XV. Time-Dependent Methods
Up to now we have focused almost
exclusively on Hamiltonians that have no
time-dependence, so H(t) = H. In this
chapter we focus on using methods where
the Hamiltonian might contain explicit time
dependence, so we have
i
H(t)
HF
HI
d
  t   H  t    t  . (15.1)
dt
t
T
In general, we would like to imagine a
Figure 15-1: Qualitative depiction of the
Hamiltonian which changes from an initial
time-dependent problem. The Hamiltonian is
Hamiltonian HI to a final Hamiltonian HF,
assumed to change only from t = 0 to t = T.
over the course of a time period T, as
illustrated in Fig. 15-1. Usually the initial
state will be some eigenstate of the initial Hamiltonian  I and we would like to know the
probability that the wave function ends in some other state  F , where
H I  I  EI  I
and H F  F  EF  F .
The probability will simply be given by
P  I  F    F  t 
2
.
Calculating this transition probability, P  I  F  will be our primary goal.
A. The Sudden Approximations
Consider first the situation where the Hamiltonian changes very quickly from its initial state
to its final state. At t = 0, it is in the initial state   0    I . For short times, we can Taylor
expand the wave function around this state, so that
  t     0  t
d
 t 
dt
t 0

t   
2
I

t
H  0  I .
i
(15.2)
If the time is short enough, then at t = T, the first term will dominate the second, and we can
approximate  T    I ; however, this state is no longer an eigenstate of the Hamiltonian.
The probability that it ends in a particular new eigenstate is therefore given simply by the
overlap, and therefore
PI  F   F I
© 2016, Eric D. Carlson
257
2
.
(15.3)
XV. Time-Dependent Methods
This is called the sudden approximation, and applies if the Hamiltonian changes quickly. What
do we mean by quickly? From Eq. (15.2) we see that the corrections will be small if TH  I
is small compared to  I . The Hamiltonian typically leads to a factor of the energy of various
eigenstates, so we would expect Eq. (15.3) to hold provided ET
, where Ei is an eigenvalue
i
of H. This turns out to be unnecessarily restrictive. Recall that adding a constant to the
Hamiltonian never affects anything, so our results can never depend on the actual energies
themselves. A more careful analysis will reveal that the restriction really has to do with the
difference in energies between eigenstates. Hence we can use Eq. (15.3) provided T E
.
As a simple example of the sudden approximation, imagine a Harmonic oscillator with a
time-dependant frequency, so that
H t  
2
1 2 1
P  2 m   t  X 2 .
2m
20
where the frequency increase from   0   0 to
0
(t)
 T   20 over a short time interval T, as illustrated in Fig.
15-2. What is the probability that the particle in the ground
state remains there as the frequency changes? The ground
state wave function is given by
 m0 x 2 
 m0 
 I  x  0  x  
 exp   2 
  


14
T
t
Figure 15-2: A Harmonic
oscillator has its frequency
double over a time interval T.
and the final state wave function is identical if we replace 0 by 20 . We therefore have
PI  F  
2

 dx  x   x 
*
F

I

m0  2
 2

  3m0
2m0

 m0 x 2 m0 x 2  
m0  

  exp  
 dx 
  
2

 
2
2
 2 2
,
 
3

where the integral was performed with the help of Eq. (A.13f). The condition of validity
T E
is just 0T 1 .
B. The Adiabatic Approximation
The complementary situation is when the Hamiltonian changes very slowly. At each
moment 0  t  T , imagine finding all the eigenstates  i  t  , satisfying
H  t   i  t   Ei  t   i  t  .
(15.4)
They will be orthonormal at all times, so in particular we see that
XV. Time-Dependent Methods
258
© 2016, Eric D. Carlson
0
d
 i  i   i  i   i  i  2 Re   i  i
dt
,
where we have dropped the explicit time-dependence of our eigenstates, and written time
derivatives with a dot. The real part of this overlap must vanish, but in fact the imaginary part is
undetermined and undeterminable; we can change it simply by redefining the phase of our
eigenstates as an arbitrary function of time. Hence we can choose the imaginary part to be
anything we want, and to make things simple, we will define our eigenstates in such a way that
i i  0 .
Taking the time-derivative of Eq. (15.4), we can also demonstrate that the overlap between
any eigenstate and the time derivative of any other eigenstate will also be small:
d
d
H  i    Ei  i  ,

dt
dt
H  i  H  i  Ei  i  Ei  i .
Let the bra  j act on both sides of this expression for some j  i :
 j H  i   j H  i   j Ei  i   j Ei  i ,
 j H  i  E j  j  i  0  Ei  j  i ,
 j i 
1
 j H i
Ei  E j
for j  i .
(15.5)
Thus our eigenstates are changing little if the Hamiltonian H is changing slowly.
If we have eigenstates at all times, we can write our wave function in terms of them, so that
  t    ci  i .
i
where both our coefficient ci and our eigenstates  i are implicit functions of time. We now
attempt to solve Schrödinger’s Eq. (15.1). We have
i
 c
i
i
i
d
 ci  i  H i ci  i ,
dt i
 i  ci  i    ci Ei  i .
i
Act on the left with  j on both sides, using Eq. (15.5) and orthonormality:
i
 c
i
i
 j  i  ci  j  i    ci Ei  j  i ,
i
i cj  i
c
i
 j i  cjEj ,
i
i c j  E jc j  
i j
© 2016, Eric D. Carlson
i ci
 j H i .
E j  Ei
259
(15.6)
XV. Time-Dependent Methods
Now, if the Hamiltonian is changing slowly, the right side will be small, and perhaps we can
ignore it. This inspires the introduction of new constants bj, defined by
i t

b j  t   exp   E j  t   dt  c j  t  ,
 0

(15.7)
Solving for the cj’s, and substituting into Eq. (15.6), we see that

 i t

 i t

i bi
i b j  E j b j  E j b j exp   E j t  dt   
 j H  i exp   Ei t  dt  ,
0
0

 i  j E j  Ei


t
i

bi
bj  
 j H  i exp    E j  t   Ei  t  dt  .
i  j E j  Ei
 0


We now simply integrate this expression over time to get the answer:
T
b j T   b j  0    
i j 0
bi  t   j H  i
Ei  t   E j  t 
i t

exp    E j  t   Ei  t  dt  dt .
 0

(15.8)
This rather messy looking expression took a lot of work, but we are about to get a big payoff.
First note that the factor  j H  i is small, because the Hamiltonian is changing slowly;
however, it is being integrated over a large time T. However, the exponential will be oscillating
rapidly, while the other factors, such as bi  t  , will be changing relatively slowly. Hence, to
leading order, the outer integral will vanish, and we can approximate b j T   b j  0  . Since
Eq. (15.7) tells us that cj and bj are the same up to a phase, we conclude that c j T   c j  0 .
2
2
Since it started in the eigenstate  I , it ends in the state  I (except for phase), and therefore
P  I  F    FI .
(15.9)
You can’t get a much simpler formula than that! It should be noted that this does not mean that
the final state is identical with the original. The original and final states are, in general,
eigenstates of different Hamiltonians. What has happened is that the eigenstates of the original
Hamiltonian have smoothly transformed into the eigenstates of the final Hamiltonian. Hence if
the particle starts in the ground state (or first excited state, or second . . .) it will end in the
ground state (or first excited state, or second. . .). Eq. (15.9) is called the adiabatic
approximation.
When is Eq. (15.9) valid? Look carefully at Eq. (15.8). Let’s assume that  j H  i is not
only small, but changing slowly, so that it is relatively constant on time scales of order T. The
exponential factor goes through its whole cycle (adding a total of zero to the integration) over
one period, which is given by P  2 E . If this time is much shorter than the characteristic
time T during which the Hamiltonian changes, then our approximation will be valid. In other
words, we need T E
. Obviously, this is just the complement of the condition for the
sudden approximation. As an illustration of the adiabatic principle, consider an example
identical to the one illustrated in Fig. 15-2, a harmonic oscillator that doubles its frequency, but
XV. Time-Dependent Methods
260
© 2016, Eric D. Carlson
this time assume T0 1 . Then the
probability of the ground state becoming the
ground state is just 1.
2
2
Using the adiabatic approximation in more
complicated situations can be rather complex.
It might seem like the n’th energy state always
goes into the n’th energy state, provided the
Hamiltonian changes slowly enough, but
sometimes this just isn’t true. In general, if you
1
1
gradually change the Hamiltonian, the various
eigenenergies will shift up and down, and
sometimes two states will approach each other
in energy (a level crossing). The most common
situation is that as the states become very close
Figure 15-3: Level mixing. Two states
in energy, they will have a great deal of mixing,
have energies that approach each other at
as is evident from the energy denominators, in
some critical time. Generally, mixing will
Eq. (12.45b). This causes the two states to
cause the lower energy eigenstate to
never really have the same energy, which
always remain the lowest energy state
generically leads to the kind of energy structure
(solid lines) so a system starting in state 1
illustrated in Fig. 15-3. In such cases, the
will smoothly evolve to state 1, and
relevant characteristic time T is the time it takes
similarly, 2 will evolve to 2. However, if
us to move through the large mixing region, and there is a symmetry preventing mixing
the energy splitting E is dominated by this
(dashed lines), the states will never mix,
mixing region, where the energy splitting is at a
and a system beginning in state 1 will
minimum. Indeed, it is even possible for the
smoothly evolve to the higher energy state
two states to be completely unmixed, and then
2, and vice versa.
there can be a true level crossing, as illustrated
in Fig. 15-3. In general, if there is absolutely no mixing between the levels, it is probably
because the mixing is prevented by some symmetry. Provided a symmetry is present, the
Hamiltonian can never change the corresponding eigenvalues, and the eigenstates will pass
through the level crossing, after which the lower energy state will have become the higher, and
vice versa.
C. Time-dependent Perturbation Theory
We now approach the subject of time-dependent perturbation theory. Suppose we have a
Hamiltonian that depends on time, but only by a small amount, so that
H  H0  W t  ,
where, in a manner similar to before, we assume W  t  is small. We will also assume that W  t 
is non-zero only for 0  t  T . We will assume that we know the exact eigenstates of H0. Let
these states n be orthonormal and have energy En, so that
H 0 n  En n .
© 2016, Eric D. Carlson
261
XV. Time-Dependent Methods
Any quantum state can be written as a linear combination of these states. In particular, the exact
state vector we are interested in can be written as
  t    cm  t  m ,
m
where our goal is to find the functions cm  t  . We start by substituting these wave functions into
the Schrödinger equation, which tells us
i
i
m
d
  H    H 0  W  t   ,
dt
d
cm  t  m   cm  t   Em  W  t   m .
dt
m
Let the bra n act on this to yield
i
d
cn  t   Encn  t    cm t  n W t  m .
dt
m
(15.10)
Now, since W is presumed small, we expect that the primary source of change for the
coefficient cn will be due to the right hand side, which suggests cn  t  eiEnt . We therefore
define the quantity bn  t  by
cn  t   bn  t  exp   iEnt
.
(15.11)
Substituting this into Eq. (15.10) and simplifying, we find1
i bn  t   En  eiEnt  Enbn  t  eiEnt   bm  t  eiEmt
n W t  m ,
m
i bn  t    bm  t  n W  t  m exp  i Ent
 i Emt

(15.12)
m
We now define the frequency difference and abbreviate the matrix element as
nm   En  Em 
,
Wnm  t   n W  t  m .
Substituting these into Eq. (15.12) , we have
bn  t    i
1  bm t Wnm t  ei
nmt
.
m
We now take this equation, and integrate it over time, to yield
1
Much of this discussion is clearer in the interaction picture (see section 11D). In this picture, the states are
changing only due to the interaction W, so that the factors ci(t) are never introduced, and Eq. (15.12) is arrived at
almost immediately. The phases in the interaction term simply represent the additional time dependence that
appears in operators when relating the interaction and Schrödinger pictures.
XV. Time-Dependent Methods
262
© 2016, Eric D. Carlson
1
bn T   bn  0  
i
T
  dtW t  e 
i
nmt
nm
bm  t  .
m 0
In general, we want to assume that we are starting in an eigenstate I of the initial Hamiltonian
H0, so that bn  0   cn  0    In . Then this equation becomes
bn T    nI 
1
i
T
  dtW t  e 
i
nmt
nm
bm  t  .
m 0
Now, substitute this equation into itself repeatedly. Each time we do so, we will get a new
term, with an extra integral. There will also be an additional sum, but one of the sums will be
handled automatically by the Kronecker delta function. We find
T
T
t
1
1
i t 
bn T    nI   dtWnI  t  einI t 
dt  dt Wnm t Wmp t  einmt e mp bp t  
2  
i 0
i  m p 0 0
.
It is clear we are producing a perturbation series in W. Since we are primarily interested in the
probability of going to a particular final state, let’s focus on n = F. We’ll give this amplitude a
special name, SFI, for which we have, to third order in W,
T
T
t
1
1
SFI  bF T    FI   dtWFI t  eiFI t 
dtWFm t  eiFmt  dt WmI t  eimI t 
2 
i 0
i  m 0
0

