Physics 406
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Physics 406
Introduction to Quantum Mechanics
J Kiefer
January 2007
© 2007
Table of Contents
TABLE OF CONTENTS ....................................................................................... 1
I.
WAVE MECHANICS ..................................................................................... 2
A.
Wave Function .................................................................................................................................... 2
B.
The Schrödinger Equation ................................................................................................................. 5
C.
Wells and Barriers—Non-Free Particles .........................................................................................10
D.
Harmonic Oscillator ..........................................................................................................................17
II.
HYDROGEN ATOM .................................................................................... 23
A.
Schrödinger Equation in Three Dimensions ....................................................................................23
B.
Angular Momentum ..........................................................................................................................25
C.
Radial Equation .................................................................................................................................30
D.
Spin......................................................................................................................................................35
III.
SELECTED QUANTUM MECHANICAL ISSUES .................................... 37
A.
Vector Spaces .....................................................................................................................................37
B.
Formal Quantum Mechanics ............................................................................................................41
C.
Time Independent Perturbation Theory ..........................................................................................43
1
I.
Wave Mechanics
A.
Wave Function
We propose to describe the motion or state of a particle by a wave function—a solution to
a wave equation.
1.
Statistical Interpretation of the Wave Function
The wave function is a complex function of position (x) and of time (t), denoted x, t .
We intend that it contain a complete description of the behavior of a particle, such as an
electron.
a.
Probability density
x, t * x, t
2
2
The quantity x, t dx is the probability that the particle is to be found in the
interval (x,x+dx) at time t.
b.
Normalization
The probability of finding the particle someplace must be 1.0, therefore,
x, t dx 1.0
2
if the wave function is to represent a physically realistic particle. That is, a wave function
must be normalizable. This boils down to the requirement that x, t 0 as x .
Some wave functions satisfy the wave equation but are not normalizable.
c.
Expectation values
Imagine a large number of identical but independent regions in space. In each region is a
particle described by the wave function, x, t . The particles and wave functions are
identical. In each of these identical systems we measure the position of the particle, x.
The average of all the independent measurements of x is
x xdx
*
x
2
dx .
This is called the expectation value of x. Note that this is not the most probable value of
2
x. That occurs where is a maximum. The variance of the measurements of x is
2 x 2 x 2 .
2
2.
Momentum
The other dynamical variable of interest is the particle momentum, p mv . We will
want the expectation value of p, namely <p>.
a.
Velocity
x * xˆdx * xdx
dx
xdx * x
dx
dt
t
t
*
x
0 . Now, the wave equation is
Note that
. Substitute this for the
t
x 2
t
time derivatives in the velocity. . .
2
2
dx
2*
*
xdx x 2 dx .
dt
x 2
x
Note that
x
1 and that 0 as
x
x .
Next, integrate by parts the first term on the right hand side (r.h.s.).
2*
*
x
dx
0
x x x dx
x 2
Integrate by parts again.
2
*
*
x dx 0 2 x dx
x x
x
dx
2
2
* 2 ( x )dx * x 2 dx .
So far, we have
dt
x
x
2
2 x
2
x
2 x
However, 2 x 2 x 2 2
and 2 0. What’s left is the
x
x
x
x x
x
expectation value of the particle velocity
dx
2 *
dx .
dt
x
Note for the future:
i
.
2m
3
b. Momentum operator
We define the particle momentum to be p m d x 2m * dx .
dt
x
We define the momentum operator to be pˆ 2m .
x
c. Uncertainty principle
The principle will be proven rigorously later; for now, we recognize that
h
p 2 .
Therefore, a spread in is inversely related to a spread in p. In terms of a localized
waveform,
We identify x with and obtain xp
.
2
Roughly stated, this is the mathematical origin of the uncertainty principle. The particle
position and momentum cannot be “known” simultaneously to arbitrary precision.
4
B.
The Schrödinger Equation
1.
Wave Equation
The one-dimensional wave equation is
2
t
x 2
a.
Solution
The wave equation has solutions of the form cos kx t , sin kx t , and
e i kxt . These are all traveling harmonic waves, where the wave number is
k
2
and the angular frequency is
2 f . (f is the frequency in Hz.)
We’ll concentrate on the complex exponential form: Ae i kx t . Then the
derivatives are
2
i and
k 2 .
2
t
x
Evidently,
i 2
2
t
k x 2
Whence we can identify i2 .
k
b.
Energy and momentum
2
For a free particle, E p , just the kinetic energy. We can express this in terms of the
2m
2 2
frequency and wave number, since E and p 2 k : k .
2m
Solving for the frequency, we obtain what is called a dispersion relation.
k 2
2m
If we substitute this for in the wave equation, we obtain the following:
ik 2 2 i 2
2
t
k x 2
2m x 2
If we multiply and divide the r.h.s. by i , we can see that the r.h.s. is just the momentum
operator squared, divided by 2m.
i i 2
1
i
2
t i 2m x
i 2 m
x
2
i
2 2
t
2m x 2
5
ˆ
The left hand side (l.h.s.) must be the total energy operator, E i
, since for a free
t
particle, the total energy is the kinetic energy.
c.
Conservative forces
For a conservative force acting on a particle, Fx
V x, t
, where V(x,t) is the potential
x
energy function.
We’ll just add it to the kinetic energy operator on the r.h.s. of the wave equation.
2 2
i
V x, t
2
t 2m x
2 2
V x, t
2
t 2m x
This equation is known as the Time Dependent Schrödinger Equation.
i
2.
Time Independent Schrödinger Equation.
Let us say that V is independent of time. The potential energy function is constant.
a.
Separation of variables
Assume that x, t x f t . Substitute into the Schrödinger equation. The partial
derivatives become
df
2
d 2
f
and
.
t
dt
x 2
dx 2
i
Divide both sides by f .
df
2 d 2
f
V x f
dt
2m dx 2
i df
2 d 2
V x
f dt
2m dx 2
The two sides must equal the same constant, namely the total energy, E.
df
2 d 2
i
Ef and
V x E .
dt
2m dx 2
b.
Time solution
i
df
Ef
dt
f t e
iEt
This is an oscillatory solution, as we always get with a wave equation.
6
c.
