)
For the y dependent part of the LaPlacian:
X(x)∂2[P(y)]/∂y2 = X(x)∂2[e-(ξ’^2)/2]/∂y2
(A15.82)
Concentrating for the moment on the derivatives:
∂2[e-(ξ’^2)/2]/∂y2 = ∂/∂y[∂[e-(ξ’^2)/2]/∂y]
(A15.83)
∂2[e-(ξ’^2)/2]/∂y2 = ∂/∂y[(e-(ξ’^2)/2)∂[-ξ’2/2]/∂y]
(A15.84)
= (-1/2)∂/∂y[(e-(ξ’^2)/2)∂[ξ’2]/∂y]
∂2[e-(ξ’^2)/2]/∂y2 = (-1/2)∂/∂y[(e-(ξ’^2)/2)2ξ’∂[ξ’]/∂y]
(A15.85)
= (-2/2)∂/∂y[ξ’(e-(ξ’^2)/2)∂[ξ’]/∂y]
∂2[e-(ξ’^2)/2]/∂y2 = -∂/∂y[ξ’(e-(ξ’^2)/2)∂[ξ’]/∂y]
(A15.86)
∂2[e-(ξ’^2)/2]/∂y2 = -∂/∂y[ξ’(e-(ξ’^2)/2)∂[(|e|η’)1/2y
(A15.87)
– {(|e|η’)1/2(E3EL – Bp3)}/(eη’2)]/∂y]
The right hand most term has no dependence on y, so:
∂2[e-(ξ’^2)/2]/∂y2 = -∂/∂y[ξ’(e-(ξ’^2)/2)∂[(|e|η’)1/2y – 0]/∂y]
(A15.88)
∂2[e-(ξ’^2)/2]/∂y2 = -∂/∂y[ξ’(e-(ξ’^2)/2)∂[(|e|η’)1/2y]/∂y]
(A15.99)
Since:
η’ = (B2 – E2)1/2
(A15.100)
h=c=1
(A15.101)
Again, this is in the:
mode. η’ is not dependent on y (in any system of units) so taking the derivative:
∂2[e-(ξ’^2)/2]/∂y2 = -∂/∂y[ξ’(e-(ξ’^2)/2)(|e|η’)1/2]
= ∂/∂y[-ξ’(e-(ξ’^2)/2)(|e|η’)1/2]
Here:
214
(A15.102)
∂[P(y)]/∂y = -ξ’(e-(ξ’^2)/2)(|e|η’)1/2
(A15.103)
∂2[e-(ξ’^2)/2]/∂y2 = ∂[-ξ’(e-(ξ’^2)/2)(|e|η’)1/2]/∂y
(A15.104)
For the second derivative:
The factor (|e|η’)1/2 is not dependent on y so:
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)1/2∂[ξ’P(y)]/∂y =
(A15.105)
-(|e|η’)1/2{(P(y)∂[ξ’]/∂y + ξ’∂[’P(y)]/∂y}
∂[ξ’]/∂y = (|e|η’)1/2
(A15.106)
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)1/2∂[ξ’P(y)]/∂y
(A15.107)
so:
= -(|e|η’)1/2{(P(y)∂[ξ’]/∂y + ξ’∂[’P(y)]/∂y}
becomes:
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)1/2∂[ξ’P(y)]/∂y
(A15.108)
= -(|e|η’)1/2{(P(y)(|e|η’)1/2 + ξ’∂[’P(y)]/∂y}
and now inserting the results for:
∂[P(y)]/∂y = -ξ’(e-(ξ’^2)/2)(|e|η’)1/2
(A15.109)
yields:
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)1/2∂[ξ’P(y)]/∂y = -(|e|η’)1/2{(P(y)(|e|η’)1/2 (A15.110)
+ ξ’(-ξ)’(e-(ξ’^2)/2)(|e|η’)1/2}
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)1/2∂[ξ’P(y)]/∂y = -(|e|η’)1/2{(P(y)(|e|η’)1/2 (A15.111)
- ξ’2(e-(ξ’^2)/2)(|e|η’)1/2}
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)1/2{(P(y)(|e|η’)1/2 - ξ’2(e-(ξ’^2)/2)(|e|η’)1/2} (A15.112)
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)(P(y) + ξ’2(e-(ξ’^2)/2)(|e|η’)
(A15.113)
∂2[e-(ξ’^2)/2]/∂y2 = -(|e|η’)(P(y) + ξ’2(|e|η’)P(y)
(A15.114)
Now inserting the results for the LaPlacian back into the Klein-Gordon equation:
215
{(E – eΘ)2/c2}Ψ = -h2∇2Ψ - Ψ(2eAp3/c) + e2A2Ψ/c2 + m2c2Ψ (A15.115)
becomes:
{(E – eΘ)2/c2}Ψ = -h2(∂2/∂x2 + ∂2/∂y2)Ψ
(A15.116)
- Ψ(2eAp3/c) + e2A2Ψ/c2 + m2c2Ψ
and:
{(E – eΘ)2/c2}Ψ = -h2(-p32/h2 - (|e|η’) + ξ’2(|e|η’)Ψ
(A15.117)
- Ψ(2eAp3/c) + e2A2Ψ/c2 + m2c2Ψ
and:
[(E – eΘ)2/c2]Ψ = p32Ψ + h2{|e|η’}Ψ – h2{ξ’2(|e|η’)}Ψ
(A15.118)
- Ψ(2eAp3/c) + e2A2Ψ/c2 + m2c2Ψ
The curly brackets enclose those parts of the equation that are in natural units:
h=c=1
(A15.119)
and mark a separation from the rest of the equation so these can be tracked
through calculations in case reinserting c and h in those parts of the equation
becomes important.
To make things a little simpler at this point set, for the time being:
eΘ = V
(A15.120)
[(E – V)2/c2]Ψ = p32Ψ + h2{|e|η’}Ψ – h2{ξ’2(|e|η’)}Ψ
(A15.121)
so:
- Ψ(2eAp3/c) + e2A2Ψ/c2 + m2c2Ψ
Since all the operators have operated it is now possible to divide by ψ:
[p32Ψ + h2{|e|η’}Ψ – h2{ξ’2(|e|η’)}Ψ
- Ψ(2eAp3/c) + e2A2Ψ/c2 + m2c2Ψ - [(E – V)2/c2]Ψ]/ψ = 0
and:
216
(A15.122)
p32 + h2{|e|η’} – h2{ξ’2(|e|η’)}
(A15.123)
- (2eAp3/c) + e2A2/c2 + m2c2 - (E – V)2/c2 = 0
This is an equation that is quadratic in p3. To make this a little easier to see:
p32 - (2eAp3/c) + h2{|e|η’} – h2{ξ’2(|e|η’)}
(A15.124)
+ e2A2/c2 + m2c2 - (E – V)2/c2 = 0
and:
p32 - p3(2eA/c) + e2A2/c2 - η1 = 0
(A15.125)
η1 = h2{|e|η’} – h2{ξ’2(|e|η’)} + m2c2 - (E – V)2/c2
(A15.126)
where:
Using the quadratic formula to solve for p3:
2p3 = -(-2eA/c) ± ([-(2eA/c)]2 – 4(e2A2/c2 - η1))1/2
(A15.127)
2p3 = -(-2eA/c) ± ([-(2eA/c)]2 – 4e2A2/c2 + 4η1))1/2
(A15.128)
2p3 = -(-2eA/c) ± (4e2A2/c2 – 4e2A2/c2 + 4η1)1/2
(A15.129)
2p3 = -(-2eA/c) ± (4η1)1/2
(A15.130)
2p3 = -(-2eA/c) ± 2(η1)1/2
(A15.131)
2p3 = 2eA/c ± 2(η1)1/2
(A15.132)
p3 = eA/c ± (η1)1/2
(A15.133)
From the results of testing the real Gaussian of the wave function of
cyclotron motion, which showed that the charge density equation is not capable
of handling a real element of the wave function correctly, the results are likely to
be no better in the case of the velocity selector since one of the exponentials,
P(y), is real.
There is, however, another step that can be taken with the results so far
217
for the velocity selector despite not being able to use the charge density
equation.
