slides on quantum mechanics

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Some Basic Notions
of Quantum
Mechanics
Prepared by Daniel Keren,
[email protected]
Sources:
1.
Feynman’s Lecture Notes
2.
Shankar, “Principles of Quantum Mechanics”
3.
Eisberg & Resnick, “Quantum Physics of Atoms, Molecules, Solids,
Nuclei, and Particles”
4.
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/Ster
nGerlach.html
5.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polar.html
6.
http://www.symonds.net/~deep/stuff/qp/chap03.php
7.
http://www.nobel.se/physics/articles/ekspong/
8.
http://physics.berea.edu/~king/Teaching/ModPhys/QM/Duality.htm
9.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/mulslid.html
10. http://physics.ucsd.edu/was-sdphul/labs/2dl/exp6/exp6-BACK.html
11. http://www.microscopyu.com/articles/formulas/formulasresolution.html
12. http://newton.phy.bme.hu/education/schrd/2d_tun/2d_tun.html#animation
13. Brandt & Dahmen , The Picture Book of Quantum Mechanics.
14. www.falstad.com
Some Classical Mechanics
Isaac Newton
• Lagrangian formulation: particle wants to minimize
the total of the Lagrangian ℑ = T − V = ℑ( x, x& , t ),
where:
T = kinetic energy
V = potential energy
x= location
x& = velocity (dx dt )
t = time
• Given initial and final location/time ( xi , ti ), ( x f , t f ),
the path x(t ) classically taken minimizes
(extremizes) the action
tf
S [ x(t )] = ∫ ℑ( x, x& )dt (here V is time independent)
• Why?
ti
????
• Via calculus of variations, this leads to the
following equation, which holds in the classical
path for every t:
∂ℑ
d  ∂ℑ 
= 

∂xi (t ) dt  ∂x&i (t ) 
• Often, the following notations are used:
x ≡q
i i
∂ℑ
F ≡
i ∂q
i
(force)
∂ℑ
p ≡
i ∂q&
i
(momentum)
• And so
dpi 
dv d (mv) dp 
=
=
Fi =
 ma = m

dt 
dt
dt
dt 
Force
• In the continuous case, often ℑ = ℑ( f , f& , f ′, t )
and one minimizes ∫∫ ℑ( f , f& , f ′, t )dxdt ⇒
∂ℑ d ∂ℑ d ∂ℑ
= 0.
−
−
&
∂f dt ∂f dx ∂f ′
f ′ Stands for the coupling of
“infinitely close particles”
(e.g. points of a string).
The Hamiltonian
• We have
∂ℑ
p=
∂q&
• Suppose we want to find an expression, call it Η ,
such that
∂Η
= q&.
∂p
So :
dq&
Η = ∫ 1 ⋅ q& dp = pq& − ∫ p
dp =
dp
∂ℑ dq&
pq& − ∫
dp = pq& − ℑ
∂q& dp
• What is Η in physical terms? Look at a particle
moving under a simple potential:
 mv 2
 mv 2
Η = pq& − ℑ = (mv ) x& − 
− mgx  =
+ mgx
2
 2

• So Η is the TOTAL ENERGY (kinetic + potential).
• What are the Hamiltonian equations? As we saw
∂Η
= q&
∂p
• And also
∂Η ∂ ( pq& − ℑ)
∂ℑ
=
= −
= − p&
∂q
∂q
∂q
• So
∂Η
= q&
∂p
∂Η
= − p&
∂q
Example: Free Fall
2
2
mv
p
Η=
+ mgx =
+ mgx
2
2m
• Note that we have to write the energy as a
function of the momentum and the coordinates,
as opposed to the coordinate and its time
derivatives (the latter is appropriate for the
Lagrangian formulation).
• The equations are (remember: q
= x, q& = v)
∂Η
p
= q& ⇒ = v (nothing new)
∂p
m
∂Η
= − p& ⇒ mg = −(mv&) = − ma ⇒ a = − g
∂q
Which is indeed the free fall equation.
Example: Harmonic Oscillator
• The point at which the spring is at zero potential
energy will be taken to be x = 0. The
Hamiltonian is then
p 2 kx 2 ∂Η
,
Η=
+
= − p& ⇒
2m 2
∂q
k
kx = −mv& = −m&x& ⇒ &x& = − x ⇒
m
 k 
 k 
x = A cos
t  + B sin 
t 
 m 
 m 
• A,B are determined by the initial conditions.
Example: Coupled Harmonic Oscillator
x1 ,x2 are the displacements from equilibrium.
[
]
p12 p22 k 2
2
2
Η=
+
+ x1 + x2 + ( x1 − x2 ) ⇒
2m 2m 2
2k
k
k
− 2k
&x&1 =
x2
x1 + x2 , &x&2 = x1 −
m
m
m
m
• Let us discuss differential equations of the form
Ωx = &x&
Where:
x is a vector of functions.
• Ω is Hermitian with negative eigenvalues − λ12 ,−λ22
and eigenvectors e1 , e2
 e1 
Denote E ≡   ( E is a 2 × 2 matrix)
e 
 2
•
t
&
&
Ωx = x ⇒ EΩE y = &y&
t
t
EΩE = D, x = E y, y = Ex
2
 − λ1
0 

D = 
, E is unitary
2
− λ2 
 0
• After diagonalizing, the system is easy to solve.
If we assume that the initial velocity is zero,
y = cos(λ1t ), cos(λ2t )
• Initial conditions:
x1 (0) = a, x2 (0) = b, x&1 (0) = 0, x&2 (0) = 0
 a   a′ 
• For y, the initial conditions are then E   =  
 b   b′ 
so the solution is
0  a′ 
 cos(λ1t )
 a′ 
  ≡ C  
y = 
cos(λ2t )  b′ 
 b′ 
 0
• To return to x,

0   a 
0  0
a
t  c11 (t )
 E   =
 + 
x = E CE   = E 
0   0 c22 (t )   b 
b
 0
t
t a
c11 (t )e1e1 + c22 (t )e2 e2   ⇒
b
x(t ) = U (t ) x(0) , U (t ) = c11 (t ) I I + c22 (t ) II II
t
[
]
Where we have taken the opportunity to
introduce a common QM notation, Dirac’s
famous bra and ket:
• ? , ? stand for column and row vectors.
•
I , II stand for Ω′s eigenvectors.
• c11 (t ), c22 (t ) are solutions to the differential
equations with one variable, &f& = ω1 f , &f& = ω 2 f
where ω1 , ω 2 are Ω′s eigenvalues.
• U (t ) is called the propagator and is of
fundamental importance in QM.
Example: Vibrating String
• Once we realize that the propagator formulation
carries over to the infinite dimensional case, we can
write down in a straightforward manner the
solution to more general problems, such as the
vibrating string:
L
• First, we have to write down the corresponding
differential equation. Think of the string as being
composed of many small springs, that is, the
limiting case of many masses coupled by springs:
Assume the
spring constant k
to be equal to 1.
qi
i −1 i i +1
N N N
L


≡ h
• The Lagrangian is the limit of  denote
N


[
]
1
1
... + mq&i2 + ... − ... + h 2 + (qi − qi −1 ) 2 + h 2 + (qi +1 − qi ) 2 + ...
2
2
qi −1 − 2qi + qi +1
hence the equations are
= q&&i
m
but in the limit, qi −1 − 2qi + qi +1 is just the wellknown approximation to the second derivative, so –
ignoring constants – the spring equation is
∂ f ( x, t ) ∂ f ( x, t )
=
2
2
∂x
∂t
2
2
∂ 2 f ( x, t ) ∂ 2 f ( x, t )
• To solve
, proceed as before:
=
2
2
∂x
∂t
• First, find the (normalize) eigenvectors/eigenvalues
∂2 f
of the operator 2 . Since they’re restricted to be
∂x
zero at 0 and L, they are
1
2
2
 mπ
I m =   sin
 L
 L
mπ

x , eigenvalue = − 2
L

2
2
• What are the equivalents of c11 (t ), c22 (t ) ?
As before, if we restrict the initial velocity to be 0,
 mπ
the solutions are cos
 L

t , and – also as

before – the general solution can be written with
a propagator:
 mπ  t
U (t ) = ∑ cos
t I m I m , f ( x, t ) = U (t ) f ( x,0) ⇒
 L 
m =1
∞

 mπ 
ψ ( x, t ) = ∑ cos
t  I m I m  ψ ( x,0) =
 m =1  L 

∞
 mπ 
cos
t  I m ψ ( x,0) I m
∑
 L 
m =1
∞
But in the limit the internal product is just an
integral, so
L


 mπ  
t   ∫ f ( x,0) I m dx  I m
f ( x, t ) = ∑ cos
 L 0
m =1

∞
Why all this?....
• Later, we will see that in QM, every single
particle has a function – the probability
amplitude – associated with it, and the main
goal is to determine how it changes over
time. It turns out that this function satisfies a
first order (in time) differential equation (as
opposed to our examples which had a
second time derivative), called
Schrödinger’s equation, and one seeks the
propagator which – given the probability
amplitude at t = 0 – computes it for every t:
ψ ( x, t ) = U (t ) ψ ( x,0)
Poisson Brackets and Canonical Transformations
Let ω ( p, q) be some function of the state variables
p,q, where we assume no explicit dependence on t.
Then its change over time is given by
 ∂ω
∂ω 
dω
= ∑ 
q&i +
p& i  =
∂pi 
dt
i  ∂qi
 ∂ω ∂Η ∂ω ∂Η 
∑i  ∂q ∂p − ∂p ∂q  ≡ {ω , Η}
i
i 
 i i
• For any two ω ( p, q), λ ( p, q) define the Poisson
bracket by
 ∂ω ∂λ ∂ω ∂λ 

{ω , λ} = ∑ 
−
∂pi ∂qi 
i  ∂qi ∂pi
• Note that
{q , q }= 0 {p , p }= 0 {q , p }= δ
i
j
i
j
i
j
ij
• We may want to define new variables
q → q ( p, q), p → p ( p, q)
A straightforward calculation yields that if we
want Hamilton’s equations to hold in the new
coordinates, we must have
{q , q }= 0 {p , p }= 0 {q , p }= δ
i
j
i
j
i
j
ij
Such a transformation is called canonical.
Note the similarity: in classical mechanics
dω
={ω ,Η}
dt
And in quantum mechanics
d Ω
−i
= [Ω,H ]
dt
h
Examples:
• Rotation is a canonical transformation (check it).
• Important: the two body problem. Assume we have
two bodies in space, with locations and masses
given by ( m1 , r1 ), ( m2 , r2 ). Assume also that there’s
a potential which depends only on their relative
distance, V ( r1 − r2 ). We can then define a
canonical transformation that reduces the analysis to
that of two independent bodies:
m1r1 + m2 r2
, mCM = m1 + m2 , VCM = 0 (the center of mass)
m1 + m2
mm
r = r1 − r2 , m = 1 2 ≡ µ , V = V ( r ) (the reduced mass)
m1 + m2
rCM =
• Thus the center of mass is a free particle, and the
reduced mass moves under the potential V. One can
then solve for them separately. One such system, for
example, is the hydrogen atom.
• We will later show how one can guess such a
transformation; it follows from the requirement that
the Hamiltonian be diagonal in the new coordinates.
The Electromagnetic Lagrangian
• The force acting on a charge q, moving at speed v,
due to an electric field E and a magnetic field B, is
v 

F = q E + ×B , with the potentials A,φ
c 

1 ∂A
satisfying ∇×A = B, E = −∇φ −
.
c ∂t
1
q
Next, define a Lagrangian ℑ = mv⋅v − qφ + v⋅A
2
c
d  ∂ℑ  ∂ℑ
We have to check that 
 = , it yields
dt  ∂vi  ∂xi




q
d
qA 
d
(
)
(
)
= − q ∇φ + ∇ v⋅A ⇒ (mv )=
mv +


dt
c
c
dt 1
24
3
 "4

canonical 
 momentum" 



q
dA
 1 ∂A 
− q(∇φ )+ −
+∇(v⋅A ) = q(−∇φ ) + q  −
+
c
dt
c
dt



{
 ∂A +( v⋅∇ )A

 ∂t



 
q
v 
(
)
(
)
q
v
A
v
A
E
∇
⋅
−
⋅
∇
=
+
×B . Note however that

c  1442443  
c 
 v× ∇×A =v×B 

 

q
qφ − v⋅A is not really a potential.
c
The Electromagnetic Hamiltonian
1
As before, H = p⋅v − ℑ= mv⋅v + qφ =T + qφ
2
But in order to write it correctly, it has to be
q
expressed in terms of p = mv + A, so
c
2
q
p− A
c
+ qφ .
H=
2m
QM was developed in order to explain
physical phenomena which were not
consistent with classical physics.
Max Planck
Werner
Heisenberg
Louis de Broglie
Albert Einstein
Erwin
Schrödinger
Paul Dirac
Niels Bohr
Max Born
Richard Feynman
Part I: Physical Intuition
and the FiniteDimensional Case
Prelude: The Nature of Light
Light was thought to be made of particles, which
move in a straight line. Later, light interference
was discovered (Young, 1801), which led to the
conclusion that light is a wave:
Two slit
experiment
The mathematics behind this: a wave can be described by
the equation e i ( kx − ω t ) . Such a wave moves through
space at a speed of ω k . ω is the frequency, k the wave
number. If another wave is ∆ behind, their sum is
e i ( kx − ω t ) + e i ( k ( x + ∆ ) − ω t ) = e i ( kx − ω t ) (1 + e ik ∆ ) = 1 + e ik ∆
which peaks/is zero when k∆ = 2nπ / k∆ = ( 2n + 1)π .
Denoting λ = 2π k (the wavelength), there is a maximum
if ∆ = nλ , which is hardly surprising – it means that if the
lag between the two waves is a multiple of the wavelength,
they contribute to each other (they are in phase).
Wave interference: as opposed to the drawing,
assume d >> a, x. Then the lag between the two
waves is d 2 + x 2 − d 2 + ( x − a ) 2 ≈ ax = ∆.
d
so, the maxima correspond to ax d = nλ , which
implies that the distance between adjacent maxima
is λd a (this can actually be used to measure the
light’s wavelength).
• Mathematically, if the (complex) wave functions
are h1 , h2 ,then the energy arriving when slit i is
2
open is hi , and the energy arriving when both are
2
2
open is NOT h1 + h2 , but
2
2
2
h1 + h2 = h1 + h2 + 2 h1 h2 cos(δ )
14
4244
3
h1
δ
interference term
So far so good – but it turns out that the wave
model fails to explain certain phenomena.
h2
When you shine light upon certain metals,
electrons are emitted (the photoelectric effect).
It turns out that
• The emitted electrons move with greater speed if the light
has a higher frequency.
• No electron is emitted until the light passes a threshold
frequency – no matter how strong the light is.
• The electrons are emitted too quickly.
All this contradicts the hypothesis about the wavelike nature
of light.
• Einstein solved this problem (and was awarded the Nobel
prize for that), by suggesting that light is composed of little
“quanta”, which impact like “baseballs”, and not like waves.
These were later termed photons, and it turned out that a
photon with frequency ω has an energy and momentum
given by
h is a very
small constant.
E =hω
p =hk
note: E = pc.
It also turns out that not only don’t the
lightwaves behave like waves – also,
particles don’t behave like particles.
Let’s go back to the two slit experiment. How will it
look with particles (electrons)? We may assume it
will look like
i.e., the number of electrons hitting any specific
point on the wall when both slits are open, will be
the sum of the numbers when only the
first/second slit is open.
But that doesn’t happen; the electrons also
interfere. That is, at some points, LESS
electrons hit when BOTH slits are open!!
This happens even if the electrons are fired
very slowly (one at a time).
There is no way to explain this – it just happens.
• Note: we can immediately realize that each
electron cannot go through one slit. However, they
don’t “break”, because when they arrive at the
wall, the energy corresponds to a “whole” electron.
• Fortunately, the simple model
that describes
2
wave interference ( h1 + h2 ) also explains particle
interference. The problem is, of course, to compute
the so-called “probability amplitudes” h1 , h2 .
There are h1 , h2 such
that Pi = hi 2 and
2
P12 = h1 + h2 .
electron gun
What happens if we try to watch the electrons? If
we see them all, there’s no interference!
To be explained
later. Intuitively, the
measurement forces
the particle to
assume a definite
position.
Some QM axioms:
• Probability is the squared absolute value of the
amplitude.
• If there are several possibilities for an event to
happen, the amplitudes for these possibilities are
summed. So, to get the total probability, you first
sum and then take absolute value squared – thus
there is interference.
• When it is possible to decide what alternative was
taken, you sum the probabilities (as usual – no
interference).
• There is uncertainty; we cannot predict what will
happen, even if we know everything in advance. This
uncertainty is reflected, for example, in the inability to
exactly describe both a particle’s position and
momentum – the famous Heisenberg uncertainty
principle. All we can do is provide probability
distributions. But, as opposed to our standard
understanding of what uncertainty is, here it’s much
deeper; it’s not “I know the particle is somewhere, I just
don’t know where” – the particle isn’t anywhere, so to
say. And if we try to nail it down, we lose its
momentum. There’s no way out of this.
It’s strange – but it works.
To quote Feynman…
“That’s how it is. If you
don’t like it, go to a
different universe, where
the laws are simpler” – R.P
Feynman, NZ, 1979.
The Heisenberg Uncertainty Principle
• Take the limiting case of a wave with amplitude
e i ( kx − ω t ) . It has exact momentum ( h k ) and
energy (h ω ) but there’s no information on its
location, because for every x the probability is 1.
• Real particles have amplitudes that are not
uniform in space. They may look like a Gaussian,
meaning that there’s a far higher probability to
find the particle near the Gaussian’s center. But in
order to get a Gaussian, you have to add many
“pure waves”. Each “pure wave” has definite
momentum – but when you mix them, the
uncertainty in the momentum increases. The
narrower the “position Gaussian”, the broader the
“momentum Gaussian” (more later).
x
Momentum certain, but zero Position certain, but zero
knowledge on position
knowledge on momentum
Momentum
uncertainty
Position
uncertainty
Trying to measure both location and momentum
The smallest distance between two points in an object that
will produce separated images is ∆ x ≈ λ sin(θ ). If a
photon has energy hω , it possesses momentum h λ . To
be collected by the lens, the photon must be scattered
through any angle between − θ to θ . So, the xcomponent of the momentum may have any value
between − h sin(θ ) λ to h sin(θ ) λ . Thus the
uncertainty in the electron's momentum is ≈ h sin(θ ) λ .
So, the product of the uncertainties is ≈ h.
Intuitively, we need a “powerful” photon (high frequency) to
get good position measurements; but the higher the energy,
the more it knocks off the electron, hence the momentum
uncertainty becomes larger.
Interlude – the Resolution of a Microscope
The resolution of an
optical microscope is
defined as the shortest
distance
between two
points on a specimen that
can still be distinguished
by the observer or
camera system as
separate entities. When
becomes close to , it’s
impossible to distinguish.
That’s why electron
microscopes are used
(much smaller
wavelength).
a
a
λ
d = 104 , a = 0.005, λ = 0.01
Distance between peaks
should be λd a = 2 ⋅ 104
d = 104 , a = 0.1, λ = 0.01
Distance between peaks
should be λd a = 103
d = 104 , a = 0.01, λ = 0.01
Distance between peaks
should be λd a = 104
d
a
λd
θ
≈ d sin(θ )
a
In order to distinguish the objects, the two
maxima should fall in the view area, so we
must have
λd
λ
d sin(θ ) ≥
⇒a≥
sin(θ )
a
Another experimental demonstration of the
uncertainty principle – shooting a particle
through a single slit (Fraunhofer diffraction):
wavelength = λ
One can prove that
θ ≈ λ W.
θ
y
slit width = W
Before the particle (wave) hits the barrier, we can assume
that its y-momentum is zero. After it passes the slit, we know
its position with an accuracy of ∆y ≈ W ,but there’s an
uncertainty in the y-momentum, ∆p y ≈ pλ W (where p is
the original momentum). So, ∆y∆p y ≈ pλ , but QM tells
us that pλ = h ( = 2πh ).
• The narrower we make the slit, the more confident we are
about the location; but then the diffraction pattern becomes
wider, and we lose confidence in the momentum.
De Broglie Waves: The Wavy Nature of Particles
In the previous slide, we adapted to particles a result
connecting the wavelength and momentum of a photon. This
is based on De Broglie’s ingenious “hunch” in 1924: “if
waves (light) behave also like particles, then particles should
also behave like waves”. It turns out that a moving body
behaves in certain ways as though it has a wavy nature. Its
wavelength λ is defined, just like
for a photon, by
h
h
λ= =
.
p mv
If so, electrons for example must diffract. This was proved
experimentally by Davisson and Germer in 1927: they shot
electrons at a crystal made of layers. An electron is scattered
from different layers, hence there’s a phase difference, hence
there’s diffraction.
λ
Note: larger objects (baseballs) also show interference, but
is so small
that the interference pattern is practically impossible to create and observe.
Application: the Radius of an Atom
• Let the “average radius” of the hydrogen atom be
denoted by a.
• The momentum p should be, on the average, h a.
this implies that the average kinetic energy is
p2
h2
=
2m 2ma 2
• The potential energy is − e 2 a , where e is the
electron charge. Hence the total energy is
h2
e2
E=
−
2
2ma
a
• It’s easy now to find the a minimizing this
expression, and it’s indeed pretty close to the true
radius.
• Why then can’t we squash an atom, and why doesn’t the
electron collapse onto the nucleus? Because then the
uncertainty in the position will go to zero, and the uncertainty
– and hence the average – of the momentum (and the energy)
will go to infinity.
More on Probability Amplitudes
Two slit experiment again:
Denote the amplitude that a particle from s will
arrive at x by Dirac’s “bra and ket” notation:
particle arrives at x particle leaves s ≡ x s
We will later see that it’s just an inner product.
We know that the probability is the square of the
amplitude. We also know that amplitudes add:
x s
both slits open
= x s
slit 1 open
+ x s
slit 2 open
Another law about amplitudes: the amplitude
for a path is the product of the amplitudes to go
part of the way and the rest of the way.
x s =
∑
xα α i i s
i =1, 2
α = a ,b , c
• We will later address amplitudes with a time
variable (e.g. what is the amplitude for a particle
to be somewhere at a given time).
• Another amplitude rule – if two particles don’t
interact, the amplitude for them to do two things
is the product of the individual amplitudes.
Applying the amplitude rules to the two-slit
experiment with a light source (trying to find
what slit the electron went through)
x
s
If there’s no light, x s = φ1 + φ2 = x 1 1 s + x 2 2 s
If there’s light, there are certain amplitudes for a photon to
arrive at D1 ( D2 ) if the electron went through 1 – call these
amplitudes a(b). Applying the principles and some symmetry
D2
considerations:
1
6photon
47at4D8
6photon
47at4
8
2
Pr( electron at x ) = aφ1 + bφ2 + aφ2 + bφ1
2
If we can detect the slit via which the electron went with total
2
2
certainty, then b=0 and the probability is φ1 + φ2 , hence no
interference at all. If we try to decrease the photon energy, we
will have to increase its wavelength, hence b will grow and
we’ll have more and more interference.
Question: how can a light source affect the
interference of baseballs?
Another Example: Scattering from a Crystal
neutron
source
crystal
θ
neutron
counter
There are some crystals for which the scattering has
both “classical” and interference components:
number of
neutrons
θ
Why does this happen?
• Sometimes the crystal nuclei have a property called
“spin” (they may be “spin up” or “spin down” – more
later).
• Usually, the neutron is scattered without “leaving a
mark” on the specific atom which scattered it. That case is
like the two slit experiment without light, and we get
interference.
• But sometimes the neutron can “leave a mark” on the
atom that it scattered from – by changing the latter’s spin
(and also its own spin). That is equivalent to the two-slit
experiment with light; in that case, we can distinguish
what path the neutron took, and therefore we should add
probabilities and not amplitudes (so no interference).
• In this crystal the neutron sometimes changes the nuclei
spin and sometimes it doesn’t – hence we have a mixture
of both distributions: interference and no interference.
Note: it makes no difference whether we try or don’t
try to find the atom from which the neutron scattered!
+
=
result
spin changed
Spin not changed
Bosons and Fermions
Let’s look at a scattering of two particles:
D1
D1
D2
D2
• If the amplitude for the left event is f (θ ), then
the probability of any particle arriving at the
detector D1 is
π 
Pr = f (θ ) + f (π − θ ) : θ = ⇒ Pr = 2 f  
2
2
2
2
π
2
• But what if the particles are identical? Then we
cannot differentiate between the left and right events,
and we have to add amplitudes before taking the
squared absolute value to find the probability.
•. Exchanging the particles does not change the
physics (probabilities), but it may change the phase.
Since exchanging again brings us back to the
original system, the amplitude must either remain
the same or be multiplied by − 1. It turns out that
there are two types of particles: for Bosons the
amplitudes are the same, for Fermions they are
inverse.
• So, for a scattering of two bosons, the probability
of one of them arriving at D1 is
π 
Pr = f (θ ) + f (π − θ ) : θ = ⇒ Pr = 4 f  
2
2
2
π
2
• And for two Fermions it is
2
Pr = f (θ ) − f (π − θ ) : θ =
π
2
⇒ Pr = 0
• So, two Fermions cannot scatter at an angle of
π
2
!
1
2
a
b
• Let’s look at a double scattering. We know that the
probability for this is
1a
2
2
2b .
• Assume that the particles are distinct. The probability for
( a → 1 ∧ b → 2) ∨ ( a → 2 ∧ b → 1)
is
1a
2
2b
2
+ 1b
2
2
2a .
• If 1 approaches 2, we can denote
1 a = 2 a = a , 1 b = 2 b = b.
2
2
so the probability for both particles to go to 1 approaches 2 a b .
• But if they are Bosons, we cannot differentiate which went
where, so we have to add amplitudes and then square:
1a 2b + 1b 2 a
2
2
→ 4a b
1→2
2
• This yields a probability two times bigger than for distinct
particles. So: Bosons like to go to the same place, Fermions
like to stay afar from each other. This is the Pauli exclusion
principle, and without it atoms would have looked very
different (electrons would stick together).
Light Polarization
Waves can be characterized by being either:
• Transverse: the thing that is "waving" is
perpendicular to the direction of motion of the
wave (light waves).
• Longitudinal: the thing that is "waving" is in
the direction of motion of the wave (sound
waves).
Z
Z
Two possible polarizations of light waves. The
arrows depict the oscillations of the
electromagnetic wave. These will be called x and y
polarizations.
Polarizers:
Polarization by reflection.
There are apparatuses that allow only light as a
certain polarization to go through (LCD, polaroid
sunglasses).
• From the QM point of view, every single photon
has a mixture of the two polarizations associated
with it, and we can only compute the probability
that it goes through a certain polarizer.
• x and y resp. refer to the state of a single
photon in an x resp. y beam which is polarized in
the classical sense.
• Filtering with a polarizer in angle θ relative to
x-y yields the state x ' = cos(θ ) x + sin(θ ) y .
If we place a second polarizer at angle 0, then the
probability that x ' goes through is cos2 (θ ) , which is
in perfect accordance with classical physics – but
here, it’s not that the wave energy is decreased by this
2
factor, but that only cos (θ ) of the photons go
through – and those which do, don’t lose any energy.
Thus – a classical, macro, continuous
phenomena is explained by a (rather
different) micro, discrete, QM phenomena.
Light is (classically) said to be RHC polarized if
its x and y components are equal but 90oout of
phase (in this case, the field is not linear as before,
but oscillates in a circle which lies in the plane
perpendicular to the light’s direction). The QM
interpretation for a single photon is
1
( x +i y
RHC =
2
)
Where does the i come from? We know that a 90o
phase shift, when repeated twice on y , yields − y .
So, we can see why it is represented by a
multiplication with i.
Interlude: Wiesner’s Quantum Money
• Bank has a list of bills; each has a serial number
and 20 photons embedded in the bill, each with
one of four possible polarizations:
• A would-be counterfeiter cannot copy the bill,
since any attempt to measure the polarization
will not only probably destroy the original
polarization, but will also leave too many
possibilities. For example, if a photon measured
with a green polarizer gives zero (the photon
doesn’t pass), it can be either a black, red, or
blue photon. As with the electron, the
measurement operation changes the state; this
is why the method works.
• Note that the bank may say “yes” even if some photons are not
correctly polarized – but the probability of the counterfeiter to
obtain this can be made arbitrarily small.
Introduction to
base states:
spin and the
Stern-Gerlach
experiment
The total amount of deflection is a function of
• The total amount and distribution of electric charge on the ball.
• The orientation and rate of spin. As the rate of spin increases, so does the deflection. As
the axis of the spin becomes more vertical, that amount of deflection also increases.
If the beam from the electron gun is directed to the magnets, as shown to the right, the
beam is split into two parts. One half of the electrons in the beam are deflected up, the
other half were deflected down. The amount of deflection up or down is exactly the same
magnitude. Whether an individual electron is deflected up or down appears to be random.
Stern and Gerlach did a version of this experiment in 1922.
This is very mysterious. It seems that the "spin" of electrons comes in only two states. If
we assume, correctly, that the rate of spin, total charge, and charge distribution of all
electrons is the same, then evidently the magnitude of the angle the spin axis makes with
the horizontal is the same for all electrons. For some electrons, the spin axis is what we
are calling "spin up", for others "spin down".
You should beware of the term "spin." If one uses the "classical radius of the electron"
and the known total angular momentum of the electron, it is easy to calculate that a point
on the equator of the electron is moving at about 137 times the speed of light! Thus,
although we will continue to use the word "spin" it is really a shorthand for "intrinsic
angular momentum.“ It has no classical counterpart.
Building a Spin Filter
The blocking mechanism is really a
“filter” – allows only one spin type to
go through.
Shorthand notation for “spin up” filter. On the average, it allows half
the electrons through. We can rotate the filter. Still, on the average, it
allows half the electrons through.
None of those which
passed the first will pass
the second (nothing passes
both of them).
All those which passed the
first will pass the second.
Half of those which passed
the first will pass the
α 
second: in general, it’s cos2  
One quarter will pass.
2
To realize that something strange is going on,
think about a filter with both paths unblocked
being here:
Correlation measurements in a radioactive substance
that emits a pair of electrons in each decay:
If the right electron passes, then its left hand companion does not
pass its filter, and vice-versa. We say that each radioactive decay
has a total spin of zero: if one electron is spin up its companion is
spin down.
Again, one-half of the right hand electrons pass through their
filter and one-half of the left hand electrons pass through their
filter. But this time if a particular right hand electron passes its
filter, then its companion left hand electron always passes its
filter. Similarly, if the right hand electron does not pass its filter,
its companion electron doesn't pass through its filter either.
1.
One-half of the right hand electrons emerge from their filter.
2.
One-half of the left hand electrons emerge from their filter.
3.
If a particular right hand electron passes its filter, one-half of
the time its companion left hand electron will emerge from its
filter, one-half of the time it will not.
It turns out that in the “world of Stern-Gerlach
apparatuses” (SGA), the states U and D (denoting
spin up and down) form something that resembles
a basis in linear algebra. What does this mean?
For example:
• If an electron is in one of these base states (B),
we can predict the probability it goes through any
SGA, regardless of its previous history.
• Every state and every transition amplitude can
be describe as a mixture of these states:
χ =∑ i χ i
i∈B
χ φ =∑ χ i iφ
i∈B
• Since the base states are complete (i.e. they span
all states), we must have that the sum of
probabilities of a state to be in any of the base
2
∗
states is 1, or
i χ =
i χ i χ =1
∑
i∈B
∑
i∈B
But the probability of a state to go to itself must
be 1, so according to , we must have
χ χ = ∑ χ i i χ = 1 ⇒ ∀ χ ,φ χ φ = φ χ
i∈B
∗
Just like in linear algebra, we have many sets of
base states; e.g. the outputs of SGAs which are
rotated in various angles, each time leaving only
one beam unblocked. And just as in linear algebra,
we can switch from description in B1 = {i} to a
description in B2 = { j}, if we know j i for all
i,j. For example, for spin ½, and for rotation
around the longitudinal axis, the transition matrix
is