Stationary states
2 d 2
V x E
2m dx 2
Et
Et
Consider the probability density. x, t * *e e x . This is
constant in time, therefore it is called a stationary state. A stationary state has a definite
total energy; the uncertainty E 0 .
2
2
2 d 2
V x . The expectation value of this
Define the Hamiltonian operator Hˆ
2
2m dx
*
operator is H Hˆ dx E
2
a constant. If the wave function, , is
normalized, then H E .
In compact form, the (Time Independent) Schrödinger equation looks like Hˆ E .
3.
Free Particles and Wave Packets
For a free particle, V(x) = 0. The wave equation is just
2 d 2
E .
2m dx 2
a.
General solution
Aeikx Be ikx (time independent)
Aei kx t Be i kx t (time dependent).
These represent harmonic waves, one traveling to the right, the other to the left along the
2
x-axis. The wave number is k 2mE and the angular frequency is k .
Notice that
2m
*
dx dx ! The free particle wave function is not normalizable.
However, this does not mean that a free particle cannot exist. It does mean that a free
particle does not have a definite energy, nor can it be considered to have a localizable
position.
b.
Wave packets
A localized wave form can be constructed by a superposition of harmonic waves.
1
x,0
k e ikx dk
2
It might look like this:
7
The wave packet travels with a speed called the group velocity, vg. The individual ripples
within the packet travel with a speed called the phase velocity, vp. The phase velocity is
the familiar wave speed.
vp
c.
k
E
2m
Group velocity
2
The angular frequency of a harmonic wave is k . This is called a dispersion
2m
relation, because it implies that waves of different frequency travel with different wave
speeds.
Let’s say that the wave packet is composed of harmonic waves with a narrow range of kvalues, centered on ko. We might expand k in a Taylor’s Series about ko, and keep
just the first two terms.
d
k o
k ko o o k ko
dk ko
Put this in the for .
1
x, t
2
x, t
1
i kx
k e
o
o
dk
k ko t
e
i o t o k o t
k e i k x t dk
o
2
1 i o t oko t
x, t
e
x o t ,0
2
Evidently, the wave packet slides along the x-axis at the speed vg
vg
d
d k 2
k
o
dk k o dk 2m k
m
o
8
d
dk
.
ko
Notes: i) v g 2v p
where k
d.
2
ii) the classical speed of a particle with momentum p is
p k
,
m m
and is the deBroglie wavelength.
Uncertainty relation
We have the wave packet and its Fourier transform:
Wave packet: x,0
1
k eikx dk
2
Transform: k
x, oe
1
2
ikx
dx
The details depend on the exact shape of . For the sake of simplicity, let’s say that
2
x,0
1
x
for
x
x
x
and zero elsewhere. Then the transform is
2
2
k
1
2
x
sin k
1 ikx
2
2
e dx
x
k
x
Take k to be the width of the central peak, where k 0 . We find that
1
k
px , since p k .
x
For more realistically-shaped wave packets we find that px .
2
9
C.
Wells and Barriers—Non-Free Particles
1.
Square Wells
a.
Infinite
x 0
V x 0 0 x a
x a
We have three (3) distinct regions, because the potential energy function (potential for
short) changes discontinuously. So, we have to solve the Schrödinger Equation three (3)
times.
Fortunately, for x 0 and x a , the solutions are trivial: x 0 , since V x .
Within the well, V x
0 , so the particle is “free.”
2 d 2
E
2m dx 2
The general solution is x A sin kx B cos kx . The parameters A, B, and k are
determined by the boundary conditions on the and by the normalization requirement.
That is, we expect that 0 a 0 .
At x = 0: A sin 0 B cos 0 0 B 0
At x = a: A sin ka B cos ka 0 sin ka 0 , since we cannot have A and B both be zero.
It follows that ka ,2 ,3 , . From this we obtain the discrete allowed energy levels
for the particle confined in the well.
2mE
k 2 2
k
E
2m
Putting in the allowed values of k yields
2 2
En n 2
, n = 1,2,3,… .
2ma2
The n is known as the principle quantum number. It labels the energy levels, or energy
states of the particle. Finally we need to normalize the wave function.
10
a
2
dx 1
0
a
A sin 2 kxdx 1
2
0
a
1
2
2
A
a
A2
Our solution is n x
2
n
sin
x for the particle inside the well.
a
a
Properties of the solution:
i) even or odd
ii) orthonormal
iii) form a complete set.
Note: While n
2
is unchanging with time, the x, t is not.
x, t
n 2 2 2
i
t
2
n
2
sin
x e 2 ma
a
a
This form is very similar to the case of a vibrating string, fastened at both ends.
b.
Finite square well
x a
0
V x Vo a x a
0
xa
We find it convenient, but not necessary, to place the origin in the center of the well.
Again, we solve the Schrödinger equation in three regions: x a , a x a , x a .
We’ll solve for the so-called bound states whose energies E < 0.
11
2 d 2
V x E
2m dx 2
The solutions will have the same form as for the infinite well, but the k and the
coefficients will differ in the three regions.
Ceikx De ikx
a x a
Aeikx Be ikx
Feikx Geikx ,
x a
xa
where k
ik
2mE Vo and
k
2mE .]
2mE . [We’ll be setting, as in the text,
The boundary conditions are that and
d
be continuous at x = - a and x = a.
dx
Ceika De ika Feik a Geik a
ikCeika ikDe ika ik Feik a ik Geik a
Ceika De ika Aeik a Be ik a
ikCeika ikDeika ik Aeika ik Be ika
Before proceeding blindly to solve these simultaneous equations for A, B, C, D, E, & F
2 mE
x
(we only have 4 equations), let us notice that since E < 0, the term e
diverges
for x < -a. Therefore the A = 0. Likewise, G = 0. So we really have only four unknowns
remaining.
Ceika De ika Feik a
ikCeika ikDe ika ik Feika
Ceika De ika Be ik a
ikCeika ikDeika ik Be ik a
Put ‘em all in the same order. . .
Ceika De ika Fea 0
ikCeika ikDe ika Fea 0
Be a Ceika De ika 0
Be a ikCeika ikDeika 0
This system of simultaneous equations will have a solution if the determinant of the
coefficients vanishes.