Boundary Conditions
For this it is necessary to have workable wave functions for incident and
reflected particles. The boundary between A = V = 0 and A, V ≠ 0 can be located
at x = 0 and in the xz plane so the incident and reflected particle can workably be
described by plane waves as is the standard in many quantum mechanics
textbooks:
ψincident = aψ1 = aei(p(1)x – E(1)t)/h
(A15.134)
ψreflected = dψ2 = de-i(p(2)x + E(2)t)/h
(A15.135)
ψtransmitted = fψ3 = fei(p(3)x + E(3)t)/he-(ξ’^2)/2
(A15.136)
For the boundary conditions the wave functions and their derivatives have
to be continuous across the boundary. Since this is a boundary in x, the
derivatives of importance are with respect to x. For the first equation, matching
the wave functions across the boundary at x = 0:
aψ1 + dψ2 = fei(p(3)x + E(3)t)/he-(ξ’^2)/2
(A15.137)
and:
aei(p(1)x – E(1)t)/h + de-i(p(2)x + E(2)t)/h = fei(p(3)x + E(3)t)/he-(ξ’^2)/2
(A15.138)
x=t=0
(A15.139)
At:
this becomes:
a + d = fe-(ξ’^2)/2
The other boundary condition equation involves the derivatives of the wave
functions with respect to x:
218
(A15.140)
∂[aei(p(1)x – E(1)t)/h]/∂x + ∂[de-i(p(2)x + E(2)t)/h]/∂x
(A15.141)
= ∂[fei(p(3)x + E(3)t)/he-(ξ’^2)/2]∂/x
Taking the derivatives:
a(ip1/h)ei(p(1)x – E(1)t)/h]/∂x – d(ip2/h)e-i(p(2)x + E(2)t)/h
(A15.142)
= f(ip3/h)ei(p(3)x + E(3)t)/he-(ξ’^2)/2
At:
x=t=0
(A15.143)
a(ip1/h) – d(ip2/h) = f(ip3/h)e-(ξ’^2)/2
(A15.144)
the result is:
Here it is possible to simplify by dividing by (i/h):
ap1 – dp2 = fp3e-(ξ’^2)/2
(A15.145)
The results for both boundary condition equations are:
a + d = fe-(ξ’^2)/2
(A15.146)
and:
ap1 – dp2 = fp3e-(ξ’^2)/2
(A15.147)
To get an idea of what the transmission coefficient looks like, solve both these
equations for d:
d = fe-(ξ’^2)/2 – a
(A15.148)
ap1 – dp2 = fp3e-(ξ’^2)/2
(A15.149)
– dp2 = fp3e-(ξ’^2)/2 – ap1
(A15.150)
For the other equation:
d = ap1/p2 – (fp3/p2) e-(ξ’^2)/2
(A15.151)
Now hooking the boundary condition equations together:
fe-(ξ’^2)/2 – a = d = ap1/p2 – (fp3/p2) e-(ξ’^2)/2
219
(A15.152)
the result is:
fe-(ξ’^2)/2 – a = ap1/p2 – (fp3/p2) e-(ξ’^2)/2
(A15.153)
fe-(ξ’^2)/2 + (fp3/p2)e-(ξ’^2)/2 = ap1/p2 + a
(A15.154)
f(e-(ξ’^2)/2 + (p3/p2)e-(ξ’^2)/2) = a(p1/p2 + 1)
(A15.155)
f/a = (p1/p2 + 1)/(e-(ξ’^2)/2 + (p3/p2)e-(ξ’^2)/2)
(A15.156)
f/a = (p1/p2 + 1)/[(1 + (p3/p2))e-(ξ’^2)/2)]
(A15.157)
Now solving for f/a:
Multiplying the right hand side by p2/p2:
f/a = (p1 + p2)/[(p3 + p2))e-(ξ’^2)/2)]
(A15.158)
f/a = [(p1 + p2)/(p3 + p2)]e(ξ’^2)/2
(A15.159)
or:
Toward a reflection coefficient, starting with the two boundary condition
equations:
a + d = fe-(ξ’^2)/2
(A15.160)
ap1 – dp2 = fp3e-(ξ’^2)/2
(A15.161)
and:
…this time solve each for f:
a + d = fe-(ξ’^2)/2
(A15.162)
f = (a + d)e(ξ’^2)/2
(A15.163)
becomes:
and for the other boundary condition equation:
ap1 – dp2 = fp3e-(ξ’^2)/2
becomes:
220
(A15.164)
f = (ap1 – dp2)(1/(p3e-(ξ’^2)/2))
(A15.165)
f = (ap1 – dp2)(e(ξ’^2)/2/p3) = (ap1/p3 – dp2/p3)e(ξ’^2)/2
(A15.166)
and:
The two boundary condition equations are now:
f = (a + d)e(ξ’^2)/2
(A15.167)
f = (ap1/p3 – dp2/p3)e(ξ’^2)/2
(A15.168)
(a + d)e(ξ’^2)/2 = f = (ap1/p3 – dp2/p3)e(ξ’^2)/2
(A15.169)
(a + d)e(ξ’^2)/2 = (ap1/p3 – dp2/p3)e(ξ’^2)/2
(A15.170)
and:
They can now link:
and:
In this case, the remaining exponential divides out:
a + d = ap1/p3 – dp2/p3
(A15.171)
d + dp2/p3 = ap1/p3 – a
(A15.172)
d(1 + p2/p3) = a(p1/p3 – 1)
(A15.173)
(d/a)(1 + p2/p3) = (p1/p3 – 1)
(A15.174)
d/a = (p1/p3 – 1)/(1 + p2/p3)
(A15.175)
and solving for d/a yields:
Multiplying the right hand side by p3/p3 yields:
d/a = (p1 – p3)/(p2 + p3)
(A15.176)
The raw coefficients are:
f/a = [(p1 + p2)/(p3 + p2)]e(ξ’^2)/2
(A15.177)
and:
d/a = (p1 – p3)/(p2 + p3)
221
(A15.178)
Since:
p3 = eA/c ± (η1)1/2
(A15.179)
f/a = [(p1 + p2)/(eA/c ± (η1)1/2 + p2)]e(ξ’^2)/2
(A15.180)
d/a = (p1 – p3)/(p2 + eA/c ± (η1)1/2)
(A15.181)
h2{|e|η’} – h2{ξ’2(|e|η’)} + m2c2 - (E – V)2/c2
(A15.182)
f/a = [(p1 + p2)/(eA/c ± (h2{|e|η’} – h2{ξ’2(|e|η’)}
(A15.183)
The two equations are:
and:
and since:
that leaves:
+ m2c2 - (E – V)2/c2)1/2 + p2)]e(ξ’^2)/2
and:
d/a = (p1 – p3)/(p2 + eA/c ± (h2{|e|η’}
– h2{ξ’2(|e|η’)} + m2c2 - (E – V)2/c2)1/2)
222
(A15.184)
APPENDIX 16
SIGN CONVENTIONS OF KEY QUANTITIES
Some of the conclusions of this thesis rest on the signs, negative or
positive, of factors in the Klein-Gordon equation, in the related charge density
and current equations or factors in boundary conditions. Most of this has to do
with the sign of e, the unit electric charge, or Θ, the scalar potential. Both these
quantities can be positive or negative.
The various elements that either have their sign determined by the sign of
e or Θ or are involved in associated terms are E, the particle energy; ρ, the
particle density; j, the particle current; eρ, the electric charge density; ej, the
electric current and eΘ, the potential energy.
There are several basic rules that determine the way the signs in all these
elements come out:
1) E, ρ and j are positive for particles and negative for antiparticles if the
usual conventions of the signs in wave functions are followed; for example:
for a free particle, localized over an arbitrary volume, with positive velocity and
momentum [2],[25]:
ψ = ei(px – Et)/h
(A16.1)
ψ* = e-i(px – Et)/h
(A16.2)
and for the counterpart antiparticle:
2) e must change sign so that eρ and ej are the sign of true charge density and
electric current instead of the sign of the energy. This is a reflection of the
fact that both antiparticles and particles, depending on the type [10],
can have positive or negative electric charges. One example of this is the
223
particle the electron, which has a negative electric charge, and its antiparticle
the positron, that has a positive electric charge. Another example is the
particle the proton, which has a positive electric charge, and its antiparticle
the antiproton, which has a negative electric charge.
3) Θ, the scalar potential, must change sign such that a repulsive potential eΘ is
really repulsive and an attractive potential eΘ is really attractive. This is really
exhibited in the factor (E - eΘ). If E and -eΘ are of opposite signs a barrier
really will be repulsive and if E and -eΘ are of the same sign the potential will
be attractive. This is perhaps easiest to see in the case of a non-relativistic
particle that has enough energy to mount a repulsive potential step barrier.
p2 = 2m(E - eΘ)
The momentum of the particle is proportional to the factor (E - eΘ). For a
particle E is positive so by the current reasoning -eΘ must be negative. If this
is so, increasing the strength of the potential step eΘ, while the magnitude of
eΘ stays below that of E, will decrease the momentum of the particle as it
passes by the barrier. The barrier is repulsive. If -eΘ is positive here, the
magnitude of (E - eΘ) will increase immediately as -eΘ increases from zero
and the barrier will have the effect of increasing the momentum of the particle.
This is an attractive potential step not a repulsive potential barrier. The same
logic is necessary for correct calculations involving Klein’s Paradox.
How do all the different combinations of factors and signs work out?