1
T
dtWFm  t  eiFmt
3 
i 
m 0
t 
t
  dtWmn t eimnt  dtWnI t einI t 
n 0
.
(15.13)
0
SFI is often called the scattering matrix, or S-matrix. It is a unitary matrix representing the
amplitude for ending in a particular state starting from an arbitrary state. The leading term is
clearly the probability that nothing happens, the first term represents the amplitude that a single
transition occurs, etc.
The probability of having a transition, then, is simply given by
P  I  F   F  T 
2
 SFI .
2
(15.14)
We are rarely interested in the probability of the
state remaining the same, so usually we will drop
the first term in Eq. (15.13). To leading order, the
transition probability is given by
PI  F  
1
2
2
T
W t  e
iFI t
FI
dt .
(15.15)
0
As a simple illustration, imagine we have a
hydrogen atom initially in the ground state 100
Figure 15-4: An electric field produces
a pulse over a characteristic time T. At
early and late times, the pulse quickly
dies away.
when a passing electromagnetic pulse produces an electric field E  E0 zˆ exp   t 2 2T 2  . This
pulse is illustrated in Fig. 15-4. We would like to calculate the probability that this pulse causes
© 2016, Eric D. Carlson
263
XV. Time-Dependent Methods
a transition to a new state nlm . An electric field E can be produced from an electric potential
U  r  with the relation E  U  r  . The easiest way to produce a constant field, then, is with a
potential U  E  r , which means that an electron with charge –e would have potential V  eU ,
so the perturbation is
 t2 
W  eE0 Z exp   2  .
 2T 
Now, in our approximations, we assumed that the perturbation is turned on at 0 and off at T, but
clearly this expression never vanishes. Nonetheless, the field dies away quickly enough that
there is no harm in integrating over all time. Hence our probability is just
PI  F  
1
2
2

 dt
nlm Z 100 eE0e
 t 2 2T 2 iFI t
e

e2 E02
nlm Z 100
2

 e2 E02
2
2

e
2
 t 2 2T 2 iFI t
dt

nlm Z 100
2
2 T 2 eT
P  I  F   2 e2 E02
2
2
FI
2
2
2
,
nlm Z 100 T 2eT FI .
2
2 2
(15.16)
The matrix elements may be computed easily from the explicit form of the hydrogen wave
functions. The operators R connect only states with l values that differ by one, so we must have
l = 1, and Z commutes with Lz, so m = 0.
It is interesting to note that Eq. (15.16) vanishes both in the limit of small and large T. Why?
In the limit of small T, we can evaluate the answer by the sudden approximation, in which the
wave function doesn’t change in the short time available, and since the final Hamiltonian is
identical with the initial Hamiltonian, the electron is still in the ground state. In the limit of large
T, the Hamiltonian is changing adiabatically, and therefore the ground state remains the ground
state. Indeed, the maximum transition probability occurs when TFI  1 , or T E  ,exactly
the boundary given between the limiting constraints T E
and T E
.
D. Harmonic Perturbations
Of particular interest, especially when studying the interaction of light with atoms, is
perturbations which are harmonic; that is, having a time dependence like a sine or cosine.
Suppose our perturbation takes the form
W  t   Weit  W †eit .
(15.17)
We will implicitly assume   0 , so we are definitely not talking about a constant perturbation.
The operators W and W † are independent of time. For example, if we pick W to be a Hermitian
(anti-Hermitian) operator, we will simply have an operator with a cosine (sine) time dependence.
Of course, to have a truly harmonic source, it should behave this way forever, from minus
infinity to plus infinity, so we will be interested primarily in situations where T is large enough
that the perturbation goes through many cycles of oscillation. To help give the “sense” of it
XV. Time-Dependent Methods
264
© 2016, Eric D. Carlson
being on forever, and to make the math a tad simpler, we will turn it on at  12 T and off again at
 12 T , changing our limits in Eq. (15.13).
Let’s calculate the scattering matrix assuming the initial and final states are distinct. We’ll
only include the first order perturbation, so that
1
SFI 
i
1T
2
 dt W
FI
 12 T
S FI 
eit  WFI† eit  eiFI t
1T
WFI eiFI  t WFI† ei FI t  2


 ,
i



i







FI
FI

  12T
1

i

sin  12 FI    T 
sin  12 FI    T  
†
W

W
 FI

FI
FI  
FI  


2
i
(15.18)
where WFI  F W I . We now wish to make an approximation. Assuming the perturbation
W is small, this expression will be small unless one of the two denominators is small. We will
therefore focus on regions where one or the other denominators is small, but not both, since we
assumed   0 . For example, suppose FI  0 and is close to  , then the first term will
dominate the second. The probability for a transition will then take the form
2
4 WFI sin  12 FI    T 
PI  F  
.
2
2
FI   
2
(15.19)
Consider for the moment the last factor. If T is
large but finite, it is a function with a very large
peak at   FI which grows taller and narrower
as T becomes larger, as illustrated in Fig. 15-5.
This suggests that in the limit of large T, it
approaches a delta function; in other words, we
suspect
lim
T 
sin 2  12 FI    T 
FI   
2
 A FI    .
Figure 15-5: A plot of the frequency
dependence of the probability of a
transition, plotted for
. As T
increases, the peak becomes narrower
and taller, increasingly resembling a
Dirac delta function.
To determine the strength of the delta function,
we simply integrate both sides. The integral on
the right is just A, and the left side can be done
with the help of Eq. (A.13j) to yield 12  T , so
A  12  T . Thus we have
lim
T 
sin 2  12 FI    T 
FI   
2
 12  T FI    .
(15.20)
We therefore have
P  I  F   2
© 2016, Eric D. Carlson
2
WFI  FI    T .
265
2
XV. Time-Dependent Methods
A couple of comments are in order here. First of all, the probability is proportional to time,
which makes sense. The longer you wait, the more chance you have for a conversion to occur.
The probability per unit time is called the decay rate and denoted  . With the help of
Eq. (A.14d), we can easily show that we can include a factor of inside the delta function if we
include a factor of outside as well, so this can be rewritten, using nm   En  Em  , as
  I  F   2
1
WFI   EF  EI    , EF  EI .
2
(15.21)
We have completed the case where the final energy is greater than the initial energy. We
now want to calculate the other case. Rather than reversing the inequality, let’s keep EF  EI
but simply reverse the role of the final and initial state, so that F is the initial state, and I
the final state. Without going through the details, we will only keep the second term in
Eq. (15.18), and we convert to a delta function as before. We find
  F  I   2
1
WIF†   EI  EF    , EF  EI .
2
(15.22)
Note that in Eqs. (15.21) and (15.22), energy is not conserved. This is to be expected when
the Hamiltonian is not time-dependant. The external source of the Hamiltonian is being treated
as an inexhaustible source of energy. Clearly, when our quantum state is increasing in energy,
from Eq. (15.21) we see that it must be absorbing an energy of  from the source of this
external perturbation, and from Eq. (15.22), it is emitting this amount of energy. When we
interpret this external source as an electromagnetic wave, we will simply interpret this as the
absorption or emission of one photon of energy. However, to make this rigorous, we will have to
quantize the electromagnetic field, a task we wish to put off until a later chapter.
Comparison of Eqs. (15.21) and (15.22) reveals a surprising fact. The delta functions differ
only by a minus sign in their arguments and are therefore identical, and for any operator W,
WIF†  WFI* , so the matrix elements are identical as well. Hence the two rates are equal. This
presents a small paradox. Suppose we place a collection of atoms in a thermal background of
electromagnetic energy (black body radiation). According to Eqs. (15.21) and (15.22), the atom
is as likely to transition from, say, a 1s state to a 2p state as vice versa. Hence, given enough
time, there will be equal numbers of electrons in the two states (actually, there would be more in
the 2p states, since there are three times as many states). We will ultimately realize that
Eq. (15.21) is basically correct, but Eq. (15.22) is incorrect. The flaw will be corrected when we
quantize the electromagnetic field in chapter seventeen, and then include interactions with atoms
in chapter eighteen.
How do we interpret the delta functions in Eqs. (15.21) and (15.22)? If our system has only
discrete energy levels, these are a bit difficult to interpret, since a delta function is either zero or
infinity. One solution in this case is to revert to forms like Eq. (15.19). We will find that if the
energy conservation conditions are nearly satisfied, the rate is large but not infinite. In contrast,
if they aren’t well matched, the transitions do not grow with time. Roughly speaking, it is easy
to show that the rate will be large provided the energy is matched with an approximate error
T E  . This is yet another uncertainty relationship from quantum mechanics, one we will
discuss later.
In other situations, the delta function is not really problematic. One that is commonly
encountered is a situation where the perturbation is not strictly speaking harmonic, but rather a
complicated function of many frequencies. In such circumstances, it is possible to break the
XV. Time-Dependent Methods
266
© 2016, Eric D. Carlson
perturbation into its Fourier components, and then add up (integrate) over all frequencies. The
delta functions will pick out only that Fourier component with the right frequency to cause a
transition. For example, in a thermal bath of electromagnetic radiation (black body radiation),
there are electromagnetic waves of all frequencies present. An integration over all frequencies is
necessary, in which case the delta function will make the computation easier.
Another common situation is when there are a continuum of final states F ; for example, if
the initial state is a bound electron and the final state a free electron. Then the final state
eigenstates might be labeled k , and have energy
H0 k  Ek k 
2
k2
k .
2m
These states will be normalized, k  k   3  k   k  . We are not generally interested in the
probability into a particular final state k , but rather the probability of the electron coming free,
which will be given by
I  k    d k
3

2
WkI   Ek  EI    
2
2
 d 

0
2
k dk WkI
2
 2k 2


 EI   
 2m

2 mk 2
2 mk
2
2
d  WkI 
d  WkI .
2
3


k
The matrix element WkI will generally depend on direction. It is also possible to find the
differential rate d  d  simply by not performing the final angular integral.
E. Electromagnetic Waves and the Dipole Approximation
We consider now a special circumstance of considerable practical importance, the effect of
electromagnetic plane waves on atoms. The full Hamiltonian will be given by Eq. (9.20),
repeated here for clarity:
N
2
e
 1 

H  
Pj  eA  R j , t   eU  R j , t   B  R j , t   S j   Va  R1 , R 2 ,

m

j 1  2m
where Va  R1 , R 2 ,
,
 represents all the internal interactions of the atom, and A and B represent
the effects of the external field, and (for this chapter and beyond) we have approximated g = 2
for the electron. We will now define the unperturbed Hamiltonian H0 as
N
1 2
Pj  Va  R1 , R 2 ,
j 1 2m
H0  
© 2016, Eric D. Carlson
267
.
XV. Time-Dependent Methods
This Hamiltonian will have atomic eigenstates m with energy Em. The perturbation will be all
the remaining terms, but since we only want to keep only leading order in the electromagnetic
fields, we will drop the A2 term. So we have1
N
e

W  t      A  R j , t   Pj  B  R j , t   S j   eU  R j , t  .

j 1  m
(15.23)
We now wish to include an electromagnetic wave. There is more than one way (“gauge
choice”) to write such a wave, which we will address more carefully in chapter seventeen, but for
a specific choice of gauge (“Coulomb Gauge”), it can be written as
U  r, t   0, A  r, t   A0  εeikr it  ε*eikr it  .
(15.24)
where ε is a unit vector (the “polarization” 2) normalized so that ε  ε*  1 , ε  k  0 , and A0
denotes the amplitude of the wave. The electric and magnetic fields are given by
E  U  r, t   A  r, t   iA0  εeikr it  ε*eikr it  ,
B   A  iA0  k  ε  eikr it   k  ε*  eikr it  .
Substituting into (15.23), we have
W t  