θ 
θ  
 cos  sin  
2
 2


θ 
θ 
 − sin 2  cos 2  
 
 

Operators and base states
Suppose we do something to a particle – for
example, make it go through a mess of SGAs.
Let’s call this mess A an operator.
• If we have such an operator, we may ask
questions such as: given two states χ , φ , what is
the amplitude for φ → A → χ , which will be
denoted χ A φ . We get another familiar result
from linear algebra (remember that it makes more
sense if you read from left to right):
χ Aφ = ∑ χ j j A i i φ
i, j
Where i,j range over base states.
• Other “linear algebra laws” follow – the
composition of operators is the matrix product of
their individual matrices, we have the usual laws
for transforming between bases, etc.
• The matrix product representation encapsulates
a rather strange result – composition of operators
(like position and momentum) is not necessarily
commutative, as opposed to classical physics.
This has profound consequences.
Dependence of Amplitude on Time: The
Hamiltonian
•
How does the amplitude change with time?
•
For a particle in a state of definite energy E0 ,
the amplitude is independent of position and is
given by ae − i ( E0 h ) t (intuition: it’s like a wave,
and it has zero uncertainty in momentum, hence
infinite uncertainty in position).
•
Note that any probability question for such a
particle is independent of time; hence such a
particle is said to be in a stationary state.
•
In order to have probabilities which change in
time, the particle has to have more than one
energy state: note that a e −i ( E1 h ) t + a e −i ( E2 h ) t
1
2
depends on the time t.
•
The amplitude of a particle in uniform motion
is ae − ( i h )( Et − p⋅ x ) , where E is the particle’s total
energy.
2
Interlude: Jumping Over a Potential Barrier
Suppose that a particle with a certain energy
somehow goes over a potential barrier which is
larger than its energy (this is, of course, impossible
in classical physics). Its energy has to be negative,
and since E = p 2 2m, its momentum has to be
imaginary. Since the amplitude behaves like
− ( i h )( Et − p⋅ x ) we may expect it to look like e −αx .
ae
,
Incredibly enough, this is true. It explains, for
example, particles getting into places which are
forbidden by classical physics. More on this later.
A Little More on Bras and Kets
• Recall that if i ranges over base states, then
χ φ =∑ χ i iφ
i
• Inspired by this, we write φ =
• And further: | =
∑i
∑i
iφ
i
i
i
• i is a bra and i is a ket (these notations are
valid not only for base states).
• If φ =
∑iC
i
i
and χ =
∑ i D , then
i
i
χ = ∑ Di∗ i and χ φ = ∑ Di∗Ci
i
i
• So it’s just like an inner product. These notions
extend naturally to an infinite number pf base
states, with the L2 structure( f , g ) = ∫ f ∗ g (plus
some functions which are not in the classical L2 ).
• For an operator A we have A = ∑ i i A j j
i, j
• It’s common in QM to denote the state vector ψ
by anything which identifies it. For example, a
state with a definite momentum p will be denoted
by p , etc.
How does the amplitude change under an
apparatus, or operator? Passage of time is also an
operator; denote by U (t2 , t1 ) the change which takes
place between time t1 and t2 . By that we mean
that if a system is at state φ at time t1 , then its
amplitude to be at state χ in time t2 is
χ U (t 2 , t1 ) φ So, it’s enough to know
i U (t 2 , t1 ) j for base states i, j.
Since U (t3 , t1 ) = U (t3 , t2 )U (t2 , t1 ), it’s enough to
know U (t + ∆t , t ). Denote as before Ci (t ) = i ψ (t ) ,
and U i , j = i U | j . All this yields
Ci (t + ∆t ) = ∑U i , j (t + ∆t , t )C j (t ).
j
Letting ∆t → 0, guess that U i , j = δ i , j + Ki , j ∆t
For reasons to be explained, this is written as
i
U i , j ( t + ∆t , t ) = δ i , j − H i , j ( t ) ∆t
h
i
so Ci (t + ∆t ) = ∑ [δ i , j − H i , j ∆t ]C j (t )
h
j
Ci (t + ∆t ) − Ci (t )
i
= − ∑ H i , jC j (t )
h j
∆t
and
dCi (t )
ih
= ∑ H i , jC j (t )
dt
j
H is called the Hamiltonian. It could also have been
called the energy matrix: look at C1 (t ) = e − i ( E0 h ) t ,
the equation we saw for the amplitude for a particle
at rest, with energy E0 . Then we have
dC1 (t )
ih
= E0C1 (t ) = H1,1C1 (t ) ⇒ H1,1 = E0
dt
This is why H represents energy, and why the − i h
factor was inserted before. In classical mechanics,
the function giving the total energy of the system is
called the Hamiltonian, hence the name here.
Interlude: the Hamiltonian is Hermitian
Recall that the adjoint of a matrix A is defined by
[ A+ ]i , j = A∗j ,i
and that A is Hermitian if A+ = A.
dC (t )
Since ih
= HC (t ),we can write the solution as
dt
(forget h for the moment,
− iHt
C (t ) = e
so
C ( 0)
it’s only a constant).
C ( t ) = C ( 0) e
iH + t
multiplying yields
1 = C ( t ) C ( t ) = C ( 0) e
i H + t − iHt
e
C ( 0)
i H + t − iHt
e e = ( I + iH + t + ...)( I − iHt + ...) =
I + it ( H + − H ) + o(t 2 )
So, we must have that H + = H .
Some useful facts: A Hermitian matrix has only real eigenvalues, and
the eigenvectors corresponding to different eigenvalues are orthogonal.
Also, if H is Hermitian, e − iH (and therefore e − iHt )
is unitary. So, the evolution of a state is actually a rotation.
A Two-State Example: the Ammonia Molecule
The molecule can flip,
and every state is a
combination of the “up”
and “down” states:
ψ = 1 1ψ + 2 2 ψ =
1 C1 + 2 C2
The question – what is the
Hamiltonian? Based on
symmetry considerations:
dC1
dC2
ih
= E0C1 − AC2 , ih
= E0C2 − AC1 ⇒
dt
dt
a −( i h )( E0 − A) t b −( i h )( E0 + A) t
C1 (t ) = e
+ e
2
2
a −( i h )( E0 − A) t b −( i h )( E0 + A) t
C2 ( t ) = e
− e
2
2
• What do these solutions mean? It’s always useful
to look for states with one frequency (or a definite
energy). These occur in two cases: if b = 0 the
energy is E 0 − A, and the amplitudes to be in 1
and 2 are the same. If a = 0 ,the energy is E 0 + A,
and the amplitudes to be in 1 and 2 have
opposite signs (but the probabilities are equal!).
• Suppose we know that when t = 0 the state is 1 .
This implies
C1 (0) = 1, C2 (0) = 0 ⇒ a = b = 1 ⇒
( i h ) At
+ e −( i h ) At 
 At 
− ( i h ) E0 t  e
− ( i h ) E0 t

 = e
C1 (t ) = e
cos 
2
 h 


( i h ) At
−( i h ) At


e
e
−
 At 
− ( i h ) E0 t
− ( i h ) E0 t


C2 ( t ) = e
sin 
= ie


2
 h 


2
So C2 (t ) = sin 2 ( At h ). Hence, if we know that
at t = 0 the state is “up”, we also know that at
t=
πh
2A
the state is “down” (!!).
Interlude: Symmetry and Conservation
A hydrogen ion can be
in one of the two states
1 and 2 .
The operator P is
defined by a reflection
with the property
P 1 = 2 , P 2 = 1.
Assume that the
physics of the ion is
symmetric with respect
to P. What does this
mean? If we reflect by
P, let the system evolve
for some time, and reflect again, it will be just like
letting the system evolve and then reflecting. If we
denote, as before, the operation of “waiting” by U, it
means that P and U commute: PU = UP. In general,
an operator Q will be said to have a symmetry if it
commutes with U.
e
− iHt
h
Recall that U =
. So Q commutes with U
iff Q commutes with H (remember that U is a
sum of powers of H).
Suppose there’s a state ψ such that ψ ′ = Q ψ
is physically the same state as ψ ; so, it differs
from ψ only by a phase factor. For example,
look at the two states
I =
1 + 2
2
, II =
1 − 2
2
: P I = I , P II = − II
so, the phase change is 0 for I and π for II .
• Another example: look at an RHC polarized
photon, and rotate it by an angle δ about the ziδ
axis. Then its state is multiplied by e .
• It’s straightforward to see that if the symmetry
operation of Q on a state ψ 0 multiplies ψ 0 by a
phase factor eiδ , then this is true forever:
Q ψ 0 = eiδ ψ 0 ⇒ Q (U (t ,0) ψ 0 ) = U (t ,0)(Q ψ 0 ) =
U (t ,0)( eiδ ψ 0 ) = eiδ U (t ,0)( ψ 0 ) ⇒ Q ψ t = eiδ ψ t
But this is conservation!
Parity
Look at the operator P ( x, y , z ) = ( − x,− y ,− z ).
iδ
P
ψ
=
e
ψ ,
Since P = I , it is immediate that if
then eiδ = 1 or eiδ = −1.
2
• P is called the parity operator and the two cases
above correspond to states having even or odd parity.
For the hydrogen ion, I has even parity, II has odd
parity, and 1 has no definite parity.
• It turns out that parity is not always conserved; it is
violated by β decay.
• If β decay is not involved, it turns out that any state with
definite energy which is not degenerate (meaning that there are
no other definite energy states that have the same energy) must
have either even or odd parity. As a matter of fact, let Q be any
symmetry operator, and let ψ be a non-degenerate state with
energy E. So H ψ = E ψ , and
H (Q ψ ) = QH ψ = QE ψ = E (Q ψ )
So, Q ψ is a state with definite energy E, but we assumed that ψ
is the only such state. So we must have Q ψ = eiδ ψ , but as
we saw before, if Q = P, then eiδ = 1 or − 1.
The conservation laws in QM
•Various conservation laws can be expressed in QM as
multiplication by a certain phase factor which depends on
time, location etc. with a constant which equals the conserved
quantity. Note: since there are states without a definite
momentum for example, we cannot always talk about
“conservation of momentum” as in classical physics (the most
we can do for such states is hope that the average momentum
is conserved). Some examples:
• Let Dx (a ) denote translation by a, and assume that
there’s a special state ψ 0 such that Dx ( a ) ψ 0 = eika ψ 0
Then kh corresponds to the classical momentum. Similarly,
there may be special states ψ 0 such that after the passage of
− iωτ
time τ , the state turns into e
ψ 0 . Then ωh
corresponds to the classical energy.
If you find this perplexing, check what happens to the wave equation
e i ( kx − ω t ) when x → x + a or t → t + τ (and look again
at the equations for a photon’s momentum and energy).
• If Rz (φ ) denotes rotation by φ around the z-axis, and
Rz (φ ) ψ 0 = eimφ ψ 0 , mh is the angular momentum for ψ 0 .
It turns out that for an RHC/LHC photon this is equivalent to the
classical notion of angular momentum, with values ± h. For linearly
polarized light, there’s equal amplitude to be in RHC or LHC, so a
single photon doesn’t have a definite angular momentum, and that’s
why (classically) a beam of linearly polarized light has zero angular
momentum.
.
Infinitesimal Transformations
As in Lie group theory, it is customary in QM to study transformations
with an infinitesimal displacement. For example, let Dt (τ ) = U (t + τ , t ).
As we saw before,
ψ (t ) = e
ψ ( ∆t ) = U ( ∆t ,0)e
− iH∆t
h
− iHt
h
ψ (0) , so for a small ∆t


ψ ( 0) ≈  1 −
i
Dt ( ∆t ) = 1 − ( ∆t ) H .
h
− iH∆t 
 ψ ( 0) ⇒
h 
But we know that for a state of definite energy E the passage of time
multiplies the state by e−iω∆t ≈ 1 − iω∆t ⇒ H = ωh = photon's energy.
So, it makes sense to refer to H as the energy operator (more on this
later).
• Similarly, for a small displacement Dx ( ∆x ),
we may assume that


the state changes linearly in ∆x : Dx ( ∆x ) ψ = 1 +
i

p x ∆x  ψ
h

Dx ( ∆x ) ψ = eik∆x ψ ≈ (1 + ik∆x )ψ ⇒ p x = hk
for a definite momentum state; so, call p x the momentum operator.
but
Note that (i h) p x behaves like dψ dx should, so we can guess
That p = −ih d ψ . That turns out to be true (more later).
x
dx
• Similarly, we can define the angular momentum operator
Jz
i


Rz ( ∆φ ) ψ = eim∆φ ψ = 1 + J z ∆φ  ψ ⇒ J z ψ = mh ψ
 h

And it immediately follows that
J z seems like xp y − yp x .
by
Part II: the Continuous
(Infinite-Dimensional)
Case: the Wave Function
The Wave Function and the Axiomatic
Foundations of QM
• So far, we have dealt with a finite number of base
states. That is, a system could be described by a
finite linear combination of probability amplitudes.
• This does not always suffice. Think for example
about the location of a particle along the x-axis.
There is a continuum of possible locations, and
they cannot be expressed as a finite linear
combination of anything.
• Same holds for momentum.
• This led to the definition of the wave function.
For a particle in a one-dimensional world, it is a
continuous function which defines the amplitude to
be, for example, at any location x.
• Next, we present the axioms of QM for what a
wave function is and how it behaves over time.
They are given side-by-side with the classical
physics axioms, to emphasize both the similarity
and difference between the two.
Classical
QM
The state of a particle at time t is
given by its position x (t ) and
momentum p (t ).
The state of a particle at time t
is given by a vector (function)
in Hilbert space, ψ (t ) .
Every dynamical variable is a
function of x and p.
The x of classical mechanics is
replaced by the operator X,
defined by X f ( x ) = xf ( x ) ,
and p is replaced by the
operator P defined by
X and P are Hermitian.
A measurement of a dynamical
variable Ω( x, p ) yields a
uniquely defined result, and the
state is unchanged by the
measurements.
P f ( x ) = −ih
df
.
dx
If a particle is in the state ψ , a
measurement of the variable
Ω yields an eigenvalue ω of
the operator Ω( X , P ) with a
2
probability Pr(ω ) ∝ ω ψ ,
and the state changes from ψ
to ω ( ψ collapses to ω ).
ψ (t ) changes according to
The system changes with time
according to Hamilton’s equations, Schrödinger’s equation:
where Η is the classical
d
i
h
ψ (t ) = H ψ (t )
Hamiltonian (total energy):
dt
∂Η
∂Η
&x =
&
, p=−
∂p
∂x
where H is the quantum
Hamiltonian, created by
substituting x → X , p → P
In the classical Hamiltonian.
Points for Discussion:
•
∞
∗
What is the wave function? Inner product: ψ φ = ∫ψ φ
−∞
•
What are the X and P operators, and what are their
eigenvectors and eigenvalues? Relation to Fourier transform!

{ X : x = δ ( x′ − x ), x},  P : p =

ipx

1
h
e , p
2πh

•
The meaning of the collapse of the state vector for the X and
P operators. What happens if an eigenstate is measured?
Reminder: two-slit experiment with measurement. Question:
how can a photon cause the wave function of a battleship to
collapse? Reminder: Stren-Gerlach experiment.
•
The Schrödinger equation for a free particle as the limit of a
finite dimensional state: where does the derivative come
from? (Later).
Complications:
1.
What does the classical xp go to: XP or PX? Answer: average.
2.
What if Ω is degenerate? Answer: Pr(ω ) ∝ ψ Pω |ψ ,
where P is the projection on the subspace of vectors with
ω
eigenvalue ω .
3.
The eigenvalue spectrum of Ω is continuous. Answer: replace all
sums by integrals, and “probability” by “probability density”.
b
Ω = X : Pr( a ≤ position ≤ b) = ∫ ψ ( x ) dx
2
a
So we should have ∫ ψ ( x ) dx = 1 (normalization).
Example:
∞
−∞
2
Expectation and Uncertainty
• We saw that the measurement process in QM is
probabilistic. Yet, we can ask what the average result of
measuring Ω in the state ψ is: following the usual
definition of expectation, it equals
Ω ≡ ∑ω ω ψ
ω
2


= ∑ ω ω ψ ψ ω = ψ ∑ ω ω ω  ψ
ω
ω

where ω , ω range over the eigenvalues/eigenvectors of Ω.
Since Ω is Hermitian, its (normalized) eigenvectors are
orthonormal. It’s easy to see that Ω can be resolved by
Ω = ∑ω ω ω
ω
(hint: verify that both sides are equal when
operating on every ω , and use the fact
that ω span the entire space).
{ }
So, the expectation is Ω = ψ Ω ψ . This is nice, since we
don’t need to find Ω′s eigenvalues and eigenvectors in order
to compute the expectation.
• The average has a meaning only when computed over an
ensemble of identical particles. It cannot be computed for a
single particle, because the first measurement changes it.
• For an eigenstate, the expectation is equal to the
corresponding eigenvalue. Thus, the Energy/Position
/Momentum measurement of a state with definite EPM yields
the “correct” and unique EPM result.
The uncertainty for Ω in a state ψ is also defined just
like the classical variance:
( ∆Ω) = ∑ Pr(ω )(ω − Ω
[
)
= ψ (Ω − Ω
2
2
ω
)ψ
2
]
1
2
Example – the Gaussian centered at a with width ∆ :
ψ ( x) =
1
(π∆ )
2
1
e −( x − a )
2
2 ∆2
⇒ Pr( x )dx =
4
1
(π∆ )
2
1
e −( x −a )
2
∆2
2
The average of the position is
∞
X = ψ X |ψ = ∫ψ ∗ ( x ) xψ ( x )dx = a
−∞
which is hardly surprising. As for the uncertainty:
( ∆X ) 2 = ψ ( X − X
)
2
(
|ψ = ψ X 2 − 2 X X + X
2
ψ X |ψ − 2 X ψ X |ψ +
2
1424
3
is just X
So we have to compute
ψ X 2 |ψ =
1
(π∆ )
2
and finally, ∆X =
1
2
∆
2
∞
expectation of a
const. is the const.
−( x − a )
e
∫
−∞
X
{
2
2
2∆
2
) |ψ
=
2
= ψ X |ψ − X
{
x 2 e −( x − a )
2
this is a 2
2
2
∆
2∆
dx =
+ a2
2
2
(also hardly surprising).
What about the momentum operator P?
∞
df
P f ( x ) = −ih
⇒ P = −ih ∫ψ ∗ψ ′dx = 0
dx
−∞
∞
2
h
h
2
∗
2
(∆P ) = −h ∫ψ ψ ′′ dx = 2 ⇒ ∆P =
2∆
2∆
−∞
as can be directly
verified by substituting
and integrating.
ψ ( x) =
1
(π∆2 )
1
e
−( x − a ) 2 2 ∆2
4
• It is instructive to see how ψ (x ) looks in momentum space.
Since the orthonormal basis corresponding to momentum space is
given by p = ( 2πh )
momentum space by
∞
−1 2 ipx h
e
, then ψ (x ) is represented in
1
4
2