224
(A16.3)
TABLE A16.1
ELEMENTS AND SIGN CONVENTIONS
E
ρ
j
e
eρ
ej
-eΘ
eΘ
Θ
1) [+,pa,rp] +
+
+
+
+
+
-
+
+
2) [-,pa,rp]
+
+
+
-
-
-
-
+
-
3) [+,an,rp] -
-
-
-
+
+
+
-
+
4) [-,an,rp]
-
-
-
+
-
-
+
-
-
5) [+,pa,ap] +
+
+
+
+
+
+
-
-
6) [-,pa,ap] +
+
+
-
-
-
+
-
+
7) [+,an,ap] -
-
-
-
+
+
-
+
-
8) [-,an,ap] -
-
-
+
-
-
-
+
+
Situation
The nomenclature of the situations in Table 1 is [electric charge, particle
(pa) or antiparticle (an), repulsive potential (rp) or attractive potential (ap)].
Situation 1: A positively charged particle encountering a repulsive potential
eΘ .
Because this is a particle, its energy is positive and so are the particle
density and particle current. The particle carries a positive charge so e must also
be positive to result in a positive charge density and electric current. The particle
is encountering a repulsive potential so the sign of -eΘ must be negative since E
is positive. This means in turn that eΘ is positive. With a positive e this yields a
positive Θ.
225
Situation 2: A negatively charged particle encountering a repulsive
potential eΘ.
Because this is a particle, its energy is positive and so are the particle
density and particle current. The particle carries a negative charge so e must be
negative to result in a negative charge density and electric current. The particle is
encountering a repulsive potential so the sign of -eΘ must be negative since E is
positive. This means in turn that eΘ is positive. With a negative e this yields a
negative Θ.
Situation 3: A positively charged antiparticle encountering a repulsive
potential eΘ.
Because this is an antiparticle, its energy is negative and so are the
particle density and particle current. The particle carries a positive charge so e
must be negative to result in a positive charge density and electric current. The
particle is encountering a repulsive potential so the sign of -eΘ must be positive
since E is negative. This means in turn that eΘ is negative. With a negative e this
yields a positive Θ.
Situation 4: A negatively charged antiparticle encountering a repulsive
potential eΘ.
Because this is an antiparticle, its energy is negative and so are the
particle density and particle current. The particle carries a negative charge so e
must be positive to result in a negative charge density and electric current. The
particle is encountering a repulsive potential so the sign of -eΘ must be positive
since E is negative. This means in turn that eΘ is negative. With a positive e this
yields a negative Θ.
226
Situation 5: A positively charged particle encountering an attractive
potential eΘ.
Because this is a particle, its energy is positive and so are the particle
density and particle current. The particle carries a positive charge so e must also
be positive to result in a positive charge density and electric current. The particle
is encountering an attractive potential so the sign of -eΘ must be positive since E
is positive. This means in turn that eΘ is negative. With a positive e this yields a
negative Θ.
Situation 6: A negatively charged particle encountering an attractive
potential eΘ.
Because this is a particle, its energy is positive and so are the particle
density and particle current. The particle carries a negative charge so e must be
negative to result in a negative charge density and electric current. The particle is
encountering an attractive potential so the sign of -eΘ must be positive since E is
positive. This means in turn that eΘ is negative. With a negative e this yields a
positive Θ.
Situation 7: A positively charged antiparticle encountering an attractive
potential eΘ.
Because this is an antiparticle, its energy is negative and so are the
particle density and particle current. The particle carries a positive charge so e
must be negative to result in a positive charge density and electric current. The
particle is encountering an attractive potential so the sign of -eΘ must be
negative since E is negative.
This means in turn that eΘ is positive. With a negative e this yields a Negative Θ.
227
Situation 8: A negatively charged antiparticle encountering an attractive
potential eΘ.
Because this is an antiparticle, its energy is negative and so are the
particle density and particle current. The particle carries a negative charge so e
must be positive to result in a negative charge density and electric current. The
particle is encountering an attractive potential so the sign of -eΘ must be
negative since E is negative. This means in turn that eΘ is positive. With a
positive e this yields a positive Θ.
228
APPENDIX 17
GENERAL FORM FOR BOUNDARY CONDITIONS FOR A
STEP POTENTIAL
The wave functions of particles approaching and reflecting from a one
dimensional step potential and being created at the barrier in the “classic” Klein
paradox and the velocity selector have the form, with motion along the x axis [2]:
Ψincident = aΨ1 = aei(p(1)x – E(1)t)/h
(A17.1)
Ψincident* = aΨ1* = ae-i(p(1)x – E(1)t)/h
(A17.2)
Ψreflected = dΨ2 = de-i(p(2)x – E(2)t)/h
(A17.3)
Ψreflected* = dΨ2* = dei(p(2)x – E(2)t)/h
(A17.4)
Ψtransmitted = fΨ3 = fei(p(3)x – E(3)t)/h
(A17.5)
Ψtransmitted* = fΨ3* = fe-i(p(3)x – E(3)t)/h
(A17.6)
These are plane waves, which have to be normalized to an arbitrary
volume for use. Despite this disadvantage, plane waves are often used to show
the behavior of particles in step potential situations such as the “classic” Klein’s
paradox.
At the boundary, which can be conveniently located at x = 0 and t = 0,
the wave functions and their first derivatives on each side must merge
seamlessly with each other. This yields two equations, one for the wave
functions and one for the first derivatives of the wave functions.
For the wave functions:
aΨ1 + dΨ2 = fΨ3
(A17.7)
aei(p(1)x – E(1)t)/h + de-i(p(2)x – E(2)t)/h = fei(p(3)x – E(3)t)/h
(A17.8)
229
Invoking the boundary and initial conditions of x = 0 and t = 0 yields
the first equation:
a+d=f
(A17.9)
To calculate the other equation, it is necessary to take the derivatives
of all three wave functions with respect to x and then relate them for continuity
across the barrier:
∂aΨ1/∂x + ∂dΨ2/∂x = ∂fΨ3/∂x
(A17.10)
Taking the constants out of the derivatives:
a∂Ψ1/∂x + d∂Ψ2/∂x = f∂Ψ3/∂x
(A17.11)
Inserting the details of the wave functions:
a∂ei(p(1)x – E(1)t)/h/∂x + d∂e-i(p(2)x – E(2)t)/h/∂x = f∂ei(p(3)x – E(3)t)/h/∂x
(A17.12)
and taking the derivatives results in:
a(ip1/h)Ψ1 + d(-ip2/h)Ψ2 = f(ip3/h)Ψ3
(A17.13)
There is no reason to retain the imaginary number, so dividing by i removes it:
(ap1/h)Ψ1 + (-dp2/h)Ψ2 = (fp3/h)Ψ3
(A17.14)
which is the second equation which states the continuity of the first derivatives of
the wave functions across the boundary. When the boundary and initial
conditions of x = 0 and t = 0 are applied the finished result is:
(ap1/h)ei(p(1)x – E(1)t)/h + (-dp2/h)e-i(p(2)x – E(2)t)/h = (fp3/h)ei(p(3)x – E(3)t)/h (A17.15)
(ap1/h)ei(p(1)●0 – E(1)●0)/h + (-dp2/h)e-i(p(2)●0 – E(2)●0)/h
(A17.16)
= (fp3/h)ei(p(3)●0 – E(3)●0)/h
(ap1/h)e0 + (-dp2/h)e0 = (fp3/h)e0
(A17.17)
ap1/h - dp2/h = fp3/h
(A17.18)
A last step eliminates the superfluous h:
230
(ap1/h - dp2/h = fp3/h)/h
(A17.19)
ap1 - dp2 = fp3
(A17.20)
The finished set of boundary condition equations derived from the wave
functions and their first derivatives is:
a+d=f
(A17.21)
and:
ap1 - dp2 = fp3
(A17.22)
For a reflection coefficient it is necessary to solve this set of equations for
d/a.To do this, solve each equation for f and then equate the two by using f as a
link. This eliminates f and it is then possible to solve the resulting equation for
d/a. Solving for f:
f=a+d
(A17.23)
ap1 - dp2 = fp3
(A17.24)
fp3 = ap1 - dp2
(A17.25)
(1/p3)(fp3 = ap1 - dp2)
(A17.26)
(1/p3)(fp3) = (ap1 - dp2)(1/p3)
(A17.27)
and:
f = (ap1 - dp2)(1/p3)
(A17.28)
f = (ap1)(1/p3) – (dp2)(1/p3)
(A17.29)
f = a(p1/p3) – d(p2/p3)
(A17.30)
The two equations, now both solved for f, are:
f = a(p1/p3) – d(p2/p3)
(A17.31)
f=a+d
(A17.32)
Equating the two equations using f:
231
a(p1/p3) – d(p2/p3) = f = a + d
(A17.33)
a(p1/p3) – d(p2/p3) = a + d
(A17.34)
a(p1/p3) – d(p2/p3) = a + d
(A17.35)
a(p1/p3) – a = d + d(p2/p3)
(A17.36)
a(p1/p3 – 1) = d(1 + p2/p3)
(A17.37)
Solving for d/a:
becomes:
and:
Multiplying by p3:
p3(a(p1/p3 – 1)) = p3(d(1 + p2/p3))
(A17.38)
a(p1 – p3) = d(p3 + p2)
(A17.39)
a[(p1 – p3)/(p3 + p2)] = d
(A17.40)
Now dividing by (p3 + p2):
and then dividing the entire equation by a:
{a[(p1 – p3)/(p3 + p2)] = d}/a
(A17.41)
{a[(p1 – p3)/(p3 + p2)]}/a = d/a
(A17.42)
(p1 – p3)/(p3 + p2) = d/a
(A17.43)
d/a = (p1 – p3)/(p2 + p3)
(A17.44)
results in:
or:
…which can be regarded as a reflection coefficient.