N
e
ik R it
 ik R j it
ε*  Pj  i  k  ε*   S j  .
A0  e j ε  Pj  i  k  ε   S j   e


m j 1
Comparison with Eq. (15.17) then tells us the perturbation with the time part factored out is
W
N
e
A0  eik Rj ε  Pj  i  k  ε   S j  .
m j 1
(15.25)
We now want to spend a moment looking at Eq. (15.25). We will consider the case where
we are using electromagnetic waves with energy  comparable to the energy level splittings of
1
an atom. These energy levels are typically of order mc2 2 , where   137
is the fine structure
constant. A wave number k for such light will be of order k mc2 2 c  mc 2 . In contrast,
the characteristic size of an atom will tend to be of order a0  2 ke e2 m  c m . As a
consequence, we have k  R j ka0  . This is a small number, so to leading order, we simply
treat it as zero, and drop the phase factor in Eq. (15.25). Similarly, if you compare the first and
second terms in Eq. (15.25), the first term will be of order P
a0  mc , while the second will
be of order kS
k
our perturbation is
mc 2 . Hence we can ignore this as well. In summary, to leading order,
1
Eq. (15.23) seems to contain an error, since we commuted A with Pi, not always a valid assumption. This
assumption turns out to work in Coulomb gauge, since the choice of A given in Eq. (15.24) has no divergence.
2
The polarization vector ε may be chosen real, yielding cosine dependence, or pure imaginary, yielding sine
dependence. More complicated possibilities exist; for example, one may make one component real and another
imaginary, resulting in circular polarization.
XV. Time-Dependent Methods
268
© 2016, Eric D. Carlson
N
e
e
WE1  A0  ε  Pj  A0ε  P ,
m j 1
m
where P is just the sum of the momenta of all the electrons. This approximation is called the
electric dipole approximation, and the E1 subscript just denotes this fact. The transition rate
between two atomic states is then given by Eq. (15.21), assuming the final energy is greater than
the initial energy.
I  F  
2 A02e2
ε  F P I
m2 2
2
 FI    .
(15.26)
A couple of observations allow us to convert this into a more convenient form. The first is to
note that the commutator of H0 with R  i Ri is given by

 H 0 , R   

k
1 2
Pk  Va  R1 , R 2 ,
2m

, R j  

j
1
i
i
Pk2 , R j     Pj   P .

2m k , j
m j
m
It follows that the matrix elements can be rewritten as
F P I 
im
F  H 0 , R  I 
im
F  EF R  REI  I  imFI F R I . (15.27)
It is also useful to rewrite the amplitude in terms of intensity. The intensity of a wave is
given by the time-averaged magnitude of the Poynting vector,1 which is given by
S 
1
0
E B 
A02
2 A2
ε   k  ε *   ε *   k  ε    0 k .

0 
0
Recalling that for light we have   ck , and finally remembering that 0  4 ke c 2 , we find
that we can write the intensity , or power per unit area, as
 S 
2 A02 2 A02 2c

.
0 c
2 ke
(15.28)
Substituting Eqs. (15.27) and (15.28) into Eq. (15.26) then yields
4 2 kee2FI2
I  F  
2
c 2
ε  F R I
2
 FI    .
The matrix element is often abbreviated rFI  F R I (the electric dipole matrix element).
The delta function assures that FI   , and we can write this more easily with the help of the
fine structure constant   ke e2
c , so this simplifies to
  I  F   4 2
1
1
ε  rFI  FI    .
2
(15.29)
See, for example, p. 259, Jackson, Classical Electrodynamics, Third Edition (Wiley, 1999).
© 2016, Eric D. Carlson
269
XV. Time-Dependent Methods
This yields an infinite rate if the frequency is appropriate. If the incoming wave has a range of
frequencies, as all real sources do, then the intensity can be described as an intensity distribution,
  with units of power per unit area per unit angular frequency, so that1
  d .

We therefore modify Eq. (15.29) by including an appropriate integral, yielding
  I  F   4 2
1
FI  ε  rFI
2
.
(15.30)
The rate will be identical for the reverse reaction.
Eq. (15.30) is appropriate for an electromagnetic wave with known polarization vector ε . If
the polarization is random, we need to average over both polarizations perpendicular to the
direction of the wave n̂ . The easiest way to do this is to sum over all three potential
polarizations and subtract the “missing” polarization n̂ . The result is
unpol  2 2
1
FI   rFI
2
 nˆ  rFI
2
.
If the incoming wave is coming in random direction, or if the atom is itself oriented in a random
direction, we need to average over all angles, which yields
random  43  2
1
FI  rFI
2
.
Let’s try applying our formulas in a simple case. Consider a hydrogen atom in the ground
state that is influenced by an electromagnetic field with sufficient frequency so that the electron
can be ionized. The initial state is 100 , but what shall we use for the final state? Far from the
atom, the electron will be approximately a plane wave, but nearby the wave function is much
more complicated. To simplify, assume the frequency is much higher than is necessary to free an
electron from hydrogen. In this case the final state will also be nearly a plane wave near the
atom, since the kinetic energy is much greater than a typical binding energy. We need to
calculate the matrix element rFI  F R I  k R 100 , which with the help of Eq. (15.27) we
can write as
rFI  k R 100 
1
k
k P 100 
k 100 .
imFI
im
We now simply put in the specific form of the wave function and take its Fourier transform to
give
rFI 
k
1
1
k
1
d 3reikr
er a0 
32 
im  2 
im 2 a03
 a03
2 k

i ma03 2
1
1
 a
1
0
 ik cos   d cos  
1
3
1

1
0
 r a ikr cos
2
 d cos  r dre 0
k
 a 1  ik 2   a 1  ik 2  ,
0
0

 2mka03 2 
A more likely experimental quantity than the intensity per unit angular frequency would be intensity per unit
frequency
 f   2   or intensity per unit wavelength      2   2 c .
XV. Time-Dependent Methods
270
© 2016, Eric D. Carlson
rFI 
2 2i ka03/2
 m 1  a02 k 2 
2
.
Substituting this into Eq. (15.29), we have
 8 2 a03 
2
2 2 4
 2 2 2  1  a0 k  ε  k  FI   
 m  
4
 2k 2

2
 32 a03 2 ε  k m2 2 1  a02 k 2   
 EI    .
 2m

 1s  k   4 2
1
The dot product is just ε  k  k cos where  is the angle between the polarization of the light
and the direction of the final electron. We would then integrate this over all possible final wave
numbers k to yield
 1s  k  

32 a03 2 k 2
m2 2 1  a02 k 2 
 2k 2
 2
2
0   2m  EI    k dk  cos  d  ,
4
32 a03k 3
d

cos2  ,
4
2
2
2
d  m 1  a k 
0
where the final wave number k is defined by
2
k 2 2m  EI   . The final angular integral, if
desired, yields an additional factor  cos2  d   43  .
F. Beyond the Dipole Approximation
Suppose the dipole moment rFI  F R I between an initial and final state vanishes.
After all, the matrix element will generally vanish unless lF  lI  1, so this will commonly be the
case. Does this mean that the decay does not occur? No, it only means that its rate will be
suppressed. Let us expand out Eq. (15.25), but this time keeping more than just the leading term.
We have
W
N
e
A0  ε  Pj  i  k  R j  ε  Pj   i  k  ε   S j  .
m j 1
(15.31)
The leading term is responsible for electric dipole transitions, which we assume vanishes. The
middle term can be rewritten, with the help of some fancy vector identities, in the form1
k  R ε  P   k  R ε  P   k  P ε  R   k  ε    R
j
1
j
1
2
j
j
j
j
1
2
j
 Pj  .
One might suspect that you have to be careful with the commutators of k  P j and ε  R j , since momentum and
position don’t commute. However, because
© 2016, Eric D. Carlson
k  ε  0 , it follows that k  Pj and ε  R j do commute.
271
XV. Time-Dependent Methods
The last term on the right contains the angular momentum Lj of the j’th electron, and it is not
hard to show that the first term can be written as a commutator, in a manner similar to before
k  R ε  P   k  P ε  R   im  H , k  R ε  R  .
j
j
j
j
0
j
j
With a bit of work, we then find the matrix elements are given by
F W  I 
iA0e
F
m
 im 

H 0 ,  k  R j  ε  R j   12  k  ε   L j   k  ε   S j  I ,


j 1
N
  2
WFI   12 FI A0e F  k  R j  ε  R j  I 
N
j 1
iA0e
 k  ε   F
m
 12 L  S  I
.
The first term is referred to as the electric quadrupole term, and the second is the magnetic
dipole term. Both terms commute with parity, unlike the electric dipole term, and therefore they
can only connect states with the same parity. Hence the states connected by these terms are
always different from the ones connected by the electric dipole term, and there is no interference
between the two. The electric quadrupole term is a rank two spherical tensor operator; as such it
changes l by either zero or two, but cannot connect two l = 0 states. The magnetic dipole term
commutes with H0, assuming we ignore spin-orbit coupling, hyperfine splitting, and external
magnetic fields. Hence it can only cause transitions for states that are split by these smaller
effects.
These expressions can then be used to compute rates, in a manner similar to the electric
dipole moment. For example, the quadrupole contribution to decay is given by
I  F    
2
1
FI   F k  R j ε  R j  I
N
2
.
j 1
This is suppressed compared to the electric dipole rate by a factor  k  R  , typically of order  2 .
2
G. Time-dependent Perturbation Theory with a Constant Perturbation
The last type of perturbation we want to consider is one that is constant in time, W  t   W .
It may seem odd to use time-dependent perturbation theory in such a circumstance, but in fact, it
is often a very good way of thinking about things. For example, if I scatter a particle off of a
potential, we intuitively imagine a particle approaching the potential (initially in a plane wave),
interacting with the potential (the interaction) and then leaving the potential again (a plane
wave). Although the potential is in fact always constant, it is easier to think of it as a temporary
perturbation.
Our starting point will once again be Eq. (15.13), but we will make a couple of
modifications. First of all, we ultimately want to consider the perturbation as constant, not
having time dependence at all, so that it is always on and always will be on. For this reason, we
will change the integration to run from  12 T to 12 T . Eq. (15.13) then will look like
XV. Time-Dependent Methods
272
© 2016, Eric D. Carlson
W
S FI   FI  FI
i

m
n
WFmWmnWnI
i 
3
1T
2
 dt e
iFI t
 12 T

m
1T
2
1T
2
1T
2
 12 T
 12 T
 12 T
WFmWmI
1T
2
1T
2
 12 T
 12 T
dt dt   t  t   e
i   
iFmt imI t 
e
2
 dt  dt  dt  t  t  t   t  e
where we have used the Heaviside function,
defined by
iFmt imnt  inI t 
e
e

,
(15.32)
Im()
1 if   0 ,
    
0 if   0 .
We now wish to prove the identity
1  ei d
.
 0 2 i    i

    lim
Re()
(15.33)
This integral will be evaluated by the method of
contour integration.
Contour integration is a method that can be
used to perform integrals in the complex plane.
As defined, the integral in Eq. (15.33) runs from
minus infinity to infinity along the real axis.
Figure 15-6: Contours for performing the
The first step of performing a contour integral is
integral (15.33). To the initial path (purple
to “close” the contour: to add an additional path
line), we add a half-circle at infinity in the
from infinity back to minus infinity to make a
upper half plane for  > 0 (red semi-circle)
complete loop. The
or the lower half plane for  < 0 (blue
trick is to choose that
Im()
semi-circle). The orange X is the pole at
path such that it
Re()
contributes negligibly   i .
to the integral. Since
we are going to be wandering into values of  that are imaginary, our
Figure 15-7: The
goal is to pick values of  such that ei will be very small. This will
contours after they
be the case provided i has a large negative real part. If   0 , this
have been shrunk.
is simply achieved by forcing Im    0 (the “upper half plane”)
The pole has been
whereas if   0 , we force Im    0 (the “lower half plane”). We
moved to   0 .
For   0 , the
can therefore add a half loop “at infinity” to make the whole integral a
contour contains the
closed loop, as illustrated in Fig. 15-6.
pole (red curve), and
We now employ the magic of contour integration, which says that
the integral is nonthe result of an integral is independent of the path chosen, provided we
zero, for   0 , the
do not cross the pole. We can therefore shrink, in each case, the
contour can be
integral to a tiny circle, enclosing the pole in the case   0 , and
shrunk to zero (blue
enclosing nothing for   0 , as illustrated in Fig. 15-7. While we are
curve), and the
at it we can take the limit   0 , which places the pole at the origin.
integral vanishes.
We now have
© 2016, Eric D. Carlson
273
XV. Time-Dependent Methods
 i

1 ei d
1   e d  if   0,
lim


i
 0 2 i    i
2

i
i

  e d  if   0.
The loop for   0 can be shrunk to zero with impunity, and we can locate it well away from the
pole if we wish, so the integral vanishes. For   0 , we shrink it to a tiny circle of radius  . In
the limit   0 we have   0 , and we can approximate ei  1 . However, we dare not make
this approximation in the denominator. We parameterize the tiny circle around the origin by an
angle  and let    ei , where  runs from 0 to 2. Then we have, for   0 ,
2

1
ei d
1
lim


 0 2 i
2 i
   i

d  ei 
0
 ei
1
2 i

2
 id  1,
0
while it vanishes for   0 , which proves Eq. (15.33).
We now wish to use this identity to simplify Eq. (15.32). We replace each of the Heaviside
functions with the messy integral Eq. (15.33), which results in
 W W
iFI t
Fm mI
dt
e

lim


2
 1T
 0
 m  i 
2
1T
2
W
S FI   FI  FI
i

m
1T
2
WFmWmnWnI

i 
3
n
1T
2
dt
 12 T

 12 T
dt 
1T
2

1T
2
1T
2
 12 T
 12 T
 dt  dt e
iFmt imI t 
e

ei t t  d
ei t t  d
 2 i   i  
2 i   i  



 12 T
ei t t  d
 2 i   i 


dt eiFmt eimnt einI t  

  

.