12
0
0
e ika
ikeika
e ika
ikeika
ea
ea
ea
ea
e ika
ikeika
e ika
ikeika
0
0
Firstly, divide the 1st and 4th columns by
0
e ika
0
ike
e a .
ika
e ika
ikeika
Secondly, expand on the 1st column.
1
e ika
ikeika
ikeika
e ika
1 ikeika
ikeika
0
e ika
ikeika
1
e ika
ikeika
0
0
0
e ika
ikeika
e ika
1
e ika
ikeika
0
e ika
1
0
0
Expand the two determinants on their 3rd columns.
1 1
e ika
ikeika
ikeika
e ika
1
ikeika
ikeika
ikeika
ikeika
1
ikeika
e ika
ikeika
e ika
e ika
e ika
e ika
0
e ika
Expand the 2x2 determinants.
k e
2 2ika
Rearrange.
ike
e
k 2 e 2ika ike2ika ike2ika
k
2
2ika
ike
2ika
2
2ika
e 2ika 0
2 e 2ika e 2ika 2ik e 2ika e 2ika 0
e i e i
e i e i
.
and cos
2i
2
Furthermore, sin 2 2 sin cos and cos 2 cos 2 sin 2 .
Now, sin
So, we can write
k
k
2
2
2
2 2i sin 2ka 4ik cos 2ka 0
2i 2 sin ka cos ka 4ik cos
2
2
Divide by cos ka .
k
2
2 4i 2 tan ka 4ik 1 tan 2 ka 0
13
ka sin 2 ka 0
The left hand side is a quadratic in tan ka .
tan 2 ka
k 2 2
tan ka 1 0
k
Use the quadratic formula:
k 2 2 1
tan ka
2k
2
k
2
2 4k
k 2
2
2
tan ka k
k
We have, after all that, two sets of solutions. We solve for k graphically.
An allowed energy level occurs where these curves intersect.
Properties of the solution:
i) always at least one solution
ii) the allowed energies are discrete
iii) the number of solutions is finite, depending on a.
iv) +/- solutions alternate in energy (odd/even, as with the infinite well).
2.
Potential Step
a.
Solution of the time-independent equation
2 d 2
V ( x) E
2m dx 2
14
In this instance, we have two regions: x < 0 and x > 0. For x < 0, V = 0, so at once we
can write Aeikx Be ikx , as usual.
For x > 0, V = Vo. We will assume the same mathematical form for the solution, but we
may or may not get oscillatory behavior.
Ceix De ix
Before applying the boundary conditions, we’ll solve for the .
d 2 2mE Vo
dx 2
2
d 2 Ceix De ix
2mE Vo ix
Ce De ix
2
2
dx
Evidently, 2mE Vo .
b.
Boundary conditions
The boundary conditions are that Aeik 0 Be ik 0 Cei 0 De i 0 and that
ikAeik 0 ikBe ik 0 iCei 0 iDe i 0 . We obtain two equations and four unknowns.
A B C D
ikA ikB iC iD
However, we have additional information. In the region x < 0, we have traveling
harmonic waves, one incident from the left, the other reflected from the step and traveling
toward the left. These cannot be normalized, so we’ll just set A = 1. On the other hand,
in the region x > 0, the D = 0, because there is no additional barrier or interface to cause
another reflection.
1 B C
ik ikB iC
Firstly, divide them.
ik ikB
i
1 B
1
k k
k
1
k
Substitute this result back into the first equation.
1 B C
k
2k
C 1
k k
B
c. Cases
Now, consider the cases E > Vo and E < Vo.
15
i)
E > Vo: is a traveling wave, but of differing wave number from the
incident wave.
ii)
E < Vo:
is imaginary. Therefore is a decaying exponential function.
2 m Vo E
x
Ceix Ce
Here’s the funny thing. The particle is not moving to the right for x > 0, but
2
yet 0 . There is some non-zero probability that the particle will be
observed at an x > 0. Classically, this is impossible.
3.
Barrier Tunneling
In this case, we have three regions. We can take advantage of our findings for the
potential step.
x0
e ikx Re ikx
0 x a Aeix Be ix
xa
Te ikx
As before, k 2mE and 2mE Vo . We apply the boundary conditions at x = o
and at x = a.
1 R A B
ik 1 R i A B
Aeia Be ia Teika
i Aeia Be ia ikTeika
Solve these for A, B, R, and T any way you can. . .
Qualitatively, we have the same two cases: E > Vo and E < Vo.
One thing we see is that even if E < Vo, the particle probability density is not zero on the
far side of the barrier.
16
D.
Harmonic Oscillator
1.
Schrödinger Equation
a.
Potential energy
V x
b.
1 2
kx
2
Hamiltonian
E T V
1
1
mv 2 kx2
2
2
Schrödinger equation
2 d 2 1 2
kx E
2m dx 2 2
The total energy operator is called the Hamiltonian: Hˆ E .
There are two ways to solve this equation.
2.
First Way—the Algebraic Method
a.
Ladder operators
2 d 2
1 d
p2
d
k
and pˆ
. Thus
.
and
2
2m dx
2m i dx
2m
i dx
m
We rewrite the Schrödinger equation as
2
1 d
2 2 2
m x E
2m i dx
Factor the thing in the square bracket.
d
1 d
im x
im x E
2m i dx
i dx
Note that the order is important. If we write out the product, we must write it thusly:
2
Recall that T
17
d d
d
im x im x d m 2 2 x 2
i dx i dx
i dx
i dx
2
d
d
im x im x d imx d m 2 2 x 2
2 2
i dx
i dx
dx
i dx
2 d 2 1
m 2 2 x 2 E
2m dx 2 2
Evidently, the operators don’t factor quite as numbers do-- u 2 v 2 u iv u iv .
Let a
1 d
1 d
im x and a
imx . Evidently,
2m i dx
2m i dx
2
2
d
1
1
m 2 x 2 a a
2
2m dx
2
2
The Schrödinger equation becomes
1
a a E ,
2
which looks neater, anyway.
b.
Raising and lowering
Suppose that
a , then
1
a a E
2
1
1
a a a a a
2
2
1
a a a a
2
1
a a a
2
Ea a
E a
E
That is, if is a solution to the Schrödinger equation with energy E, then
a
is also a
solution, with energy E . Similarly, a produces the solution with energy
E . Hence the terms raising and lowering operators.
c.