For a transmission coefficient it is necessary to solve this set of equations:
a+d=f
and:
232
(A17.45)
ap1 - dp2 = fp3
(A17.46)
for f/a. To do this, solve each equation for d and then equate the two by using d
as a link. This eliminates d and it is then possible to solve the resulting equation
for f/a. Solving for d in the wave function equation:
a+d=f
(A17.47)
d=f–a
(A17.48)
Solving for d in the first derivative equation:
ap1 - dp2 = fp3
(A17.49)
-dp2 = fp3 - ap1
(A17.50)
-1(-dp2 = fp3 - ap1)
(A17.51)
dp2 = ap1 - fp3
(A17.52)
(dp2 = ap1 - fp3)/p2
(A17.53)
(dp2)/p2 = (ap1 - fp3)/p2
(A17.54)
d = ap1/p2 - fp3/p2
(A17.55)
With both equations solved for d they can now be linked by that variable:
f – a = d = ap1/p2 - fp3/p2
(A17.56)
f – a = ap1/p2 - fp3/p2
(A17.57)
(f – a = ap1/p2 - fp3/p2)p2
(A17.58)
(f – a)p2 = (ap1/p2 - fp3/p2)p2
(A17.59)
fp2 – ap2 = ap1 - fp3
(A17.60)
fp2 + fp3 = ap1 + ap2
(A17.61)
Eliminating d:
Multiplying by p2:
Re-arranging the terms:
233
f(p2 + p3) = a(p1 + p2)
(A17.62)
(f(p2 + p3) = a(p1 + p2))/a
(A17.63)
(f(p2 + p3))a = a(p1 + p2)/a
(A17.64)
(f/a)(p2 + p3) = (a/a)(p1 + p2)
(A17.65)
Dividing by a:
(f/a)(p2 + p3) = (p1 + p2)
(A17.66)
(f/a)(p2 + p3)/(p2 + p3) = (p1 + p2)/(p2 + p3)
(A17.67)
f/a = (p1 + p2)/(p2 + p3)
(A17.68)
Finally, dividing by (p2 + p3):
which can serve as a transmission coefficient.
The results now for reflection and transmission coefficients are,
respectively:
d/a = (p1 – p3)/(p2 + p3)
(A17.69)
f/a = (p1 + p2)/(p2 + p3)
(A17.70)
What are the results for antiparticles, i.e., for the complex conjugates of
the wave functions?
The procedure is the same:
For the wave functions:
aΨ1* + dΨ2* = fΨ3*
(A17.71)
ae-i(p(1)x – E(1)t)/h + dei(p(2)x – E(2)t)/h = fe-i(p(3)x – E(3)t)/h
(A17.72)
Invoking the boundary and initial conditions of x = 0 and t = 0 yields
the first equation:
a+d=f
To calculate the other equation, it is necessary to take the derivatives
234
(A17.73)
of all three wave functions with respect to x and then relate them for continuity
across the barrier:
∂aΨ1*/∂x + ∂dΨ2*/∂x = ∂fΨ3*/∂x
(A17.74)
Taking the constants out of the derivatives:
a∂Ψ1*/∂x + d∂Ψ2*/∂x = f∂Ψ3*/∂x
(A17.75)
Inserting the details of the wave functions:
a∂e-i(p(1)x – E(1)t)/h/∂x + d∂ei(p(2)x – E(2)t)/h/∂x = f∂e-i(p(3)x – E(3)t)/h/∂x
(A17.76)
and taking the derivatives results in:
a(-ip1/h)Ψ1 + d(ip2/h)Ψ2 = f(-ip3/h)Ψ3
(A17.77)
There is no reason to retain the imaginary number or h, so multiplying by h/i
removes it:
(-ap1)Ψ1* + (dp2)Ψ2* = (-fp3)Ψ3*
(A17.79)
which is the second equation which states the continuity of the first derivatives of
the wave function complex conjugates across the boundary. When the boundary
and initial conditions of x = 0 and t = 0 are applied the finished result is:
(-ap1)e-i(p(1)x – E(1)t)/h + (dp2)ei(p(2)x – E(2)t)/h = (-fp3)e-i(p(3)x – E(3)t)/h
(A17.80)
(-ap1)e-i(p(1)●0 – E(1)●0)/h + (dp2)ei(p(2)●0 – E(2)●0)/h = (-fp3)e-i(p(3)●0 – E(3)●0)/h (A17.81)
(-ap1)e0 + (dp2)e0 = (-fp3)e0
(A17.82)
-ap1 + dp2 = -fp3
(A17.83)
(-ap1 + dp2 = -fp3)(-1)
(A17.84)
ap1 - dp2 = fp3
(A17.85)
Multiplying by -1:
At this point, the two wave functions stemming from the requirement that
the complex conjugate wave functions and their first derivatives must be
235
continuous at the barrier:
a+d=f
ap1 - dp2 = fp3
(A17.86)
(A17.87)
are of the same form as those derived from the wave functions:
a+d=f
(A17.88)
ap1 - dp2 = fp3
(A17.89)
so the developmental process for the complex conjugate wave function reflection
and transmission coefficients can be truncated here with the statement that the
general form of these coefficients:
d/a = (p1 – p3)/(p2 + p3)
(A17.90)
f/a = (p1 + p2)/(p2 + p3)
(A17.91)
is the same regardless of whether the wave functions or their complex
conjugates are used as long as they are not mixed. This also means, in effect,
that the coefficients are of the same form for particles and antiparticles that are
linked by the relevant set of wave functions and their complex conjugates.
236
APPENDIX 18
DERIVATION OF THE QUANTIZATION OF CYCLOTRON ENERGY
The quantized energies of the velocity selector are, in the:
h=c=1
(A18.1)
mode [15]:
(E – ELpx/B )2 = (η’/B)2[(2n + 1)|e|η’ + m2]
(A18.2)
where:
pz = 0
(A18.3)
η’ = (B2 – EL2)1/2
(A18.4)
and:
The big difference between the velocity selector and cyclotron motion is the
presence of an electric field EL in the former and not in the latter. What happens
to the energy quantization of the velocity selector when EL → 0? What should
turn up is the quantization of energy for relativistic cyclotron motion:
Starting with:
(E – ELpx/B )2 = (η’/B)2[(2n + 1)|e|η’ + m2]
(A18.5)
η’ → (B2 – EL2)1/2
(A18.6)
and inserting:
the result is:
(E – ELpx/B )2 = ((B2 – EL2)1/2/B)2[(2n + 1)|e|(B2 – EL2)1/2 + m2]
(A18.7)
Now taking EL → 0:
(E – 0)2 = ((B2 – 0)1/2/B)2[(2n + 1)|e|(B2 – 0)1/2 + m2]
(A18.8)
E2 = ((B2 )1/2/B)2[(2n + 1)|e|(B2)1/2 + m2]
(A18.9)
E2 = (B/B)2[(2n + 1)|e|B + m2]
237
(A18.10)
E2 = (2n + 1)|e|B + m2
(A18.11)
This is still in the h = c = 1 mode. The terms m2 and (2n + 1)|e|B must have units
of energy squared. That means:
E2 = (2n + 1)|e|B + m2c4
(A18.12)
What has to happen to the (2n + 1)|e|B term is not much of a stretch. Formally:
|e|B must have units of energy squared. The actually units are:
|e|Bunits = Coulomb(Nt-sec/Coulomb-m) = Nt-sec/m
(A18.13)
This is force divided by a velocity. Sorting this out:
Nt-sec/m = (kg-m/sec2)(sec/m) = kg/sec
(A18.14)
In terms of what must be multiplied against |e|B, some combination of h and c, to
transform the units of kg/sec to energy squared:
(kg/sec)W = kg2m4/sec4
(A18.15)
Solving for W:
{(kg/sec)W = kg2m4/sec4}(sec/kg)
(A18.16)
(sec/kg)(kg/sec)W = (kg2m4/sec4)(sec/kg)
(A18.17)
W = kg-m4/sec3 = hc2units
(A18.18)
The energy quantization is now:
E2 = (2n + 1)|e|Bhc2 + m2c4
(A18.19)
for relativistic cyclotron motion. Since eBh units are momentum units squared:
E2 = (2n + 1)p2c2 + m2c4
(A18.20)
which is a quantized version of the relativistic momentum-energy relation.