 −ipa h − p 2 ∆2
∆
∗
p ψ = ∫ p ψ dx =  2  e
e
 πh 
−∞
2 h2
So, the narrower the position distribution (small ∆), the broader
the momentum distribution. This not only agrees with the
physics discussed before, it is also no surprise to anyone who
studied the Fourier transform! Note: ∆P ⋅ ∆X = h 2 .
144244
3
uncertainty principle
• This happens because P and X don’t commute,
and don’t have common eigenvectors
(“eigenkets”). Since only the eigenkets of an
operator are unaffected by a measurement of the
operator, we cannot have a system for which we
know both the position and momentum.
• If two operators commute, there is a basis of
common eigenkets, and the order of the
measurements doesn’t make a difference.
• The commutator of two operators is defined as
[Ω, Λ ] ≡ ΩΛ − ΛΩ
• It is straightforward to see that [X , P ] = ih
Interlude: Proof of Heisenberg’s Uncertainty Theorem
We want to obtain a lower bound on
(∆X ) (∆P) = ψ ( X − X
2
2
)
2
ψ ψ (P − P
)
2
ψ
The trick is to reduce this to an expression depending on
[X , P] = XP − PX = ih
define A =
X − X , B = P − P ⇒ AB − BA = XP − PX
So we should bound
Since A,B are Hermitian
ψ A2 ψ ψ B 2 ψ = Aψ Aψ Bψ Bψ
From the Cauchy-Schwartz inequality, this is larger than
Aψ Bψ
2
= ψ AB ψ
2
Now comes the trick. Denote
AB − BA
AB + BA
C=
,D =
⇒ AB = C + D
2
2
So we should bound
ψ C + Dψ
2
Now:
ψ C + Dψ
2
= ψ Cψ + ψ Dψ
2
ψ Cψ
+ ψ Dψ
+
∗
∗
ψ Dψ
ψ
and since
which is of course positive, and prove that
∗
2
ψ C |ψ
is normalized,
To get the inequality, we will throw away
ψ Cψ ψ Dψ
=
2
ψ Cψ ψ Dψ + ψ Cψ
ih
But C =
,
2
2
h2
= .
4
2
ψ D |ψ ,
∗
+ ψ C ψ ψ D ψ = 0!
1444444424444444
3
Call this G
Ignoring the constant factor of 2, this last expression equals
G ≡ ψ AB − BA ψ ψ AB + BA ψ
ψ AB − BA ψ
∗
Now we have to remember that
And that
A+ = A, B + = B,
∗
+
ψ AB + BA ψ
∀ψ ,Ω : ψ Ω ψ
and that ∀ Λ ,Ω
∗
= ψ Ω+ ψ
: (ΩΛ ) + = Λ+ Ω +
And it immediately follows that G is indeed 0. Thus the proof is
complete, and after taking roots:
( ∆X )( ∆P ) ≥ h 2
• Is the minimum attained? We can assume X = 0.
The Cauchy-Schwartz inequality is not strict only if
the two vectors are linearly dependent, so we must
have
i
(P − P )ψ = cX ψ ⇒ ψ ′ = ( P + cx )ψ ⇒
h
ψ ( x ) = ψ ( 0) e

i  cx 2


+
P
x

h  2

But to minimize, we must also have
ψ (P − P )X + X (P − P )ψ = 0
but
(P −
P )ψ = cX ψ , ψ (P − P ) = c∗ ψ X ⇒
ψ (P − P )X + X (P − P )ψ = (c + c∗ ) ψ X 2 ψ = 0
So c must be pure imaginary,
correcting by shifting with X
ψ ( x ) = ψ ( X )e
c = c i and so (after
) we get
2
 x− X 
(
)
x
−
X


i P 

h
 −c

e
2h
That is, a Gaussian with a phase shift proportional
to the momentum saturates the uncertainty.
Interlude: Finite-Dimensional Example: the effect of
measurement (Shankar Ex. 4.2.1)
Consider the following operators on C 3 :
 0 1 0
1 0 0 
1 
Lx =
1 0 1 , Lz = 0 0 0 



2
0 1 0
0 0 − 1
• What are the possible values obtained if Lz
Answer: the
Lz
is measured?
eigenvalues/normalized eigenvectors are
  1     0     0   So the possible results when
            is measured are 0,1,-1.
1,  0  , 0,  1  , − 1,  0  
  0    0    1 
        
• In the state in which
Lz
Lz = 1, what are Lx , L2x , ∆Lx ?
Answer: if Lz = 1, the state must be (1,0,0)t . According to what
we proved before:
Lx = (1,0,0) Lx (1,0,0)t = 0
L2x = (1,0,0) L2x (1,0,0)t = 1 2
∆Lx =
L2x − Lx
2
=1
2
Note: we should take
conjugate transpose,
but everything here’s
real.
• Find the eigenvalues and normalized eigenkets of
basis.
Lx
in the
Lz
Answer: the Lz basis is the standard one. An immediate
computation yields that the
eigenvectors are
Lx
eigenvalues/normalized

 − 1    1   
 1 

1
 1    1   
0,
 0  , 1,  2  , − 1,  − 2  
2
2    2   


1
1
1

     


• If a particle is in the state with Lz = −1, and Lx is measured,
what are the possible outcomes and their probabilities?
Answer: if Lz = −1, the state must be
ψ = (0,0,1)t . According
to the axioms, the possible outcomes are the Lx eigenvalues,
with probabilities equal to the square of the inner product of ψ
with the corresponding eigenket. A straightforward calculation yields
that we get 0 with probability 1 2, 1 with probability 1 4, and − 1
with probability 1 4 .
(
)t
2
• Consider the state ψ = 1 2,1 2,1 2 in the Lz basis. If Lz
is measured in this state and the result is 1, what is the state after the
measurement? If Lz is then measured, what are the possible
outcomes and respective probabilities?
Answer:
1 0 0
L2z = 0 0 0


0 0 1
So, there’s a degenerate eigenvalue, 1, with an
eigenket subspace V = {(a ,0, b)},and a 0
eigenvalue with an eigenket (0,1,0). If 1 was
measured, the result is the (normalized) projection
of ψ on V, which equals 1 3 1,0, 2 t .
(
)(
)
Next, Lz is measured. The possible results, and their probabilities,
are: 0 is measured with probability 0, 1 is measured with
probability 1 3 , and –1 is measured with probability 2 3.
NOTE: the measurement of
measurements of Lz !
L2z
does not determine the subsequent
• A particle is in a state for which the probabilities are
Pr( Lz = 1) = 1 4, Pr( Lz = 0) = 1 2, Pr( Lz = −1) = 1 4.
note that the most general state with this property is
eiδ1
eiδ 2
eiδ 3
ψ =
Lz = 1 +
Lz = 0 +
Lz = −1
2
2
2
does this mean that one can choose arbitrary values of
without affecting the physical state?
δ1,δ 2 ,δ 3
Answer: no. The physical state is immune only to global phase
shifts: ψ and eiδ ψ are physically equivalent in the sense that
they generate the same probability distribution for any observable.
However, if for example δ 1 = δ 2 = δ 3 = 0, then ψ = Lx = 1 ,
and if
δ 1 = δ 3 = 0 and δ 2 = π , then ψ = Lx = −1 .
A Note on the Position, Momentum,
and Energy Operators
• We saw that for a state of definite position (delta function)
the operator X indeed returns the position, and that for a state
of definite momentum (a pure wave, ei ( kx −ωt ) )
the operator P indeed returns the momentum.
• What about the other states? The most we can ask for is
that the average of X (P) will be what we consider intuitively
to be the average position (momentum). That is indeed the
case.
• Let ψ be a state. The probability density to be at the
2
position x is ψ ( x ) = ψ ∗ ( x )ψ ( x ), so the average position
is defined by
∗
x
ψ
∫ ( x )ψ ( x )dx, but the average of the X
operator is defined by
X = ψ X ψ = ψ Xψ = ψ xψ = ∫ xψ ∗ ( x)ψ ( x )dx
And they are indeed equal.
Note: if the integration limits are unspecified, they are assumed
to be
−∞
to
∞.
What about the momentum? Remember that the momentum
operator is defined by
and that its
ipx
dψ eigenkets, with
1
h
P ψ = − ih
p =
dx eigenvalues p, are
2πh
defined by
e
So the probability for the state ψ to be with momentum p is
pψ
2
= pψ ψ p
ψ ψ p dp.
∫1p4p44
2444
3
So the average momentum is
I
We want to prove that this equals the average of the
operator P, which is
P = ψ P ψ = (−ih ) ∫ψ ∗ψ ′dx
14
4244
3
II
To prove that I=II, note that p ψ =
1
2πh
∫e
− ipx
h
ψ ( x )dx
Which is just ψ ′s Fourier transform (up to the h factor).
Now, remember that (up to constants which we won’t bother
with here), the product of the Fourier transform with p is the
transform of the derivative; this reduces p p ψ
in I to
the transform of the derivative. Lastly, remember that the
Fourier transform is unitary, so the inner product of the
transforms equals the inner product of the respective
functions. This immediately reduces I to II.
In very much the same way, the Hamiltonian H is the energy
operator, and H = ψ H |ψ is the system’s average energy.
ψ =
• Example: Gaussian
wave packet
e
ik0 ( x + a ) − ( x + a ) 2 2 ∆2
e
(π∆ )
2 14
A straightforward calculation yields
• Average position =
ψ X ψ = ∫ xψ ∗ψdx = − a
• Average momentum =
• Average energy =
(for a free particle)
ψ P ψ = (− ih )∫ψ ∗ψ ′dx = k0 h
h2
∗
ψ Hψ =−
ψ
ψ ′′dx =
∫
2m
k02 h 2
h2
+
2m 4m∆2
• Question: what happens to the average energy when ∆ → 0 ,
and why? It looks odd that the mass m is in the denominator –
can you explain it?
Schrödinger’s Equation
h 2 ∂ 2ψ
d
d
ih ψ (t ) = H ψ (t ) ⇒ ih ψ (t ) = −
dt
dt
2m ∂x 2
14444244443
For a free particle
( H =P2 2m)
• Tells us how a state changes over time, hence is
very important. Where does it come from? There are
various heuristics to justify the equation; for
example, suppose a free electron is moving through
a crystal, in which the atoms are close to each other:
δ
0
1
2
C0
C1
C2
Let Ci be the amplitude to be at atom i at time t.
Now look again at the equations for the ammonia
molecule. Here, we have three states (if we assume
that, as a very short time passes – remember, we’re
looking at the time derivative – the change in C1
depends only on it and on C0 , C2 ) . So we can
guess the following equation (up to constants):
dC1
ih
= E0C1 − AC0 − AC2
dt
Now, use the well-known approximation for
the second derivative at x:
f ′′( x ) ≈
f ( x − δ ) − 2 f ( x) + f ( x + δ )
δ2
And thus the right-hand side equals, up to
constants, the second derivative of the wave
function by x.
• The simplest (and very common) case for a non-free
particle has the Hamiltonian
h2 d 2
−
+ V ( x)
2
2m dx
Where V is some potential.
• It is impossible to prove Schrödinger’s
equation. It does not follow from the
axioms.
Solving Schrödinger’s Equation
d
ih ψ (t ) = H ψ (t ) , ih ψ& = H ψ
dt
• This is a differential equation, which resembles equations
in classical physics.
• The goal is to find the propagator
U (t ) which satisfies
ψ ( t ) = U ( t ) ψ ( 0) .
• As often happens, a good direction to proceed is to first
find the (normalized) eigenkets and eigenvalues of the
Hermitian operator H.
• Since H is the energy operator, we will often denote the
eigenkets/eigenvalues with the letter E:
H E =EE
• Suppose we have these eigenkets/eigenvalues. Since they
are orthonormal and span the space, then for every state
ψ (t ) = ∑ E E ψ (t ) ≡ ∑ E a E (t )
E
E
• Plugging this last form into the equation yields
iha& E (t ) = Ea E (t ) ⇒ a E (t ) = a E (0)e − iEt h
E ψ (t ) = a E (t ) = a E (0)e − iEt h
• So,
• And
ψ (t ) = ∑ aE (t ) E = ∑ a E (0)e − iEt h E =
∑
E
E
E
E ψ (0) e −iEt h E = ∑ E ψ (0) E e −iEt h =
E

−iEt h 
−iEt h
(
0
)
(
)
E
E
e
ψ
⇒
U
t
=
E
E
e
∑
∑

E
E

• So we have a closed-form expression for the propagator. In
the infinite dimensional case, the sum is replaced by an
integral.
− iEt h
• The so-called normal modes, E ( t ) = E e
,
are also called stationary modes, because they change only
by a phase factor, hence have the same probabilities to be in
a certain state regardless of the time. They correspond to
systems which start off in an eigenket of H (definite energy
state).
• Another expression for the propagator is given by
U (t ) = e
− iHt h
But it is not always manageable. Note that it implies that U (t )
is a unitary operator.
• Schrödinger’s equation on the line is usually solved in the
X-basis.
• Note that
U (t ) is an operator. By ψ (t ) = U (t ) ψ (0)
∞
ψ ( x, t ) = ∫ U
( x, t; x′,0) ψ ( x′,0) dx′
14243 123
we mean that
−∞
where
propagator
initial conditions
U ( x, t; x′,0) = ∑ x E E x′ e − iHt h
E
Or, in the infinite dimensional case,
U ( x, t; x′,0) = ∫ x E E x′ e − iEt h dE
E
• How does one derive such equations? It’s straightforward
from U ( t )
= ∑ E E e − iEt h :
E
the left and by
x′
on the right.
multiply by
x
on
• The propagator as the evolution of simple states: let’s
look at the equation
∞
ψ ( x, t ) = ∫ U
( x, t; x′,0) ψ ( x′,0) dx′
14243 123
−∞
propagator
initial conditions
And take ψ (x′,0) to be a state with definite position, that is,
a delta function at a. Then
∞
ψ ( x, t ) = ∫ U ( x, t; x′,0)δ ( x′ − a )dx′ = U ( x, t; a ,0)
−∞
taking
x = b yields
U (b, t; a ,0) = The amplitude to be in b, at time t, when at
t = 0 the state is a delta function at a.
• And in general, U ( x, t ; x′,0) is the factor weighing the
contribution of the amplitude to be in x′ at time 0 to the
amplitude to be in x at time t.
Solving Schrödinger’s Equation
for a Free Particle
• If there’s no potential, the equation is relatively simple:
P2
ih ψ& = H ψ =
ψ
2m
• Is it trivial? Not at all; even for relatively simple initial
states, the development of the wave function over time
may be surprisingly complicated.
• To solve the equation, first find the eigenkets of H:
P2
H E =
E =EE
2m
• The kets p , , which are eigenstates of P, suggest
themselves. Remember that
ipx
1
p =
eh,
2πh
P p =pp
and so
2
2
P
p
p =
p = E p ⇒ p = ± 2mE
2m
2m
Hence for every E there are two corresponding eigenstates:
E ,+ = p = 2mE , E ,− = p = − 2mE
• Physically, this means that a particle with energy E can be
moving to the left or right. But in QM, the following state –
which has no meaning in classical mechanics – also exists:
E = α p = 2mE + β p = − 2mE
Note that this is a single particle
which can be caught moving either
left or right!
• The next step is to compute the propagator. In the
position eigenkets it equals
∞
U ( x , t ; x ′, 0 ) =
1
2 πh
∫e
ipx h −ipx ′ h −ip 2t 2 mh
e
e
dp
−∞
U ( x, t; x′,0) = ∫ x E E x′ e − iEt h dE
E
The integral evaluates to
1
2
 m


 2πhit 
e
im ( x − x ′ ) 2 2 ht
and so for an initial state
1
2
 m

ψ ( x, t ) = 
 2πhit 
ψ ( x′,0),
∞
∫e
the state at time t is
im ( x − x ′ ) 2 2 ht
ψ ( x ′ , 0 ) dx ′
−∞
• What happens if t → 0 ? We get the integral
2
of a function multiplied by things like cos( Ax ),
where A is very large:
− x2
Plot of e
multiplied by cos( Ax 2 ). If we
integrate, only the value at 0 remains.
Interlude: The Free Particle Propagator
in the Momentum (Fourier) Domain
• For a free particle, Schrödinger’s equation is
dψ
dψ
ih d 2ψ
h 2 d 2ψ
ih
=−
⇒
=
2
dt
2m dx
dt 2m dx 2
• Switching to the Fourier domain yields
dΨ (u, t ) ih 2
=
u Ψ ( u, t )
dt
2m
• Which is solved by
Ψ ( u, t ) ∝ t e
ihtu 2
2 m Ψ ( u,0)
• This observation highlights the “convolution
nature” of the free particle propagator, and the
manner in which it blurs spatially, and sharpens in
the frequency (momentum) domain, as time goes
on. The convolution kernels are like those in the
previous slide.
Interlude: The Density Matrix
• Expresses the expectation as a linear function of the
operator being measured.
• Allows to include “usual” uncertainty about the state.
• Closely related to Boltzmann distribution and quantum
statistical mechanics.
Let un be an orthonormal basis. The probability of finding
ψ (t) in the state un is cn (t ) , where ψ (t ) =∑cn (t ) un . If
2
A is an operator, and A u p =∑ Anp un , then
n
n
A t = ψ (t ) Aψ (t ) =∑ A c (t )c p (t )=Tr ( ρ (t ) A),
∗
np n
n, p
where ρ (t )=ψ (t ) ψ (t ) is the density matrix.
Note: ρ (t ) is a positive operator, and Tr( ρ (t ))=∑cn (t )cn∗ (t )=1.
Also, ρ = ρ , and ρ = ρ ( ρ is Hermitian).
2
+
n
Note: ρ (t ) does not have an arbitrary phase factor, hence it
captures the physically relevant properties.
From the Schroedinger equation, it follows immediately that
1
ρ& (t )= [H ,ρ (t )].
ih
• Reminder – what is the probability to obtain a
value an when measuring A (at time t)?
Answer: Let Pn be the projection on the subspace
of eigenvectors with eigenvalue an . It is immediate
that the probability to obtain an is
ψ (t ) Pn ψ (t ) =Tr (ρPn ).
• Incorporating “classical” uncertainty:
suppose the starting conditions of a physical
system are not exactly known, and that it has
probability pk to be in state ψ k ( pk ≥0, ∑ pk =1).
Then, the density matrix can be naturally k
extended by defining
ρ =∑ p k ψ k ψ k
k
And linearity immediately proves that the
expected value of an operator A is Tr( ρA).
• Some further properties:
•
•
ρ is Hermitian.
Tr( ρ )=1.


2
2
ρ
• = ∑ pk ψ k ψ k  =∑ pk pl ψ k ψ l ψ k ψ l ,
 k
 k ,l
and since {u n }is an orthonormal basis, Tr( ρ 2 )=
2
ρ
u
∑ n un = ∑ pk pl ψ k ψ l unψ k ψ l un =
n
∑p
∑p
n ,k ,l
k
k ,l
pl ψ k ψ l ψ l ψ k =∑ pk pl ψ k ψ l
2
≤
k ,l
k
pl =1, with equality holding only if all
k ,l
of the ψ 's are identical (i.e. a "pure state").
• The meaning of ρ nn is the probability to
obtain u n in a measurement. It is termed
“population”, while ρ np are termed
“coherence”.
• If we’re working in the basis of eigenstates
of the Hamiltonian H, then from the
equation for the development of ρ np over
time, it follows immediately that
ρ np (t )=exp (E p − En )t  ρ np (0).
i
h


• Relation to quantum statistical mechanics: in
thermodynamic equilibrium, it turns out that
1
Z
 H 
 , where k is Boltzmann’s
 kT 
ρ = exp −
constant and T the temperature. Z is chosen so
to normalize the trace to 1 (it is the partition
function). If En is the basis of H’s eigenstates,
then
1
1
 En 
 H 
ρ nn = un exp −  un = exp − ,
Z
Z
 kT 
 kT 
and n≠ p⇒ ρ np =0
(Boltzmann's distribution).
Path Integrals: Feynman’s
Formulation of QM
U ( x, t; x′,0) ∝ ∑ eiS [ x ( t )] h
Where the summation ranges over all paths connecting ( x ′, 0 )
to ( x , t ), and S is the classical action (the integral over x ( t )
of the kinetic minus potential energy).
• What does this mean?
• The classical path is the one with the smallest action. In
Feynman’s formulation, the propagator is the sum of all
paths, each weighted by the exponent of i times the action
over h . What happens for objects with a large mass, for
example, is that due to the h in the denominator, the
contributions around any path but the classical one are in
very different phases and therefore cancel out.
• This formulation is equivalent to the one we had shown, but
highlights the connection to classical physics. Also, in some
cases (but not usually) it is easier to use it to calculate the
propagator.
• Proof of the path integral formulation: look at the
propagator with a fixed potential V. Proceeding
much like as in the free particle case, this time p
is an eigenstate with energy p 2 2m + V , so the
propagator is
∞
U (t ) =
∫
p p
e
−iE ( p )t h
dp =
−∞
∞
∫
p p
e

2
−i p 2m + V t h





dp =
−∞
e − iV t h
∞
∫
p p
e





2
−i p 2m t h

dp =
−∞
 m 


 2πhit 
1
2
e
−iVt h im( x − x′) 2 2ht
e
since the integral is exactly the one for the free particle
propagator. Now, if the elapsed time is very small, ∆t
(this is the justification for assuming a fixed potential),
and v( x) stands for speed in the interval between x and
x′, this equals
1
2
 −iV( x)(∆ t ) h imv 2 ( x)(∆t ) 2h

m
 e

e
 2πhi (∆t ) 
Using the postulates in Part I, we compute the
amplitude of getting from ( x′,0) to ( x, t ) as the
sum of the amplitudes of all different paths
(think of more and more screens with more
and more slits). For every path, the amplitude
is the product of amplitudes to go over its sub paths. In the limit the sub - paths go to zero, and
we get the product of the above, following the
postulate about the amplitude of events in
succession. so the powers in the exponents add

 mv 2 ( x)
to (i h )∫ 
− V( x) dt = (i h )S [ x(t )] ,
2

0
which completes the proof * . It turns out that
t
this formulation is equivalent to Schrodinger' s.
* up to a scale factor - see following.
• Derivation of the free particle propagator via
path integrals:
∞
ψ ( x, T ) = ∫ U ( x, T ; x′,0)ψ ( x′,0)dx′ =
−∞
i m 2

∞
∞
i
x& ( t ) dt 
∫
S [ x ( t )]
h 2

ψ ( x′,0)dx′ = 
ψ ( x′,0)dx′ =
0
h
e
e
∑
∫−∞ x (∑
∫



0 ) = x′
− ∞ x ( 0 ) = x′
 x (T ) = x

 x (T ) = x

T
2
mi 
x − x′ 

∞
x& ( t ) +
dt 


∫
2h 
T 

ψ ( x′,0)dx′
0
e
∑
∫−∞ x (0)=0

 x (T ) =0

 using the fact that the transformation x (t ) → x (t ) + x′ + x −T x′ t maps 




the loops starting and ending at 0 to the paths from x′ to x





T
2
 
mi  2
 x − x′ 
 x − x′ 
 dt 
x& ( t ) + 2 
 x& ( t ) + 


∫


2
h
T
T

4424
 4
1424
∞
1
3
3
0
 


independent 
integrates
e
ψ
( x′,0)dx′ =

∫−∞ x (∑
of t , hence
to 0, since
 x (T0 ))==00

x(0) = x(T ) integrates to


( x − x′ )
T


T
mi 2
2

x& ( t ) dt  ∞  mi ( x − x′ ) 
∫
2h

 e 2 hT ψ ( x′,0)dx′
e 0
∑
 x ( 0 )=0
 ∫

∞
 x (T ) =0
 −1
4
4
4
4
2
4444
3
1442443
Compare to slide 96
T
2
Propagator from 0 to 0
Path Integral Normalization Factor
• The question arises as to how to normalize the
summation over all paths.
• Approximate each path by a piecewise linear
function:
xN = b
.
.
.
.
xi +1
xi
. .
x
. .
x0 = a
ta
t
ti ti +1
tb
Denote ε = ti +1 − ti , so Nε = tb − t a ≡ T .
The free particle propagator is then
( A is a normalizing factor)
 i N xj m 2 
A lim ∫ ∫ ... ∫ exp ∑ ∫
x& dt dx1...dx N −1 ≅
ε →0
 h j =1 x j−1 2

− ∞− ∞ − ∞
∞ ∞
∞
∞ ∞
∞
 mi N
2
−
(
)
A lim ∫ ∫ ... ∫ exp
x
x
∑ j
dx1...dx N −1
j −1
ε →0
 2εh j =1

− ∞− ∞ − ∞
which follows immediately by replacing x& with
x j − x j −1
ε
The Gaussian integral evaluates to
−
N −1
2
N −1 mi
( x N − x0 ) 2
 m 
π 2 e 2hT
AN 