Ground state
The state having the lowest energy is called the ground state. We label it o . Since it is
the state of lowest energy, evidently a o 0 .
18
1 d o
imx o 0
2m i dx
d o
m
o
dx
d
m
oo xdx
ln o
o Ao e
m 2
x a constant
2
m 2
x
2
The ground state energy is obtained from the Schrödinger equation.
a a
o Eo o
2
0
o Eo o
2
1
Eo ( 0 ! )
2
Notice that not only is the energy discrete, but the ground state energy is not zero, but
1
.
2
d.
Energy levels
Ei 1 Ei , i 0,1,2,3,
1
En n , n 0,1,2,3,
2
The energy levels are uniformly spaced, and n is the principle quantum number.
The energy state functions are obtained from the raising operator:
3.
Second Way—the Analytical Method
a.
Asymptotic behavior of
i 1 a i .
2 d 2 1
2m
m 2 x 2 E 2
2
2
2m dx
d 2 m 2 2 x 2
2mE
2
2
2
dx
2
2 2
d m 2 2mE
x 2
dx 2 2
If
x
d 2 m 2 2 2
, then 2 2 x , which has solutions of the form
dx
19
( x) Ae
B 0 because that term diverges as x
m2 2
m2 2
x2
x2
.
Be
. So, let us propose that
2
2
( x) h( x)e
2 2
m2 2
22
x2
.
b.
Sturm-Liouville equation
The Sturm-Liouville equation is one of those “well known” differential equations for
which people worked out the solutions before television was invented. By substituting
the assumed into the Schrödinger equation, and making a change of variable, we’ll
produce a differential equation for that unspecified function,
h(x) .
m
y2
m
Let y
x . Then ( y) h( y)e 2 .
d 2 m 2 2 2 2mE
x 2
dx 2 2
m d 2 m 2 2mE
y 2
dy 2
The first and second derivatives of with respect to y are
y
d
d
he 2
dy dy
2
d 2
e
2
dy
y2
2
y
y
y
e 2 dh hye 2 dh hy e 2
dy
dy
2
2
2
2
d 2h
dh
y
dh
2 y h yh ye 2
dy
dy
dy
2
d 2h
y
dh
2 2 y ( y 2 1)h e 2
dy
dy
The
e
y2
2
divides out, so we have an equation for the h(y).
m 2 2mE
m d 2 h
dh
2 2 y
( y 2 1)h
y 2 h
dy
dy
d 2h
dh
2E
2y
( y 2 1)h y 2 h
h
2
dy
dy
d 2h
dh
2y
( 1)h 0
2
dy
dy
2E
Where we have set
.
c.
Solution of the Sturm-Liouville equation
The standard tack is to assume a series solution:
20
h( y ) ao a1 y a2 y 2 a3 y 3 .
Plug this into the S-L equation, and we obtain a recursion relation for the coefficients.
We’d like to have all the sums start at the same place, as well.
dh
j 1
ja j y ( j 1)a j 1 y j
dy j 1
j 0
d 2h
j 2
j
(
j
1
)
a
y
( j 2)( j 1)a j 2 y j
j
2
dy
j 2
j 0
j 1
j 1
y ja j y j 1 ja j y j ( j 1)a j 1 y j 1
j 0
j 0
j 0
( j 2)( j 1)a j2 y j 2 ( j 1)a j1 y j1 ( 1) a j y j 0
j 0
j
Collect the coefficients of y and set equal to zero.
( j 2)( j 1)a j 2 2 ja j ( 1)a j 0
Solve for
2 j 1
a j 2
aj .
( j 2)( j 1)
This result is a recursion relation for two sequences of coefficients. Namely
a0 , a2 , a4 , and a1 , a3 , a5 , .
However, an infinite series is not normalizable. So the series must terminate at some
finite number of terms; say at jmax n . That is an2 0 and all succeeding a j 0
also. The h(y) is a finite polynomial of degree n. Of course, n is going to be the principle
quantum number.
Through the , the n is related to the total energy, E.
Set 2 j 1 0 , solve for E when j = n.
n 2n 1
1
En (n )
2
These are the energy levels for the harmonic oscillator. We have to solve finally for the
h(y).
d.
Hermite polynomials
Of course, we have to get ao and a1 in order to start the recursions for the rest of the
coefficients. For that we use the normalization condition. After that, with 2n 1 ,
2(n j )
a j 2
aj .
( j 2)( j 1)
The results are the coefficients of the Hermite Polynomials, Hn. The first few Hermite
polynomials are shown in Table 2.1 in the text. These form a complete orthogonal set of
21
functions. This means that an arbitrary function can be expanded as a series of Hermite
polynomials, just as any vector can be expanded in terms of the Cartesian unit vectors.
However, there are an infinite number of Hermite polynomials; they may be regarded
mathematically as unit vectors in some infinite-dimensional space.
e.
Harmonic oscillator wave functions
Finally, we put the pieces together to form the harmonic oscillator wave functions.
m
n ( x)
See Fig. 2.5 in the text.
1
4
m
H n
2 n n!
1
22
m x2
x e 2
II. Hydrogen Atom
A.
Schrödinger Equation in Three Dimensions
1.
Hamiltonian Operator
a.
Momentum operators
Hˆ i
t
pˆ x
, pˆ y
, pˆ z
i x
i y
i z
b.
Commutation relations
A commutator is a particular combination of two operators:
[ Aˆ , Bˆ ] f Aˆ Bˆ f Bˆ Aˆ f .
In general, [ Aˆ , Bˆ ] f 0 , that is Aˆ Bˆ f Bˆ Aˆ f . If [ Aˆ , Bˆ ] f 0 , then the two operators are
said to commute.
Take a look at
xˆ, pˆ x f
We say that xˆ, pˆ x i .
Now, try
xˆ pˆ x f pˆ x xˆf
f xf
x
i x i x
f f
x
x
f
i x i x
f
i
i f
xˆ, pˆ f xˆ pˆ f pˆ xˆf
y
y
y
f xf
x
i y i y
f f
x
x
0
i y i y
0
23
Repeat for all possible combinations of
xˆ , yˆ , zˆ, pˆ x , pˆ y , pˆ z .
c.