In the non-relativistic limit, the energy quantization is [15]:
E = [(2n + 1)|e|B + pz2]/2m + ELpx/B + m
with:
238
(A18.21)
pz = 0
(A18.22)
E = [(2n + 1)|e|B]/2m + ELpx/B + m
(A18.23)
this reduces to:
For non-relativistic cyclotron motion:
EL → 0
(A18.24)
E = [(2n + 1)|e|B]/2m + ELpx/B + m
(A18.25)
E = [(2n + 1)|e|B]/2m + m
(A18.26)
which turns:
into:
This is still in the h = c = 1mode. To go back to h = h and c = c:
E = (2n + 1)|e|Bh/2m + mc2 = (|e|Bh/2m)(2n + 1) + mc2
(A18.27)
ω = (|e|B/2m)
(A18.28)
Since:
the non-relativistic energy relation is:
E = ωh(2n + 1) + mc2
where ω is the non-relativistic cyclotron frequency.
239
(A.18.29)
APPENDIX 19
QUANTITATIVE ANALYSIS OF TERMS
This appendix contains the calculations that show the relative magnitude
of the suppressing effect that the vector potential in the symmetric gauge (with
cyclotron motion) has on particle pair production in Klein’s paradox.
Starting with equation (A13.59):
p = ±{(1/c2)(E – V)2 – ehBmL/c – e2B2r2/4c2 – m2c2}1/2
(A19.1)
This equation is in Gaussian units. Equations in the mks system will be
involved in the analysis of the terms so it is convenient now to convert (1) to mks
units. For practical purposes, this amounts to B/c → B so, writing (1) in B/c terms:
p = ±{(1/c2)(E – V)2 – eh(B/c)mL – e2(B2/c2)r2/4 – m2c2}1/2
(A19.2)
and taking B/c → B yields:
p = ±{(1/c2)(E – V)2 – ehBmL – e2B2r2/4 – m2c2}1/2
(A19.3)
Pulling a 1/c out of the radical:
p = ±{(1/c2)(E – V)2 – (1/c2)ehBmLc2 – (1/c2)e2B2r2c2/4 – (1/c2)m2c4}1/2 (A19.4)
p = ±{(1/c2)[(E – V)2 – ehBmLc2 – e2B2r2c2/4 – m2c4]}1/2
(A19.5)
p = ±(1/c){(E – V)2 – ehBmLc2 – e2B2r2c2/4 – m2c4}1/2
(A19.6)
and multiplying the whole equation by c:
[p = ±(1/c){(E – V)2 – ehBmLc2 – e2B2r2c2/4 – m2c4}1/2]c
(A19.7)
pc = ±{(E – V)2 – ehBmLc2 – e2B2r2c2/4 – m2c4}1/2
(A19.8)
The terms ehBmLc2, e2B2r2c2/4 and m2c4 all have the same sign and all
have the same suppressing effect on Klein’s paradox, i.e., each of these terms
represents a proportional amount that V will have to increase, over and above
240
V > E , for the particle pair production that characterizes Klein’s paradox to start.
The corresponding equation with no vector potential (the “classic”
Klein’s paradox) is:
pc = ±{(E – V)2– m2c4}1/2
(A19.9)
The suppressing effect that the vector potential has on particle pair
formation in Klein’s paradox, then, is due to the terms ehBmLc2 and e2B2r2c2/4.
How significant is the effect of these terms, if significant is defined as the
magnitude of the rest energy term, m2c4?
This is not entirely straightforward since the quantities mL, the angular
momentum quantum number; B, the magnetic field and r, the radius of particle
rotation, are all related. The equations of relation, for circular motion, are:
r = Mv/eB
(A19.10)
v = particle velocity
(A19.11)
where:
and:
M = relativistic mass = γm
(A19.12)
γ = 1/(1-v2/c2)1/2
(A19.13)
m = rest mass
(A19.14)
where:
a relativistic factor, and:
Also:
Lz = r x p
(A19.15)
where:
Lz = the z component of angular momentum
241
(A19.16)
and:
p = momentum
(A19.17)
Lz = rpsin90o = rp = rMv
(A19.18)
and:
and finally:
Lz = mLh
(A19.19)
r = Mv/eB
(A19.20)
rB = Mv/e
(A19.21)
Starting with:
…this can be re-arranged:
Since m and e are constants, rB must also be a constant if v is constant.
Another way of saying this is to remark that r and B are inversely proportional to
each other if v is constant.
This can also be stated in terms of angular momentum:
Lz = rMv
(A19.22)
v = Lz/rM
(A19.23)
rB = Mv/e
(A19.24)
rB = (M/e)(Lz/rM) = Lz/er
(A19.25)
Lz/e = r2B
(A19.26)
so:
Inserting this into:
yields:
and:
which says that for a given angular momentum that r2B is a constant. Finally, in
242
terms of mL:
Lz = hmL
(A19.27)
Lz/e = r2B
(A19.28)
hmL/e = r2B
(A19.29)
so:
becomes:
which says that for a given mL that r2B is a constant.
The implications of this, particularly the relationship of inverse
proportionality between r and B if v is constant, is greatest for the e2B2r2c2/4 term,
where such a relationship between r and B will tend to limit the magnitude of the
overall term. This implies that the ehBc2mL term may be more important of the
two. Additional calculations show that this is indeed the case:
Starting with e2B2r2c2/4 and:
r = Mv/eB
(A19.30)
then:
e2B2r2c2/4 = (e2B2c2/4)r2 = (e2B2c2/4)(Mv/eB)2
(A19.31)
e2B2r2c2/4 = (e2B2c2/4)(M2v2/e2B2) = M2v2c2/4
(A19.32)
Since:
Lz = mLh = rMv
(A19.33)
v = mLh/rM
(A19.34)
e2B2r2c2/4 = m2v2c2/4
(A19.35)
then:
and if this is inserted into:
the result is:
243
e2B2r2c2/4 = M2v2c2/4 = (M2/4)v2c2 = (M2/4)(mLh/rM)2c2 = mL2h2c2/4r2 (A19.36)
To make further progress:
r = Mv/eB
(A19.37)
v = mLh/rM
(A19.38)
and:
are needed, so that:
r = Mv/eB = (M/eB)v = (M/eB)(mLh/rM)
(A19.39)
r = mLh/eBr
(A19.40)
mL = r2eB/h
(A19.41)
This results in:
or:
Going back to:
e2B2r2c2/4 = mL2h2c2/4r2
(A19.42)
e2B2r2c2/4 = (mLh2c2/4r2)mL
(A19.43)
and:
Now using:
mL = r2eB/h
(A19.44)
then:
e2B2r2c2/4 = (mLh2c2/4r2)mL = (mLh2c2/4r2)(r2eB/h) = (hec2/4)mLB
The reason for all this algebra is that the terms ehBc2mL and e2B2r2c2/4
can be each be written as some constants multiplied by mLB, i.e., as
(hec2)mLB and (hec2)mLB/4.
The (hec2)mLB does indeed turn out to be dominant over (hec2)mLB/4
(which has been transformed from e2B2r2c2/4). To evaluate the effect of the
244
(A19.45)
vector potential on the particle pair formation in Klein’s paradox the terms can
merely be added:
(hec2)mLB + (hec2)mLB/4 = (5/4)(hec2)mLB
(A19.46)
which combines the two terms from the Klein-Gordon equation that originally
contained the vector potential A.
Condensing the constants in the expression will make it easier to evaluate
the whole term quantitatively:
(5/4)(hec2)mLB
(A19.47)
= (5/4)(1.055 x 10-34 J-sec)(1.602 x 10-19 Coul)(2.998 x 108 m/sec)2mLB
= (1.899 x 10-36 J-coul-m2/sec)mLB
Since the units of B are (Nt-sec/coul-m), the overall units of the expression are:
(J-coul-m2/sec)(Nt-sec/coul-m)
(A19.48)
= (J-m2/sec)(Nt-sec/m) = (J-m/sec)(Nt-sec)
= (J-m)(Nt) = J-Nt-m = J2 = Joules2 = energy squared
So:
(1.899 x 10-36 J-coul-m2/sec)mLB = (1.899 x 10-36)mLB Joules2 (A19.49)
where B is unitless now.
How significant is this term compared to the rest energy term m2c4?
Well, for a positively charged pion,
m = 2.489 x 10-28 kg
(A19.50)
For this mass the rest energy term has a magnitude:
m2c4 = (2.489 x 10-28 kg)2(2.998 x 108 meters/sec)4
(A19.51)
= 5.005 x 10-22 Joule2
For relatively minimal mL and B, the rest energy term m2c4 is many orders
245
of magnitude larger than the vector potential term. For an mL = 100 and B = 10-2
Tesla (about 100 times the average magnitude of the Earth’s magnetic field):
m2c4/[(5/4)(hec2)mLB]
(A19.52)
= 5.005 x 10-22 Joule2/[(5/4)(1.899 x 10-36)(100)(10-2)
Joule2] = 2.109 x 1014
With this ratio, mL and B over many orders of magnitude each still result in a
(5/4)(hec2)mLB term lower in magnitude than the rest energy term.