We now proceed to perform all but one of the time integrals. We will do so simply by
working in the infinite time limit, so that the integrals turn into trivial expressions like



eimI  t dt  2 mI    . Using this on all the integrals except for dt, we obtain

W
S FI   FI  FI
i

m
1T
2
 dt e
iFI t
 12 T
1T
2
WFmWmnWnI
i 

3
n
 W W
 lim  Fm 2mI
 0
 m  i 
 12 T
1T
2
e
 12 T
it

e
dt
iFmt

   mI  d
i   i 
eit mn       d   nI     d 
 i   i  
i   i 



eiFmt dt 

.

We now use the Dirac delta functions to do all the  -integrations.
W
S FI   FI  FI
i

m
n
T
2

WFmWmI
iFI t
eiFm mI t dt
2
 1T dt e  lim

 
0
 m  i  i mI  i   12 T
2
1T
2
1
WFmWmnWnI
1T
2
 i  i nI  i  i mn  nI  i  T
3
1
2
eiFm mn nI t dt 

.

We rewrite ij  Ei  E j , and write this in the form
XV. Time-Dependent Methods
274
© 2016, Eric D. Carlson
1
S FI   FI 
i


WFmWmI
iFI t 
dt
e
W

lim

FI
 1T
 
 0
 m EI  Em  i

2
1T
2

m
n
We now define the transition matrix
FI
 
 
  
WFmWmnWnI
 EI  En  i  EI  Em  i
as
FI


WFmWmI
WFmWmnWnI
 WFI  lim 
 
 ,
 0
 m EI  Em  i m n  EI  En  i  EI  Em  i 

(15.34)
where we have changed    since we are taking the limit   0 anyway. The pattern is
clear: at each order in perturbation theory, we get one more factor of W in the numerator,
summed over all possible intermediate states, while in the denominator we always get factors of
the initial energy minus any intermediate energy, together with an i term that will bear some
discussion later. Then our scattering matrix will be given by
1
S FI   FI 
i
1T
2

FI
dt eiFI t   FI 
 12 T
2sin  12 FI T 
1
i
FI
FI
.
If the final state is different from our initial state, we ignore the first term and find
PI  F  
4
2
sin 2  12 FI T 
FI2
FI
2
.
In the limit of large T, we can then use Eq. (15.20) to obtain
P  I  F   2
2
2
FI
T  FI  .
We again let the rate be the probability divided by time, and conventionally place one factor of
inside the delta-function, to obtain Fermi’s golden rule:1
  I  F   2
1
2
FI
  EF  EI  .
(15.35)
This, together with Eq. (15.34), allows us to calculate transition rates to arbitrary order.
To understand how to apply Fermi’s golden rule in a simple situation, consider a plane wave
scattering off of a weak potential. We will let the unperturbed Hamiltonian be the free particle
Hamiltonian, H 0  P 2 2m , and the potential V be the perturbation. Then our eigenstates of H0
will be plane waves, and to leading order,
FI
FI
will be
 VFI  k F V  R  k I  
d 3r
 2 
3
eik F rV  r  eik I r
The rate Eq. (15.35) is then given by
1
Note that the very similar Eq. (15.21) is also called Fermi’s golden rule.
© 2016, Eric D. Carlson
275
XV. Time-Dependent Methods
I  F  
1
 2 5
2
 2 2
3 i  k F k I r
2 
d
r
e
V
r



 2m  kF  kI  .



There are two problems with this formula: first, neither our initial nor our final state plane waves
are normalized properly, and second, the delta function is a bit difficult to interpret. We can get
rid of one and one-half of these problems by “summing” over the final state momenta, which
then becomes an integral.
I  F  

1
 2 
5
3
 d kF
2
 2 2
 i  k F k I r
3
2 
d
r
e
V
r



 2m  k F  k I  




2
 2 2
 i  k F k I r
2
2 
3
k
dk

k

k
d

d
r
e
V
r
,




F
F 
F
I 

 2 5 0
 2m

2
d
mk

d 3reik F k I rV  r  .
5 3 
d   2 
1
(15.36)
The one remaining difficulty is that the incident plane wave cannot be normalized. In a manner
similar to last chapter, we note that the incident wave has a probability density    2 
2
3
and
is moving with classical velocity k m , implying a flux of    k m  2  . This allows us
3
to convert Eq. (15.36) into a differential cross-section.
d 1 d 
m2

 2
d   d  4
 d re
3
4
V r  .
i  k F k I r
2
This is none other than the first Born approximation, Eq. (14.19).
Problems for Chapter 15
.
1. An isolated tritium (hydrogen) atom 3H has its electron in the ground state when it suddenly
radioactively decays to 3He, (helium) but the nucleus stays in the same place (no recoil).
What is the probability that the atom remains in the ground state? What is the probability
that it goes into each of the n = 2 states 2lm ?
2. A neutral boron atom has a total angular momentum l = 1 and spin s = ½. In the absence of a
magnetic field, the lowest energy states might be listed as l , s, j, m j  1, 12 , j, m j , with the
j  32 state having higher energy. The atom is placed in a region of space where a magnetic
field is being turned on in the +z direction. At first, the spin-orbit coupling dominates, but at
late times the magnetic interactions dominate.
(a) Which of the nine operators L, S and J will commute with the Hamiltonian at all times?
Note that the state must remain an eigenstate of this operator at all times.
(b) At strong magnetic fields, the states are dominated by the magnetic field. The eigenstates
are approximately l , s, ml , ms  1, 12 , ml , ms . For each possible value of m j  ml  ms ,
XV. Time-Dependent Methods
276
© 2016, Eric D. Carlson
deduce which state has the lower energy. Atoms in strong magnetic fields are discussed
in chapter 9, section E.
(c) If we start with a particular value of l , s, j, m j (six cases), calculate which states
l , s, ml , ms it might evolve into, assuming the magnetic field increases (i) adiabatically
(slowly) or (ii) suddenly. When relevant, give the corresponding probabilities. The
relevant Clebsch-Gordan coefficients are given in Eq. (8.17).
3. A particle of mass m is initially in the ground state of a harmonic oscillator with frequency .
 t
At time t = 0, a perturbation is suddenly turned on of the form W  t   AXe . At late times (
t   ), the quantum state is measured again.
(a) Calculate, to second order in A, the amplitude
Sn 0
that it ends up in the state n , for all
n (most of them will be zero).
(b) Calculate, to at least second order, the probability that it ends up in the state n . Check
that the sum of the probabilities is 1, to second order in A.
4. A particle of mass m is in the ground state 1 of an infinite square well with allowed region
0  X  a . To this potential is added a harmonic perturbation W  t   AX cos t  , where A
is small.
(a) Calculate the transition rate  1  n  for a transition to another level. Don’t let the
presence of a delta function bother you. What angular frequency  is necessary to make
the transition occur to level n = 2?
(b) Now, instead of keeping a constant frequency, the frequency is tuned continuously, so
that at t = 0 the frequency is 0, and it rises linearly so that at t = T it has increased to the
value  T   2 2 ma 2 . The tuning is so slow that at any given time, we may treat it as
a harmonic source. Argue that only the n = 2 state can become populated (to leading
order in A). Calculate the probability of a transition using P 1  2    1  2 dt .
T
0
5. A hydrogen atom is in the 1s ground state while being bathed in light of sufficient frequency
to excite it to the n = 2 states. The light is traveling in the +z direction and has circular
polarization, ε  12  xˆ  iyˆ  .
(a) Calculate all relevant dipole moments rFI for final states 2lm .
(b) Find a formula for the rate at which the atom makes this transition.
(c) What is the wavelength required for this transition? Assume at this wavelength the
power is     100 W/m2 /nm . Find the rate at which the atom converts. (Note the
footnote on p. 270)
.
© 2016, Eric D. Carlson
277
XV. Time-Dependent Methods
6. A hydrogen atom is in interstellar space in the 1s state, but not in the true ground state (F =
0), but rather in the hyperfine excited state (F = 1), specifically in the state
I  n, l , j, F , mF  1,0, 12 ,1,0 . It is going to transition to the true ground state
F  n, l , j, F , mF  1,0, 12 ,0,0 via a magnetic dipole interaction.
(a) Write out the initial and final states in terms of the explicit spin states of the electron and
proton ,  . Find all non-zero components of the matrix F S I , where S is the
electron spin operator.
(b) Show that the rate for this transition for a wave going in a specific direction with a
definite polarization is given by   4 2 m2 2
k  ε   SFI   EF  EI    .
2
(c) Show that for a wave going in a random direction with random polarization, this
2
simplifies to   I  F   43  2 m2c2 S FI  E f  Ei   .


(d) For low frequencies, the cosmic microwave background intensity is
   kBT 2  2c2 where kB is Boltzman’s constant and T is the temperature.
Integrate the flipping rate over frequency. Find the mean time -1 for a hydrogen atom to
reverse itself in a background temperature T = 2.73 K for FI  2 1.420 GHz  .
7. A spin ½ particle of mass m lies in a one-dimensional spin-dependant potential
H 0  P 2 2m  12 m 2 X 2   . The potential only affects particles in a spin-up state.
(a) Find the discrete energy eigenstates for spin-up ( ,i ) and the continuum energy
eigenstates for spin-down ( ,  ). Also, identify their energies.
(b) At t = 0, a spin-dependant perturbation of the form V   x , where  x is a Pauli matrix,
is turned on. Calculate the rate  at which the spin-up ground state “decays” to a
continuum state.
XV. Time-Dependent Methods
278
© 2016, Eric D. Carlson
XVI. The Dirac Equation
We turn our attention to a discussion of a relativistic theory of the electron, the Dirac
Equation. Although the approach we use is now considered obsolete, it does provide important
insights into relativistic quantum mechanics, and ultimately it was an important step on the way
to a modern theory of particle physics.
As a first step, consider the procedure we used to produce the free particle Hamiltonian. We
started with the non-relativistic formula for energy, namely, E  p2 2m , multiplied by a wave
function   r,t  and then made the substitutions E  i  t and promoted p to an operator,
p  P  i  to produce the free particle Schrödinger equation,
i

1 2

P .
t
2m
Now, the corresponding relativistic equation for energy is E 2  c2p2  m2c4 . It is then easy to
derive a corresponding Schrödinger-like equation by the same prescription:
 