 2hεi 
comapring this with the free particle propagator, it must
equal
1
−
2
mi
N
2
−
(
)
x
x
 m  2hT N 0
 m 2
. So, A = 

 e
 .
 2πihT 
 2πhiε 
It is customary to write the path integral as
1
2
∞ ∞
∞
1
2
dx N −1
1
 2πhiε 
(i h )S cl [ b , a ] dx1 dx2
lim ∫ ∫ ... ∫ e
...
,B=
 .
ε →0 B
B B
B
 m 
− ∞− ∞ − ∞
.
Equivalence to Schrödinger’s Equation
For an infinitesimal time interval ε , the path
integral propagator U ( x, ε ; x′,0) is approximately
1
2
  m( x − x′) 2
 m 
 x + x′   
− εV 
,0   h 

 expi 
2ε
 2
 
 2πhiε 

where the approximations are standard to the first order in
ε . Note that there is no external normalizing factor, since
there is no midpoint to integrate over. So, ψ ( x, ε ) should be
1
∞
2
 imη 2 
 iε  η  


exp
exp
∫−∞  2εh  − h V  x + 2 ,0 ψ ( x + η ,0)dη
imη 2
≥ π , the exponent oscillates
where η = x′ − x. When
2εh
 m 


h
π
ε
2
i


1
2
 2εhπ 
very rapidly, so we' ll concentrate on the region η ≤ 
 .
 m 
This means that if we' re working to first order in ε , we
should work to second order in η .
The integral to compute for ψ ( x, ε ) is
∞
 imη 2 
 iε  η  
∫−∞exp 2εh  exp− h V  x + 2 ,0 ψ ( x + η ,0)dη
Expanding to first order in ε
∂ψ η 2 ∂ 2ψ
+
ψ ( x + η ,0 ) = ψ ( x ,0 ) + η
2 ∂x 2
∂x
 iε  η  
iε
exp − V  x + ,0  = 1 − V ( x,0 ) (we neglect
h
2 
 h 
terms or order ηε , since they' re O(ε 3 2 ) ). The integral
1
∞
2
 imη 2  
iε
∫−∞exp 2εh ψ ( x,0) − h V (x,0)ψ ( x,0) +
∂ψ η 2 ∂ 2ψ 
η
dη (again, ηε etc. terms are dropped).
+
2 
2 ∂x 
∂x
 m 
is 

 2πhiε 
This is a Gaussian integral in η . Eventually we get

− iε  − h 2 ∂ 2
+ V ( x,0)ψ ( x,0)
ψ ( x, ε ) −ψ ( x,0) =

2
h  2m ∂x

which indeed agrees with Schrodinger' s equation
prediction.
Calculating the Path Integral
Usually, it' s impossible. For potentials of the form
V = a + bx + cx 2 + dx& + ex&x, the propagator can be
computed by expanding around the classical path.
Write the path x(t ) as xcl (t ) + y (t ). Since x(t ) and
xcl (t ) agree at the endpoints, y (0) = y (T ) = 0. So
0
i


U ( x, t ; x′,0) = ∫ exp S [xcl (t ) + y (t )]D[ y (t )]
h

0
When expanding S , the linear terms vanish, since xcl (t )
i

is an extremum. Also, exp S cl  factors out. So, the
h 
i

integral equals exp S cl  ×
h 
i T1 2
 
2
∫0 exp h ∫0  2 my& (t ) − cy (t ) − ey(t ) y& (t ) dt D[ y(t )]
144444444424444444443
Depends only on T
0
Note that the probabilistic information about the wave
function can still be extracted, since for a given T we
know the function up to a scale factor.
How to compute
i T1 2
 
2
∫0 exp h ∫0  2 my& (t ) − cy (t ) − ey(t ) y& (t ) dt D[ y(t )] ?
For example, for the harmonic oscillator it equals
0
i T m 2

2 2
∫0 exp h ∫0 2 y& (t ) − ω y (t ) dt D[ y(t )] (where y(0) = y(T )
∞
 nπt  ∞
= 0). Represent y (t ) as ∑ an sin 
 ≡ ∑ an yn . It is easy to
 T  n =1
n =1
[
0
]
 π 2 n 2 ω 2T 
−
see that ∫ y& (t ) − ω y (t ) dt = ∑ a 
. Going to
2 
n =1
 2T
0
this representation, and forgetting for the moment about the
T
[
2
2
]
2
∞
2
n
Jacobian and other constants, the integral is the infinite
 i 2  π 2 n 2 ω 2T  
−
product of the Gaussian integrals ∫ exp an 
 dan ,
2 
−∞
 h  2T
∞
π n
2 
−ω T 
or (forgetting i, h, π ...) ∏ 
n =1  T

∞
 π n

∏
n =1  T
∞
2
2
 ω T
1 − 2 2
 π n
2
−
1
2
2



2
−
1
2
2
−
1
2
=
 ωT
= K (T )∏ 1 − 2 2
π n
n =1 
∞
2
2



−
1
2
=
 sin(ωT ) 
K (T ) 
. Normalization is possible since we know

 ωT 
the limit as ω → 0 (it' s the free particle propagator).
Perturbation Theory and Feynman Diagrams
 i T m 2
 
′
&
U ( x, T ; x ,0) = ∫ exp ∫  x (t ) − V [ x(t )]dt Dx[t ]
 
x′, 0
h 0  2
expand the exponent to write the integrand as
x ,T
i T m 2  i
 i T m 2 T
exp  ∫ x& dt  − exp  ∫ x& dt  ∫ V ( x)dt + ...
h 0 2
 h
0
h 0 2
Integrating over all paths, the first summand is just the free
particle propagator. To calculate the second summand, look
at the discrete approximation (ignoring ε etc.) :
∞
∞
[
 mi
2
2
...
exp
(
x
x
)
...
(
x
x
)
−
+
+
−

N
N −1
∫ ∫  2h 1 0
−∞ −∞
[V ( x0 ) + ... + V ( xN )] dx1...dxN −1
] ×

To compute the summand with V ( x j ), note that in the limit, if we
assume bounded variation, then since the time between any two
successive xi goes to zero, xi +1 − xi → 0 as well, and we can drop
[
]

 mi
the terms with exp
( x j − x j −1 ) 2 + ( x j +1 − x j ) 2  which goes

 2h
to 1. We' re left with the product of free particle propagators from 0
to t j and from t j to T . When this consideration is extended to all
summands, we have two variables to integrate over : the time τ and
the location at that time, denote it y. So, the integral equals
T ∞
U1 ( x, T ; x′,0) ≡ ∫ ∫ U 0 ( x, T ; y ,τ )V ( y )U 0 ( y ,τ ; x′,0)dydτ
0 −∞
where U 0 is the free particle propagator.
The physical meaning is : the particle propagates
freely from ( x′,0) to ( y,τ ), is then acted upon
(scattered) by the potential, and then propagates
freely to ( x, T ). In the same manner, the third
summand corresponds to a particle scattered twice
by the potential (represented by a quadruple integral),
etc. U1 ( U 2 ) (a " first (second) order perturbation" ) is
represented by the following Feynman diagram:
( x, T )
( x, T )
time
( y 2 ,τ 2 )
( y ,τ )
(x′,0)
V
( y1 ,τ 1 )
(x′,0)
V
V
The second order perturbation equals
 T

 i T m 2  T
1 i
 −  exp  ∫ x& dt   ∫ V [ x( s ), s ]ds   ∫ V [ x( s′), s′]ds′
2!  h 
h 0 2
 0
 0

which equals
2
 i
− 
 h
2 T ∞ τ2 ∞
∫ ∫ ∫ ∫U
0
( x, T ; y2 ,τ 2 )V ( y2 )U 0 ( y2 ,τ 2 ; y1 ,τ 1 )V ( y1 )
0 −∞ 0 −∞
U 0 ( y1 ,τ 1 ; x′,0)dy1dτ 1dy2 dτ 2
Note that the 1 2! factor (and, in general, the 1 n! factor)
vanished, because the integral is carried over the set
τ 1 ≤ τ 2 . It is customary to assume that τ 2 < τ 1 ⇒
U (?,τ 2 ; ?,τ 1 ) = 0, which allows to replace τ 2 in the
quadruple integral above by T , and the same for higher
orders.
In General
T ∞
U k ( x, T ; x′,0) = ∫ ∫ U 0 ( x, T ; y,τ )V ( y )U k −1 ( y,τ ; x′,0)dydτ
0 −∞
and it is easy to deduce from the expansion of the exponent
the integral equation : U ( x, T ; x′,0) =U 0( x, T ; x′,0) −
T ∞
i
U 0 ( x, T ; y,τ )V ( y )U ( y,τ ; x′,0)dydτ
∫
∫
h 0 −∞
A Path Integral Formulation for the
Density Matrix
Noting the similarity between the quantum mechanics
propagator and the density matrix
 −i

U ( x2 ,t 2 ;x1 ,t1 )=∑exp Ek (t 2 −t1 )  x2 Ek Ek x1
h

k
ρ ( x2 ,x1 )=∑exp(− βEk ) x2 Ek Ek x1
k
One sees that the procedure for evaluating a path
integral for the propagator can be used to evaluate
a path integral expression for the density matrix:
 1 βh m 2


ρ ( x2 ,x1 )= ∫exp − ∫  x& (u )+V ( x(u )) du Dx(u )
 
 h 02
where the integration is over all "paths" such that
x( β h )= x2 , x(0)= x1.
Evolution of a Gaussian Packet
If the initial state is a Gaussian wave function
ψ ( x′,0 =
e
ip0 x ′ h − x 2 2 ∆2
e
(π∆ )
2 14
then a direct integration yields
ψ ( x ,t =
[ ( )] e
π1 2 ∆+
iht
m∆
1
2
 − ( x − p0 t m ) 2 
 2
2 
 2 ∆ (1+ iht m∆ ) 
e
 ip0 
p0 t  
x
−



h
2
m



with the probability density
Pr( x, t ) =
1
π (∆2 + h 2t 2 m 2 ∆
• The mean position at time t is
case).
p0t m
to
1
2
at time t.
e
(like the classical
• The uncertainty in position grows from ∆
∆ 
h 2t 2 
1 + 2 4 
2 m ∆ 
)
2 12
− [( x − ( p0t m ]2
∆2 + h 2t 2 m 2 ∆2
2
at time 0
Position uncertainty =
1
2
∆ 
h 2t 2 
ht
1 + 2 4  ≈
for large t
2 m ∆ 
2 m∆
This can be viewed as t times the initial uncertainty
in the velocity.
• Note also that the uncertainty in position grows
very slowly – especially for a heavy particle.
• Note that there is a non-zero probability for the
particle to “go backwards”, that is, be captured at the
opposite direction in which it is “generally moving”.
• The Gaussian packet has an “average” velocity, associated
with the motion of its center. This is the so-called “group
velocity”.
• Let a (stationary) packet be defined by
∞
ikx
ψ ( x,0) = ∫ g (k )e dk ≡ ∫ g (k ) ei (α (k ) + kx) dk
∞
−∞
−∞
where α (k ) is the phase of g (k ). Assume that g (k ) has a
strong peak at k 0 , then the integral can be approximated by
k 0 + ∆k
∫
g (k ) e
i (α (k0 ) + (k − k0 )α ′(k0 ) + kx)
dk
k 0 − ∆k
This wave packet has a peak at the location in which the
phases of the exponentials change the slowest, since then they
interfere the least. This occurs for the x at which the derivative
of the phase by k is zero, that is,
x0 = − α ′( k0 ).
• If the packet is moving, we have
∞
ψ ( x, t ) = ∫
+
−
i
(
(
k
)
kx
(
k
)
t
)
α
ω
g (k ) e
dk
−∞
So the packet’s center is at
and its “group velocity” is
x0 (t ) = − α ′(k 0 ) + ω ′(k 0 )t
ω ′(k0 ).
Operator Power Series for the Propagator
• We saw that the propagator for a free particle can be
expressed as
e
−iHt h
2n
∞
1  iht  d 2 n
1  iht  d ψ ( x,0
=∑ 
⇒ ψ ( x, t = ∑ 


2n
2n
!
2
!
2
n
m
dx
n
m
dx




n =0
n =0
∞
n
n
• Sometimes, this can be used to compute ψ ( x, t .
An apparent counterexample
  πx 
L
x≤
sin 
ψ ( x,0 =   L 
2
 0
otherwise
It appears as if the spread of the wave function is not growing
over time, because all the derivative are nonzero only at the
interval x ≤ L 2 . How is this possible?
The operator power series doesn’t
converge (at the boundaries).
Look more carefully at
2n
1  iht  d ψ ( x,0
ψ ( x, t = ∑ 

2n
!
2
n
m
dx


n =0
∞
n
You can try and sum a few of the first terms,
and obtain a reasonable approximation, if the
higher derivative don’t diverge and if t is
small enough.
• But one has to be cautious: look at a particle
moving under the influence of a force f. The
Hamiltonian is then (assuming we fix all
constants to obtain a simple form):
n
d

d
n
H = 2 + fx ⇒ H ψ =  2 + fx  ψ
dx
1dx

42
4 43
4
2
2
The calculation of this operator is
tricky because the operators of
second derivative and multiplication
by x do not commute!
Solving Schrödinger’s Equation
for a Particle in a Box
• For a particle under the influence of a potential V (x ),
the states of definite energy satisfy
 P2

2m( E − V ( x ))

ψ
+ V ( x )  ψ = E ψ ⇒ ψ ′′ = −
2
h
 2m

• Next, we solve it for a particle bounded by an infinite
potential (“walls”) on both sides:
L

0
x≤
V ( x) = 
2
 ∞ otherwise
I
III
II
−L 2
L2
• In regions I and III,
2m( E − V ( x ))
ψ ′′ = −
ψ ⇒ ψ ′′ ≈ V0 ψ , V0 → ∞
2
h
• So the solutions are
ψ = Ae
V0 x
+ Be
− V0 x
V x
• In region III, for example, the Ae 0 part is not
− V x
physically allowed, as it diverges to infinity. Only the Be 0
part is allowed, but as V → ∞ , the solution is zero. Same
0
holds for region I.
• So the solution can be non-zero only in region II. The
equation there is
2mE
ψ ′′ = − 2 ψ ≡ ψ ′′ = −k 2 ψ ⇒
h
ψ = A cos(kx ) + B sin(kx )
• Boundary conditions: ψ ( L
2) =ψ ( − L 2) = 0.
• It immediately follows that the (normalized) solutions are
ψn
1

2
2


   cos nπx 
 L 
L 

=
1
 2  2  nπx 

  sin
 L 
 L 
n = 1,3,5...
n = 2,4,6...
• There is no solution corresponding to
equals 0.
• The energy (E) corresponding to
ψn
n = 0,
since it
h 2π 2n 2
is En =
.
2
2mL
First (green) and second (red) states: left=amplitude, right=probability.
• There
is no state with definite energy 0!
• A state with energy 0 would violate the uncertainty
principle:
• The particle is bound between − L
2 and L 2, so
the position uncertainty is bounded by
L 2.
2
L
h
h
h
2
(
)
∆X ≤ , ∆ X∆ P ≥ ⇒ ∆P ≥ ⇒ ∆ P ≥ 2
2
2
L
L
• We must have
P = 0, else the particle will drift
away to infinity, but that’s impossible, as it is bound
in the box. But
(∆P )2 =
P− P
2
= P
2
h2
≥ 2
L
• What about the average energy? We know that
2
E
2
= H
2
P
h2
=
≥
2m 2mL2
While the lowest energy level is obtained when n = 1
and it equals h 2π 2 2mL2 (the ground state
energy). The uncertainty principle can sometimes
be applied to find a rough estimate of the ground
state energy.
Example Question (Shankar, 5.2.1):
• A particle is in the ground state in a box of length
L. Suddenly the box expands to twice its size.
What is the probability to find the particle in the
ground state of the new box?
• Answer: the
two ground
state are
1


2
 g1 =  2  cos πx  small box 


 L
 L


1


 πx 
 1 2
big box 
 g 2 =   cos 
 L
 2L 


And the probability to find g1 in g 2 is
2
2


8

∗
 g1 g 2dx  =   ≈ 0.72
 ∫
  3π 
 −L 2

L2
SHOW MAPLE FILE HERE
Development in time of probability
amplitude, at t = 0 the state is (ψ 1 + ψ 2 )
(ordering is raster)
2
Quantization of Energy
• We saw that the energy levels of the particle in a
box are quantized.
• A particle which is bound by a potential from
escaping to infinity is said to be in a bound state.
Formally, lim ψ ( x ) = 0.
x → ±∞
• It turns out that the energy levels of a bound
particle are always quantized.
• Why? Look again at Schrödinger’s Equation:
2m( E − V ( x ))
ψ ′′ = −
ψ
2
h
• Assume that V (x ) is everywhere bounded. It
follows that both ψ and ψ ′ are continuous.
Assume also that there’s a bounded region (a
“well”) in which the particle’s energy is larger than
the (fixed) potential, but that outside the well the
potential is greater than the energy:
I
2m( E − V ( x ))
ψ ′′ = −
ψ
2
h
II
V >E
ψ ′′ = cI ψ , cI > 0
V <E
Solution is sum of ψ ′′ = − cII ψ , cII > 0
rising and
Solution is sum of
decreasing
sines and cosines
exponentials
III
V >E
ψ ′′ = cIII ψ , cIII > 0
Solution is sum of
rising and
decreasing
exponentials
But one of them has to be zero
because the rising exponentials go to
infinity as x → ∞.
• So we’re left with one parameter (degree of freedom) for each
of regions I and III, and two d.o.f for II. We have four constraints
– the continuity of the function and its derivatives at each of the
borders (just like splines). It appears that we can solve for every
E; but note that there is a redundancy for multiplying by a scalar,
hence we really have 3 d.o.f and four constraints – so there will
be a solution only for “special” cases. This is why the energy
levels for a bound state are quantized. This argument can be
extended to any finite potential. If the potential is infinite
(particle in a box) we have to enforce only the continuity of the
function, not the derivative.
Finite Potential Well
• Like particle in a box, but the walls are lowered
to a finite height: we’ll assume the energy to be E0 ,
which is smaller than V0 .
x = −a
x=a
I
2m(V0 − E0 )
k2 =
h2
• The solution
has the form
}
V0
II
2mE0
k1 =
h2
III
2m(V0 − E0 )
k2 =
h2
 region I :
Ae k2 x

ik1 x
−ik1 x
+
region
II
:
Be
Ce

−k2 x
region III :
De

• As noted, enforcing continuity of the function and
its derivative yields four equations in A,B,C,D, but
these can be solved only up to a scale factor, hence
the system is overdetermined and can be solved only
for special values of energy. The resulting equations
are transcendental and don’t have a closed-form
solution.
The Probability Current
• For particles, or classical flux, the current is the
amount of something crossing, say, a unit area
during unit time.
• In QM, it makes sense to talk about the
probability current. If many particles are present,
it will be equivalent to the classical definition.
• The probability current is defined for a onedimensional particle as the amount of probability
crossing over: how did the probability to find the
particle to the right of x0 change between time t0
and t0 + ∆t ?
t0 + ∆t
t0
x0
x0
• The probability change is
∞
∫ Pr( x, t
0
x0
∞
+ ∆t )dx − ∫ Pr( x, t0 )dx =
x0
 ∞ ∂ Pr( x, t ) 
0
[
]
+
∆
−
≈
∆
Pr(
,
)
Pr(
,
)
x
t
t
x
t
dx
t
dx 

0
0
∫x
∫
∂t
 x0

0
∞
• Assume that there’s a function J such that
∂ Pr
∂J
=−
∂t
∂x
Then the probability change is J∆t , and J is
therefore the probability current.
• A straightforward application of Schrödinger’s
Equation yields
h  ∗ dψ
dψ ∗ 

ψ
J=
−ψ
2mi 
dx
dx 
State at
t=2
Probability
current at t = 3
State at
t=3
• Initial state: Gaussian packet centered at 0,
moving to the right at unit speed.
• Why is the current negative at certain points?
• Because the packet is moving to the right, but is
also expanding! So there’s an area in which the
probability to the left of x is increasing.
• For the right moving wave packet
Ae
i
p2 
 px −
t 

h
2m 
= Ae
i
( px − Et )
h
• The probability current is
p
J= A
m
2
• Intuitively, this is true because p m is the
speed of the packet, and the packet is uniform,
so the current depends only on the speed.
• It can also be immediately verified that
current for two such wave packets moving at
the opposite directions is
(
)
p
J= A −B
m
2
2
Where A,B are the coefficients of the right and left
moving packets.
Introduction to Scattering
When a classical particle hits a potential barrier,
it either passes it (if its energy is higher than the
barrier’s energy), or is reflected (if its energy is
lower). The situation in QM is rather different.
• Start by considering a step
incoming
ψI
}
barrier
V0
t=0
x=0
transmitted
reflected
ψR
ψT
t >> 0
An incoming packet hits the barrier and
disintegrates into a left and right going packets.
Straightforward Solution
• Find the basis of the normalized eigenstates of
the step potential Hamiltonian.
• Project the initial wave function on that basis.
• Propagate each of the basis functions in time
(with the corresponding coefficient).
• As the time tends to infinity, identify the right
and left moving components.
It turns out that, under certain
assumptions, a much simpler method can
be used – which is rooted in classical
optics, and in the wave-particle duality.
• The problem can be solved in principle for any
packet, but it is impossible to obtain a closed-form
expression for the solution.
• Often, one is interested in cases in which the
incoming wave is very wide, and therefore has a
sharp peak in momentum space.
• In that case, it is possible to derive the ratio of the
reflected and transmitted parts with relative ease by
essentially reducing it to a stationary problem,
dependent only on the energy.
• The main trick is to look at all the components –
incoming, reflected, and transmitted – as coexisting,
and to measure the ratio of their probability currents.
The justification is: if the incoming wave is very
wide, it is impossible to tell when it hits the barrier.
• We’ll assume that k0 is the wave number of the
incoming packet, hence its energy and momentum are
energy = E0 = hk02 2m
momentum = hk0
And its stationary state
behaves like e
ik0 x
.
• The reflected part has the same wave number (and
frequency, of course) as the incoming wave. The transmitted
part has a wave number k1 which must satisfy
2mV0
E0 − V0 = hk 2m ⇒ k1 = k − 2
h
2
1
2
0
• So, we have the following situation:
Ae
ik0 x
Ceik1x
Be − ik0 x
x=0
• We can find A,B,C up to a scale factor because the function
and its derivatives are continuous at 0:
B k0 − k1 C
2k0
=
,
=
A k0 + k1 A k0 + k1
• The currents associated with incoming, reflected, and
2
2
2
transmitted are A hk m, B hk m, C hk m.
0
2
0
2
So, the
C k1 C
E0 − V0
T
=
=
probabilities
2
2
E0
A
k
A
0
are:
for transmitted
1
R=
B
2
A
2
= 1− T
for reflected
Tunneling
• What happens if V0 > E0 ?
ψ
energy =
}
I
E0
x=0
V0 “the forbidden zone”
II
• The solution to Schrödinger’s Equation in the
forbidden zone is
2m( E0 − V0 )
ψ ′′ +
ψ =0⇒
2
h
2m(V0 − E0 )
− k2 x
ψ ∝e
, k2 =
h2
ik1 x
− ik1 x