In order to compare terms over a wide variety of mL, B, v and r some more
calculations are necessary:
r = (mLh/eB)1/2
(A19.53)
Checking the units:
(mLh/eB)1/2units = {(J-sec)/[(Coul)(Nt-sec/Coul-m)]}1/2
(A19.54)
(mLh/eB)1/2units = {(J-sec)/(Nt-sec/m)}1/2
(A19.55)
(mLh/eB)1/2units = {(J-sec)/[(kg-m/sec2)(sec/m)]}1/2
(A19.56)
(mLh/eB)1/2units = {(J-sec)/[(kg/sec2)(sec)]}1/2
(A19.57)
(mLh/eB)1/2units = {(J-sec)/(kg/sec)}1/2
(A19.58)
(mLh/eB)1/2units = {(J)(sec2/kg)}1/2
(A19.59)
(mLh/eB)1/2units = {(kg-m2/sec2)(sec2/kg)}1/2
(A19.60)
(mLh/eB)1/2units = {(kg/kg)(m2)(sec2/sec2)}1/2
(A19.61)
(mLh/eB)1/2units = {m2}1/2
(A19.62)
runits = (mLh/eB)1/2units = m = meters
(A19.63)
and the units do work out. Quantitatively, in terms of mL and B:
r = (mLh/eB)1/2 = (2.566 x 10-8)(mL/B)1/2 meters
Another parameter is particle velocity. This can also serve as another
246
(A19.64)
check on whether or not the calculations represented in
figures 1-20 are realistic since superluminal velocities should not occur. The
velocity derivation starts out with the three basic equations
used to make the calculations in this appendix:
r = Mv/eB
(A19.65)
where:
M = γm = m/(1 – v2/c2)1/2
(A19.66)
Lz = mLh = rMv
(A19.67)
is the relativistic mass. Also:
Starting with:
r = Mv/eB
(A19.68)
r = (m/eB)(v/(1 – v2/c2)1/2
(A19.69)
[r = (m/eB)(v/(1 – v2/c2)1/2]2
(A19.70)
r2 = (m2/e2B2)(v2/(1 – v2/c2)
(A19.71)
r2(1 – v2/c2) = v2(m2/e2B2)
(A19.72)
(r2e2B2/m2)(1 – v2/c2) = v2
(A19.73)
r2e2B2/m2 - (r2e2B2/m2)(v2/c2) = v2
(A19.74)
r2e2B2/m2 = v2 + (r2e2B2/m2)(v2/c2)
(A19.75)
r2e2B2/m2 = (v2/c2)(c2 + r2e2B2/m2)
(A19.76)
Squaring the equation:
to get rid of the radical yields:
Now some algebra:
Factoring out v2/c2:
and now solving for v2/c2:
247
(r2e2B2/m2)/(c2 + r2e2B2/m2) = v2/c2
(A19.77)
Since none of the factors involved in this equation are negative:
v2/c2 < 1
(A19.78)
and particle velocities of v ≥ c do not occur.
For deriving the velocities as a function of only mL and B it is necessary to use:
r = Mv/eB
(A19.79)
hmL/e = r2B
(A19.80)
r2 = hmL/eB
(A19.81)
M2v2/e2B2 = hmL/eB
(A19.82)
and:
Continuing:
Substituting for r2:
Multiplying by e2B2:
M2v2 = hmLEb
(A19.83)
γ2m2v2 = hmLeB
(A19.84)
γ2v2 = hmLeB/m2
(A19.85)
Dividing by the rest mass m:
and expanding γ:
v2/(1 – v2/c2) = hmLeB/m2
(A19.86)
v2 = (1 – v2/c2)hmLeB/m2
(A19.87)
v2 = hmLeB/m2 – (v2/c2)hmLeB/m2
(A19.88)
v2 = hmLeB/m2 – (v2)hmLeB/m2c2
(A19.89)
v2 + (v2)hmLeB/m2c2 = hmLeB/m2
(A19.90)
v2(1 + hmLeB/m2c2) = hmLeB/m2
248
(A19.91)
v2 = (hmLeB/m2)/(1 + hmLeB/m2c2)
(A19.92)
[v2 = (hmLeB/m2)/(1 + hmLeB/m2c2)]1/2
(A19.93)
v = [(hmLeB/m2)/(1 + hmLeB/m2c2)]1/2
(A19.94)
As for units, due to the 1 the denominator should be unitless and the units
of the numerator, before the square root is taken, should be m2/sec2. First the
numerator:
(hmLeB/m2)units = [(J-sec)(Coul)(Nt-sec/Coul-m)]/kg2
(A19.95)
(hmLeB/m2)units = [(J-sec)(Nt-sec/m)]/kg2
(A19.96)
(hmLeB/m2)units = [(J-Nt)(sec2/m)]/kg2
(A19.97)
(hmLeB/m2)units = [(kg-m2/sec2)(kg-m/sec2)(sec2/m)]/kg2
(A19.98)
(hmLeB/m2)units = [(m2/sec2)(m/sec2)(sec2/m)]/[(kg2/kg2)]
(A19.99)
(hmLeB/m2)units = (m3/m)(sec2/sec4)
(A19.100)
(hmLeB/m2)units = (m2)(1/sec2)
(A19.101)
(hmLeB/m2)1/2units = (m2/sec2)1/2
(A19.102)
(hmLeB/m2)1/2units = m/sec
(A19.103)
Because of this result and the unitless 1 in the denominator, the other
term in the denominator, hmLeB/m2c2, must also be unitless:
(hmLeB/m2c2)units = (m2/sec2)/c2units = (m2/sec2)(sec2/m2) = 1 (A19.104)
The denominator of:
v = [(hmLeB/m2)/(1 + hmLeB/m2c2)]1/2
(A19.105)
is indeed unitless and the entire fraction has units of v = m/sec.
Now to reduce this to a point where the only values left to enter in the
expression are mL and B:
he/m2 = (1.055 x 10-34)(1.602 x 10-19)/(2.489 x 10-28)2
249
(A19.106)
he/m2 = 272.8
(A19.107)
So now:
v = [(272.8mLB)/(1 + 272.8mLB/c2)]1/2 m/sec
(A19.108)
v = [(272.8mLB)/(1 + 272.8mLB/(2.998 x 108)2)]1/2 m/sec
(A19.109)
v = [(272.8mLB)/(1 + 3.035 x 10-15mLB)]1/2 m/sec
(A19.110)
Note: for situations where velocities are very close to c a greater accuracy of
numbers was necessary (more significant digits) and the formula used is:
v = [(272.880782mLB)/(1 + 3.03620817 x 10-15mLB)]1/2 m/sec (A19.111)
Now the derivation and checking of equations with which to graphically
plot data are finished and, in total, are:
Magnitude of vector potential term = (5/4)(hec2)mLB Joules2 (A19.112)
Magnitude of vector potential term = (1.899 x 10-36)mLB Joules2 (A19.113)
also with the pair:
r = (mLh/eB)1/2
(A19.114)
r = (2.566 x 10-8)(mL/B)1/2 meters
(A19.115)
v = [(hmLeB/m2)/(1 + hmLeB/m2c2)]1/2 m/sec
(A19.116)
v = {272.8mLB/(1 + 3.035 x 10-15mLB)}1/2 m/sec
(A19.117)
and also the pair:
Following are a series of figures that compare these two terms for various
mL and B. Each of those figures is paired with a second figure that indicates the
radii of rotation for cyclotron motion for those mL and B using the relationship and
a third figure that shows the relevant particle velocities for the same mL and B:
Synchrotron radiation will be a significant energy loss, especially at the
higher mL. No such energy loss is reflected in the following calculations so they
250
must be regarded as neglecting synchrotron radiation or assuming the energy
loss is made up by the cyclotron or synchrotron in which the particle is
circulating.
251
Figure A19.1. The (5/4)(hec2)mLB Term, with mL to 104 and B to 104 Tesla, versus
the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 1 shows that even with a magnetic field millions of times the
average strength of the Earth’s magnetic field that the vector potential term
((5/4)(hec2)mLB) is still much lower in magnitude than the rest energy term
(m2c4).
252
Figure A19.2. Radii of Cyclotron Motion (in meters) for mL to 104 and B to = 104
Tesla.
Figure 2 shows that rotation is occurring over scales << 1 meter. For
cyclotron motion, mL (and the angular momentum) and B are inversely
proportional to each other.
253
Figure A19.3. Velocities Associated with mL to 104 and B to 104. The Vertical Axis
is in Meters/Second.
Velocities here are still non-relativistic.
254
Figure A19.4. The (5/4)(hec2)mLB Term, with mL to 107 and B to 107 Tesla, versus
the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 4 shows that the magnitude of the vector potential term
((5/4)(hec2)mLB) at the higher mL and B is approaching the magnitude of the rest
energy term.
255
Figure A19.5. Radii of Cyclotron Motion (in Meters) for mL to 107 and B to 107
Tesla.
Figure 5 shows that rotation is still occurring over radii << 1 meter.
256
Figure A19.6. Velocities Associated with mL to 107 and B to 107 Tesla. The Vertical
Axis is in Meters/Second.
Some relativistic velocities are now occurring.