Total energy
1
p2
Classically, E mv v V ( x, y, z )
V ( x, y, z ) . The corresponding operator is
2
2m
2 2
2
2
2 2
ˆ
V x, y, z
H
V .
2m x 2 y 2 z 2
2m
This is referred to as the Hamiltonian operator.
2.
Spherically Symmetric Potential Energy Function(s)
V r V r , with no dependence on or .
a.
b.
Spherical polar coordinates
1 2
1
1
2 2
r
2
sin
2 2
r r r r sin
r sin
2
2
Separate the variables
2 2
r , , V (r ) r , , E r , ,
2m
Assume that r, , R(r )Y ( , ) and substitute that into the Schrödinger equation.
2 Y 2 R
R
Y
R
2Y
r
sin
VRY ERY
2m r 2 r r r 2 sin
r 2 sin 2 2
2mr 2
Divide both sides by RY and multiply both sides by 2 , and collect terms.
1 2 R 2mr 2
1 1
Y
1 2Y
r
2 V (r ) E
sin
2
0
sin 2
R r r
Y sin
The quantities in the
must separately be equal to a constant,
won’t be valid for all r , , . Therefore, we have two equations,
Angular equation:
Radial equation:
1
Y
1 2Y
CY
sin
2
sin
sin 2
d 2 dR 2mr 2
r
2 V (r ) E R CR
dr dr
These have to be solved separately.
24
C , else the equation
B.
Angular Momentum
Classically,
L r p.
1.
Angular Momentum Operators
a.
Cartesian components of
L
Lx yp z zp y
L y zp x xpz
Lz xpy yp x
Writing the momentum components as operators, we obtain the angular momentum
operators:
Lˆ x ( y z )
i
z
y
Lˆ y ( z x )
i x
z
Lˆ z ( x y )
i y
x
b.
Commutation relations
[ Lˆ x , Lˆ y ] f iLˆ z f ; [ Lˆ y , Lˆ z ] f iLˆ x f ; [ Lˆ z , Lˆ x ] f iLˆ y f
You can verify these by writing them out.
The other angular momentum operator we will need is the total angular momentum
squared, L2.
Lˆ2 f ( Lˆ2x Lˆ2y Lˆ2z ) f Lˆ x ( Lˆ x f ) Lˆ y ( Lˆ y f ) Lˆ z ( Lˆ z ) f
2.
Angular Momentum Ladder Operators
2
Let us assume that f is an eigen function of both the L̂ and
Lˆ2 f f and Lˆ f f ,
L̂z
operators. That is,
z
.where and are numbers, known as the eigen values.
a.
Raising & lowering
Define the angular momentum ladder operators as Lˆ Lˆ x iLˆ y .
Claim: If Lˆ2 f f , then Lˆ2 Lˆ f Lˆ f also.
Proof: Consider the following—
25
[ Lˆ z , Lˆ ] f [ Lˆ z , Lˆ x ] f i[ Lˆ z , Lˆ y ] f
iLˆ y f i (iLˆ x ) f
( Lˆ x Lˆ y ) f
Lˆ f
Now, consider
[ Lˆ2 , Lˆ x ] f [ Lˆ2x , Lˆ x ] [ Lˆ2y , Lˆ x ] [ Lˆ2z , Lˆ x ] f
0 0 Lˆ (iLˆ ) (iLˆ ) Lˆ
Lˆ x [ Lˆ x , Lˆ x ] [ Lˆ x , Lˆ x ]Lˆ x Lˆ y [ Lˆ y , Lˆ x ] [ Lˆ y , Lˆ x ]Lˆ y Lˆ z [ Lˆ z , Lˆ x ] [ Lˆ z , Lˆ x ]Lˆ z f
y
0
z
z
y
Lˆ z (iLˆ y ) (iLˆ y ) Lˆ z f
!!
Similarly, [ Lˆ2 , Lˆ y ] 0 and [ Lˆ2 , Lˆ z ] 0 . It follows that [ Lˆ2 , Lˆ ] 0 , as well.
Therefore, Lˆ2 ( Lˆ f ) Lˆ ( Lˆ2 f ) Lˆ (f ) ( Lˆ f ) . Tah dah. Or q.e.d.
At the same time, ( Lˆ f ) is an eigenfunction of
L̂z
with eigenvalue ( ) .
[ Lˆ z , Lˆ ] f Lˆ f
Lˆ z ( Lˆ f ) Lˆ ( Lˆ z f ) Lˆ f
Lˆ ( Lˆ f ) Lˆ ( Lˆ f ) Lˆ f
z
z
Lˆ ( f ) Lˆ f
( )( Lˆ f )
The remaining task is to evaluate the and the .
b.
Angular momentum quantum numbers
As a vector component, Lz cannot exceed L . Therefore, there are both upper and
lower limits on . Let’s call the maximum, or top . Then,
Lˆ f 0
Lˆ z f f
Lˆ2 f f
What is in terms of ? It’s quickest if we wrote L̂2 in terms of the L̂ , because we
“know” Lˆ2 f and Lˆ f and Lˆ z f , but not Lˆ x f , etc.
Lˆ Lˆ f ( Lˆ x iLˆ y )( Lˆ x iLˆ y ) f
Lˆ
Lˆ
Lˆ2x Lˆ2y i ( Lˆ x Lˆ y Lˆ y Lˆ x ) f
2
Lˆ2z [ Lˆ x , Lˆ y ] f
2
Lˆ z Lˆ z f
26
Solve for Lˆ2 f and plug in f .
Lˆ2 f Lˆ Lˆ Lˆ2z Lˆ z f
Lˆ2 f Lˆ ( Lˆ f ) Lˆ z ( Lˆ z f ) Lˆ z f
0 2 2 f 2 f
( 1) 2 f
Thus, we can identify ( 1) 2 . By a similar token, there is a minimum value of
; call it for the time being.
Lˆ f 0
Lˆ z f f
Lˆ2 f f
Lˆ2 f Lˆ Lˆ Lˆ2z Lˆ z f
0 2 2 f
f
Evidently, ( 1) ( 1) . Therefore, we must have .