region
I
:
Ae
Be
+
If k1 = 2mE0 h ,

−k2 x
the solutions are
region
II
:
Ce

2
 B k1i + k2
B
=
⇒
=1


A k1i − k2
A
where 
2
2
C
2
k
i
C
4
k
1
 =
⇒
= 2 1 2
 A k1i − k2
A
k1 + k2
2
• It is immediate to see that the probability
current going into the right half is zero; yet –
some particles manage to escape into it.
• How can this happen? It seems to violate
energy conservation!
• From the uncertainty principle, conservation of
energy is violated only if we can measure this
violation. The position uncertainty for the particle
in the forbidden zone is very small (a decreasing
exponential), hence the uncertainty in momentum
is large enough – and it doesn’t allow the
measuring process to determine that energy
conservation was violated.
• So, a particle can break into the forbidden zone
(although it cannot go too far into it). If the potential
drops again to zero, the particle can escape through
the barrier and then it is free (like some particles
escaping from a nucleus).
A 1D particle hitting a step with a potential higher
than the particle’s energy. Note short penetration
into the forbidden zone.
Tunneling: a 2D particle hitting a barrier. The
probability amplitude is depicted.
Potential Barrier
V0 > E0
Incoming
(energy = E0 )
I
}
II
V0
III
x=a
x = −a
• In I,
ψ = Aeik x + Be − ik x , k0 = 2mE0 h 2
• In II,
ψ = Cek x + De − k x , k1 = 2m(V0 − E0 ) h 2
0
1
0
1
ik x
• In III, ψ = Ee 0
• As before, enforcing the function and its
derivative to be continuous at x = ± a yields four
equations, which can be solved up to a scale
factor, and then
2
E
1
T=
=
A
cosh 2 ( 2k1a ) + α 2 sinh 2 ( 2k1a )
V0 − 2 E0
α=
4 E0 (V0 − E0 )
Scattering off a
potential barrier.
Horizontal axis
is E V0 ,
vertical axis is
T. Scales are
chosen so that
2mV0 a 2 = 9h 2
(following
Eisberg &
Resnick, p.
202). Lower is
detail of upper;
note that T is
equal to 1 at
certain energies
(strange,
because we can
have more
transmission at a
lower energy!).
Xmaple code for potential
barrier, tunneling
V0 > E
assume(a,positive); assume(k1,real); assume(k2,real);
psi1:=A*exp(I*k1*x)+B*exp(-I*k1*x); # (-infinity,-a/2) k1=sqrt(2*E*m/h^2)
psi2:=C*exp(k2*x)+DD*exp(-k2*x); # (-a/2,a/2) k2=sqrt(2*(V0-E)*m/h^2)
psi3:=F*exp(I*k1*x); # (a/2,infinity)
# now force continuity and first derivative continuity
eq1:=subs(x=-a/2,psi1-psi2);
eq2:=subs(x=-a/2,diff(psi1,x)-diff(psi2,x));
eq3:=subs(x=a/2,psi2-psi3);
eq4:=subs(x=a/2,diff(psi2,x)-diff(psi3,x));
sol:=solve({eq1,eq2,eq3,eq4},{F,C,B,DD});
Fsol:=subs(sol,F)/A; numF:=numer(Fsol);
help1:=expand(simplify(numF*conjugate(numF)));
denomF:=denom(Fsol);
help2:=expand(simplify(denomF*conjugate(denomF)));
T:=help1/help2;
T1:=subs({k1=sqrt(2*m*E/h^2),k2=sqrt(2*m*(V0-E)/h^2)},T);
# Following Eisberg & Resnick p. 202
T2:=simplify(subs(h=sqrt(2*m*V0*a^2/9),T1));
T3:=simplify(subs(E=V0*r,T2));
plot1:=plot({T3},r=0..1,thickness=3,axes=boxed,axesfont=[TIMES,BOLD,16],n
umpoints=1000):
V0 < E
#################################### end V0>E, start E>V0
assume(a,positive); assume(k1,real); assume(k2,real);
psi1:=A*exp(I*k1*x)+B*exp(-I*k1*x); # (-infinity,-a/2) k1=sqrt(2*E*m/h^2)
psi2:=C*exp(I*k2*x)+DD*exp(-I*k2*x); # (-a/2,a/2) k2=sqrt(2*(E-V0)*m/h^2)
psi3:=F*exp(I*k1*x); # (a/2,infinity)
# now force continuity and first derivative continuity
eq1:=subs(x=-a/2,psi1-psi2);
eq2:=subs(x=-a/2,diff(psi1,x)-diff(psi2,x));
eq3:=subs(x=a/2,psi2-psi3);
eq4:=subs(x=a/2,diff(psi2,x)-diff(psi3,x));
sol:=solve({eq1,eq2,eq3,eq4},{F,C,B,DD});
Fsol:=subs(sol,F)/A; numF:=numer(Fsol);
denomF:=denom(Fsol);
T:=simplify(abs(numF/denomF)^2);
T1:=subs({k1=sqrt(2*E*m/h^2),k2=sqrt(2*(E-V0)*m/h^2)},T);
#following Eisenberg&Resnick p. 202
T2:=simplify(subs(h=sqrt(2*m*V0*a^2/9),T1));
T3:=simplify(subs(E=V0*r,T2));
plot2:=plot({T3},r=1..10,thickness=3,axes=boxed,axesfont=[TIMES,BOLD,16],
numpoints=1000):
with(plots):
display({plot1,plot2});
plot(T3,r=2..10,thickness=3,axes=boxed,axesfont=[TIMES,BOLD,16],numpoints
=200);
A Delta Function Potential
• Assume the same situation as before, but instead
of a step potential we have a “Delta function”
potential – that is, a very narrow infinite wall.
Aeik0 x
Be
Potential = aδ (x )
− ik0 x
Ce
ik0 x
x=0
• Note that the wave number on the right of the
barrier is the same as on the left, because there is no
region in which is loses energy.
• As before, there are two equations: the function
has to be continuous at 0 (hence A+B=C). The first
derivative equation is a bit more tricky, because
Schrödinger’s Equation is
2m( E0 − V0 )
2mE0
2maδ ( x )
ψ ′′ = −
ψ =− 2 ψ +
ψ
2
2
h
h
h
• What happens to ψ ′ at x = 0 ? Since it is the
integral of ψ ′′, which contains a product of the
Delta function, it jumps by 2maC h 2 when
crossing the barrier from left to right. So, the
equation for ψ ′ is
2maC
ik (C − A + B ) =
h2
• Solving this together with A+B=C yields
2
2
2
B
ma
R=
= 2 2
, T = 1− R
2 4
A
m a + k0 h
• As the particle is heavier, and its wave number
smaller, it will tend more to behave classically (that
is, reflect).
Arbitrary Barrier
• Express as the limit of rectangular barriers of
infinitesimal width, and calculate the limit of the
product of the corresponding transmission
coefficients.
Some results on Degeneracy
• Recall that an eigenvalue ω is degenerate if there
are two linearly independent states with eigenvalue ω .
Theorem: there is no degeneracy in one dimensional
bounded states.
Proof: assume
h2
h2
−
ψ 1′′ + Vψ 1 = λψ 1 , − ψ 2′′ + Vψ 2 = λψ 2
2m
2m
Multiply by ψ 2 and ψ 1 and subtract to get
′ 2 − ψ 1ψ 2′′ = 0 ⇒ (ψ 1′ψ 2 − ψ 1ψ 2′ )′ = 0 ⇒
ψ 1′ψ
ψ 1′ψ 2 − ψ 1ψ 2′ = const ., take x → ∞ to get const . = 0
ψ 1′ ψ 2′
So,
=
⇒ log(ψ 1 ) = log(ψ 2 ) + const . ⇒ ψ 1 = e const .ψ 2
ψ1 ψ 2
But a difference in scale factor means that the states
represent the same physics.
Theorem: If parity is conserved (that is, the
Hamiltonian commutes with the parity
operator), then every non-degenerate state of
definite energy has definite parity.
Proof: Let Q be the parity operator.
H ψ = E ψ ⇒ H (Q ψ
Q(E ψ ) = E (Q ψ ) ⇒
) = Q(H ψ ) =
non -degeneracy
Q ψ = e iδ ψ
but Q 2 = 1 ⇒ eiδ = ±1, so there' s definite parity.
There can be degeneracy in two dimensional
systems – such as a particle in a 2D box:
• The states of definite energy are products of the 1D
states for x and y, and their energy is the sum of
energies of the corresponding 1D states.
• So, if we know for example that the energy is
10h 2π 2
, then the state is not uniquely determined,
2
2mL
but instead it can be any normalized linear
combination of
ψ 1,3
ψ 3,1
2
 πx   3πy 
= cos  cos
, or
L
 L  L 
2
 3πx   πy 
= cos
 cos 
L
 L   L
That is, the state has to be of the form
2
2
α ψ 1,3 + β ψ 3,1 , α + β = 1
The Harmonic Oscillator
• Very important – models the behavior of a
system around an equilibrium point, even for
many particles (as follows by diagonalizing the
Hamiltonian).
• The Hamiltonian is the same one as in classical
mechanics, with the usual substitution p → P, x → X :
P 2 mω 2 X 2
H=
+
2m
2
• Since X and P are Hermitian, it follows that H
cannot have negative eigenvalues.
• To find the definite energy states, we must solve
d 2ψ 2m 
mω 2 x 2 
ψ = 0
+ 2  E −
2
dx
h 
2 
• Define constants and a variable change
1
2
E
 h 
b=
, x = by
 ,ε =
hω
 mw 
2
′
′
• To obtain ψ + ( 2ε − y )ψ = 0
• To solve it, look at the asymptotic behaviors:
y2
±
m
2
y → ∞ ⇒ ψ ′′ − y 2ψ ≈ 0 ⇒ ψ ≈ Ay e
m
y2
2
• But Ay e is ruled out from physical
considerations.
• At the other end
y → 0 ⇒ ψ ′′ + 2εψ ≈ 0 ⇒
ψ ≈ A cos( 2ε y ) + B sin( 2ε y ) =
2
A + cy + O ( y ), c = B 2ε
• Next, guess a solution
ψ ( y ) = u ( y )e
y2
−
2
Where
u ( y ) → A + cy + (higher powers)
y →0
u ( y ) → y m + (lower powers)
y →∞
u′′ − 2 yu′ + ( 2ε − 1)u = 0
• Guess further u ( y ) =
∞
n
C
y
∑ n
n =0
• And obtain Ck + 2
2k + 1 − 2ε
2
=
Ck → C k
( k + 2)( k + 1) k →∞ k
• This in itself would not be very bad, but it’s
y2
trivial to verify that u ( y ) behaves like e
when y → ∞, which implies that
y2
2
ψ ( y ) ≈ e , but that is physically untenable.
y →∞
• Fortunately, there’s a way out: look at the recursion
Ck + 2
2k + 1 − 2ε
Ck
=
( k + 2)( k + 1)
This has to stop somewhere in order to prevent the
solution from diverging, so we must have
2k + 1
ε=
, k = 0,1,2...
2
• So, energy quantization somehow came (again)
into the picture. The solution for energy E is


mω
ψ E ( x ) = 

2n
2 
 πh2 ( n! ) 
1
4
1

 − mω x 2
2
 mω   2 h

Hn 
 x e
 h  


1

Where E =  n + hω
2

Hermite, not
Hamiltonian!
• The H’s are the Hermite polynomials. The first are
H 0 ( y ) = 1 H1 ( y ) = 2 y
H 2 ( y ) = −2(1 − 2 y 2 )
2 3
4 4


2
H 3 ( y ) = −12 y − y  H 4 ( y ) = 121 − 4 y + y 
3 
3 


A few examples.
The upright arrows
mark the classical
turning point for an
oscillator with the
given energy; note
that, much like as
with the potential
barrier, the oscillator
enters the classically
forbidden zone with
non-zero probability.
Note also that as the
energy increases, the
behavior starts to
resemble the
classical behavior.
Interlude: Finding the Oscillator’s Ground
State from the Uncertainty Principle
P 2 mω 2 X 2
+
H=
2m
2
• Let’s search for the minimal energy state, that is,
the normalize ψ which minimizes ψ H ψ :
1
1
2
2
2
ψ Hψ =
ψ P ψ + mω ψ X ψ
2m
2
Recall that ψ X ψ = (∆X ) + X (just as
with ordinary average and variance), and the same for
P. Since we want a minimum, take X = P = 0,
and then
2
2
2
1
1
2
2
2
(
)
(
)
ψ Hψ =
∆P + mω ∆X
2m
2
We know that
(∆P ) (∆X )
2
2
2
h2
h
2
≥
⇒ (∆P ) ≥
2
4
(
)
4 ∆X
And therefore
h2
1
2
2
ψ Hψ ≥
+ mω (∆X )
2
8m(∆X ) 2
The minimum is obtained when
(∆X )
2
hω
h
, for which H =
=
2 mω
2
• We proved that the minimum is obtained for a
Gaussian; since ∆X = ∆P = 0, the Gaussian is
centered at 0 and has no phase (momentum) shift,
which after normalizing equals
 mω 
ψ min ( x ) = 

 πh 
1
4
mωx 2
−
e 2h
• Due to the non-degeneracy of one dimensional
bounded states, ψ min must indeed be the ground
state (there’s only one state with the minimal energy
equal to its eigenvalue). Thus, to quote Shankar, we
have found the ground state by “systematically
hunting in Hilbert space”.
• Example question: a potential barrier is added, so
that the potential for x ≥ 0 is ∞. What is the ground
state energy?
• Answer: no even solutions are possible, since
they are non-zero at x = 0. So the ground state
energy is (3 2 )hω .
Analysis of the Oscillator with the
“Raising” and “Lowering” Operators
• An important and particularly attractive technique
due to Dirac for solving the oscillator in the energy
basis.
• In order to solve
 P 2 mω 2 X 2 
 E = E E
H E = 
+
2 
 2m
• Define the two operators
1
2
1
2
 mω 
 1 
a=
 X + i
 P
 2h 
 2 mω h 
1
2
1
2
 mω 
 1 
a =
 X − i
 P
 2h 
 2 mω h 
+
• It readily follows that their commutator is
[a,a ] =aa
+
+
+
−a a =1
• And they “nearly factor” H (“nearly” because
X,P do not commute):
(
)
H
1
+
ˆ
=a a+
H = a a + 1 2 hω , so define H =
hω
2
+
• Let’s try and solve
Ĥ ε = ε ε
• The action of a + and a on the eigenvectors of Ĥ
is very simple, because
[a, Hˆ ] = a, [a , Hˆ ] = −a
= ε ε , Hˆ a ε = (aHˆ − [a, Hˆ ]) ε
(aHˆ − a ) ε = (ε − 1)(a ε ).
+
Hˆ ε
+
Similarly, Hˆ a + ε = (ε + 1)(a + ε )
=
• So, operating with a, a + lowers and raises the eigenvalues
of Hˆ . But since H (and therefore Hˆ ) are positive definite, the
operation by a has to stop somewhere:
∃ε 0 a ε 0 = 0 ⇒ a + a ε 0 = 0 ⇒
Hˆ − 1 2 ε 0 = 0 ⇒ Hˆ ε 0 = 12 ε 0
(
)
• Due to the non-degeneracy of one dimensional bounded
states, every such “decreasing sequence” has to end with the
same ε 0 , denote it 0 .
+
• Label the successive operations of a on
ε0
by
n.
( )
It follows that a n = n n − 1 , a + n = n + 1 n + 1 , because : n is a +
n
0
normalized. Since Hˆ 0 = (1 2) 0 and every a + raises by 1, we have
(
)
Hˆ n = (n + 1 2) n . Now, Hˆ a n = aHˆ − a n = (n − 1 2 )(a n ) and since Hˆ is
non - degenerate and Hˆ n − 1 = (n − 1 2) n − 1 there' s a constant C such that
a n = C n − 1 . Take adjoints to obtain n a + = n − 1 C ∗ , and multiply to get
n a + a n = C . But n a + a n = n H − 1 2 n = (n + 1 2 ) − 1 2 = n.
2
(
a )
n =
+ n
So:
(n!)
12
0
• So, the representations of a , a + , H in the n
basis are very simple, and so are those of X and P:
1
2
1
2
 h 
 mω h  +
+
(
)
X =
 a + a , P = i
 (a − a )
 2 
 2 mω 
Example question:
• What are X , X 2
in the state n ?
1
2
• Up to a constant c = (h 2mω ) , X = a + a + .
Ignoring c for the moment,
X = n a + a+ n = n a n + n a+ n = 0 + 0 = 0
Since
X
2
a, a + lower and raise
= n (a + a
)
+ 2
n and since n m = δ m ,n
n = n a n + n (a
2
)
+ 2
n +
n a + a n + n aa + n = 0 + 0 + n a + a n + n aa + n
Now, use the identities
a n = n n − 1 , a+ n = n + 1 n + 1
To obtain (after multiplying by c 2 )
X
2
1 h

= n + 
2  mω

Passage from the Energy Basis to
the Coordinate Basis
1
2
1
2
1
2
 mω 
 1 
a=
 X + i
 P⇔
 2h 
 2 m ωh 
1
2
 h  d
 mω 
12
denote y = (mω h ) x


 x+
 2mω  dx
 2h 
1
1
− 
− 
d 
d  +
2
2
so a = 2  y + , a = 2  y − 
dy 
dy 


• To find the ground state in the coordinate basis:
a ψ0

d 
= 0 ⇒  y +  ψ 0 = 0 ⇒ the ground state is
dy 

ψ 0 = A0e
y2
−
2
1
4
 mω 
=
 e
 πh 
mωx 2
−
2h
(after normalization)
And the other eigenstates can be derived (up to a
normalizing factor) by
n
ψn
(
a )
=
+ n
(n!)
12
−
ψ0
n
d 
2 
=
y −  ψ 0
12 

(n!)  dy 
2
Coherent States
• Seek the “classical limit” of the oscillator,
i.e. states with a relatively small uncertainty
in energy etc.
• Time development of classical oscillator:
x& (t )=(1 m ) p (t ), p& (t )=− mω x(t )
Define xˆ (t )= β x(t ), pˆ (t )=(1 hβ ) p (t )
β = mω h
xˆ& (t )=ωpˆ (t ), pˆ& (t )=−ωxˆ (t ).
Define α (t )= 1 2 [xˆ (t )+ipˆ (t )]
−iωt
α& (t )=−iωα (t )⇒α (t )=α (0)e
ˆx(t )= 1 2 α 0 e −iωt +α 0*e iωt
−iωt
* i ωt
pˆ (t )= −i 2 α 0 e −α 0 e
2
(
)
(
(
(
)[
)[
)
]
]
The energy (for all t ) = hω α 0
2
• There is a lot of similarity between these
relations and the average behavior of the
quantum oscillator:
(
)(
)
)(
+
ˆ
X =βX = 1 2 a+a
+
ˆ
P=(1 hβ )P= −i 2 a −a
(
(
H =hω a a +1 2
+
)
( 2 )[ α (0)e
Pˆ (t )=(−i 2 )[ α (0)e
Xˆ (t )= 1
−iωt
)
*
+ α ( 0) e
*
]
]
− α ( 0 ) e iω t
This is almost identical to the classical
oscillator equations, with α 0 playing
the role of α 0 . We therefore ask that
ψ (0) aψ (0) =α 0 . For the energy to
behave similarly, we should have
2
+
H =hω a a (0)=hω α 0 -- but
[
H =hω 1 2+ α 0
2
that α 0 >>1 2.
2
−iωt
i ωt
]. Assume then
• This analysis suggests to require
ψ (0) aψ (0) =α 0 , ψ (0) a aψ (0) = α 0
+
2
The simplest way to enforce this is to
require aψ (0) =α 0 ψ (0) .
How do the eigenvectors of a look like?
Let a α =α α , denote α =∑cn (α ) n .
n
So a α =∑cn (α )a n =∑cn (α ) n n−1 =
n
n
α ∑cn (α ) n ⇒cn (α )=
n
cn (α )=
α
n
c0 (α )
αcn−1 (α )
n!
And after normalization:
α =e
−α
2
2
∑
n
α
n
n!
n.
n
⇒
• It follows that:
4
2 1
 2 1
2
2 2
H α =h ω  α +  , H = h ω  α + 2 α + 
α
4
2


∆H α
And so ∆H α =hω α ⇒
<<1.
H α
Since X ,P are expressible in terms of α ,α + , it
is straightforward to derive:
2h
X α=
Re(α ), P α = 2mωh Im(α )
mω
h
h
mωh
, ∆Pα =
∆X α =
⇒∆X α ∆Pα =
2mω
2
2
Time evolution:
n
2
α
−α
2
e 0 ∑ 0 e −iEnt h n =
n!
n
−iωt n
2
(
α 0 e ) −iEnt h
−iωt 2 − α 0 2
−iωt 2
−iωt
e
e
e
n
=
e
α
e
∑n n!
0
So, the state remains an eigenvector of a. This also means
that ∆X ,∆P don't change. The wave function is a "small"
α
α
Gaussian which moves back and forth.
The Operator D(α ) and the Wave Function ψ α (x)
αa + −α *a
Define D(α )=e
, D(α ) is unitary.
2
*
+
Since αa ,α a =− α , it follows from Galuber's
formula for two operators which commute with
[
]
their commutator that D(α )=e
e
−α *a
αa +
0 = 0 , and e
D(α ) 0 =e
−α 2
∑
αn
0 =∑
− α 2 αa + −α *a
αn
n!
e
e
.
n , so
n =α , which allows to
n!
calculate the wave function ψ α ( x)= xα = x D(α ) 0 .
The result is
2
14




x
X
−
m
ω
x




α
ψ α ( x ) =e i θ α 
 exp −
 +i P α 
2∆X α 
h
 πh 




Field Quantization
• For two coupled oscillators, new variables can
be defined which decouple them. This can also
be done for an infinite chain of oscillators, a
continuous string, and the electromagnetic field.
• The Hamiltonian for a pair of coupled
oscillators equals
p12 p22 1
2 1
2
2
2
+ + mω ( x1 −a ) + mω 2 ( x2 + a ) +λmω 2 ( x1 − x2 )
2m 2m 2
2
where λ controls the strength of the coupling.
• Defining new variables and new masses
x (t )+ x2 (t )
x (t )− x2 (t )
m
,µG =2m,µ R =
,xR (t )= 1
xG (t )= 1
2
2
2
decouples the resulting differential equations,
and the Hamiltonian can be written as
2
p
p
1
1
2a 

2 2 4λ
m
a
+
ω
+
+ µGωG2 xG2 + µ Rω R2  xR −

2µG 2µ R 2
2
1+ 4λ 
1+ 4λ

p1 (t )− p2 (t )
where pG (t )= p1 (t )+ p2 (t ), p R (t )=
,ωG =ω ,
2
ω R =ω 1+4λ
2
G
2
R
• The quantum mechanics case proceeds similarly:
new observables are defined by
X1+ X 2
P1 − P2
XG =
,PG = P1 + P2 , X R = X 1 − X 2 ,PR =
2
2
• From the commutation relations of the original
observables
[X 1 ,P1 ]=[X 2 ,P2 ]=ih (and the other commutators = 0)
It follows that the same relations hold for the new
observables. It also follows that the Hamiltonian
P12 P22 1
2 1
2
2
2
+ + mω ( X 1 −a ) + mω 2 ( X 2 + a ) +λmω 2 ( X 1 − X 2 )
2m 2m 2
2
equals
2
P 1
P 1
2a 

2 2 4λ
+ µGωG2 X G2 +
+ µ Rω R2  X R −
+
m
a
ω

2µG 2
2µ R 2
+ 4λ 
1+ 4λ
 4414
4442
43
14
42443 14
2
G
2
R
HG
HR
Note that this is true only because of the
commutation relations between the new
observables.
A basis for the system’s wave functions is given by
the tensor product of the corresponding
eigenvectors of H G and H R , which are given by
the respective raising and lowering operators:

1  µ G ωG
i
a = 
XG −
PG 
h
2 
µGωG h 
+
G

1  µ Rω R
2a
i
a = 
X R′ −
PR , X R′ = X R −
h
1+ 4λ
2 
µ Rω R h 
+
R
the eigenvectors of H G are given by
( )
1 + n G
ψ =
aG ψ 0 ,with an eigenvalue of
n!
 1
R
 n+ hωG , and similarly for ψ n .
 2
G
n
The eigenvectors of the system are given by
( )( )
1
+ n
+ p
aG a R ψ 0, 0 ,with an energy of
ψ n, p =
n!p!
 1
 1
2 2 4λ
 n+ hωG + p + hω R + mω a
1+ 4λ
 2
 2
• In the (0,0) state, both the average and the distance
between the two masses are Gaussians around zero.
• In the (1,0) state, the average’s distribution has two
peaks away from zero, and the distance is a Gaussian
around zero.
• In the (0,1) state, the average’s distribution is a
Gaussian around zero, and the distance’s distribution
has two peaks away from zero.
• In the (1,1) state, the distributions of both the
average and the distance have two peaks away from
zero.
(1,0)
(0,0)
(0,1)
(1,1)
Continuous String
 1 ∂2
F
∂2 
 2 2 − 2 u ( x,t )=0, v=
µ
∂x 
 v ∂t
2
L
F  ∂u ( x,t ) 
Potential energy = ∫
 dx
2 0  ∂x 
µ  ∂u ( x,t ) 
L
Kinetic energy =

∫
2 
2
 dx
∂t 
0
2  πx 
Fourier expansion with f k ( x)= sin  k :
L  L
∞
u ( x,t )=∑qk (t ) f k ( x)⇒qk (t ),q& k (t ) determine
k =1
the state of the string at time t.
• In Fourier domain, the equations are
kπv
2
decoupled: q&&k (t )+ωk qk (t )=0, ωk ≡
L
• It’s straightforward to see that the
µ
[
2
2 2
&
Lagrangian equals L= ∑ qk −ωk qk
2 k =1
]
• The momentum conjugate to qk is
∂L
pk =
= µq& k
∂q& k
µ
 pk2 µ 2 2 
• The Hamiltonian is H = ∑ + ωk qk 
2 k =1  2 µ 2

1
pk
Denoting qˆ k ≡ β k qk , pˆ k =
βkh
1
2
2
ˆ
ˆ
yields H =∑ hωk qk + pk
2
[
]


µω
k
 βk ≡



h


A schematic proof that qk , pk =q& k are
canonical (most constants ignored, L=2π ):
[
2π
]
Lagrangian=L= ∫ (∂u ( x,t ) ∂t ) −(∂u ( x,t ) ∂x ) dx,
2
2
0
and the momentum conjugate to u ( x) is ∂L ∂u& ≈u&.
Fourier expansion with f k ( x)=sin (kx ):
∞
∞
k =1
k =1
u ( x,t )=∑qk (t ) f k ( x)⇒u& ( x,t )=∑q& k (t ) f k ( x), so
2π
2π
qk ≈ ∫u ( x,t ) f k ( x)dx, pk ≈ ∫u& ( x,t ) f k ( x)dx.
0
0
Commutation relations: {qk ,ql }=


2π
 ∂qk ∂ql
∂qk ∂ql 
∫0  ∂u ( x) ∂u& ( x) − ∂u& ( x) ∂u ( x) dx=0, same for {pk , pl }.