257
Figure A19.7. The (5/4)(hec2)mLB Term, with mL to 108 and B to 107 Tesla, versus
the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 7 shows that it is possible for the magnitude of the vector potential
term ((5/4)(hec2)mLB) to exceed the magnitude of the rest energy term.
258
Figure A19.8. Radii of Cyclotron Motion (in Meters) for mL to 108 and B to 107
Tesla.
Figure 8 shows that cyclotron motion is still occurring only at radii far less
than 1 meter.
259
Figure A19.9. Velocities Associated with mL to 108 and B to 107 Tesla. The Red
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
The higher relativistic velocities are now approaching the speed of light.
260
Figure A19.10. The (5/4)(hec2)mLB Term, with mL to 1015 and B to 1 Tesla, versus
the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 10 shows some additional combinations of mL and B for which the
magnitude of the vector potential term ((5/4)(hec2)mLB) exceeds the magnitude
of the rest energy term.
261
Figure A19.11. Radii of Cyclotron Motion (in meters) for mL to 1015 and B to= 101
Tesla.
Figure 11 shows that radii of cyclotron motion is on the room size scale.
This means that figure 7 shows values of mL and B for which the vector potential
term exceeds the magnitude of the rest energy term for radii on the laboratory
scale.
262
Figure A19.12. Velocities Associated with mL to 1015 and B to 101 Tesla. The Red
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
263
Figure A19.13. The (5/4)(hec2)mLB term, with mL to 1019 and B to 10-4 Tesla, versus
the m2c4 term (blue). The vertical axis is in units of Joule2.
Figure 13 shows values of mL for which the magnitude of the vector
potential term ((5/4)(hec2)mLB) exceeds that of the rest energy term for magnetic
fields in strength up to about the average surface strength of the Earth’s
magnetic field.
264
Figure A19.14. Radii of Cyclotron Motion for mL = 1019 and B = 10-4 Tesla.
Figure 14 shows that rotation is now taking place with radii on the order of
several tens of kilometers.
265
Figure A19.15. Velocities Associated with mL to 1019 and B to 10-4 Tesla. The Blue
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
266
Figure A19.16. The (5/4)(hec2)mLB Term, with mL to 1014 and B to 104 Tesla, versus
1000 times the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 16 shows that for some values of mL and B the magnitude of the
vector potential term ((5/4)(hec2)mLB) can exceed 1000 times the value of the
rest energy term.
267
Figure A19.17. Radii of Cyclotron Motion (in meters) for mL = 1014 and B = 104
Tesla.
Figure 17 shows that the radii of circulation are again below 1 meter.
268
Figure A19.18. Velocities Associated with mL to 1014 and B to 104 Tesla. The
Purple “Ceiling” is the Speed of Light. The Vertical Axis is in Units of
Meters/Second.
The higher velocities are now getting close to the speed of light.
269
Figure A19.19. The (5/4)(hec2)mLB Term, with mL to 1017 and B to 10 Tesla, versus
1000 times the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Compared to figure 16, in figure 19 the strength of the magnetic field has
been lowered and the angular momentum boosted.
270
Figure A19.20. Radii of Cyclotron Motion (in meters) for mL = 1017 and B = 10
Tesla.
Figure 20 shows rotation radii now occurring on a room scale size.
271
Figure A19.21. Velocities Associated with mL to 1017 and B to 10 Tesla. The Dark
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
272
Figure A19.22. The (5/4)(hec2)mLB Term, with mL to 1019 and B to 102 Tesla, versus
2,000,000 times the m2c4 Term (blue). The Vertical Axis is in Units of Joule2.
Figure 22 shows that the strength of the vector potential term
((5/4)(hec2)mLB) can exceed 2 million times the magnitude of the rest energy
term.
273
Figure A19.23. Radii of Cyclotron Motion (in meters) for mL = 1019 and B = 102
Tesla.
Figure 23 shows radii of rotation from room scale to somewhat large than
room scale.
274
Figure A19.24. Velocities Associated with mL to 1019 and B to 102 Tesla. The Blue
“Ceiling” is the Speed of Light. The Vertical Axis is in Units of Meters/Second.
Most velocities in this plot are very close to the speed of light.
275
Figure A19.25. The (5/4)(hec2)mLB Term, with mL to 1015 and B to 109 Tesla, versus
2 Billion times the m2c4 Term (Blue). The Vertical Axis is in Units of Joule2.
Figure 25 shows that for some values of mL and B that the magnitude of
the vector potential term ((5/4)(hec2)mLB) can exceed 2 billion times the
magnitude of the rest energy term. The strength of B is now on the order of that
found around neutron stars with particularly strong magnetic fields.
276
Figure A19.26. Radii of Cyclotron Motion (in Meters) for mL = 1015 and B = 109
Tesla.
Figure 26 shows that with the ratio of mL to B decreasing that the radii of
rotation decrease as well to << 1 meter.
277
Figure A19.27. Velocities Associated with mL to 1015 and B to 109 Tesla (Green
and Red). The Blue “Ceiling” is the Speed of Light (the Scale is now fine enough
that some Truncation Error is Showing up). Particle Velocities are now a Very
Large Fraction of the Speed of Light for most mL and B. The Vertical Axis is in
Units of Meters/Second.
Almost all velocities now are very close to the speed of light.
278
Figure A19.28. The (5/4)(hec2)mLB Term, with mL to 1023 and B to 109 Tesla, versus
1017 times the m2c4 term (Blue). The Vertical Axis is in Units of Joule2.
Figure 28 shows that for some mL and B the strength of the vector
potential term ((5/4)(hec2)mL) can massively exceed the magnitude of the rest
energy term.
279
Figure A19.29. Radii of Cyclotron Motion (in meters) for mL = 1023 and B = 109
Tesla.
Figure 29 shows that the radii of cyclotron motion are on the room size
scale for figure 28.
280
Figure A19.30. Velocities Associated with mL to 1023 and B to 109 Tesla (Red). The
Blue “Ceiling” is the Speed of Light (the scale is now fine enough that some
truncation error is showing up). Particle Velocities are now a very large fraction of
the Speed of Light for most mL and B. The Vertical Axis is in Units of
Meters/Second.
Most velocities are very close to the speed of light. Only the lower
velocities are visible here as red.
281
These data show that for conditions of low angular momentum and
relatively weak magnetic fields, the suppressive effect of the vector potential is
very weak compared to the rest energy term. However, there is no upward bound
(ignoring energy loss to synchrotron radiation) on angular momentum or
magnetic field strength so if the angular momentum
and magnetic field strength are high enough the suppressive effect of the vector
potential becomes much stronger than the magnitude of the rest energy term. In
such a case, V may be many times larger than mc2 without any particle pairs with
real momenta being produced.
282
APPENDIX 20
GAUSSIAN VERSUS MKS UNITS
For most of this thesis the equations are written using the Gaussian
system of units. For actual calculations with numerical values, however, this
thesis will occasionally use the mks system (this will be clearly noted). Here’s
why:
In Gaussian units the minimal coupling of the vector potential to the
momentum is given as:
p – eA/c
(A20.1)
Aunits = kg-m/coul-sec
(A20.2)
(eA/c)units = (coul)(kg-m/coul-sec)(sec/m) = kg
(A20.3)
The units of A are:
Including the units for c and e:
This term should have units of momentum to be able to couple to the
momentum p. There is an apparent error in units.
This apparent error is actually resolved in the formulation of Gaussian
units so that there is no contradiction. However, this does show why
computations involving actual numbers, not just variables, are sometimes easier
in the mks system.
For the coupling of the vector potential to the momentum the transition
from Gaussian to mks units mainly involves manipulation of c. For example, in
Gaussian units the Lorentz Force Law is:
F = e(EL + (1/c)v x B) = e(EL + v x B/c)
In the mks system:
283
(A20.4)
F = e(EL + v x B)
(A20.5)
As another example, in the Gaussian system Faraday’s Law is:
∇ x EL + (1/c)∂B/∂t = 0
(A20.6)
∇ x EL + ∂B/∂t = 0
(A20.7)
and in the mks system:
The transition from Gaussian to mks units is not always as easy but in the
case of p – eA/c, the it can be effected by:
B/c → B
(A20.8)
B=∇xA
(A20.9)
Because:
holds in both systems and c is a scalar, A transforms as does B:
A/c → A
(A20.10)
which means the transformation of p – eA/c from Gaussian to mks units is:
p – eA/c → p – eA
(A20.11)
(eA)units = (coul)(kg-m/coul-sec) = kg-m/sec
(A20.12)
Checking the units of eA:
and the units work out in a more straightforward way.
284
APPENDIX 21
MAPLE GRAPHICS
plot(exp(-1.266*10^4*r^2),r=0..(.03),y=-1..2,title =
"Gaussian Exponential, B = 10^(-2)
Tesla",titlefont=[HELVETICA, BOLD, 16], labels=["Rotation
Radius", "Gaussian
Magnitude"],labeldirections=[horizontal,vertical]);
This graphic is used in figure 11.19.