2 2
2 4
2 2 2
2 4
i
   c P   m c   c    m c  ,

t


2
(16.1)
the Klein-Gordon equation. The problem with Eq. (16.1) is that it is second order in time. As
such, to predict the wave function   r,t  at arbitrary time, we would need to know not just the
initial wave-function   r, 0  , but also its first derivative   r, 0  , contradicting the first
postulate of quantum mechanics. We need to find a way to convert Eq. (16.1) into a first order
differential equation for the wave function.
A. The Dirac Equation
Our goal is to replace Eq. (16.1) with an equation that is first order in time. For example, if
you start instead with the equation E  c 2p2  m2c 4 , one would obtain
i

  c 2 P 2  m2 c 4  .
t
The problem is, it is difficult to see exactly what the square root on the right side of this equation
means. The momentum operator P is an operator. If one were so foolish as to replace the square
root with something naïve, like cP  mc2 , not only would there be cross-terms, but the
expression doesn’t even make sense, since P is a vector while mc2 is not.
Dirac’s brilliant idea was to include matrices, so that the square root becomes an expression
like cα  P  mc2  where α is a set of three Hermitian matrices and  is one more, collectively
known as Dirac matrices, carefully selected so that
 cα  P  mc  
2
© 2016, Eric D. Carlson
2
 c 2 P 2  m2 c 4 .
279
(16.2)
XVI. The Dirac Equation
If you expand out the left side of Eq. (16.2), you will obtain
c2  x2 Px2   y2 Py2   z2 Pz2   ax y   y x  Px Py   ax z   z x  Px Pz   ay z   z y  Py Pz 
mc3  x    x  Px   y    y  Py   z    z  Pz   m2c 4  2  c 2 P2  m2c 4
This can be satisfied provided you assure that all four matrices anticommute with each other,
and have square equal to the identity matrix, so that we have
i j   ji  2 ij , i    i  0,  2  1.
(16.3)
Dirac’s (free) equation then, in terms of these matrices, is given by
i

   cα  P  mc2     ic α   mc 2 
t
(16.4)
The free Dirac Hamiltonian is just
H  cα  P  mc2   ic α   mc2 
(16.5)
Of course, it remains to demonstrate that there exist four matrices that satisfy Eqs. (16.3).
Dirac was able to demonstrate that the minimum size for such matrices was 4  4. For example,
one way of writing them, known as the chiral representation, is
σ 0 
 0 1 
α
and   

,
 0 σ 
 1 0 
(16.6)
where the σ ’s are Pauli matrices, given explicitly by Eqs. (7.17). Note in particular that the 0’s
and 1’s in Eqs. (16.6) stand for 2  2 matrices. Although it is occasionally useful to have explicit
forms such as Eqs. (16.6), it is far more often useful to avoide writing them explicitly, and use
the anti-commutation relation Eqs. (16.3) whenever possible.
There are other ways of writing the Dirac matrices. Let U be any 4  4 unitary matrix, and
define
  U , α  UαU † , and    U U † .
Then by multiplying Eq. (16.4) by U on the left and inserting factors of 1  U †U appropriately, it
is easy to show it is equivalent to
i

  cα  P  mc2   .
t
This is obviously the Dirac Eq. (16.4), but with α and  redefined. The new matrices α  and
  will still be Hermitian and still satisfy the relations Eqs. (16.3). In particular, for
U
1  1 1 

,
2 1 1 
the Dirac matrices will take the form of the Dirac representation, given by
0 σ
1 0 
α  
and    

.
σ 0
 0 1 
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280
(16.7)
© 2016, Eric D. Carlson
You can work in any representation you prefer. We will stick with the Chiral representation
because it is slightly simpler for our purposes.
B. Solving the Free Dirac Equation
We would like to solve the time-independent Dirac equation, which is simply
E  H   cα  P  mc 2   .
We expect to find plane wave solutions of the form   u  k  eik r , where u is independent of
r. Substituting this expression in, we find
 cα  P  mc   u k  e
2
ik r
 c σ  k mc 2 
ik r
ik r

 u  k  e  Eu  k  e .
2
c σ  k 
 mc
(16.8)
We are thus trying to solve this eigenvalue equation. As a first step, we consider the 2  2 matrix
σ  k , which when squared yields  σ  k   k 2  k 2 , and therefore has eigenvalues  k . Let us
2
define the two eigenstates of this matrix  , so that  σ  k     k . In explicit form, if you
need them, one can show that if k is in the spherical direction denoted by  and , we have
 cos  12   
  sin  12   
  
i 
1
 and    cos  1   ei  .
2
 sin  2   e 


Then we will guess that we can find solutions of Eq. (16.8) of the form
 a 
u k      .
 b 
(16.9)
Substituting this into Eq. (16.8), we find
2
 a   c σ  k mc2   a    bmc  ac k   
E
.



2
c σ  k   b    amc2 bc k   
 b   mc


This yields the two equations

a E c k
mc 2


.
b
mc 2
E c k
(16.10)
Cross-multiplying leads to the equation E 2  c2 2k 2  m2c4 , exactly what we want. Up to
normalization, we can solve Eq. (16.10) for a and b, and then substitute into Eq. (16.9), yielding
two solutions for each value of k: 1
1
If you wish to use wave functions of amplitude 1, these should be divided by
proves to be inconvenient, though it is irrelevant for us.
© 2016, Eric D. Carlson
281
2E .
For technical reasons this
XVI. The Dirac Equation
 E  c k   ik r
e ,
 E c k  


 u  r   u  k  eik r  
(16.11)
where E  c 2 2 k 2  m2c 4 .
What are these two solutions? Obviously, they have the same momentum p  k . It turns
out that the spin operator in the Dirac equation is given by
σ 0
S  12 Σ, where Σ  
.
0 σ
(16.12)
If we measure the spin along the direction the particle is traveling we will find
 kˆ  σ 0 
kˆ  S    12 
   12   .
 0 kˆ  σ  


 
Thus it is in an eigenstate of spin in this direction. Of course, linear combinations of the two
solutions given by Eqn. (16.11) are also allowed. Hence the Dirac equation predicts two positive
energy solutions for every value of the wave number k corresponding to the two spin states.
The eigenvalue Eq. (16.9) involves a 4  4 matrix, and hence should have four eigenvectors,
not two. This suggests that we have somehow missed two solutions. Indeed, we have, and the
problem is they have negative energy. They work out to be
  E c k  ik r
e
,
  E  c k 


 v  r   v  k  eikr  
where E   c 2 2 k 2  m2c 4 , as can be verified by direct substitution into the time-independent
Dirac equation. It has momentum  k and spin along the direction k̂ of 12 (which makes it
look like we mislabeled these solutions). But what do these solutions mean?
Dirac came up with the following solution. Dirac reasoned that because these states had
negative energy, the ground state would be a state where these particles were present, not for just
one momentum p, but for all momenta. He reasoned that since electrons were fermions, they
satisfied the Pauli exclusion principle. Hence once you “filled” these states, there was no lower
energy state. He suggested that what we normally term the “vacuum” is not truly empty, but
rather is filled with electrons sitting in all the negative energy states. We cannot notice this “sea”
of states because it is just the normal state of affairs.
Does this mean these negative energy states are irrelevant? Not at all. Suppose we applied a
perturbation, say an electromagnetic pulse, of sufficient energy to take one of these negative
energy states and promote it to a positive energy state. We would instantly “see” an electron that
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282
© 2016, Eric D. Carlson
had not been there previously, as illustrated in
Fig. 16-1. We would also simultaneously “see” a
missing state from the negative energy sea, much as
a bubble underwater indicates an absence of water.
This “hole,” as he called it, would be perceived as
an absence of momentum  k , (or in other words,
a momentum k ), an absence of spin angular
momentum 12 in the direction of k̂ (spin  12 ),
E
p
an absence of energy  c 2 2 k 2  m2c 4 (or a
presence of enerby E  c 2 2 k 2  m2c 4 ) and an
absence of charge – e (or q  e ). In other words,
we would see a particle identical to the electron,
with the same mass and spin, except having the
opposite charge. Dirac assumed this particle was
the proton (and was puzzled why it didn't have the
same mass), but we now recognize it as a separate
particle, the positron, also known as the antielectron. Dirac’s arguments can be generalized to
other spin-1/2 particle, and, in fact, we now expect
every elementary particle to have an anti-particle,
and, indeed, all particles do apparently have antiparticles (though some particles, like the photon,
are their own anti-particle).
Figure 16-1: The Dirac equation has
both positive (blue curve) and negative
(red) energy solutions. Dirac assumed
all the negative energy states were
“filled” with an undetectable sea of
electrons (black dots). Sufficient
energy can move (green arrow) one of
these electrons from a negative energy
state to a positive energy state,
producing a visible hole (open red
circle) and an electron (blue dot).
C. Electromagnetic Interactions and the Hydrogen Atom
We now wish to include electromagnetic interactions in our Hamiltonian. This is an easy
matter: we simply change P to π  P  eA and then add the electrostatic potential energy eU
to yield
H  cα  P  eA  R, t   eU  R, t   mc 2  .
(16.13)
So that the Dirac equation becomes
i

  cα  i   eA  R, t    eU  R, t    mc 2 
t
(16.14)
Let’s try solving a particularly difficult and interesting problem in this case, namely,
hydrogen, with potential U  ke e R . Then our Hamiltonian Eq. (16.13) is
ke e2
 c
H  cα  P 
 mc 2   cα  P 
 mc 2  ,
R
R
© 2016, Eric D. Carlson
283
XVI. The Dirac Equation
where we have used the fine structure constant   ke e2 c (not to be confused with ) to
rewrite the potential term. We will be trying to find eigenstates of this equation, solutions of the
time-independent Schrödinger equation
 c


E   cα  P 
 mc 2   ,
r


 c

2
E
   cα  P  mc   .
r 

We now let the operator on the right act a second time on both sides to yield
 cα  P  mc    E  r c    cα  P  mc    .
2
2
2
On the right hand side, we can use Eq. (16.2) to simplify. On the left side, we note that if the
factor of cα  P  mc2  were allowed to act on the wave function, we would only get another
factor of  E   c r  . However, the derivative P can also act on the potential  c r , yielding
another term. The result is
 c
i c

 α  rˆ    c2P2  m2c4  .
E

2
r
r


2
2 2
We now notice something interesting: The only matrix in this whole equation is α , which is
block-diagonal. Therefore, if we write our wave function  as two pieces, each of which
themselves have two components, so that
 
  ,
 
then the equations for  and   completely decouple.1 Hence we can focus on one at a time. It
is not hard to show that they have the same energy eigenvalues, so we will focus exclusively on
 , for which our equation is
2 2
 2 2 cE

c
 2  2  i  σ  rˆ     c 2 P2  m2c 4   .
E 
r
r


(16.15)
Our problem is spherically symmetric, so we should probably switch to spherical
coordinates. With the help of Eqs. (A.31d) and (7.23), it is not hard to show that in spherical
coordinates,
P2  
2
L2
r


  2.
r r 2
r
2
Substituting this into Eq. (16.15) and rearranging a bit, we have
1
This is why we chose the chiral representation, the decoupling is far less obvious in the Dirac representation.
XVI. The Dirac Equation
284
© 2016, Eric D. Carlson
E
2
 m2c4   
2 2
c 2
2 cE
c
r







2
r r
r
r2
2 2
2
L2   2  i σ  rˆ  .
(16.16)
We note that all the angular and spin-dependence of this expression is contained in the last
factor, which we define as A, so that
A
L   2  iσ  rˆ .
2 2
It makes sense to try to factor  into an angular and radial part, just like we did for nonrelativistic Hydrogen. We let
  r   R  r    ,   ,
where R  r  is an ordinary function, but   ,   has two components, so it has both the angular
and spin dependence. Substituting this form into Eq. (16.16), we have
c 2
2 cE
R  r    ,  
rR  r     ,   
2 
r r
r
2 2
c
 2 R  r  A  ,   .
r
 E 2  m2c4  R  r    ,   
2 2
(16.17)
We’d like our angular functions   ,   to be eigenstates of A. This is the equivalent of finding
the spherical harmonics, with the added complication of spin, though our primary interest here is
only finding the eigenvalues of A.
To find the eigenvalues of A, consider first that the original Hamiltonian must be rotationally
invariant, and hence commutes with all components of J, and specifically with J2 and Jz. It
therefore makes sense to try to work with eigenstates of these two operators. We already know
that the angular and spin states of a spherically symmetric problem can be written in the basis
l , j, m j (we have suppressed the spin label s  12 ) . Since J2 and Jz commute with the
Hamiltonian, it is hardly surprising that they commute with A, and therefore A can only connect
states with the same values of j and mj. Thus the only non-zero matrix elements of A in this basis
are
l , j, m j A l , j, m j  l , j, m j