2
2
In conclusion, we have the eigenvalues of L̂ z differing by and running from to
. The eigenvalues of L̂2 are equal to ( 1) 2 .
We’ll label the eigenstates according to the quantum numbers m and :
Lˆ2 f m ( 1) 2 f m
Lˆ f m mf m
z
The azimuthal quantum number is m ; the orbital quantum number is .
The next task is to solve the Schrödinger equation for the f m .
3. The Angular Solutions
1
Y
1 2Y
sin
CY
sin 2 2
sin
We need to solve for Y and evaluate the C.
a.
Separate the angles
Assume that Y ( , ) ( )( ) . Substitute into the differential equation and divide
both sides by Y.
27
2
d
d
1
1 d
(sin
)] C sin 2
0
[sin
2
d
d
d
As usual, the two terms must separately be equal to a constant, D .
d 2
D
d 2
d
d
sin
(sin
) (C sin 2 D) 0
d
d
It can be shown that in fact C ( 1) and D m 2 . This done is the following
paragraph b, which may be skipped if time is short.
b.
Polar coordinates
We are using spherical polar coordinates at this point because in the Hydrogen atom the
potential energy function is spherically symmetric. Let us write the angular momentum
vector in terms of polar coordinates.
L [ r ]
ˆ
[r (iˆ ˆj
k )]
x
y
z
1
1
[r (rˆ ˆ
ˆ
)]
r
r
r sin
1
[r (rˆ rˆ) (rˆ ˆ)
(rˆ ˆ )
]
r
sin
1
[ˆ
ˆ
]
sin
We will require also the operators Lˆ , Lˆ , Lˆ , so we need to write the angular unit
x
y
z
vectors in terms of the Cartesian unit vectors, as well.
ˆ iˆ cos cos ˆj cos sin kˆ sin
ˆ iˆ sin ˆj cos
Now, substitute these into L and collect the iˆ , ˆj , kˆ terms. We obtain the polar versions
of the angular momentum operators.
Lˆ x f ( sin
cos cot
)f
i
Lˆ y f (cos
sin cot
)f
i
Lˆ z f
f
i
In terms of these, the raising and lowering operators become in polar coordinates
28
Lˆ f ( Lˆ x Lˆ y ) f
[( sin i cos )
(cos i sin ) cot
]f
i
e i ( i cot
)f
f
Now, Lˆ z f
mf , so evidently f g ( )e im . The f is also an eigenfunction of
i
the operator
L̂2 ,
Lˆ2 f ( Lˆ Lˆ Lˆ2z Lˆ z ) f ( 1) 2 f .
Substitute the polar forms for the operators, as well as f g ( )e im . After some algebra,
we obtain
d dg
Lˆ2 ( ge im ) 2 [e im
(
mg cot )
d d
d
e im (m 1) cot (
mg cot ) m 2 ge im mgeim ]
d
2 ( 1) ge im
2 im
We divide through by e
.
d dg
d
(
mg cot ) (m 1) cot (
mg cot ) m(m 1) g ]
d d
d
( 1) g
Take those derivatives on the left hand side, and after some more algebra,
d 2g
dg
sin cos
m 2 g ( 1) sin 2 g
2
d
d
d
dg
sin
(sin
) [( 1) sin 2 m 2 ]g 0
d
d
sin 2
That last line is identical to the -part of the Schrödinger equation. We can identify that
D m 2 and that C ( 1) .
c.
Spherical harmonics
We have e im . What remains is to solve for the ( ) . [Known as g ( ) in the
preceding paragraph b.]
d
d
sin
(sin
) [( 1) sin 2 m 2 ] 0
d
d
This is a “well known” differential equation, whose solutions are the Associated
Legendre Functions, Pm (cos ) . These are defined in turn in terms of the Legendre
Polynomials, P (x) , with x cos .
29
P ( x)
1 d 2
( x 1)
2 ! dx
m
m
d
P ( x) (1 x ) P ( x)
dx
m
2
2
See Table 4.1 in the text.
So, we do not have to solve for all over from scratch. What is left to do is to
normalize the angular solution. Ym ( , ) e im Pm (cos )
2
Y
m 2
sin d d 1
0 0
This has been done already, so we just look up the result:
Ym ( , )
(2 1) ( m )! im m
e P (cos )
4 ( m )!
(1) m if m 0
where
1 if m 0
See Table 4.2 in the text.
The Ym are called Spherical Harmonics, and comprise a complete set of orthogonal
functions. That means that any function of and can be expanded as a sum of
C.
Ym .
Radial Equation
d 2 dR 2me r 2
V (r ) E R ( 1) R
r
dr dr
2
Since we have m for the azimuthal, or magnetic, quantum number, we’ll start using me
for the mass of the electron. In addition, we’ll find it convenient to make a change of
variable, letting u (r ) rR (r ) .
2 d 2u
2 ( 1)
u Eu
V
(
r
)
2me dr 2
2me r 2
Classically, we do the same thing when solving Newton’s “Law” for orbits in a
gravitational field.
1.
Potential Energy
a.
Coulomb
For the Hydrogen atom, the force acting on the electron is the electrostatic Coulomb
e2
force: The potential energy function is V (r )
. The radial equation therefore
4o r
becomes
30
2 d 2u 2 ( 1)
e2
u Eu
2me dr 2 2me r 2
4o r
b.
Another change of variable
To make it more confusing, but to simplify the writing, we’ll change the variable again.
2me E
me e 2
Firstly, let
. Secondly, let r and o
. Now, the radial
2o 2
equation at least has a cleaner appearance:
d 2u o ( 1)
1
u
d 2
2
2.
Radial Solution
The construction of a solution is broken into two parts. The first part is called the
asymptotic behavior, referring to the solution at very large distance from the proton or
very close to the proton.
a.
Asymptotic behavior
d 2u
i) as , 2 u , which has the solution u ( ) Ae .
d
d 2u ( 1)
u , which has the solution u ( ) C 1 .
d 2
2
Therefore, we propose that the radial solution has the form u ( ) 1e v( ) . We’ll
plug this into the radial equation and assume that v ( ) is a power series in .
ii) as 0 ,
b.