123 123
0
0




2π
 ∂qk ∂pl
∂qk ∂pl 
{qk , pl }= ∫ 
−
dx=
∂u ( x) ∂u& ( x) ∂u& ( x) ∂u ( x) 
0
123 123
0
0


2π
∫f
0
k
( x) f l ( x)dx=δ kl .
• Now, quantize by promoting qˆ k , pˆ k to
Operators Qˆ k ,Pˆk satisfying [Qˆ k1 ,Pˆk 2 ]=iδ k1k 2 .
(note: as for one particle, the classic variables q& k ,qk are timedependent, but the operators are not).
(
)
1
2
2
ˆ
ˆ
H
=
h
ω
Q
+
P
• Denote k
k
k
k . It has
2
eigenstates and eigenvalues
1

H k nk = nk + hωk .
2

• Since the H k commute, a general state can
be defined by n1 n2 ... nk , with energy
∑n hω
k
k
above the ground state.
k
As before, define creation and annihilation
operators: ak = 1 2 Qˆ k +iPˆk ,
+
ak = 1 2 Qˆ k −iPˆk .
(
)(
(
)
)(
)
ak n1...nk ... = nk n1...nk −1...
+
k
a n1...nk ... = nk +1 n1...nk +1...
(
a )
...
(
a )
n ...n ... =
1
k
+ nk
k
+ n1
1
n1!
nk !
... 0
The observable u ( x,t ) is promoted to an
operator U ( x)=∑ f k ( x)Qk =
(
k
1
1
+
f
x
a
a
+
(
)
∑
k
k
k
2 k βk
) (β
k
≈ k
)
And it is not hard to show that U ( x) (t )
satisfies the classical equation.
• What is the ground state 0 ? Its k’th Fourier
coefficient is an oscillator with frequency
proportional to k, at its ground state – i.e. a
Gaussian with variance ≈ 1 k . So a sample of
the ground state is a function with slowly
decreasing Fourier coefficients. This decrease
reflects the coupling in the spatial domain,
and the samples are not random noise, but
have some measure of smoothness associated
with them.
• The ground state is a fractal of dimension 2.
• An excited state is the superposition of
sinusoids with the ground state.
Quantization of the EM Field
Maxwell's Equations :
∂ρ
∇⋅E = 4πρ ∇⋅B = 0 ∇⋅J = −
∂t
1 ∂B
4π 1 ∂E
∇×E = −
∇×B = J +
c ∂t
c c ∂t
1 ∂A
Potentials : ∇×A=B, E=−∇φ −
c ∂t
1
∂
Λ
Gauge transform: A′=A−∇Λ, φ ′=φ +
c ∂t
Coulomb gauge: ∇⋅A=0, (φ =0 if ρ =0, J=0).
1 ∂φ
⇒
Lorentz gauge: ∇⋅A= −
c ∂t
2
π
4
J
2
φ =−4πρ
A= −
2
1 ∂
≡∇ − 2 2
c ∂t
2
2
c
 v 
Force on a charged particle : q E + ×B 
 c 
1
Energy of the EM field : ∫∫∫ E 2 + B 2 d 3r
8π
[
]
• Experimental
motivation for
quantizing the EM field: spontaneous
decay of hydrogen atoms when no
field is present.
Recipe for quantization: find canonical
coordinates Ai and momenta Π j such that
Π i =∂L ∂A&i , { Ai ,Π j }=δ ij , and promote
them to operators with similar commtation
relations.
•
• The EM Lagrangian is
[
]
1
2
2
3
L= ∫∫∫ E − B d r =
8π
2

1
1 ∂A
−∇φ − ∇×A
−
∫∫∫
8π  c ∂t
2
 3
d r

and indeed it yields Maxwell’s equations when
varied with respect to the potentials.
• But since φ& does not appear in L,
conjugate momentum.
φ has no
• Since we’re working in free space (no
charges/currents), this problem can be alleviated
by working in Coulomb gauge with φ =0.
• The Lagrangian is then simpler, but it doesn’t
yield the equation ∇⋅E=0 any more. This
condition now has to be enforced.
• This will result in one less degree of freedom for
the vector potential A.
• The coordinates are A
(
r
)
i
, and the
−E i ( r )
conjugate momenta are Π ( r )=
.
i
4πc
• Alas, they are not canonical. We’d like to have
{A i (r ),Π j (r′)}=δ ijδ (r −r′)
but taking the divergence with respect to r
yields zero on the left-hand side, but not the righthand side.
• The way out of this is to parameterize A by
independent variables, so that ∇⋅A=0 will hold.
As in some cases in classical mechanics, this is
easier to do in the Fourier domain, in which the
condition is algebraic and not differential, and in
which the coordinates can be decoupled.
• Since A (r ) and Π (r ) are real, there is
a complex vector function a(k) satisfying:
[
A(r )= ∫ a(k)e +a*(k)e
ik⋅r
− ik⋅r
]d k
3
[
]
1
ik⋅r
−ik⋅r
3
Π ( r )=
k
a(k)
e
−
a*
(k)
e
d
k
4πic ∫
• The existence of a(k) is guaranteed by
the redundancy in the Fourier transforms
of the real functions A and Π : for a real
f, F (− k)= F * (k).
• The choice of a(k) is guided by the
following identities for the oscillator:
12
 h 
X =

 2mω 
12
 mωh 
a + a , P =i

 2 
(
+
)
(a
+
−a
)
∇⋅A=∇⋅Π =0 are
expressed in the Fourier domain by
• The conditions
k⋅[a(k)+a∗(-k)]=k⋅[a(k)−a∗(-k)]=0⇒k⋅a(k)=0
which suggests to expand a(k) in a basis
orthogonal to k.
• Define for every k an orthonormal basis
ε (k1),ε (k2 ),ε (k3), such that ε (k3) k.
• For every k expand a(k) as follows:
3
(
a(k)=∑ c 4π ω
λ =1
2
2
)
12
a (kλ )ε (kλ ) (ω =kc)
This yields
12
 c 
A(r)=∑∫ 2 
λ  4π ω 
2
[a(kλ )ε (kλ )e
12
1 ω 
Π (r)=∑∫ 
4 
i
64
π


λ
ik⋅r
[a(kλ )ε (kλ )e
]
+ a∗(kλ )ε (kλ )e −ik⋅r d 3 k
ik⋅r
]
−a∗(kλ )ε (kλ )e −ik⋅r d 3k
• Before imposing transversality
A (r),A (r′)=Π (r),Π (r′)=0,
 i
  i

j
j

 

A (r),Π (r′)=δ δ (r-r′)
 i
 ij
j


And it follows that
{a(kλ ), a(k′λ ′)} = {a ∗ (kλ ), a ∗ (k′λ ′)} = 0,
{a(kλ ), a ∗ (k′λ ′)} = − iδ λλ′δ 3 (k - k′)
Outline of proof : ignoring all constants,
[
]
a (kλ ) = i ∫ A(r)e −ik⋅r d 3 r + ∫ Π (r)e −ik⋅r d 3 r ⋅ ε (kλ )
[
]
a ∗ (kλ ) = − i ∫ A(r)e ik⋅r d 3 r + ∫ Π (r)e ik⋅r d 3 r ⋅ ε (kλ )
Now continue as usual, with the derivatives by
A j (r0 ), Π j (r0 ) given by e −ik⋅r0 ε j (kλ ) , etc.
• Next, discard a (k 3) which must be
zero due to the transversality.
•
The EM Hamiltonian is
[
]
1
2
2
3
H = ∫∫∫ E + B d r
8π
and it equals
2
3
ω
a
∗
(
k
λ
)
a
(
k
λ
)
d
k
∑∫
λ =1
• Guided by the oscillator equations, define
1
[a(kλ )+a∗(kλ )]
q ( kλ ) =
12
(2ω )
12
ω 
p (kλ )=−i  [a(kλ )−a∗(kλ )]
2
it can be verified that they satisfy the
canonical Poisson bracket relations.
• It immediately follows that
[
]
1
H = ∑ ∫ p 2 ( kλ ) + ω 2 q 2 ( kλ ) d 3 k
2 λ
so the field is a sum of a continuous
family of oscillators, two at each
frequency.
• Next, promote p ( kλ ), q (kλ ) to operators
P (kλ ), Q (kλ ) satisfying the canonical
commutation relations
[P(kλ ), Q(k′λ ′)] = ihδ λλ′δ 3 (k - k′)
and define raising and lowering operators
12
12
ω 
 1 
a ( kλ ) =   Q ( kλ ) + i 
 P ( kλ )
 2h 
 2ωh 
12
12
ω 
 1 
a ( kλ ) =   Q ( k λ ) − i 
 P ( kλ )
 2h 
 2ωh 
which satisfy
+
[a(kλ ), a
+
]
(k ′λ ′) = δ λλ ′δ 3 (k - k ′)
A and Π are also promoted to operators:
12
 hc 
A(r)=∑∫ 2 
λ  4π ω 
2
[a(kλ )ε (kλ )e
12
1 hω 
Π (r)=∑∫ 
4 
i
64
π


λ
ik⋅r
[a(kλ )ε (kλ )e
]
+a + (kλ )ε (kλ )e −ik⋅r d 3 k
ik⋅r
]
−a + (kλ )ε (kλ )e −ik⋅r d 3 k
To find the Hamiltonian, symmetrize a ∗ a
by (1 2 )(aa ∗ + a ∗ a ) to obtain
1
 +
3
H = ∑ ∫ a (kλ )a (kλ ) + hωd k
2

λ
• In the ground state, ALL the oscillators
are in their ground state, so for all k, λ
a ( kλ ) 0 = 0.
• Formally, the energy of the ground state
(“vacuum”) is E0 ≡
∑λ ∫ (1 2)hωd
3
k = ??
• It can be verified from the definition of H
and from the commutation relations that
(
)
H a (kλ ) 0 =hωa (kλ ) 0 =(hkc )a (kλ ) 0
+
+
+
(this is computed “above” E0 ).
• The momentum operator of the field can
be computed by promoting the classical EM
1
3
momentum
(
)
E
×
B
d
r to
∫
4πc
[
]
P = ∑ ∫ a ( kλ ) a ( kλ ) hkd k
λ
+
3
And it can be verified that the momentum
of a + ( kλ ) 0 ≡ kλ is hk. Since
E = p c + m c it follows that kλ is
2
2 2
2 4
a massless particle – the photon. Its wave
function is
ε ( kλ ) e i k ⋅ r
kλ →
32
(2π )
• The state of the field is given by the
number of photons at each k, λ , that is, by
successive application of creation operators
(Fock state):
 a + (k1λ1 )n1 a + (k 2 λ2 )n2


....  0 =


!
!
n
n
1
2


ni photons with wave number ki and
polarization λi , and linear combinations
of such states.
• Expected value and variance of vacuum
electric field in location r:
0 E (r) 0 = 0, but 0 E (r) 0 = ∞
2
• Why? It’s enough to prove for
Π (r) :
• For simplicity’s sake, write
[
]
Π (r) ≈ ∫ k 1 2 a(k)e ik⋅r − a + (k)e − ik⋅r d 3k
So 0 Π (r) 0 =
[
]
0 ∫ k 1 2 a (k)e ik⋅r − a + (k)e −ik⋅r d 3 k 0
but for all k : a (k) 0 = 0, 0 a + (k) = 0.
Now to the variance :
0 ∫∫ k k
12 12
1
2
(a(k )e
2
ik 2 ⋅ r
[(a(k )e
1
+
ik 1 ⋅ r
− a (k 2 )e
+
− a (k1 )e
− ik 2 ⋅ r
− ik 1 ⋅ r
)]d k d k
3
3
1
2
)
0=
0 ∫∫ k11 2 k 21 2 ei ( k1 − k 2 )⋅r a (k1 )a + (k 2 )d 3 k1d 3 k 2 0 =
[
0 ∫∫ k11 2 k 21 2 ei ( k1 − k 2 )⋅r a + (k 2 )a (k1 ) + δ 3 (k1 − k 2 )
]
d 3 k 1d 3 k 2 0 =
0 ∫ k1d 3 k1 0 =O  k04  for a frequency cutoff k0 .
The Classical Limit
• We should hope that, as mass and energy scales
increase to “everyday scale”, QM laws should
converge to Newtonian mechanics laws. Look for
example at the harmonic oscillator when the energy
increases – it resembles the behavior of a
“Newtonian” oscillator.
• It makes sense, in this regard, to look at the
average, or expectation, of position and momentum.
Therefore let us look at how the average changes
over time: let Ω be a function of X and P. Then
d
d
Ω =
ψ Ωψ =
dt
dt
& ψ + ψ Ω ψ&
ψ& Ω ψ + ψ Ω
& = 0.
• Assuming Ω has no explicit time dependence, Ω
We then obtain, using Schrödinger’s Equation
d
−i
Ω =   [Ω, H ]
dt
 h 
Ehrenfest’s theorem
 dω


= {ω , Η}
 dt

Remember!
• For example
P
∂H
d
−i
=
X =   [X , H ] =
∂P
m
dt
 h 
2


P
 assuming H =
+ V ( X ) 
2m


• and
d
∂H
dV ( x )
=−
P =−
dt
∂X
dx
• which look just like Hamilton’s equations. Alas,
we cannot argue that the latter hold for average
quantities, since in general
∂ H
∂H
≠
, etc.
∂P
∂ P
• However, it’s still a good approximation.
Rotational Invariance
and Angular Momentum
Motivation: Central Potential (Atoms)
2D Translations
• Translations in 2D are given by
∂
∂
Px → −ih , Py → −ih
∂x
∂y
and linear combinations of these. Much like in the
1D case, translation by a vector a is given by
T (a) = e
− ia ⋅ P h
, P = Px i + Py j
dψ d
i
,
= P,
dx dx h
U a (ψ ( x)) = ψ ( x − a ) =
In 1D : P ψ = −ih
a2
ψ ( x) − aψ ′( x) + ψ ′′( x) − ... =
2


d a2 d 2
I
a
...
−
+
−

ψ =
2
dx 2! dx


e
−a
d
−iaP
dx = e h .
One can also see this via infinitesimal translations :
U a +ε (ψ ) − U a (ψ )
 dU a 
=

ψ = lim
ε
0
→
ε
 da 
ψ ( x − a − ε ) −ψ ( x − a )
dψ ( x − a )  d

lim
=−
=  − U a ψ
ε →0
ε
dx
 dx 

d 

−iaP
a − 
 dx 
d
d

so, U a = − U a ⇒ U a = e 
=e h
da
dx
• Let a 2D rotation be described by
 x   cos(φ0 ) − sin(φ0 )  x 
 
  → 
 y   sin(φ0 ) cos(φ0 )  y 
• The operator that rotates the 2D vectors is
denoted by R (φ0 k ), and the corresponding
Hilbert space operator is denoted
U [R (φ0 k )].
For an infinitesimal rotation,
i ε z Lz
− iφ 0 L z
, U [R (φ0 k )] = e
U [R (ε z k )] = I −
h
∂
Lz = XPy − YPx =
{ − ih
∂φ
in polar
coordinate
[
]
• Physical interpretation: if Lz , H = 0, H is
rotationally invariant, and experiments will yield
the same results in the coordinates are rotated.
Also, there is a common basis for H and Lz .
h
Lz ' s Eigenvalues
L z l z = l z l z ⇒ −i h
∂ψ l z ( ρ , φ )
∂φ
= l zψ l z ( ρ , φ ) ⇒
∞
ψ l ( ρ , φ ) = R( ρ )eil φ h ( R has to be normalizable w.r.t ∫ ρdρ )
z
z
0
∗
Imposing Hermiticity, ψ 1 Lz ψ 2 = ψ 2 Lz ψ 1 , implies
ψ ( ρ ,0) = ψ ( ρ ,2π ) ⇒ e 2πil
z
h
= 1 ⇒ l z = mh, m = 0,±1,±2...
m is called the magnetic quantum number. Why can we
impose Hermicity? It' s a postulate.
• There are still an infinite number of degrees of
freedom for choosing R( ρ ). This is solved by
demanding that the eigenvectors of Lz are also the
eigenvectors of, say, a rotationally invariant
Hamiltonian. Physically, this means that the state
is determined not only by its angular momentum,
but also by its energy.
Define Φ m (φ ) = (2π ) −1 2 eimφ

 2π
 ∫ Φ m (φ )Φ m′ (φ )dφ = δ mm′ 



0
Examples (Shankar p. 315):
• A particle’s wave function is
ψ ( ρ , φ ) = Ae
− ρ2
2 ∆2
cos 2 (φ ).
Prove that
1
2
P(l z = 0) = , P (l z = 2h) = P (l z = −2h) = .
3
6
Solution : to calculate P(l z = mh ), the ρ part is only
a normalizing factor. So, up to normalization,
2

π
m=0
2π

P(l z = mh ) = ∫ψ ∗ ( ρ , φ )Φ m (φ )dφ = π 2 4 m = ±2
0
 0
otherwise

2
• A particle’s wave function is
ρ

ψ ( ρ , φ ) = Ae
 cos(φ ) + sin(φ ) .
∆

1
Prove that P (l z = h ) = P (l z = −h ) = .
2
−ρ 2
2 ∆2
Solution: the same as above – note that
2
2π
const. m = ±1
∫0 ( A cos(φ ) + sin(φ ))Φ m (φ )dφ =  0 otherwise
Rotationally Invariant Problems
Assume that V ( ρ , φ ) = V ( ρ ). Then the eigenvalue
equation for H is ( µ denotes the mass) :
 − h2  ∂2

1 ∂
1 ∂2 
 + V ( ρ )ψ E ( ρ , φ ) = Eψ E ( ρ , φ )
 2 +
+ 2

2 
ρ ∂ρ ρ ∂φ 
 2µ  ∂ρ

• Note that in this case H commutes with Lz ,
since Lz is represented by ∂ ∂φ , and we must
have
 ∂ψ
V 
 ∂φ
 ∂
(Vψ )
 =
 ∂φ
which will hold if V is a function of ρ only.
• We seek common eigenvectors of H and
the form (separation of variables):
Lz of
ψ Em = REm ( ρ )Φ m (φ ) ⇒ The radial equation :
 − h2  d 2

1 d m2 
 2 +
− 2  + V ( ρ ) REm ( ρ ) = EREm ( ρ )

ρ dρ ρ 
 2 µ  dρ

Angular Momentum in 3D
Three (Hermitian) generators : Lx = YPz − ZPy ,
Ly = ZPx − XPz , Lz = XPy − YPx , which satisfy :
[L , L ] = ihL , [L , L ] = ihL , [L , L ] = ihL
x
y
z
y
z
x
z
x
x
or L × L = ihL or
[L , L ] = ih∑ ε
3
i
j
k =1
ijk
Lk , ε ijk = antisymmetric tensor.
• The total angular momentum operator (squared)
is defined by
[
] [
] [
]
L2 = L2x + L2y + L2z ⇒ L2 , Lx = L2 , Ly = L2 , Lz = 0
2
L
The square root of
is regarded as the total
angular momentum of the system. The goal
is to characterize a “pure” state by its
(definite) total angular momentum, energy,
and magnetic quantum number (the last being
the angular momentum around the z-axis).
Finite Rotations in 3D
• Rotations are represented by
θ = θθˆ, θ = rotation angle,
θˆ = rotation axis (unit vector)
• Turns out that U
[R(θ)] = e
− iθθˆ⋅L h
=e
− iθ⋅L h
Proof: when a vector r is rotated by an infinitesimal angleδθ,
it moves to r + δθ × r (because the addition to the vector is
perpendicular both to the vector and the rotation axis). So, the
infinitesimal change induced by U [R(δθ)] is
ψ (r) → ψ (r + δθ × r) = ψ (r) + (δθ × r ) ⋅ ∇ψ
Putting aside h, i for the moment, we should show that this
change in ψ (r) is also induced by I + (δθ )( L ⋅ θˆ). the
vector product term is
θˆx
θˆy
θˆz and the change induced byI + (δθ )( L ⋅θˆ)
ˆ L + θˆ L + θˆ L )ψ
is
(
δθ
)(
θ
x x
y y
z z
r
δθ rx ry
z
ψ x ψ y ψ z To prove that the two changes are
equal, compare e.g. the coefficients
of θˆz (and remember that Lz = XPy − YPx ).
If H is invariant under rotation (U [ R ]H = HU [ R ]),
then, since Lx , Ly , Lz are represented by infinitesimal
rotations, [ H , Li ] = 0, and so [ H , L2 ] = 0. Hence, all
the three components of L, and the total angular
momentum, are conserved.
However, one cannot find a base which diagnolaizes
H , Lx , Ly , Lz , since the Li don' t commute. But [ L2 , Li ] = 0,
so one usually tries to simultaneously diagonalize H , L2 , Lz .
Common Eigenvalues for L2 and Lz
Following the paradigm used for the oscillator to
common eigenvectors L2 αβ = α αβ , Lz αβ = β αβ
Here, too, there are lowering and raising operators :
[
]
L± = Lx ± iL y ⇒ [Lz , L± ] = ± hL± . Note L± , L2 = 0.
Lz ( L+ αβ ) = (L+ Lz + hL+ ) αβ = ( β + h )( L+ αβ ).
L2 ( L+ αβ ) = α ( L+ αβ ). L+ αβ = C+ (α , β ) α , β + h
And similarly L− αβ = C− (α , β ) α , β − h
So, given αβ , the action of L± on it creates α , β ± kh
This has to end somewhere, since for a given total angular momentum
(α ), the angular momentum around z ( β ) has to be bounded.
αβ L2 − L2z αβ = αβ L2x + L2y αβ , but L2x + L2y
is positive definite (since Lx , Ly are Hermitian), so
we must have α ≥ β 2 . So there' s an αβ max that
cannot be raised : L+ αβ max = 0. Applying L− to
both sides and using L− L+ = L2 − L2z − hLz , we get
α = β max ( β max + h) , similarly α = β min ( β min + h)
= − β max . Since we got from
It follows that β
min
αβ min to αβ max in k steps of size h, we must have
β max
hk

2  k  k
=
, k = 0,1,2... , α = h   + 1.
2
 2  2 
k β max
is called the angular momentum of the state.
=
2
h
Unlike in classical mechanics, it' s smaller than the total
angular momentum.
Angular
k
 
Momentum  2 
β max
α
0
0
0
1/2
αβ
0,0
h 2  1  3 h 2
 2  2 
3 2 h
 h ,
4 2
3 2 h
 h , −
2
4
2h 2 , h
2h 2 , 0
h
(1)(2)h 2
3/2
..
..
..
:
:
:
:
1
But we proved that
multiplies of h ??
2h 2 , − h
l z can only have integer
L×
L7=4ihL
6
44
4
8
• Answer: here we only used the commutation rules,
not the operators’ specific shapes.
• There are particles whose wave function is
more complicated (non-scalar). For these
particles, there are rotation-like operators J
which satisfy the commutation rules
J × J = ihJ
• For example:
ψ( x, y, z ) = ψ x ( x, y, z )i +ψ y ( x, y, z ) j +ψ z ( x, y, z )k
• Rotations of such a function involve rotating
both the components and the coordinates: under
an infinitesimal rotation ε k,
z
ψ x → ψ x ( x + yε z , y − xε z ) − ε zψ y ( x + yε z , y − xε z ) = ψ ′x
ψ y → ε zψ x ( x + yε z , y − xε z ) +ψ y ( x + yε z , y − xε z ) = ψ ′y
ψ ′x   1 0 iε z  Lz 0  iε z  0 − ih   ψ x 
 
−
−
ψ ′  =  





 ψy
0
L
0
1
h
i
0
h
h
y




z


  
 