2
L2   2  i σ  rˆ  l , j, m j .
(16.18)
Since j  l  12 , there are only two possible values for l, and hence finding the eigenstates and
eigenvalues of A is reduced to the problem of diagonalizing a 2  2 matrix.
The first two terms in Eq. (16.18) are obviously diagonal in this basis, so we see that
l , j, m j

2
L2   2  l , j, m j   ll '  l 2  l   2  .
The last term is more problematic. If you rewrite σ  rˆ  σ  r r , and then recall that the operator
R (which becomes r) connects only states which differ by one value of l, you will realize that the
on-diagonal components of this term must vanish, so
l , j, m j σ  rˆ l , j, m j  0 .
© 2016, Eric D. Carlson
285
XVI. The Dirac Equation
On the other hand, we know that  σ  rˆ   1 , the identity matrix. It follows that
2
1  l , j, m j  σ  rˆ  σ  rˆ  l , j, m j 


l ' j  12
l , j, m j σ  rˆ l , j, m j
2

l ' j  12
l , j, m j σ  rˆ l , j, m j l , j, m j σ  rˆ l , j, m j
,
where we have inserted a complete set of intermediate states. However, there are only two terms
in the sum, and one of them is zero, so the other one must be a number of magnitude one.1 In
other words,
j  12 , j, m j σ  rˆ j  12 , j, m j  j  12 , j, m j σ  rˆ j  12 , j, m j
*
 , where   1 .
2
We now know every matrix element of A; in the basis where we put l  j  12 first and
l  j  12 second, A takes the form
  j  12  j  12    2
A  
i *


i
.
2
 j  12   j  32    
It is a straightforward matter to find the eigenvalues of this matrix, which turn out to be
 j  12 
2
 2 
 j  12 
2
 2 .
It is very helpful to define the quantity
j 
 j  12 
2
 2 .
(16.19)
Then our eigenvalues are  j2   j . Hence for every value of j and mj we find two solutions
 j ,j  ,   satisfying
m
 L2
 m
m
m
A j ,j  ,     2   2  i σ  rˆ   j ,j  ,      j2   j   j ,j  ,  .


Substituting this into Eq. (16.17), we see that the angular portion cancels out, and we are left
with two very similar radial equations,
2
 E  m c  R  r    rc drd 2 rR  r   2 r cE R  r  
2 2
2
2 4
c   j2   j 
2 2
r2
R r  .
Solving this will prove to be quick, because of its remarkable similarity to the standard
hydrogen atom. To make the comparison as obvious as possible, first divide by 2E.
2
2 2
2 2
E 2  m2c 4
c d2
 c
c j  j
R r   
rR
r

R
r

R r  .







2E
2 Er dr 2 
r
2E
r2
1
(16.20)
Actually, it is one, but we won’t need this fact.
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286
© 2016, Eric D. Carlson
Compare this to the radial equation for hydrogen, Eq. (7.41), which we reproduce here, rewritten
slightly:
ER  r   
2 2
d2
 c
l l
rR
r

R
r

R r  .






2 

2 r dr
r
2 r 2
2
(16.21)
Comparing Eqs. (16.20) and (16.21), we see that the latter can be transformed into the former if
we identify
E
E 2  m2 c 4
E
,   2 , l 2  l   j2   j .
2E
c
The last transformation must be treated as two separate cases, but it is pretty easy to see that the
two cases are
l  j
or l   j  1.
The point is that we can now instantly use every equation we derived from hydrogen, we just
need to make the corresponding transformations. For example, our formula for the radial wave
function will take the form (similar to Eq. (7.50a))
 1
Rnl  r   er a  f p r p
p J
or Rnl  r   er a
 1


p
J
1
f pr p ,
(16.22)
where a is given by (similar to Eq. (7.42)):
2 2
E 2  m2 c 4
c
c

, or a 
.
2
2
4
2E
2 Ea
m c  E2
Eqs. (16.22) require a bit of discussion. Note that the lower limit on the sum is not an
integer, and therefore the powers p of r are non-integer as well. The upper limit  will therefore
also not be an integer, but it will differ from the lower limit by an integer; that is   j  k
,where k is a positive integer in the first case, and a non-negative integer in the second case.
The energies are given by an equation analogous to Eq. (7.51)
Ek 2e4
E 2  m2 c 4
E 2
  2e2 2  
,
2
2E
2 c
2 j  k 
2
E  mc2 1   2   j  k  


1 2
.
(16.23)
This, together with Eq. (16.19), gives us the energies. It should be noted that in Eq. (16.23), k
can be zero or a positive integer. When it is zero, we only have one case, but when k is positive,
m
we will have two distinct cases, corresponding to the two angular functions  j ,j  ,   .
If we are dealing with a hydrogen-like atom, rather than true hydrogen, the only change is to
substitute   Z . So the more general version of Eqs. (16.19) and (16.23) are
j 
© 2016, Eric D. Carlson
 j  12 
2
287
 Z 2 2 ,
(16.24a)
XVI. The Dirac Equation
2
E  mc2 1  Z 2 2   j  k  


1 2
.
(16.24b)
It is helpful to expand these in powers of the fine structure constant to understand what the
various choices correspond to. We find, to fourth order in  ,
E  mc2 
mc2 2 Z 2
2  j  12  k 
2

mc2 4 Z 4  3 j  12  k 


4 
 j  12  k   8 2 j  1 
.
Looking at these terms order by order, it is easy to identify them. The first term is the rest
energy of an electron. The second is the leading order binding energy. Indeed, since j is halfinteger, the combination j  12  k is an integer, which we name n, and the last term is simply the
first relativistic correction. We therefore have
E  mc 2 
mc 2 2 Z 2 mc 2 4 Z 4  3
n 

 

2
4
2n
n
 8 2 j 1 
,
(16.25)
with now the restriction that n  j  12 . Indeed, the same restriction occurs for non-relativistic
hydrogen. When n  j  12 , there are two different possibilities with exactly the same energy
(corresponding to l  j  12 ), while for n  j  12 , there is only one (since l  j  12 is no longer
allowed). The final term in Eq. (16.25) is simply the first relativistic correction. The jdependence is reflecting the conventional spin-orbit coupling, though other effects are included
as well. Note, as expected, the corrections grow as powers of  Z , which means that larger Z’s
will have much larger effects. In most practical situations, such atoms will in fact be neutral (or
nearly neutral), and hence they won’t be hydrogen-like at all.
It is interesting to note that the energy Eq. (16.24b) depends only on j and k, or in
conventional notation, j and n. This means that, for example, the 2s1/2 and 2p1/2 levels are still
exactly degenerate, if we ignore hyperfine interactions. This is surprising, because there is no
evident symmetry protecting them. There are interactions that split this apparent degeneracy,
but they come about by interactions with the quantized electromagnetic field, the subject of the
next chapter.
Despite the successes of the Dirac equation, the awkward way it deals with anti-particles and
other problems ultimately led to its abandonment as an approach to developing a fully relativistic
theory of particles. Due to its complications, we will henceforth ignore it, and use nonrelativistic electrons whenever possible. But electromagnetic waves, which always move at the
speed of light, must be quantized in a fully relativistic formalism.
Problems for Chapter 16
1. Starting with the free Dirac Eqn. (16.4), multiply it on the left by  † . Then take the
Hermitian conjugate of Eqn. (16.4) and multiply it on the right by  and subtract them. If
we define the probability density as   † , show that we can obtain a conservation
equatioin of the form  t   j  0 for a suitable choice of the probability current j. What
is j? Now repeat with the electromagnetic version of the Dirac Eqn. (16.14).
XVI. The Dirac Equation
288
© 2016, Eric D. Carlson
2. In the class notes, we claimed that the spin was defined by S  12 Σ , Eq. (16.17). To make
sure this is plausible:
a) Demonstrate that S satisfies proper commutations relations  Si , S j    k i  ijk Sk .
b) Work out the six commutators  L, H  and S, H  for the free Dirac Hamiltonian.
c) Show that  J, H   0, where J  L  S .
© 2016, Eric D. Carlson
289
XVI. The Dirac Equation
XVII. Quantizing Electromagnetic Fields
When Max Planck first “discovered” quantum mechanics, it was by conjecturing that
electromagnetic waves come in quantized packets of energy E   . And yet here we are,
nearly 300 pages later, and we have yet to quantize the electromagnetic field. This is because
electromagnetism presents some special challenges. Fortunately, these challenges have been
generally overcome, and we have a fully relativistic theory of the electromagnetic field.
Ultimately, we will want to interact electromagnetic waves with electrons or other particles, but
for now, let us try to quantize the pure EM field.
A. Gauge Choice and Energy
One complication that is especially annoying about electromagnetism is that it will turn out
to be necessary to work with the vector and scalar potential, A  r,t  and U  r, t  , not the electric
and magnetic fields E and B directly. As we discussed in chapter 9, we must make a gauge
choice; that is, we have to decide which of several possible specific forms for A and U we will
use to describe the electric and magnetic fields. A gauge transformation is given by Eqs. (9.11),
repeated here:
U   r, t   U  r, t  

  r, t  and A  r, t   A  r, t     r, t  .
t
(17.1)
Though several different gauges are useful for different purposes, we will choose, in this case, to
select the Coulomb gauge, defined by the constraint
  A  r, t   0 .
Can such a gauge choice be made? The answer is yes. Suppose, for example, that this were not
the case; that is, suppose   A  0 . Then define   r,t  to be the solution of the equation
2   r, t     A  r, t 
(17.2)
We know, in general, that such a solution always exists; indeed, it isn’t hard to show that it is
explicitly given by
  r, t   
  A  r, t  3
d r .
4 r  r
Then it follows from Eqs. (17.1) and (17.2) that   A  r, t   0 .
In terms of the vector and scalar potential, the electric and magnetic fields are
given by Eqs. (9.6) and (9.7), repeated here
E  r, t   A  r, t   U  r, t  , B  r, t    A  r, t  .
XVII. Quantizing Electromagnetic Fields
290
(17.3)
© 2016, Eric D. Carlson
The first of Maxwell’s equations (Coulomb’s Law) states that1
  r, t    0 E  r, t    0  A  r, t   U  r, t    02U  r, t  .
Note there are no time derivatives in this equation, so the scalar potential is determined, in this
gauge, by the instantaneous charge density
U  r, t   
  r, t  3
d r ,
4 0 r  r
the same formula we have been using all along. All of the dynamics, and interesting quantum
mechanical effects, will come from the vector potential A. In the absence of charges, the scalar
potential will be trivial, U  r, t   0 .
We now want to write the classical Hamiltonian, which is just the energy formula.
The energy density of the electromagnetic fields is2
u  r, t   12  0E2  r, t   12 01B2  r, t   12  0 E2  r, t   c 2B2 r, t  .
(17.4)
so the energy of the fields is given by


E  12  0  E2  r, t   c 2B2  r, t  d 3r  12  0  A 2  r, t   c 2  A  r, t  d 3r .
2
(17.5)
Though it is not obvious, Eq. (17.5) is nothing more than a formula for an infinite number of
coupled harmonic oscillators. Think of A  r,t  as an infinite number of independently varying
“positions”; then the time derivative term is just like a kinetic energy term for a particle. The curl
term is a derivative, which we can think about as the difference between A  r,t  at “adjacent”
points, and hence this term is a coupling term.
B. Fourier Modes and Polarization Vectors
To make our work easier, we proceed to work in finite volume. We will imagine that the
universe has volume V = L3 , possessing periodic boundary conditions in all three directions.
Therefore, any function   r  will have the property
  x  L, y, z     x, y  L, z     x, y, z  L     x, y, z  .
Any such function can written as a sum of Fourier modes,
A  r, t    Ak  t  eik r .
(17.6)
k
1
In previous chapters, we have avoided using
 0 , and preferred Coulomb’s constant ke.
The two are related by
4 ke 0  1 .
2
See, for example, Jackson, Classical Electrodynamics, third edition (Wiley, 1999) equations (4.89), p. 166 and
(5.148), p. 213
© 2016, Eric D. Carlson
291
XVII. Quantizing Electromagnetic Fields
Because of the periodic boundary conditions, our k-values in the sum are not continuous, but
discrete, so that
k
2
2
n
 nx , ny , nz  .
L
L
where n is a triplet of integers. The mode functions Ak  t  are further restricted by the demand
that they must lead to a real vector potential A  r,t  , which requires that A k  t   A*k  t  . Also,
because we are working in Coulomb gauge, we must have   A  0 , which implies
k  Ak  t   0 .
Substituting Eq. (17.6) into Eq. (17.5), we see that the energy takes the form
2
2
 

ik r 
2 
ik r   3
E   0    Ak  t  e   c  ik  Ak  t  e   d r

k
 
  k
1
2


 12  0   Ak  t   Ak   t   i 2c 2 k  Ak  t   k   Ak   t  eik r ik r d 3r .
k
k

The space integral is easy to do,  eikr ik r d 3r  V  k ,k , which yields


E  12  0V  Ak  t   A k  t   c 2k 2 Ak  t   A k  t  ,
(17.7)
k
where we have used k  Ak  t   0 to simplify the dotted cross products. We now need to take
advantage of the restriction A k  t   A*k  t  . This allows us to simplify the terms in Eq. (17.7).
It is also apparent that the sums in Eq. (17.7) contains pairs of identical terms. To avoid
confusion later, we will divide the k values in half, treating half of them as positive; for example,
we can define k > 0 as those values for which the first non-zero component of k is positive.1
Then Eq. (17.7) becomes