Series solution
We’ll need du e [ 1 v dv ] and
d
d
d 2u
( 1)
dv
d 2v
.
e
[
2
2
]
v
2
(
1
)
2
d 2
d
d
In the radial equation, the common factors of e will divide out.
[2 2
( 1)
]v 2( 1 )
dv
d 2v
( 1)
2 [1 o
]v
d
d
2
Collect the like terms.
d 2v
dv
2( 1 )
[ o 2( 1)]v 0
2
d
d
At this point, we assume that v( ) a j j and substitute it into the differential
j 0
equation.
31
dv
ja j j 1 ( j 1)a j 1 j
d j 1
j 0
d 2v d
j
(
j
1
)
a
j ( j 1)a j 1 j 1
j 1
d 2 d j 0
j 0
j 0
j 0
j 0
j ( j 1)a j 1 j 1 2( 1 ) ( j 1)a j 1 j [ o 2( 1)] a j j 0
The coefficient of each power of must vanish separately.
j ( j 1)a j 1 2( 1)( j 1)a j 1 2 ja j [ o 2( 1)]a j 0
Solve for
a j 1 .
o 2( 1) 2 j
j ( j 1) 2( 1)( j 1)
2( j 1) o
aj
( j 1)( j 2 2)
a j 1 a j
The first coefficient is determined by the normalization condition, a0 A for the
moment, and the rest follow from the formula (called a recursion relation) above.
However, the series must terminate, else v ( ) will diverge at large . That is, the series
cannot be infinitely long, for physical reasons. There must be some maximum value of j
such that a jmax 1 0 . In the recursion relation
2( jmax 1) o 0
o
jmax 1
2
We define n jmax 1 to be the principle quantum number, because it labels the
discrete energy levels (through the o ).
3.
Radial Wavefunctions
Now, we work backward from v ( ) to R (r ) . We had set r and
u 1e v( ) .
1
1
Rn (r ) u 1e vn ( )
r
r
a.
Ground state wave function
The ground state wave function is
32
R ru and
R10
1
1
e ao 1 re r ao aoe r
r
r
mee 2
me e 2 4 o 2 r
ao
e
4o 2
r
Here, the a
ao a
e
a
4o 2
is the Bohr Radius. To evaluate the a o we normalize the R10.
me e 2
R10 r dr
2
0
2
ao
a2
2
e
2r
a
ao
r dr
2
a2
0
ao
2
a
R10 (r )
2
a
3
e
2
a3
1
4
r
a
2
b.
Excited state wave functions
For n 1, 0, 1, 2, 3, , n 1 and m , , 0, , . The vn ( ) is a polynomial of
degree n 1 (that is, jmax ) whose coefficients are given by the recursion relation
2( j 1 n)
a j 1
aj
( j 1)( j 2 2)
and
ao
is determined by normalizing the function.
As they have been defined, the vn ( ) are also “well known” as the Laguerre and
Associated Laguerre Polynomials:
q
x d
Lq ( x) e e x x q
dx
p
d
Lqp p ( x) (1) p Lq ( x)
dx
For us, x
2r
na
.
After all the normalization, we end up with the wave functions for the energy states of the
electron in the Hydrogen atoms:
2 (n 1)! na 2r 2 1 2r m
nm
e Ln1 ( ) Y ( , )
3
na
na 2n[( n )! ]
na
See Tables 4.5, 4.6, and 4.7 as well as Figure 4.4 in the text.
3
r
33
c.
Energy levels
The o is in the total energy, E: E
m e4
2 2
2 e2 2 2 .
2me
8 o o
m e2 2
, n 1,2,3,. . .
e2
2 4o
These are exactly the same energy levels obtained for the classical Bohr model of the
Hydrogen atom. However, Bohr’s model does not have and m .
m e4
me e 2
1
En 2 2e 2 2
2
2 2 2 2
8 o 4n
32 o n
n
The ground state wave function is
100 (r , , ) R10 (r )Y00 ( , )
2
a
3
e
r
a
1
4
2
1
e
r
a
a3
The corresponding ground state energy is E1 = -13.6 eV. The excited states have the
energies En
E1
.
n2
d.
Degeneracy
Recall that for any n , 0, 1, 2, 3, , n 1 and m , , 0, , . Consequently, there
may be several states whose energy is the same. This condition is called degeneracy.
The energy levels are said to be degenerate. For instance, let n 3 . Then there are three
possible values of . For each value of , there are 2 1 possible values of m .
Therefore, the degeneracy of the n 3 level is 1 + 3 + 5 = 9. There are nine wave
functions that have the same energy, E3 .
34
e.
Unbound states
If E 0 , then 0 and is real. But if E 0 , then is imaginary and we obtain freeparticle oscillating solutions, with continuous energy values rather than discrete. We
speak of the continuum of energy states lying above the discrete bound states.
D.
Spin
A property of subatomic particles that Bohr did not put into his model is something called
intrinsic spin. This is a built-in property of particles, just as is mass or electric charge. It
is manifest in the magnetic moment a subatomic particle has, even neutrons. Recall the
Stern–Gerlach experiment.
1.
Spin Angular Momentum Operators
We will define the spin operators by analogy with the orbital angular momentum.
However, the eigenstates will not be functions of angles.
a.
Spin quantum numbers
Define s and ms and such that
Sˆ z ms
Sˆ 2 2 s ( s 1)
where
1 3
s 0, , ,
2 2
ms s, s 1,,0,, s 1, s
b.
Commutation relations
Just like the angular momentum operators,
[ Sˆ x , Sˆ y } iSˆ z
[ Sˆ y , Sˆ z ] iSˆ x
[ Sˆ z , Sˆ x ] iSˆ y
[ Sˆ 2 , Sˆ z ] 0
2.
Spin
1
2
For electrons, s
1
1
so ms only.
2
2
35
a.
Spinors
1
only, there are only two (2) eigenstates of Ŝ 2 and Ŝ z : and . These
2
are sometimes called spinors.
11
Sˆ 2 2 1
22
1
Sˆ z
2
1
Sˆ z
2
Since ms
A general spin state would expressed as a sum of spinors, but there is only two—
2
2
a b where a b 1 .
If we measure the spin of the particle, we’ll get either
getting
or . The probability of
2
2
2
2
is a , is b .
2
2
b.