J z = L(z1) ⊗ I ( 2) + I (1) ⊗ S z( 2) = Lz + S z . S z = e.g. spin.
I ( 2) = 2 × 2 identity matrix w.r.t the vector components,
I (1) = identity operator w.r.t x, y.
• Just as before:
1 3
2
2
J jm = j ( j + 1)h jm , j = 0, ,1, ...
2 2
J z jm = mh jm , m = j , j − 1,...0,... − j
As before, J ± jm = C± ( j , m) j , m ± 1
(J
±
≡ J x ± iJ y )
J + jm = C+ ( j , m) j , m + 1 ⇒
{
adjoint
jm J − = C+∗ ( j , m) j , m + 1 ⇒
2
C+ ( j , m) = jm J − J + jm = jm J 2 − J z2 − hJ z jm =
j ( j + 1)h 2 − m 2 h 2 − mh 2
1
2
so J ± jm = h[( j m m )( j ± m + 1)] j , m ± 1
J+ + J−
J+ − J−
Since J x =
,Jy =
, it is straightforward
2
2i
to write their matrices in the jm basis as well. These
matrices are block diagonal in the fixed j blocks. This
makes the calculation of e −iθ⋅J h more feasible :
Finite Rotations
• In order to compute the transformation induced by
finite (as opposed to infinitesimal) rotations, one
has to calculate − iθθˆ⋅J
U [R (θ ) ] = e
h
=e
− iθ ⋅ J
h
.
• The calculation is facilitated by the fact that the
matrices for J x , J y , J z are all block diagonal in
the constant j blocks.
• Also, since the eigenvalues of θˆ ⋅ J (in the
constant-j block) are
(
) (
− j , − j + 1,... j , then J ⋅ θˆ + jI ⋅ ⋅ ⋅ J ⋅ θˆ − jI
)= 0
(since obviously its action on any vector is 0).
( )
( )( ) ( )
2 j +1
ˆ
• This allows to express θ ⋅ J
as a linear
2
2j
ˆ
ˆ
ˆ
combination of I , θ ⋅ J , θ ⋅ J ... θ ⋅ J , so
the powers in the exponential usually have a
simple periodic behavior which allows to write it
down as simple functions of θ (see also
discussion here on Stern-Gerlach apparatuses).
The Matrices of Angular Momentum Operators
Spin half
j ′m′
J
jm
(0,0)
1 1
 , 
2 2
2 1 1
 ,− 
2 2
(1,1)
(1,0)
(1,−1)
j ′m′
J
jm
(0,0)
1 1
 , 
2 2
1 1
z  ,− 
2 2
(1,1)
(1,0)
(1,−1)
Spin one
1 1 1 1
(0,0)  ,   ,− 
2 2 2 2
0
0
0
(1,1) (1,0) (1,−1)
0
0
0
0
(3 4)h 2
0
0
0
0
0
0
(3 4)h 2
0
0
0
0
0
0
2h 2
0
0
0
0
0
0
2h 2
0
0
0
0
0
0
2h 2
1 1 1 1
(0,0)  ,   ,− 
2 2 2 2
0
0
0
(1,1) (1,0) (1,−1)
0
0
0
0
(1 2)h
0
0
0
0
0
0
− (1 2 )h
0
0
0
0
0
0
0
0
0
h
0
0
0
0
0
0
0
0
0
0
−h
Spin half
jm
j ′m′
J
(0,0)
1 1
 , 
2 2
1 1
 ,− 
2 2
x (1,1)
(1,0)
1 1 1 1
(0,0)  ,   ,− 
2 2 2 2
0
0
0
(1,−1)
j ′m′
J
(1,0)
(1,−1)
0
(1,1)
(1,0) (1,−1)
0
0
0
0
0
h 2
0
0
0
0
h 2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
jm (0,0)
(0,0)
1 1
 , 
2 2
1 1
 ,− 
y 2 2
(1,1)
Spin one
1 1 1 1
 ,   ,− 
2 2 2 2
0
0
2
h
0
2
h
0
2
h
2
h
0
0
(1,1)
(1,0)
(1,−1)
0
0
0
0
0
− ih 2
0
0
0
0
ih 2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
ih
− ih
2
0
2
− ih
0
ih
0
2
0
2
Angular Momentum Functions in the Coordinate Basis
Start with the " topmost" state ll , satisfying L+ ll = 0.
 ∂
∂ 
± i cot (θ ) 
In the coordinate basis, L± = ± he ±iφ 
∂φ 
 ∂θ
(note that the r part is left out, and the φ part is taken care
of by multiplying with exp(ilφ ) ).
This allows to find ll , and the others can be found by
successive operations with L− . As before, we separate the
r part, and search for eigenfunctions with definite total
angular and z − momentum, Yl m (θ , φ ) (spherical harmonics ).
They are orthonormal under the inner product
1 2π
∫ dΩ = ∫ ∫ d (cos(θ ))dφ .
−1 0
∞
If ψ ( r , θ , φ ) = ∑
(
l
m
m
C
(
r
)
Y
∑ l l (θ , φ ), where
l =0 m = − l
)
∗
C ( r ) = ∫ Yl (θ , φ ) ψ ( r , θ , φ ) dΩ, then
m
l
m
∞
2
Pr( L = l (l + 1)h , Lz = mh ) = ∫ C ( r ) r 2 dr
2
2
m
l
0
That is, Clm ( r ) is the amplitude to find the particle
at radius r and with momentum(s) l , m.
Y00 = (4π ) −1 2
Y1±1 = m (3 8π )1 2 sin(θ )e ±iφ
Y10 = (3 4π )1 2 cos(θ )
Y2± 2 = (15 32π )1 2 sin 2 (θ )e ± 2iφ
Y2±1 = m (15 8π )1 2 sin(θ ) cos(θ )e ±iφ
(
)
Y20 = (5 16π )1 2 3 cos 2 (θ ) − 1
1
2
 (2l + 1)(l − m)! imφ m
Yl (θ , φ ) = (−1) 
e Pl (cos(θ ))

 4π (l + m)! 
m
m
l +m
1
d
2
l
Pl m ( x) = l (1 − x )
(
x
−
1
)
dx l + m
2 l!
m
2 2
• Example question (Shankar, 12.5.13): a
− ar
particle’s state is proportional to ( x + y + 2 z )e .
Prove that Pr(l z = 0) = 2 3, Pr(l z = h ) = Pr(l z = −h ) = 1 6.
• Answer: It’s immediate to see that
4π −1
π 0
1
z = 2r
3
Y1 , x = r
6
(Y
1
)
− Y1 , y = ri
4π −1
Y1 + Y11
6
(
So all that’s left is to compare the coefficients of
Y10 , Y1−1 , Y11 in x + y + 2 z.
)
Note that as m grows,
there is more of the
momentum in the z
direction, and the
probability to have large z
values decreases.
Solution of Rotationally Invariant Problems
Note: µ is mass (to avoid confusion with m).
If V (r ,θ ,φ )=V (r ), the Schr&o&dinger equation becomes
 −h 2  1 ∂ 2 ∂
∂
∂
1

+
+
θ
sin(
)
r

2
2

∂θ
 2 µ  r ∂r ∂r r sin(θ ) ∂θ

∂2 
1
+V (r )ψ E (r ,θ ,φ )= Eψ E (r ,θ ,φ )
2
2
2 
r sin (θ ) ∂φ 

Since in this case [H ,L ]=0, seek simultaneous eigenkets
of H ,L2 ,Lz :ψ Elm (r ,θ ,φ )= RElm (r )Yl m (θ ,φ ). As in 2D,
get a radial equation:
 −h 2  1 ∂ 2 ∂ l (l +1) 

− 2  +V (r ) REl = EREl (no m).
r

2


 2 µ  r ∂r ∂r r 
U El  d 2 2 µ 
l (l +1)h 2  
U =0
REl =
⇒  2 + 2  E −V (r )−
2   El
r
2 µr  
 dr h 
which is equivalent to
 h2 d 2
l (l +1)h 2 
U ≡ Dl (r )U El = EU El
+V (r )+
−
2
2  El
2 µr 
 2 µ dr
Imposing Hermiticit y of Dl with regard to
*
∞ ∗
∞ ∗
U El , ∫ U 1 Dl U 2 dr = ∫ U 2 Dl U 1 dr ⇒
0
0
∞ ∗
integratio n
∗ ″
′
′
∫ U 1 U 2 − U 1 U 2 dr = 0⇒
by parts
0
∞
∗
 ∗ dU 2
dU 1 
 U1
 =0
−U 2
dr
dr 

0
Next, address the normalizat ion of U El ,R El : to integrate R El
∞
∞
one must take ∫ r 2 R dr = ∫U dr . For this to be normalizab le,
El
El
0
0
ikr
we must have either U El →0 (bound state) or U El →e
r →∞
r →∞
(unbound state). In both cases, the upper limit in
∞
∗
 ∗ dU 2
dU 1 
 U1
 vanishes and the Hermiticit y depends on
−U 2
dr
dr 

0
∗
 ∗ dU 2
dU 1 
( 0 ) = 0. This is satisfied if
−U 2
whether  U 1

dr
dr 

(
(
( )
)
)
[
(
)
U El →c = const . If c is not zero, R ∝
r →0
]
U
r
∝
c
r
Diverges at the origin, which is impossible since
c
does not satisfy Schr&o&dinger' s equation at the
r
origin, as ∇ 2 (1 r ) = −4πδ 3 (r). So, we must have
U El r
→ 0.
→0
General Properties of
U El
• The goal is to further study the asymptotic behavior
of U El as r tends to zero or infinity, depending on
the potential.
As r → 0, and assuming that V (r ) is less
−2
singular than r , then the equation is
dominated by the so - called centrifugal
l (l + 1)
barrier : U l′′ ≅
U l ( E becomes
2
r
inconsequential in the limit).
Guessing a solution U l ≅ r α yields
 r l +1 (regular)
U l ≅  −l
r (irregular, ruled out following previous analysis)
• Intuitively – as before, as the angular momentum
increases, the particle avoids the origin more and
more.
• Next, let us study the behavior at infinity.
Assume that the potential does not dominate the
behavior at infinity; a condition which will be
justified later is
rV (r ) r

→
0
→∞
d 2U E
2µE
For large r , the equation becomes
= − 2 UE
2
dr
h
(l does not influence the behavior at infinity).
There are then two cases :
E > 0 : Particle may escape to infinity, we expect
U E to oscillate at infinity.
E < 0 : The region r → ∞ is classicaly forbidden,
we expect U E to exponentially decrease as r → ∞.
Assume E > 0. The solutions are
U E = Ae + Be
ikr
−ikr
(
, k = 2 µE h
Let us now write U E = f (r )e
)
2 12
± ikr
, and check
whether f → const. as r → ∞. To do that,
plug the above form into
 d 2 2µ 
l (l + 1)h 2  
 2 + 2  E − V (r ) −
 U El = 0
2
2 µr  
h 
 dr
To obtain at infinity
2 µV (r )
f ′′ ± (2ik ) f ′ −
f =0
2
h
Assuming the second derivative is small :
df
i µ
V (r )dr
=m
2
f
kh
  iµ  r

f (r ) = f (r0 ) exp m  2  ∫ V (r ′)dr ′
  kh  r0

If rV (r ) r
→ 0, the integral tends to
→∞
e2
a constant as r → ∞. But if e.g. V (r ) = − ,
r
as in the Coulomb potential , then
  iµe 2   r 
f (r ) = f (r0 ) exp ±  2  ln 
  kh   r0 
 

µe 2
U E (r ) ∝ exp ± i kr + 2 ln (r )
kh

 
So, the potential influences the particle
no matter how far it is from the origin.
For the case E < 0, the results carry over with
(
the change k → iκ , κ = 2 µ E h
)
2 12
,
κr
−κr
U E r
Ae
Be

→
+
, and B = 0 as usual.
→∞
As before, this is true only if rV (r ) → 0.
In the coulomb case (Hydrogen atom) we expect
as before

2
2
± µe κh 




 µe 2
 m kr
 m kr
e
U E ∝ exp ± 2 ln(r ) e
=r
 κh

For E < 0, the energy eigenkets are normalizable to
unity. The operator Dl (r ) is non - degenerate (see
proof in the discussion on degeracy), hence
U El v are orthogonal for different E , and so
REl (r )Yl m (θ , φ ) are orthogonal for different {E , l , m}.
This means that the energy, the total angular
momentum, and the z − axis angular momentum
fully determine a state.
The Free Particle in Spherical Coordinates
As before: start withψ Elm =REl (r)Yl m (θ ,φ )
DefineU El as before,and sinceV =0 we get
 d 2 2 l (l +1) 
2 2µE
 2 +k − 2 U El =0, k = 2
r 
h
 dr
 d 2 l (l +1) 
ρ =kr⇒− 2 + 2 Ul =Ul
ρ 
 dρ
This resembles the harmonic oscillator,
so define lowering and raising operators:
d l +1 + d l +1
+
dl = + ,dl =− + , dl dl Ul =Ul
dρ ρ
dρ ρ
+
+
+
+
+
dl dl dl Ul =dl Ul ,dl dl =dl+1dl+1⇒
+
+
+
+
dl+1dl+1 (dl Ul )=dl Ul ⇒dl Ul =clUl+1
So dl+ is a raising operator. We need
to start with U0: U0A (ρ )=sin(ρ ) or
U0B (ρ )=cos(ρ ).
(
(
)
)
Taking cl to be unity, and re - introducing Rl , it' s
easy to prove that
l
1 ∂ 
 R0 , which generates solutions
Rl = (− ρ ) 
 ρ ∂ρ 
l
l
 1 ∂   sin( ρ ) 
 (spherical Bessel)
 
R ≡ jl = (− ρ ) 
 ρ ∂ρ   ρ 
A
l
l
l
 1 ∂   − cos( ρ ) 
 (Neumann)
 
R ≡ jl = (− ρ ) 
ρ

 ρ ∂ρ  
So, free particle solutions which are regular in the
B
l
l
entire space are given by
2 2
h
k
m
ψ Elm (r ,θ , φ ) = jl (kr )Yl (θ , φ ), E =
2µ
• Example (Shankar 12.6.8): Find the energy levels
of a particle in a sphere of radius r0 , given l=0.
• Solution: for l=0, we must have sin( ρ ) = 0 at
the boundary, hence
n 2π 2 h 2
kr0 =nπ ⇒ En =
2 µr02

2 2 µE 
 since k = 2 
h 

( note that the quantization is finer as the radius increases)
Relation to Cartesian Coordinates Solution
In Cartesian coordinates, ψ E ( x, y, z ) =
1
ip⋅r h
e
,
32
(2πh )
p 2 h 2k 2
=
E=
. If the particle is going up the z − axis with
2µ
2µ
1
ikr cos(θ )
e
, so it must be
momentum p, ψ E (r ,θ , φ ) =
32
(2πh )
possible to express it as a sum of the eigenstates computed in
spherical coordinates :
e
ikr cos(θ )
∞
l
= ∑ ∑ Clm jl (kr )Yl m (θ , φ ). But only φ = 0 is relevant,
l =0 m = − l
as the left - hand side doesn' t depend on φ (physically, a
particle going up the z − axis has no angular momentum in
12
 2l + 1 
that direction). Since Yl 0 (θ ) = 
 Pl (cos(θ )), then
 4π 
12
 2l + 1 
e ikr cos(θ ) = ∑ Cl jl (kr )Pl (cos(θ )), Cl = Cl0 
 . It can be
 4π 
l =0
shown that Cl = i l (2l + 1), so that
∞
e
ikr cos(θ )
∞
= ∑ i l (2l + 1) jl (kr )Pl (cos(θ ))
l =0
The Isotropic Oscillator
Px2 + Py2 + Pz2 1
H=
+ µω 2 X 2 + Y 2 + Z 2
2µ
2
U El m
Yl (θ , φ ) ⇒
ψ Elm =
r
 d 2 2µ 
1
l (l + 1)  
2 2
U =0
 2 + 2  E − µω r −
2   El
2
2 µr  
h 
 dr
(
r → ∞ ⇒U ≈ e
− y2 2
)
1
2
 µω 
,y =
 r
 h 
So assume U ( y ) = v( y )e
− y2 2
⇒

l (l + 1) 
E
v′′ − 2 yv′ + 2λ − 1 −
v = 0, λ =

2
hω
y 

Incorporating the previous analysis about the
behavior near the origin, guess that
∞
v( y ) = y l +1 ∑ Cn y n
n =0
Substituting this into the equation yields
a recursion relation between Cn and Cn−2 ,
1
which implies that λ =n+l − . However,
2
l
it's easy to verify that the coefficient of y is
C1 , which has to be zero (since all is zero),
hence there are only even powers in the
expansion. The case n=0 is ruled out because
we start with C2 , and therefore the quantized
energy levels are
3

n=2(k +1)⇒ E = 2k +l + hω ,k =0,1,2...
2

(one could guess that the ground state energy is
(3 2)hω by separation of variables, and since
GSE for a 1D oscillator is (1 2)hω. But this
solution ties the state directly to the angular
momentum).
The principle quantum number is defined
by n=2k +l , and it determines the energy,
E =(n+3 2 )hω. At each n, the allowed
l values are l =n−2k =n,n−2,n−4...1 or 0.
The first eigenstates are
n=0 l =0 m=0
n=1 l =1 m=±1,0
2
,
1
,
0
n=2 l =0(k =1),2(k =0) m= 0{ ; ±
±
1
424
3
l =0
l =2
1
,
0
;
3
,
2
,
1
,
0
n=3 l =1(k =1),3(k =0) m=±
±
±
±
{ 14243
l =1
l =3
The degeneracy in l will be explained later. It
is a result of symmetries in the Hamiltonian.
The Hydrogen Atom
• Assume in the meanwhile that the proton’s mass is
infinitely larger than the electron’s, so only the
electron’s wave function is sought. The potential is
induced by the attraction between the proton and
electron, and the equation has the form
 d 2 2m  e 2 l (l +1)  
U =0, and as usual
 2 + 2 E+ −
2   El
r 2mr  
 dr h 
U El (r ) m
m
ψ Elm (r ,θ ,φ )= REl (r )Yl (θ ,φ )=
Yl (θ ,φ )
r
It was proved before that under such a potential,
then up to a polynomial factor, at infinity we have
[
−(2 mW h 2 )r ]
U El (r )≈e
, where W =− E is the binding
energy (which it would take to liberate the electron),
and as r →0 we have U El (r )≈r l +1.
(
Define ρ = 2mW h
)
2 12
r ,U El (r )=e −ρ vEl . Considering
∞
the behavior at 0, assume vEl = ρ l +1 ∑Ck ρ k . This yields
k =0
(as opposed to the other cases), a successive recursion
d 2v
dv
+
relation for the equation 2 − 2
dρ
dρ
 e 2 λ l (l + 1) 
12
2
−

 v = 0, λ = 2m h W :
2
ρ 
 ρ
Ck +1
− e 2 λ + 2(k + l + 1)
=
,
Ck
(k + l + 2)(k + l + 1) − l (l + 1)
(
)
as usual the series has to terminate and we get
the energy quantization levels defined by
− me 4
E = −W = 2
, k = 0,1,2.., l = 0,1,2..
2
2h (k + l + 1)
In terms of the principle quantum number,
n = k + l + 1, the allowed energies are
− me 4
En = 2 2 . At each n the allowed l values are
2h n
n − 1, n − 2...0, and for each of those there are the
appropriate 2l + 1 z − axis angular momentum values.
So, for a given n which specifies an energy level, the
total degeneracy is 2(n − 1) + 1 + 2(n − 2) + 1 + ... = n 2
The series terminates at k = n − l − 2, and there' s
a ρ l+1 factor outside. So the solutions are polynomials
of degree n − 1 : Rnl ( ρ ) ≈ e − ρ L2nl−+l1−1 (2 ρ ), where the
latter are the associated Laguerre polynomials :
(
L =e d
0
p
x
p
)
(
−x
p
k
p
k
When r → ∞, Rnl ≈ r n −1e −cr .
Define a0 = h 2 me 2 .
ψ 1,0, 0
ψ 2, 0, 0
ψ 2,1, 0
)
dx (e x ), L = (− 1) d k dx k L0p + k
p
1
2
 1  − r a0
=  3  e
 πa0 
 1 

= 
3 
 32πa0 
1
2

r  − r 2 a0
 2 − e
a0 

1
2
 1  r − r 2 a0

e
cos(θ )
= 
3 
 32πa0  a0
1
2
 1  r − r 2 a0
± iθ

ψ 2,1, ±1 = 
e
sin(
θ
)
e
3 
32
π
a
0  a0

For l = n − 1,ψ n ,n −1,m ∝ e − r na0 r n −1Ynm−1 (θ , φ )
Therefore, the probability to be in a spherical
shell of radius r and thickness dr is ∝ e − 2 r na0 r 2 n .
This is a maximum when r = n 2 a0 , hence a0 is
the average distance of the electron from the
nucleus in the ground state.
[
a0
2
In general, r =
3n − l (l + 1)
2
]
Multi-Electron Atoms
• Due to the Pauli exclusion principle, two
electrons cannot occupy the same state if their
parameters (energy + angular momentum + z-axis
momentum + spin) are identical.
• Hence, first the lowest energy level, n = 1, is
filled: Hydrogen and Helium.There are two n = 1,
electrons there (two spins). Then, for n = 2, the
degeneracy is 4, multiplied by two spins given 8,
and this accounts for the following 8 elements
(Lithium to Neon). For n = 3, first the l = 0
are filled (two electrons: there’s a 2l + 1 = 1
degeneracy, multiplied by two spins). Same for
l=1, for which there are six electrons (3 “l levels”
multiplied by two spins). By then we are at
2+8+8=18 (Argon). However, the next electron
(in Potassium) does not choose n=3&l=2, but
n=4&l=0. WHY?
• The answer lies in the interaction of the
electron’s charges with each other (which
we have neglected). The n=4&l=0 is farther
away from the nucleus then the n=3&l=2
electron (see equation for average distance).
So, it is partially screened from the
Coulomb potential by the charge of the other
electrons.
Spin
• Spin has no analogue in classical physics.
• An electron is prepared in a state of zero linear
momentum (i.e. space-independent) state. Since all
the angular momentum operators involve
derivatives w.r.t x,y,z, we expect that its angular
momentum(s) will all be zero. However, upon
measurement along, say, the z-axis, the results of
measuring Lz are ± h 2 .
• So, the electron has some kind of “intrinsic”
angular momentum. Same holds for other particles.
• The wave function of a particle with spin has a
few components, and the spin operators linearly
mix these components – as opposed to the orbital
momentum operators, which mix the coordinates.
• The number of components is determined by the
number of values which the “intrinsic” angular
momentum can attain. The wave function is then
just a direct product unifying the “ordinary” wave
function with these different “intrinsic” values.
The generalized rotation operator is then J = L + S,
where L is orbital angular momentum and S the spin
operator. Under an infinitesimal rotation around the
z − axis, the wave function changes according to
 ψ 1 
0
ψ 1′  

− ih ∂ ∂θ ...
 
 .  


.
.
.


 
 
 iε
iε 


.
− Sz  . 
.
 . = I− 

h 
h  
 .  

.
.
.


ψ ′  


 ψ 
h
θ
i
0
...
−
∂
∂


 n 
 n 
where S z is also an n × n matrix.
iε 

ψ ′ =  I − J z  ψ , J z = Lz + S z
h 

Now the number of components has to be recovered, as
well as the form of the operators S x , S y , S z . Since the
number of possible results for measurement of S z is 2,
we assume the dimension is 2. We also assume that J
[
]
satisfies the commutation relations J i , J j = ih ∑ ε ijk J k ,
k
and since L and S commute, these relations are also
satisfied for S x , S y , S z .
The commutation rules determine the S operators,
just as they did in the general analysis in the
general discussion on angular momentum:
h 0 1 
h 0 −i 
h 1 0 
Sx = 
, Sz = 
, Sy = 


2 0 −1
2i 0 
2 1 0 
So the electron is described by a two-component
wave function:
ψ + ( x, y,z ) 
1 
0 
ψ =
 =ψ + ( x, y,z )   +ψ − ( x, y,z )  
0 
1 
ψ − ( x, y,z ) 
This is called a spinor. It has two complex components.
For the case of zero momentum, since the momentum
operator P yields zero, we can deduce that ψ + ,ψ − are
independent of x, y,z.
0
2
2
2
2 3 2 1
The total spin is given by S = S x + S y + S z = h 
4 0 1
and so it yields a value of (3 4 )h 2 on any state. Unlike the
total orbital angular momentum, the total spin cannot be
changed and it is intrinsic to the particle. The electron
spin is defined as 1 2.
Hereafter the notation ψ ± ( x, y, z ) will
combine the spin and position amplitudes,
and the normalization condition is therefore
∫∫∫ ( ψ
2
+
+ψ−
2
)dxdydz = 1.
Assume for the moment that the orbital wave
function evolves independently from the spin.
We can then limit the discussion to a 2D Hilbert
space, whose kets are s, s z = s, mh ≡ s, m ,
where s is the total spin (1 2 in our case), and m
the spin along the z − axis. A basis is given by
1 1
s, m = ,
2 2
1
→  
S z basis 0
 
0 
1 1
s, m = ,−
→  
S
2 2 z basis 1
and a general state by
1 1
1 1
χ = α , + β ,−
2 2
2 2
Remember that we’re
assuming that
ψ ( x, y, z , s , t ) = ψ ( x, y, z , t ) χ (t )
z
where χ (t ) is a two
component spinor
independent of the
position.
α  2
2
→  , α + β = 1
S z basis β
 
Define S=(S x ,S y ,S z )=S x i+ S y j+ S z k
If nˆ ,± are the eigenstates of nˆ⋅S, for a
unit vector nˆ , nˆ ,± S nˆ ,± =±(h 2)nˆ (can
be proved by direct computation) and
these states are said to have spin up/down
the nˆ direction. Denote (nˆ x ,nˆ y ,nˆ z )=
(sin(θ )cos(φ ),sin(θ )sin(φ ),cos(θ ) ), then
nˆ⋅S=nx S x + n y S y + nz S z =
e −iφ 2 sin(θ )
h  cos(θ )
,with
 iφ 2
2 e sin(θ )
−cos(θ ) 
eigenvectors (and eigenvalues ±h 2 ):
e −iφ 2 cos(θ 2)
nˆ up ≡ nˆ + =  iφ 2

 e sin(θ 2) 
−e −iφ 2 sin(θ 2)
nˆ down ≡ nˆ − =  iφ 2

 e cos(θ 2) 
• Note that knowing S determines the
state. Actually, instead of specifying a state
by α and β , we can specify a unit vector n̂
such that the state is an eigenstate of nˆ ⋅ S
with eigenvalue h 2 .
most direct way to see this is the following:
let there be two complex numbers α ,β such
2
2
that α + β =1. We can multiply by a phase
factor (physically meaningless) and assume
that their angles are opposite: α = ρ1e −iφ ,
β = ρ 2 eiφ . Since ρ12 + ρ 22 =1, we can take
ρ1 =cos(θ ),ρ 2 =sin(θ ), and (up to the division
by a factor of 2) this is the general form of the
eigenvectors of nˆ⋅S .
Note: this indicates that any state can be prepared
by a Stern -Gerlach appartus, with its axis at the
corresponding polar angle. This is because we're
measuring the angular momentum at that direction.
•The
The form of the spin operators leads to the
definition of the Pauli matrices σ, S = (h 2 )σ :
0 1 
0 − i 
1 0 
σx = 
,σ y = 
,σ z = 