E   0V  Ak  t   A*k  t   c 2k 2 Ak  t   A*k  t  .
k 0
(17.8)
We have successfully decoupled the different Fourier modes, but we still have vector
quantities to deal with. Because of the restriction k  Ak  t   0 , Ak  t  has only two
independent components. We define two orthonormal polarization vectors εk where   1, 2 ,
chosen so that
k  εk  0 and ε*k  εk   .
For example, if we let k be defined by the spherical coordinate angles  and , so that
k  k  sin  cos  ,sin  sin  , cos   , we could pick
εk1   cos  cos  , cos  sin  ,  sin   , εk 2    sin  , cos  , 0  .
1
This might leave you concerned about the special case k = 0; however, it is easily demonstrated that a gauge
change can always eliminate such a space-independent mode.
XVII. Quantizing Electromagnetic Fields
292
© 2016, Eric D. Carlson
We then write our modes Ak  t  as
Ak  t  
 A  t  ε  .
 1,2
k
(17.9)
k
Substituting this into Eq. (17.8), we have
E   0V   Ak  t   Ak*  t   c 2k 2 Ak  t  Ak*  t   .
(17.10)
k 0 
We are now prepared to quantize our theory.
C. Quantizing the Electromagnetic Fields
Eq. (17.10) is nothing more than a sum of complex harmonic oscillators. We therefore can
take advantage of all of the work of section 5F and quickly quantize the theory. The classical
theory we quantized in section 5F had an energy given by Eq. (5.26), E  m zz*   2 zz* .


Comparison with Eq. (17.10) tells us that we can take over all our old formulas if we make the
following associations:
m   0V ,   k  ck , z  Ak .
We then simply copy equations like Eq. (5.30) without doing any additional work:
H   k  a†,k , a,k ,  a†,k , a,k ,  1 .
k 0 
Our notation is getting a bit unwieldy. At the moment, there are three indices on a, one
denoting which of the two annihilation operators we are talking about for our complex variables
Ak , one denoting the vector k (which is restricted to be positive), and one denoting the
polarization. We can combine the first two into a single index k which is no longer restricted to
be positive by defining a ,k ,  ak , and a,k ,  ak , . The commutation relations for these
operators are
ak , , ak†,    kk  , ak , , ak,    ak†, , ak†,    0 .
(17.11)
Then, splitting the constant term in half, we can rewrite the Hamiltonian in the more compact
form
H   k  ak† ak  12  where k  ck .
(17.12)
k ,
Note we have abbreviated ak , as ak , dropping the comma. The sum is no longer restricted to
positive k-values. We call ak , and ak† , annihilation and creation operators respectively.
The analog of Eqs (5.31) can then be written in terms of creation and annihilation operators
Ak 
© 2016, Eric D. Carlson
2 0V k
a
k
 a†k ,  , Ak  i
293
k †
 ak ,  ak  .
2 0V
XVII. Quantizing Electromagnetic Fields
This allows us to write our vector potential with the help of Eqs. (17.6) and (17.9):
A r   
k ,
2 0V k
a
k
 a†k ,  εk eikr .
Our analysis has not produced a manifestly Hermitian expression for A  r  , which is
surprising, since its classical analog is always real. The complication has to do with the way the
sums were temporarily restricted to k  0 , with the restriction A k  t   A*k  t  . The effect is that
we are forced to choose our polarization vectors such that ε k ,  ε*k . With this restriction, it is
more useful to rewrite the final term by taking k  k , yielding the manifestly Hermitian
expression
A r   
2 0V k
k ,
ε
k
eikr ak  ε*k eikr ak†  .
(17.13)
We can then take the curl of this expression to get the magnetic field. Furthermore, we can find
an expression for the electric field by using our expression for Ak for the time derivatives.
Skipping many steps, the results are
E r   
k ,
B r   
k ,
k
i  εk eikr ak  ε*k eikr ak†  ,
2 0V
2 0V k
(17.14a)
ik   εk eikr ak  ε*k eikr ak†  .
(17.14b)
In most circumstances, we will work with real polarizations.
Note that the time dependences in Eqs. (17.13) and (17.14) have been dropped. This is
because A, E, and B are now all to be interpreted as operators, which act on some sort of state
vector. In particular, E  r  and B  r  are quantities that can be measured experimentally, whose
values will be uncertain, due to the spread in whatever state vector you might be interested in
(A(r) is not measurable, since it is not gauge invariant), just as the position x of a particle gets
replaced by an operator X when we quantize the harmonic oscillator. The quantum states 
are, however, far more complicated, and therefore require some additional comment.
D. Eigenstates of the Electromagnetic Fields
We focus for the moment on the Hamiltonian, given by Eq. (17.12), which is clearly just a
sum of harmonic oscillators. We start by guessing that the ground state will be the state 0 , the
state annihilated by all the lowering operators, so ak 0  0 . This state has energy
E0   12 k   , thanks to the infinite number of terms in the sum. One might hope that this
might have something to do with the finite volume, and that in the limit V   this infinity
might be spread throughout all of space, thereby rendering a relatively innocuous finite energy
XVII. Quantizing Electromagnetic Fields
294
© 2016, Eric D. Carlson
density, but a more careful analysis will show that in this limit, the energy density is infinite.
How can we deal with this infinite energy?
The answer is deceptively simple. As always, adding a constant to the Hamiltonian has no
effect on the physics. We always measure differences in energy, not absolute energy values, and
hence we can simply rewrite Eq. (17.12) as
H   k ak† ak
k ,
where k  ck .
(17.15)
The ground state now has energy zero.
An arbitrary eigenstate of H could be listed by simply listing the eigenstates of all the number
operators Nk  ak† ak , but since we have an infinite number of such operators, such a listing
would become unwieldy. We therefore list only those states which have non-zero values for
these operators, so we denote our states as
n1 , k1 , 1; n2 , k 2 ,  2 ; ; nM , k M ,  M ,
(17.16)
with energy
E
 n11  n22 
 nM M  .
We will describe such a quantum state as having n1 photons of momentum k1 and polarization
1 , n2 photons of momentum k 2 and polarization  2 , etc. These states can be built up from the
ground state (or vacuum state) in the usual way, so that
n1 , k1 , 1; n2 , k 2 ,  2 ;
; nM , k M ,  M 

1
ak†11
n1 !n2 ! nM !
 a 
n1
†
k 2 2
n2
a
†
kM M

nM
0 .
The order of the triplets in Eq. (17.16) is irrelevant. Creation and annihilation operators act as
usual on the states Eq. (17.16):
ak n, k ,  ;
 n n  1, k ,  ;
ak† n, k ,  ;
 n  1 n  1, k ,  ;
,
.
We will sometimes abbreviate our states still further. For example, if we know we will not be
discussing more than one photon at a time, we might write the state as k .
E. Momentum of Photons
Classically, a photon of energy E   should have a momentum p  E c  k . Will this
work quantum mechanically as well? In electromagnetism, the momentum density of the
electromagnetic field is given by1 g   0E  B . We need to integrate this over our volume V, so
that Pem   0  E  B d 3r , and using our explicit form of the electric and magnetic fields,
Eqs. (17.14):
1
Jackson, Classical Electrodynamics, third edition (Wiley, 199) equation (6.123), p. 262.
© 2016, Eric D. Carlson
295
XVII. Quantizing Electromagnetic Fields


i2
Pem 
k k   d 3r  eikr εk ak  eikr ε*k ak†   k    eik r εk  ak   eik r ε*k  ak†  

2V k k 

 
 12
k
k 

k k  ak εk   k ,k  k   εk   ak    k ,k  k   ε*k   ak†  

ak† ε*k   k ,k  k   εk   ak    k ,k  k   ε*k   ak†   .
We use the Kronecker delta functions to rewrite this sum, also using ε k ,   ε*k  , k  εk ,  0 ,
and Eq. (A.7b) to expand the triple cross products, to yield
Pem  12
ε    k  ε   a  a


 
k

1
2

1
2
*
k 
k

k  a  a

 
k


 k  a  a
k
k
 k ,
 k , 
k
 k , 
 ak ak†    ε*k   k  εk    ak† ak   ak† a†k ,  
 ak ak†   ak† ak   ak† a†k ,  
 ak ak†  ak† ak  ak† a†k ,  .
k
For each of the first terms kak ak , , there will be another term kak ak , which exactly
cancels it. The same applies to the last term. The middle two terms can be combined with the
help of Eqs. (17.11) to yield
Pem  12
 k 1  2a  a  .
†
k
k
k
The term that is just a sum of k will now cancel between the positive and negative allowed
values of k, and the remaining term simplifies to
Pem   kak† ak ,
k
which we simply interpret as photons having momentum k , as expected.
F. Taking the Infinite Volume Limit
We will commonly want to take the limit where the volume of the universe is increased to
infinity. Generally, our strategy will be to keep the volume finite as long as possible, and then
apply the limit only when necessary. Two formulas make this easy.
Consider first the following integral, in both the finite and infinite volume limit.
V  k ,k 
if V   ,

3
ik r ik r
d
r
e


3

3
 2    k  k   if V   .
This leads to our first rule: We want the former expression to turn into the latter as V increases to
infinity, or in other words,
lim V k ,k    2   3  k  k .
3
V 
XVII. Quantizing Electromagnetic Fields
296
(17.17)
© 2016, Eric D. Carlson
The other expression comes from considering carefully any sum over momentum states.
Consider first a one-dimensional sum of k-values. In 1D, in the finite volume limit, k can only
take on the discrete values k  nk where n is an integer, and k  2 L . In the large size
limit, this spacing becomes very small. Indeed, the one-dimensional integral
 f  k  dk is
defined more or less as
2
L L
 f  k  dk  lim k  f  nk   lim
k 0
n
 f k  .
k
In one dimension, therefore, we see that the appropriate limit can be written as
dk
1

lim   f  k    
f k  .
L  L
2
 k

In three dimensions, this generalizes to
d 3k
1

lim
 f k     2 3 f k  .
V  V
 k
(17.18)
Indeed, Eqs. (17.17) and (17.18) are flip sides of each other. If you take Eq. (17.17) and
substitute it into Eq. (17.18), both sides become simply one.
G. The Nature of the Vacuum
Consider an experiment where we measure the electric and magnetic field of empty space.
Surely, we would expect the result to be zero. However, in quantum mechanics, things can get
more complicated. What would be the expectation value for these fields in the vacuum? Could
they fluctuate from zero? To answer these questions, we would need the expectation values of
the fields E  r  and B  r  and their squares E2  r  and B2  r  . We start by computing
E  r  0 and B  r  0 , which work out to
k * ikr
εk e
1, k ,  ,
2 0V
E  r  0  i 
k ,
B  r  0  i 
2 0V k
k ,
k  ε*k eik r 1, k ,  .
We then immediately find 0 E  r  0  0 B  r  0  0 . In contrast, we have
0 E2  r  0  E  r  0
2