Spin operators
The operators and spinors can be represented by matrices. Let the spinors be represented
1
0
by column matrices: and . The operators are 2x2 square matrices,
0
1
which can be figured out, knowing the eigenvalues.
c d 1 3 2 1
c d 0 3 2 0
and Sˆ 2
. Carrying
Suppose Sˆ 2
e f 0 4 0
e f 1 4 1
3 1 0
1 0
. In a similar vein, Sˆ z
.
out the multiplications yields Sˆ 2 2
4 0 1
2 0 1
There are also raising and lowering operators, such that Sˆ and Sˆ .
0 1
0 i
and Sˆ
.
Sˆ
2 1 0
2 i 0
36
III. Selected Quantum Mechanical Issues
A.
Vector Spaces
1.
Properties
A vector space consists of a set, or group, of mathematical entities, called vectors and
scalars, which obey certain rules.
a.
Addition
The sum of any two vectors is also a vector. That is, the sum of two members of the
group is another member of the same group.
Vector addition obeys the associative law,
,
and is communtative
b.
Scalar multiplication
The product of a scalar and a vector is also a vector in the vector space. Scalar
multiplication is associative and distributive.
a
ab ab
a b
a b
a a a
c.
Null vector & inverse vector
There exists a null vector, such that 0 . For every vector, there exists an
inverse vector, such that 0 .
d.
Basis vectors & components
A vector in the vector space it is said to be linearly independent of a set of other vectors
if it cannot be written as a linear combination of that set of vectors. A basis set is a
collection of linearly independent vectors that spans the vector space. That means that
every vector can be written as a linear combination of the members of this set. The
dimension of the vector space is the number of vectors in the basis set, namely n.
n
Given a set of basis vectors, ei , an arbitrary vector can be written as ai ei .
i 1
The {ai} are said to be components of the vector with respect to the basis ei .
37
2.
Inner Products
a.
Properties
The inner product of two vectors and is a complex number, such that:
*
0
0 0
b c
b c
b.
Orthonormal vectors
A unit vector is a vector whose norm is equal to one:
1. On the other
hand, two vectors are orthogonal if their inner product is zero: 0.
A set of mutually orthogonal, and normalized, vectors is called an orthonormal set.
i j ij
c.
Schwarz inequality
2
3.
Linear Transformations
A transformation, Tˆ , acts on a vector in the vector space to produce another vector.
T̂
In the case of linear transformations,
Tˆ a b a Tˆ b Tˆ aTˆ bTˆ
a.
Matrix representation
n
Let be some arbitrary vector (in the vector space). a j e j . Further, let a
j 1
n
linear transformation be defined by Tˆ e j Tij ei , (j = 1, 2, 3, 4, . . ., n).
i 1
a T
ei Tij a j ei ai ei
j
i
i j
Parenthetically, there is a form of notation in which the summation signs are not used; the
summation is implied by a repeated index. I.e., Tˆ a j Tˆ e j a j Tij ei ai ei
Tˆ a j Tˆ e j
j ij
The elements of Tˆ are defined with respect to the orthonormal basis set: Tij ei Tˆ e j .
38
A linear transformation acting on a vector can be represented by the multiplication of a
square matrix and a column vector (a matrix with one column), thusly
T11 T12 T1n a1
T
T
T
a 2
21
22
2
n
.
Tˆ
T
T
T
n2
nn a n
n1
b.
Doing things to matrices
transpose
complex conjugate
hermitian conjugate
commutator
inverse matrix
unitary matrix
similarity transformation
4.
Eigenvectors
a.
Eigen value equations
Tˆ , where is a complex number, an eigen value. The is an eigen vector,
or eigen function or eigen state of the operator, Tˆ . Commonly, the transformation is
known, and the tasks are to solve for the eigen values and the eigen vectors. As
1
0 1
.
illustration, consider the spin operator, Sˆ x Sˆ Sˆ
2
2 1 0
Sˆ x x
x
Sˆ
x
Iˆ x 0
det Sˆ x Iˆ 0
2
2 0
2
2
2
ˆ
ˆ
The equation det S x I 0 is called the characteristic equation. Its roots are the
eigen values of the operator in the equation written in matrix form.
2
39
The eigen vectors are obtained by plugging in the known eigen values.
Sˆ x x x
2
0 1 a
a
2 1 0 b
2 b
b
a
a
b
1 1
(normalize d)
x
2 1
b.
Hermitian transformations
40
B.
Formal Quantum Mechanics
1.
Postulates of Quantum Mechanics
a.
One
For a system consisting of a particle in a conservative field there is an associated
wavefunction, . This wavefunction determines everything that can be known about the
system, and is single valued in coordinates and time. In general, it is a complex function,
and may be multiplied by a complex number without changing its physical significance.
b.
Two
A Hermitian operator is associated with every physical observable. The eigen vectors of
such an operator form a complete orthonormal basis set in Hilbert space. An arbitrary
wavefunction can be expanded as a linear combination of eigen vectors.
c.
Three
The development of the wavefunction with time, assuming the system is undisturbed, is
governed by the Schrödinger Equation.
Hˆ i
t
2.
Statistical Interpretation
a.
Measurement
Let Q̂ be the Hermitian operator corresponding to an observable physical quantity, Q.
The eigen value equation for that operator would be Qˆ f q f .
n
n
n
Now consider a particle in a state . c n f n
n
If a measurement of Q is made, one of the eigen values, qn will be the result. The
2
probability obtaining a particular qn is c n , where c n f n f n* dx . When the
measurement is made, the wavefunction collapses to the eigen function, f n . If the
measurement were immediately repeated, the same qn.would be the result.
The expectation value of Q is the weighted sum over all the eigenvalues.
Q qn cn
2
n
This is the expected average of measurements performed on many independent identical
systems.
3.
Uncertainty Principle
a.
Compatible/incompatible
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b.
Uncertainty
Aˆ A Aˆ A Bˆ B Bˆ B
Aˆ A Bˆ B
A2 Aˆ A Aˆ A
A2 B2
1
2i
2
Aˆ A Bˆ B Bˆ B Aˆ A
1 ˆ ˆ
A, B
2i
2
42
2
C.
Time Independent Perturbation Theory
43