−
1
0
i
0
0
1






• They anti - commute with each other.
• They satisfy σ xσ y = iσ z , and cyclic
permutations which can be immediately
[
]
[
derived : σ i , σ j = (2i )∑ ε ijkσ k , σ i , σ j
k
]
+
= 2 Iδ ij
• They are traceless.
• Their
squares equal the identity, and more
2
ˆ
(
)
generally n ⋅ σ = I : since each such matrix
nˆ ⋅ S has eigenvalues ± (h 2 ), it follows that
The eigenvectors span the
space, and if the operator is
zero on them then it’s zero on
all the space.
hI 
hI 

 nˆ ⋅ S +  nˆ ⋅ S −  = 0
2 
2

2
2
2
ˆ
ˆ
so (n ⋅ S) = h 4 I ⇒ (n ⋅ σ ) = I .
(
•
)
If we add the identity I , we get an orthonormal
basis under the inner product (1 2 )Tr ( AB).
Since (nˆ ⋅ S) = I , it' s relatively easy to
calculate the powers of nˆ ⋅ S, which helps in
2
calculating the propagator and the rotation
operator. For the latter, recall that
−iθ⋅S h
−iθ⋅σ 2
=e
=
U [R(θ )] = e
 θ  ˆ 
θ 
θ  ˆ
exp − i θ ⋅ σ  = cos  I − i sin  θ ⋅ σ
2
2

 2
(follows immediately by equating coefficients).
Spin Dynamics (How Spin Behaves
in a Magnetic Field)
µ
B
I
• The torque induced by the magnetic field B on
the current carrying loop is to align its normal
with the field. The torque is given by T = µ × B,
I⋅ A
e⊥
Where µ =
c
is the magnetic moment,
A the loop area, and e ⊥ a unit vector
perpendicular to the loop’s plane. See slides 27-8
in electrodynamics file.
• The interaction energy between the loop and B
is given by − µ ⋅ B (so that the gradient of this
expression is the force acting on the loop).
Carry the preceding analysis with the loop replaced
by a particle of mass m and charge q, moving at speed
v in a circle of radius r. The " current" is then
qv
I=
, and the magnetic moment has a magnitude
2πr
which is the product of the current and the area :
qv πr 2  q 
µ=
⋅
= 
 l , l = mvr = the magnitude of
2πr c
 2mc 
gyromagnetic
ratio
γ
angular momentum. Since µ and l are parallel, µ = γ ⋅ l.
The effect of T here is not to align µ with B; instead, it
results in a precession of µ around B.
The equation of
B
l
dl
motion is T = =
dt
∆φ
l + ∆l
µ × B = γ (l × B)
⇒ ∆l = γ (l × B) ∆t
⇒ ∆l = γlB sin(θ ) ∆t
But ∆l is perpendicular
θ
to l and B (it' s
proportional to their
Explanations:
vector product), so the
1. The
particle
tip of the vector l moves by an
angle (relative to B) of
rotates
around l
 − ∆l 
2. T is torque
 = (−γB)∆t ,
∆φ = 
(two slides
 l sin(θ ) 
earlier).
so it precesses at a frequency
The “real” reason this happens is the colinearity
ω0 = −γB .
of the magnetic and angular moments.
• What is the equivalent of these ideas in QM?
• The Hamiltonian of a particle of mass m and
charge q in a magnetic field is
(
P − qA c )
H=
2
2m
2
2
2
q A
q
(P⋅A+A⋅P )+
= −
2
2
mc
2m 1
2mc
444
424444
3
interaction Hamiltonian
P
where A is the vector potential satisfying
∇×A =B. Taking A =(B 2 )(− yi+ xj)⇒
∇×A =B= Bk. Dropping the quadratic
term in H , the interaction part turns out to be
q
H int = −
L⋅B≡−µ⋅B, where µ is a
2mc
conjectured magnetic moment operator, and
the analogy is adopted from the classical case
in which the interaction Hamiltonian is given
by the same expression.
q
So we take µ ≡
L.
2mc
• The spin magnetic moment is assumed to
be of the same shape:
H int =−µ⋅B∝σ ⋅B
The related constant turns out to satisfy (as
justified both by experiment and theory):
H =−γS⋅B, γ =−(e mc)
which is twice the orbital case.
The evolution of the electron state ψ (0) is
−i Ht h
ψ (0) =
given as usual by ψ (t ) =e
−iθ⋅S h
iγt (S⋅B) h
ψ (0) , but the e
operator
e
rotates by θ − just like we proved for J, which
shares the same properties as S (slides207-8)−
so the effect on the electron is to rotate the state
by an angle θ(t )=−γtB. So, S precesses around
B at a frequency ω0 =−γB.
For example, consider B to be along the
iγtS z B h
z −axis, B= Bk. So U (t )=e
=
iω 0 tσ z 2
,
e
and since σ z is diagonal the exponent equals
eiω0t 2

0
. If an electron
U (t )=

ω
2
i
t
−
0
 0

e
cos(θ 2 )e −iφ 2 
starts in ψ (0) = nˆ ,+ →
iφ 2 

 sin (θ 2 )e
cos(θ 2 )e −i (φ −ω0t ) 2 
then ψ (t ) =
i (φ −ω0t ) 2 
 sin (θ 2 )e

For example, if it starts with spin along the x−
axis, it will change to spin along y, etc:
(1
θ =π 2,φ =0)→(θ =π 2,φ =π 2)→...
42
4 43
4 1442443
eigenstate of S x
eigenstate of S y
Which is precession around the z-axis.
Energy Levels of an Electron in a Magnetic Field
• As noted before, if the Hamiltonian separates
into a spin and orbital parts, we have twice the
number of orbital states – each one will now have
a ± (1 2) spin.
• Now assume there’s a magnetic field B=Bk
present. For the hydrogen atom, the coupling of
the proton to B can be ignored because its mass to
charge ratio is much higher than the electron’s.
For the electron, the combined Hamiltonian is
 − eB 
 − eB 
H = H Coulomb − 
S z
 Lz − 
 mc 
 2mc 
(remember that the coupling coefficient of the
spin is twice as large). Since H commutes with
H Coulomb , Lz , L2 , and S z , it is diagonalized by
the same states as before : nlmms (ms is the spin).
However, the eigenvalues are different :

 − R y eBh
H nlmms =  2 +
(m + 2ms ) nlmms
2mc

 n
 − R y eBh

H nlmms =  2 +
(m + 2ms ) nlmms
2mc
 n

This results in the reduction of the 2n 2 degenracy.
For example, the first level (n = 1), which had a
degenracy level of 2, now has two levels; m = 0,
but ms = ±(1 2 ), and therefore the energy levels are
eBh
now En =1 = − R y ±
(so no degeneracy). The
2mc
n = 2 level, which was eightfold degenerate, now
splits into five levels : the only degeneracy remains
in states in which m + 2ms are equal :
En = 2
 2 (m = 1, ms = 1 2)

1 (m = 0, m = 1 2, l = 0,1)

s


= − R y ± c1 ⋅  0 (m = 1, ms = − 1 2 or m = −1, ms = 1 2) 


−
=
=
−
=
m
m
l
1
(
0
,
1
2
,
0
,
1
)
s


 − 2 (m = −1, ms = − 1 2)

eBh 

c
=
 1

2mc 

This splitting is named the Zeeman effect.
The Stern-Gerlach Experiment Revisited
• As in classical electrodynamics, an inhomogenous
magnetic field creates a force on a dipole, which is
∂Bz
k (assuming the field is in the zgiven by µ z
∂z
direction). So particles with positive spin will go
upwards, and those with negative spin downwards.
By blocking one of the paths, we can ensure that the
particle leaving the apparatus will have a definite
and known spin in the z-direction (i.e. we know that
it is an eigenstate of S z , and we know the
eigenvalue). This results in a space quantization of
the emerging particles (they’re not continuously
scattered, but emerge only at distinct points
corresponding to the different spins; spin half will
concentrate at two points etc). Note, especially, that
not only the total spin is quantized, but the spin at
every axis as well.
• Suppose we have a sequence of SG apparatuses.
We may ask what percentage of those leaving
apparatus i will leave apparatus i+1 (see slides 4851). We can now solve this problem rigorously,
since the equations for the behavior of spin under
rotation of the axis have been derived.
 1
 0

• The electron is in a mixed spin state  α   + β   .
1
 0
Due to the heavy proton, the behavior is semi-classical, so the
spin up part induces an upwards moving wave-packet, and
similarly for the spin down part. So the wave packet has really
two components – upwards and downwards. The blocking is
like a measurement, and it forces the spin into a single state; if
the lower beam is blocked, we get “pure” spin up.
Examples
• Spin half particles are moving along the y-axis (see
drawing) enter two collinear SGAs, the first with its
field along the z-axis and the second’s field is along
the x-axis (i.e. it is the first, rotated by π 2). What
portion of those leaving the first will exit the second,
if the lower beams in the two SGAs are blocked?
• Answer: we can think of the particles leaving the
first being rotated by π 2 and then entering an SGA
identical to the first. The equation for rotation of spin
half particles was developed before (slide 210):
U [R(θ)] = e
−iθ⋅S h
=e
−iθ⋅σ 2
=
 θ  ˆ 
θ 
θ  ˆ
exp − i θ ⋅ σ  = cos  I 2 − i sin  θ ⋅ σ
2
2
 2

In our case
θˆ = (0,1,0),θ = π 2 ⇒ U [ R(θ)] =
1
1 1 − 1
(I 2 − iσ y ) = 1 1 
2
2

Note: all
calculations
are carried
out in the
Sz
basis
• Since the lower path of the first SGA is blocked,
1
those emerging from it are in a state  , and after
 0
U [ R (θ )] acts on them they become
1 1
  .
2 1
The probability for them to exit the second SGA
(which after the rotation also has its field pointing
up the z-axis), equals their probability to yield 1
when their z-spin is measured, i.e. the square of the
projection of the state vector on the eigenvector of
1
S z with eigenvalue 1, but this eigenvector is  ,
0

so the probability is clearly 1 2 .
• Now, assume that another SGA is placed after
the second, which transmits only spin down along
the z-axis. How many of those entering it will exit?
We know that those leaving the second have a spin
up along x, so their state vector is the eigenvector
of
S
1  1
 .
x with eigenvalue 1, which equals 2 1
The probability of this particle going through a zspin down SGA is clearly 1 2, and so 1 4 of those
leaving the first will exit the third.
• Note that if the middle SGA has both beams
unblocked (i.e. all those entering it leave it), then
no particle which leaves the first will exit the third
(they leave the first with spin up and all are
therefore blocked by the third). So, the blocking in
the middle one increases the percentage of those
leaving the third.
• Other way to compute: to see what happens with
the particle after it leaves the first (and is in a (1,0) t
state in the z-basis) upon entering the rotated
SGA: simplest way to understand it is to rotate the
entire system by − π 2 , since then we’re back with
an SGA with its field along z, calculate there, and
rotate back by π 2 .
SGA’s Acting on Spin One Particles
• The computation proceeds very much like for
spin half. Suppose that the first SGA passes only
z-spin up particles, and the second one is rotated
around the y-axis by an angleθ . What percentage
of those exiting the first will leave the second?
• Answer: as before, we compute what such a
rotation does to the particle’s state. Just like for
spin half, the rotation operator is defined by
U [R (θ)] = e
−iθ⋅S h
To compute the exponents of spin one operators,
note that, since the eigenvalues are 0,± h then
(nˆ ⋅ S + hI )(nˆ ⋅ S)(nˆ ⋅ S − hI ) = 0 ⇒ (nˆ ⋅ S)3 = (nˆ ⋅ S)
The Pauli matrices are defined just like for spin
half, but here we shall denote S = hσ (not h 2).
for example,
0 − i 0 
See

1 
σy =
slides
 i 0 −i
2
207-8

0
i
0
So in our case, θ = (0,θ ,0), and U [R (θ)] =
−iθσ y
1 2 2
−iθ⋅S h
=e
= I 3 − iθσ y − θ σ y +
e
2!
1 3
1 4 2
i θ σ y + θ σ y ... (remember : σ y3 = σ y )
3!
4!
Collecting terms : I 3 + [cos(θ ) − 1]σ y2 − i sin(θ )σ y =
 1 + cos(θ )

2

 sin(θ )

2
 1 − cos(θ )

2

−
sin(θ )
2
cos(θ )
1 − cos(θ ) 

2

sin(θ ) 
−
2 
1 + cos(θ ) 

2

sin(θ )
2
So, since we started with a particle with z − spin
1
 1 + cos(θ ) sin(θ ) 1 − cos(θ ) 
up,  0  , it transforms to 
,
,

2
2
2


0
and its probability to be in z − spin up is
2
1
cos(
)
+
θ



 , and that portion will exit the second
2


SGA.
Addition of Angular Momenta
• Question: how does a system with two spin half
particles appear (in terms of spin) when viewed as a
single object?
• Such systems are naturally described in QM by the
direct (tensor) product. For example, the system
with two spin half particles is the vector space of
dimension 4 spanned by + + , + − , − + , − −
S1z acts as usual on the first component and is equal to the identity
on the second component, etc.
Define S z = S1z + S 2 z .
and the matrix of S2 = (S1 + S2 ) ⋅
Then the matrix of S z
(S1 + S2 ) = S12 + S22 + 2S1S2 is
123
in this product basis is
++ +− −+ −−
S S +
++ +− −+ −−
1x 2 x
0
0
0
1
0
0
0
2
S S
+


y
y
1
2




0
0
0


S S
0
0
1
1
0
1z 2 z


h

2
h 
 As can be
0
0
0
0
0
1
1
0 verified by


 directly



0
0
− 1
 computing the

0
0
0
2 tensor product.
0
However, the following basis diagonalizes
S2 , S z , S12 (= S12x + S12y + S12z ), S22 :
s = 1, m = 1 :
++
s = 1, m = 0 : 2 −1 2 ( + − + − +
)
s = 1, m = −1 : − −
s = 0, m = 0 : 2 −1 2 ( + − − − +
)
Note that s, m stand for the total spin and total
spin in the z − axis. Thus, s = 1 in + + because
when S2 operates on it, it multiplies the state by
2h 2 . But according to the formalism developed
for general angular momentum operators, the
total momentum operator has eigenstates with
eigenvalue of l (l + 1)h 2 , so here clearly the total
spin is 1.
So the system of two spin half states is spanned by :
s = 1, m = 1 , s = 1, m = 0 , s = 1, m = −1 , s = 0, m = 0 .
• This means that the tensor product of two spinhalf particles behaves like the direct sum of a spin
one and spin zero particle – since each state can be
viewed as a linear combination of the three spin one
states (l=1,m=1,0,-1) and the spin zero state
(l=m=0) (careful – the spin one particles requires
three, not one, scalars to describe its state!).
• Formally, we write
1 1
⊗ = 1⊕ 0
2 2
• And in general, for two particles of spin j1 and j2 :
j1 ⊗ j2 = ( j1 + j2 ) ⊕ ( j1 + j2 − 1) ⊕ ... ⊕ ( j1 − j2 )
As follows by counting arguments and from
considering the upper and lower bounds for the total
spin. It remains to calculate the coefficients of the
basis transformation. All this carries over to systems
with more than two particles, e.g.
1 1 1 3 1 1
⊗ ⊗ = ⊕ ⊕
2 2 2 2 2 2
The Hyperfine Interaction in Hydrogen
• In addition to the Coulomb interaction, there’s an
interaction Hamiltonian of the form H hf = AS1S2
between the proton and electron, due to their
magnetic moment. It splits the ground state to two
levels: E+ = − R y + Ah 2 4, − R y − 3 Ah 2 4
Proof : compute the matrix of
S1 ⋅ S2 = S1x ⋅ S2 x + S1 y ⋅ S2 y + S1z ⋅ S2 z
++
+−
1 0

So up to constants it is
 0 −1
σ x ⊗σ x + σ y ⊗σ y + σ z ⊗σ z = 
0 2

0 0

which has eigenvalues and eigenvectors
 0 1  0
0
     
 
1  0  0
1
1,   1,   1,   − 3,  
−1
1
0
0
     
 
 0  0 1
0
     
 
+− + +−
++
−−
144444
42444444
3
Triplets
+− − +−
14243
Singlet
−+
−−
0 0

2 0
−1 0 

0 1 
A Very Short Introduction to Scattering
• Assume that a plane wave exp(ikz ) hits a potential
centered around the origin, and which decreases faster
than 1 r (so the potential has no effect at infinity).
• Asymptotically (r →∞ ) the scattered wave will behave
like a free particle. The solutions are known to be of the
form
 Al sin( kr −lπ /2) Bl cos(kr −lπ /2)  m
+

Yl (θ ,φ )
∑
kr
kr

l ,m 
for the wave to be purely outgoing, this must equal
exp(ikr )
f (θ ,φ ), and the total wave is therefore
r
exp(ikr )
exp(ikz ) +
f (θ , φ ), that is, the sum of the incident
r
and reflected parts. This expression can be guessed from
physical considerations.
• The stationary states satisfy (up to a choice of
[
]
constants) ∆ +k 2 −U (r) ψ (r)=0 ( k 2 is proportional to
the energy).
• The cross-section (amount of particles scattered per
2
angle) is proportional to f (θ , φ ) . It also turns out that,
asymptotically, the scattered current is only radial.
The Integral Scattering Equation and Green’s Function
[∆ + k
2
]
[
]
− U (r) ψ (r) = 0 ⇒ ∆ + k 2 ψ (r) = U (r )ψ (r)
• Try to solve with a Green function for the operator:
[∆ + k ]G(r) = δ (r). The idea : given an operator A
2
and equation Aψ (r ) = U (r )ψ (r ), find G(r) such that
AG (r) = δ (r), and then a solution also satisfies
ψ (r ) = ψ 0 (r ) + G (r ) ∗U (r )ψ (r ), whereψ 0 (r ) is a solution
to the homogenous equation, Aψ 0 (r ) = 0 :
Aψ (r ) = A[ψ 0 (r ) + G (r ) ∗ U (r )ψ (r )] = A(G (r ) ∗ U (r )ψ (r ) ) =
δ (r) ∗ U (r )ψ (r ) = U (r )ψ (r ). The integral equation allows an
iterative solution in powers of U (r ). Meaningful solutions
for the scattering problem are ψ 0 = exp(ikz ) and
G± (r ) (up to a constant) - exp(±ikr ) r ), G+ (r ) is the outgoing
wave in which we' re interested here.
The equation is therefore
exp(ik r − r′ )
ψ (r ) = exp(ikz ) − ∫
U (r′)ψ (r′)d 3 r′
r − r′
Since the potential is spatially limited, and we' re interested in
the behavior when r → ∞, we can assume that in the
meaningful range of the integrand, r >> r ′. It is easy to see that
then r − r′ ≈ r − (u , r′), where u is a unit vector in the direction
of r; in the denominator we can replace r − r′ with r , which
yields the asymptotic approximation
exp(ikr )
3
′
′
′
exp(
−
ik
u
r
)
U
(
r
)
ψ
(
r
)
d
r′
ψ (r) ≈ exp(ikz ) −
∫
r
exp(ikr )
which is indeed in the form exp(ikz ) +
f (θ , φ ),
r
since r depends only on θ , φ .
• It is customary to use the following notations:
The incident wave vecto r , k i , is the vector of modulus
k which is directed along the the axis of the beam, so
exp( ikz ) = exp( ik i ⋅ r). Similarly, the scattered wave
vector in the direction θ , φ is defined by k d = ku. The
scattering (or transferr ed) wave v ector in the
direction θ , φ is K = k d − k i .
The Born Approximation
ψ =exp(ik i r)+ ∫G+ (r −r′)U (r′)ψ (r′)d 3r′=
exp(ik i r)+ ∫G+ (r −r′)U (r′)exp(ik i ⋅r′)d 3 r′+
3
3
′
′
′
′
′
′
′
′
′
′
−
−
G
G
U
U
ψ
d
d
(
r
r
)
(
r
r
)
(
r
)
(
r
)
(
r
)
r
r′′+...
+
∫∫ +
This is the Born expansion, which is useful if the
potential is small. Plugging ψ =exp(ik i ⋅r)+
exp(ikr )
f (θ ,φ ) into the first order Born expansion
r
and using ∫G+ (r −r′)U (r′)exp(ik i ⋅r′)d 3 r′≈
exp(ikr )
3
′
′
′
(
)
−
−
⋅
⋅
ik
U
i
d
exp
u
r
(
r
)
exp(
k
r
)
r′ =
i
∫
r
exp(ikr )
3
′
′
′
(
)
−
−
⋅
⋅
i
U
i
d
exp
k
r
(
r
)
exp(
k
r
)
r′⇒
d
i
∫
r
f (θ ,φ )=− ∫exp(−iK⋅ r′)U (r′)d 3 r′, which is the Born
approximation. It relates the Fourier transform of the
potential to the scattering amplitude. Higher order
expansion relate to "secondary scatterings" scattered
again from the potential, etc.
Some examples of the Born approximation (left – the
potential, right – dependence of f (θ ) on θ . This is 2D,
the incoming wave is at the x-direction).
Partial Waves
• If the potential is rotationally symmetric (central), the
scattered wave can be studied by separately analyzing its
components which have fixed angular momentum.
∞
exp(ikz )=∑i l 4π (2l +1) jl (kr )Yl 0 (θ ) (no φ dependence).
l =0
∞
Similarly, we can write f (θ ,k )=∑(2l +1)al (k ) Pl (cos(θ ))
l =0
1



 4π  2 0 
 Pl (cos(θ ))= 2l +1  Yl (θ ) . al (k ) are called the lth partial






sin( kr −lπ 2)
wave amplitude. Since jl (kr ) →
, we get for the
r →∞
kr
free particle (no potential) exp(ikz ) →
r →∞
1
exp(ikr )−exp(i (kr −lπ ))
Pl (cos(θ )); thus for each l
(2l +1)
∑
r
2ik l =0
there are incoming and outgoing waves of the same amplitude,
with a phase difference due to the centrifugal barrier. The two
waves induce opposite probability currents which cancel out
(since there should be no probability flow in the steady state).
If a potential is present, then at infinity the only difference
relative to the free particle can be in the phase; so denote
∞
exp(i (kr −lπ /2+δ l )−exp(i (kr −lπ /2+δ l ))
ψ k (r ) → ∑ Al
Pl (cos(θ ))
r →∞
r
l =0
∞
Comparing with the free particle, the incoming wave must
2l +1
be the same, so Al =
exp(i (lπ /2+δ l )), and
2ik
∞
1
ψ k(r ) →
(2l +1)[exp(ikr )exp(2iδ l )−exp(−i (kr −lπ )]×
∑
r →∞ 2ikr
l =0
exp(2iδ l )−1
∞

Pl (cos(θ ))=exp(ikz )+ ∑(2l +1)
Pl (cos(θ )×
2ik
 l =0

exp(ikr )
. Recalling how f (θ ,φ ) was defined,
r
exp(2iδ l )−1
al (k )=
. The effect of the potential is therefore
2ik
to attach a phase shift exp(2iδ l ) to the outgoing wave. The
cross section can be simply expressed as a function of the δ l :
σ =(4π k
2
)∑(2l +1)sin
∞
l =0
2
(δ l )
To find the phase shift, one imposes continuity between the
solution at infinity and the solution of the steady state at
the vicinity of the origin.
The main motivation to use partial waves is when l ,k , and
the range of the potential are such that only a small number
of al (k ) need to be considered, since for l >lmax the particle
doesn't come closer to the origin more than lmax k, and if
this is larger than the extent of the potential such l 's don't
scatter (since the momentum is ≈ k and therefore the angular
momentum is roughly kr0 , where r0 is the particle's distance
from the origin).
Example: scattering of l=1 from a hard sphere
The potential is V (r )=0 for r >r0 , V (r )=∞ for r <r0 .
The general method of solution is as follows: look at the
outgoing wave as r →∞. It is of the form exp(ikz ) (the
homogenous part) plus the solution of the radial equation
in u (r )=rRk ,l (see slides of "Solution of Rotationally
Invariant Problems"). The equation is
 h 2 d 2 l (l +1)h 2

h 2k 2
+
+V (r )uk ,l (r )=
uk ,l (r ).
−
2
2
2µr
2µ
 2µ dr

Since uk ,l (r ) includes already the divison by r (see previous
slides on scattering), one only has to compute the phase
shift in uk ,l (r ) relative to the free particle shift, lπ 2:
uk ,l (r ) ≈ Csin( kr −lπ 2+δ l ). δ l is the desired shift.
r →∞
Let's compute the phase shift for the hard sphere and l =1.
Inside the sphere, the wave function is zero (since the
potential is infinite). In the outside, the radial equation
 d 2u 2 
reduces to − 2 + 2 u =k 2u⇒
 dr r 
 sin(kr )
 cos(kr )

u (r )=C 
−cos(kr )+a
+sin(kr )  which,
 kr

 kr
as r →∞, should equal (up to a constant factor)
sin( kr −π 2+δ1 )=−cos(δ1 )cos(kr )+sin (δ1 )sin( kr ), so
a =tan(δ1 ). Now it remains to enforce continuity at r =r0 :
The wave function has to be zero there, so we must have

 cos(kr0 )
sin(kr0 )
−cos(kr0 )+ a
+sin(kr0 ) =0⇒
kr0

 kr0
sin(kr0 )
−cos(kr0 )
kr0
a =−
, which in very low energies (k →0)
cos(kr0 )
+sin(kr0 )
kr0
3
is of the order (r0 k ) . Since for small angles θ ≈tan(θ ), this
is also the order of δ1 (k ). Using asymptotic properties of
the spherical Bessel and Neumann functions (see slide 222)
2 l +1
it turns out that for every l , δ l (k )≈(r0 k ) .
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