285
plot(exp(1.266*10^4*r^2)*(1+1.266*10^4*r^2+(1/2)*(1.266*10^4*r^2)^2+
(1/6)*(1.266*10^4*r^2)^3+(1/24)*(1.266*10^4*r^2)^4+(1/120)*
(1.266*10^4*r^2)^5),r=0..(.04),y=-1..3,title = "B = 10^-2
Tesla, M=1",titlefont=[HELVETICA, BOLD,
16],labels=["Rotation Radius", "WhittakerM
Function"],labeldirections=[horizontal,vertical]);
This is the graphic used in figure 11.21.
286
> plot([exp(1.266*10^4*r^2),1+1.266*10^4*r^2,(1/2)*(1.266*10^4*r^2)^2,(1/6)*
(1.266*10^4*r^2)^3,(1/24)*(1.266*10^4*r^2)^4,(1/120)*(1.266*10^4
*r^2)^5],r=0..(.03),y=-1..5,title = "B = 10^-2 Tesla,
M=1",titlefont=[HELVETICA, BOLD, 16], labels=["Rotation Radius",
"Gaussian and Hypergeom
Terms"],labeldirections=[horizontal,vertical],color=[red,green,b
lue,plum,cyan,navy]);
This is the graphic used in figure 11.20.
287
plot([1/(2*1.266*10^4*r^2),(1/(2*1.266*10^4*r^2)^2),(2/(2*1.266*
10^4*r^2)^3)],r=0..(.05),y=0..4,title = "B = 10^-2 Tesla,
M=3",titlefont=[HELVETICA, BOLD, 16], labels=["Rotation Radius",
"KummerU Terms
"],labeldirections=[horizontal,vertical],color=[red,green,blue])
;
This graphic is used in figure 11.23:
288
plot([exp(-1.266*10^4*r^2),exp(1.266*10^4*r^2)*(1+1.266*10^4*r^2+(1/2)*(1.266*10^4*r^2)^2+
(1/6)*(1.266*10^4*r^2)^3+(1/24)*(1.266*10^4*r^2)^4+(1/120)*
(1.266*10^4*r^2)^5)],r=0..(.04),y=-1..3,title = "B = 10^-2
Tesla, M=1",titlefont=[HELVETICA, BOLD,
16],labels=["Rotation Radius",
"Function"],labeldirections=[horizontal,vertical]);
This graphic is used in figure 11.22:
289
plot([exp(-1.266*10^4*r^2),exp(1.266*10^4*r^2)*(1/(2*1.266*10^4*r^2)+(1/(2*1.266*10^4*r^2)^2)
+(2/(2*1.266*10^4*r^2)^3))],r=0..(.025),y=0..4,title = "B =
10^-2 Tesla, M=3",titlefont=[HELVETICA, BOLD, 16],
labels=["Rotation Radius",
"Function"],labeldirections=[horizontal,vertical]);
This graphic is used in figure 11.21:
290
fieldplot([y,-x],x=-6..6,y=6..6,grid=[8,8],thickness=2,color=red,arrows=thick);
This graphic is used in figures 6.2, 10.3 and A15.5:
291
fieldplot([y,0],x=-6..6,y=6..6,grid=[8,8],thickness=2,color=red,arrows=thick);
This graphic is used in figures 10.2 and A15.2:
292
fieldplot([0,-x],x=-6..6,y=6..6,grid=[8,8],thickness=2,color=red,arrows=thick);
This graphic is used in figure A15.3:
293
fieldplot([-y,x],x=-6..6,y=6..6,grid=[8,8],thickness=2,color=red,arrows=thick);
This graphic is used in figure A15.4:l
294
plot3d([1.899*10^(-36)*m*B,5.005*10^(-22)],
B=0..10^4,m=0..10^4,shading=[xyz,z],labels=["Magnetic Field
(T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used for figure 11.1 and figure A19.1.
295
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^4,m=0..10^4,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.2 and figure A19.2.
296
plot3d(((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2), m =
0..10^4,B = 0..10^4,labels=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figures 11.3 and A19.3.
297
plot3d([1.899*10^(-36)*m*B,5.005*10^(-22)],
B=0..10^7,m=0..10^7,shading=[xyz,z],labels=["Magnetic Field
(T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.4.
298
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^7,m=0..10^7,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.5.
299
plot3d(((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2), m =
0..10^7,B = 0..10^7,labels=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure A19.6.
300
plot3d([1.899*10^(-36)*m*B,5.005*10^(-22)],
B=0..10^7,m=0..10^8,shading=[xyz,z],labels=["Magnetic Field
(T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.7.
301
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^7,m=0..10^8,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.8.
302
plot3d([((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2),
2.998*10^8], m = 0..10^8,B =
0..10^7,shading=[xyz,z],labels=["Quantum No.","Magnetic
Field (T","meters/second"],axes=BOXED);
This graphic is used in figure A19.9.
303
plot3d([1.899*10^(-36)*m*B,5.005*10^(-22)],
B=0..1,m=0..10^15,shading=[xyz,z],labels=["Magnetic Field
(T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.4 and figure A19.10.
304
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..1,m=0..10^15,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.5 and figure A19.11.
305
plot3d([((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2),
2.998*10^8], m = 0..10^15,B =
0..1,shading=[xyz,z],labels=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure 11.6 and figure A19.12.
306
plot3d([1.899*10^(-36)*m*B,5.005*10^(-22)], B=0..10^(4),m=0..10^19,shading=[xyz,z],labels=["Magnetic Field
(T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.7 and figure A19.13.
307
plot3d(2.566*10^(-8)*(m/B)^(1/2), B=0..10^(4),m=0..10^19,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.8 and A19.14.
308
plot3d([((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2),
2.998*10^8], m = 0..10^19,B = 0..10^(4),shading=[xyz,z],labels=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure 11.9 and figure A19.15.
309
plot3d([1.899*10^(-36)*m*B,5.005*10^(-19)],
B=0..10^4,m=0..10^14,shading=[xyz,z],labels=["Magnetic
Field (T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.10 and figure A19.16.
310
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^4,m=0..10^14,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.11 and figure A19.17.
311
plot3d([((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2),
2.998*10^8], m = 0..10^14,B =
0..10^4,view=2.99*10^8..3*10^8,shading=[xyz,z],labels=["Qua
ntum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure 11.12 and figure A19.18.
312
plot3d([1.899*10^(-36)*m*B,5.005*10^(-19)],
B=0..10,m=0..10^17,shading=[xyz,z],labels=["Magnetic Field
(T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.19.
313
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10,m=0..10^17,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.20.
314
plot3d([((272.8*m*B)/(1 + (3.035*10^(-15))*m*B))^(1/2),
2.998*10^8], m = 0..10^17,B =
0..10,view=2.99*10^8..3*10^8,shading=[xyz,z],labels=["Quant
um No.","Magnetic Field (T)","meters/second"],axes=BOXED);
This graphic is used in figure A19.21.
315
plot3d([1.899*10^(-36)*m*B,2*5.005*10^(-16)],
B=0..10^2,m=0..10^19,shading=[xyz,z],labels=["Magnetic
Field (T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.13 and figure A19.22.
316
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^2,m=0..10^19,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED)
This graphic is used in figure 11.14 and figure A19.23.
317
plot3d([((272.880782*m*B)/(1 + (3.03620817*10^(15))*m*B))^(1/2), 2.99792458*10^8], m = 0..10^19,B =
0..10^2,view=2.9979*10^8..2.99793*10^8,shading=[xyz,z],labe
ls=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure 11.15 and figure A19.24.
318
plot3d([1.899*10^(-36)*m*B,2*5.005*10^(-13)],
B=0..10^9,m=0..10^15,shading=[xyz,z],labels=["Magnetic
Field (T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.25.
319
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^9,m=0..10^15,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure A19.26.
320
plot3d([((272.880782*m*B)/(1 + (3.03620817*10^(15))*m*B))^(1/2), 2.99792458*10^8], m = 0..10^15,B =
0..10^9,view=2.997923*10^8..2.9979246*10^8,shading=[xyz,z],
labels=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure A19.27.
321
plot3d([1.899*10^(-36)*m*B,5.005*10^(-5)],
B=0..10^9,m=0..10^23,shading=[xyz,z],labels=["Magnetic
Field (T)","Quantum No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.16 and figure A19.28.
322
plot3d(2.566*10^(-8)*(m/B)^(1/2),
B=0..10^9,m=0..10^23,labels=["Magnetic Field (T)","Quantum
No.","Energy Squared"],axes=BOXED);
This graphic is used in figure 11.17 and figure A19.29.
323
plot3d([((272.880782*m*B)/(1 + (3.03620817*10^(15))*m*B))^(1/2), 2.99792458*10^8], m = 0..10^23,B =
0..10^9,view=2.997923*10^8..2.9979246*10^8,shading=[xyz,z],
labels=["Quantum No.","Magnetic Field
(T)","meters/second"],axes=BOXED);
This graphic is used in figure 11.18 and figure A19.30.
324