2 V  
0

© 2016, Eric D. Carlson
2 0V
k
k 
k
k 
c
 0V
kk   εk  ε*k    eik r ik r 1, k ,  1, k ,  
k 
k
297
c
0
d 3k
  2 
3
k,
(17.19a)
XVII. Quantizing Electromagnetic Fields
0 B r  0  B r  0
2
2


2 V  
k,
0

2 0V

k,
k  εk
k
eikr ikr
kk 
k , 
2

k2
2 0V
 
k,
k
k  εk   k  ε*k   1, k,  1, k,  

k 
 0Vc
k
d 3k
 0c   2 3
k,
(17.19b)
where at the last step, we went to the infinite volume limit. Surprisingly, even though the
average value of the electric field and magnetic fields are finite, the expectation values of the
squares are infinite, suggesting that the fields have large fluctuations. In retrospect, we probably
should have anticipated this, since these two terms contribute to the energy density, and we
already know that is infinite.
Experimentally, how come we don’t notice this infinity? One reason is that it is impossible
to measure the electric or magnetic field at exactly one point. Suppose that we use a probe of
finite size, which measures not E(r), but rather
E f  r    f  s  E  r  s  d 3s, where
 f s  d s  1,
3
and f(s) is concentrated in the neighborhood of zero.. Then we have
E f  r  0  i   f  s  d 3s eikr sε*k
k
k
1, k,  .
2 0V
Performing the s integral will then yield a Fourier transform of f, which yields a factor of 1 at
long wavelength, but it will suppress the short wavelength contribution. Since real probes are
always finite in size, this will reduce 0 E2f  r  0 and 0 B 2f  r  0 and produce a finite value
for the fluctuations. But it is always possible, at least in principle, to measure finite fluctuations
in the field due to finite sized probes. Without going into too many details, the infinity also goes
away if we average over time; the infinite contribution comes from very fast fluctuating fields,
and realistic probes can simply not measure fluctuations that fast.
H. The Casimir Effect
Before moving onwards, it is worth commenting again on the infinite energy density of the
vacuum, which we can obtain easily from Eqs. (17.19) substituted into Eq. (17.4) to yield
0 u r  0  c
d 3k
 2 
3
k.
(17.20)
As we stated before, we can argue that since the only thing that matters is differences in energies,
and this energy is there even in the vacuum, you can normally ignore this effect. There is,
however, one important effect which requires that we consider this more carefully. Eq. (17.20)
was obtained by considering the limit of infinite volume. What if we don’t have infinite
volume? Consider, for example, a parallel plate capacitor, consisting of two conductors very
closely spaced together. We will assume the plates are each of area A = L2, where L is very large
compared to the separation a, as illustrated in Fig. 17-1. The area is assumed to be so large that
it is effectively infinite, but the separation is small enough that the modes in this direction will
XVII. Quantizing Electromagnetic Fields
298
© 2016, Eric D. Carlson
take on only distinctly discrete values. Therefore the energy density
between the plates will not be given by Eq. (17.20). We must redo the
work of section B, and quantize subject to these restrictions.
Considerable work is required to figure out what modes are allowed
subject to our constraints. It is forbidden to have electric fields parallel to
a conducting surface, which suggests that we should in general have
A  x, y,0, t   A  x, y, a, t   0 . Since virtually all modes have some
A
component of A parallel to the conducting plates, this suggests we write
A  r, t    Ak  t  e
ik x x ik y y
sin  k z z  ,
a
(17.21)
k
The wave numbers are now restricted to be of the form
 2 nx 2 ny  nz 
k 
,
,
,
L
a 
 L
(17.22)
Figure 17-1: A
parallel plate
capacitor
consisting of two
conducting
plates of area A
separated by a
small distance a.
where nx and ny are arbitrary integers, but nz must be a positive integer.
However, there is one other type of mode that we have missed in
Eq. (17.21), which occurs because if k is parallel to the capacitor plates,
we can choose our polarization ε  zˆ , perpendicular to the capacitor plates. This leads to
additional terms of the form Ak  t  zˆ e
ik x x ik y y
, where
kz  0 .
Hence each value of k given in
Eq. (17.22) with positive nz will have two possible polarizations, but there will be a single
additional polarization with nz = 0.
Now, our Hamiltonian will still be given by Eq. (17.12), and therefore the ground state
energy will be E0  a   12 ck , k . In the usual manner, we can turn the kx and ky sums into
integrals by taking the limit L   , but the kz and polarization sum must be dealt with explicitly
and we find
E0  a   L2  
dkx dk y
k z ,
 2 
2
1
2
ck .
Not surprisingly, this number is infinite. However, we are interested in the difference between
this value and the empty space vacuum value, which can be found by multiplying the energy
density Eq. (17.20) by the volume L2a, so we have
E  E0  a   E0 
1
2
cL2  
dk x dk y
k z ,
 2 
2
k  L2 a c 
d 3k
 2 
3
k.
(17.23)
Now, we have a problem in that both expressions are infinite, and we don’t know how to
subtract them. Fortunately, this infinity is not physical, because real conductors do not work at
arbitrarily high frequencies. Hence we should put in some sort of cutoff function f(k) that is one
for small k and vanishes for large k. The exact form used can be shown not to matter, so I will
 k
use the simple formula f  k   e , where  is some small number, governing when the
conductor effectively becomes transparent. Substituting this into Eq. (17.23), we have
© 2016, Eric D. Carlson
299
XVII. Quantizing Electromagnetic Fields
E  12 L2 c  
dk x dk y
k z ,
 2 
2
ke  k  L2 a c 
d 3k
 2 
3
ke  k .
2
2
2
On the first term, we switch first to cylindrical coordinates k  k x  k y and angle  , perform the
 integral, and then switch to the spherical coordinate k  k  k z . In the latter term, we switch
2
2
2
to spherical coordinates and perform all integrals. The result is
E 
L2 c 
L2 a c  2   k
L2 c  2   k
3L2 a c
 k
k
dk
ke

k
ke
dk

k
e
dk



4 kz , 0
2 2 0
4 kz , kz
 2 4
L2 c  kz2 kz
1    kz 3L2 a c

   e   2 4 .
2 kz ,  2  2  3 
It’s time to do the sums on kz. As argued above, kz   n a , where n is an integer. There will
be two polarizations when n is positive, and one when n = 0, so we have
L2 c  1    2 n2 2 n 2   n a  3L2a c
E 

 e
  
 2 4 .
2   3 n1   a2  2 a  3 
  
(17.24)
All the sums can be performed exactly using the identities

x
n 1
n

x
,
1 x

 nx
n 1
n

x
1  x 

n x
2
,
2
n 1
n

x  x2
1  x 
3
,
valid for x  1 . The first is just the geometric series, and the other two can be easily derived by
taking the derivative of the previous series with respect to x. Applying these equations to
Eq. (17.24), we see that
 2  a  e2 a 
L2 c    e
2 e a
E 

2  a 2  1  e a 3
a 2 1  e

a


2e a
 3 1  e
a



1  3L2 a c

,
 3   2 4

Now, the actual frequencies at which the cutoff occurs tend to correspond to wavelengths
much shorter than the experimental separation we can achieve, so  a . We therefore expand
E in this limit. Rewrite our expression in terms of w   2a .
E 
 2 L2 c  cosh w
1
cosh w
3 
 2
 3
 4 .
3 
3
2
16a  w sinh w w sinh w w sinh w w 
We now expand in powers of w. We’ll let Maple do the work for us:
> series(cosh(w)/w/sinh(w)^3+1/w^2/sinh(w)^2+cosh(w)/w^3/sinh(w)
- 3/w^4,w,11);
E 
 2 L2 c 1 4 2 1 4
  45  315 w  315 w 
16a3
XVII. Quantizing Electromagnetic Fields
300

 2 L2 c 
1  O  w2  .
3 
720a
© 2016, Eric D. Carlson
We see that the in the limit w  0 the energy difference is finite and non-zero. Not
surprisingly, the energy is proportional to the area, so we actually have a formula for the energy
per unit area. Note that the result is negative, so there is less energy as the plates draw closer.
This means there is an attractive force between them, or rather, a force per unit area (pressure)
given by
P
1 d
2 c
1.30 mN/m2

E




.


4
L2 da
240a 4
 a m
This force is attractive, pulling the two plates of the capacitor towards each other. Though small,
it has been measured.
Problems for Chapter 17
1. In class, we quantized the free electromagnetic field. In homework, you will quantize the
free massive scalar field, with classical energy


E  12  d 3r  2  r, t   c 2   r, t    2 2  r, t 
2
This problem differs from the electromagnetic field in that: (i) there is no such thing as gauge
choice; (ii) the field   r,t  is not a vector field; it doesn’t have components, and (iii) there is
a new term
 2 2 , unlike anything you’ve seen before.
a) Write such a classical field   r,t  in terms of Fourier modes k  t  . What is the
relationship between k  t  and k  t  ?
b) Substitute your expression for   r,t  into the expression for E. Work in finite volume V
and do as many integrals and sums as possible.
c) Restrict the sum using only positive values of k. Argue that you now have a sum of
classical complex harmonic oscillators. What is the formula for k , the frequency for
each of these oscillators?
d) Reinterpret H as a Hamiltonian, and quantize the resulting theory. Find an expression for
the Hamiltonian in terms of creation and annihilation operators.
2. How do we create the classical analog of a plane wave quantum mechanically? Naively, you
simply use a large number of quanta.
a) Suppose the EM field is in the quantum state n, q, , where n is large. Find the
expectation value of the electric E  r  and magnetic fields B  r  for this quantum
state. For definiteness, choose q  qzˆ and ε q  xˆ
b) Now try a coherent state, given by
 c  e c
2
2
 c

n 0
n

n! n, q, , where c is an
arbitrary complex number. Once again, find the expectation value of the electric and
magnetic field. Coherent states were described in section 5B.
© 2016, Eric D. Carlson
301
XVII. Quantizing Electromagnetic Fields
3. Suppose we measure the instantaneous electric field using a probe of finite size, so that we
2 2
actually measure E f  r    E  r  s  f  s  d 3s , where f  s    3 2 a 3es a , where a is the
characteristic size of the probe. For the vacuum state, find the expectation value of E f  r 
and E2f  r  . You should take the infinite volume limit, and make sure your answer is
independent of V.
4. Define, for the electric and magnetic field, the annihilation and creation parts as
E  r   i 
k ,
B  r   i 
k ,
k
εk eikr ak ,
2 0V
2 0V k
E  r   i 
k
k  εk eikr ak , B  r   i 
k ,
k * ikr †
εk e ak ,
2 0V
2 0V k
k  ε*k eik r ak† .
It should be obvious that E  r   E  r   E  r  and B  r   B  r   B  r  .
(a) Define the normal-ordered energy density as


u  r   12  0 E2  r   2E  r   E  r   E2  r   c2 B2  r   2B r   B r   B2 r 
Prove that this normal-ordered energy density differs from the usual definition by a
constant, i.e., that the difference contains no operators (the constant will be infinite).
(b) Prove that the expectation value of this operator for the vacuum is zero.
(c) Consider the quantum state   38 0  13 2, q, ; i.e., a quantum superposition of the
vacuum and a two photon state with wave number q and polarization  . To keep things
simple, let the polarization εq be real. Work out the expectation value u  r  for this
quantum state.
(d) Sketch u  r  as a function of q  r . Note that it is sometimes negative (less energy than
the vacuum!). Find its integral over space, and check that it does, however, have total
energy positive.
XVII. Quantizing Electromagnetic Fields
302
© 2016, Eric D. Carlson
XVIII. Photons and Atoms
We have quantized the electromagnetic field, and we have discussed atoms as well in terms
of quantum mechanics. It is time to put our knowledge together so that we can gain an
understanding of how photons interact with matter. Our tool will be primarily time-dependent
perturbation theory, in which we divide the Hamiltonian H into two pieces, H  H0  W , where
H0 will be assumed to be solved, and W will be small. The rate at which an interaction occurs is
then given by Fermi’s golden rule, Eq. (15.35), and the transition matrix Eq. (15.34), given by
FI


WFmWmI
WFmWmnWnI
 WFI  lim 
 
 ,
 0
 m  EI  Em  i  m n  EI  En  i  EI  Em  i 

2
  I  F   2 1 FI   EF  EI  ,
(18.1a)
(18.1b)
where Wnm  n W m . Our first step will be to break the Hamiltonian up into a main part and
a perturbation, and find the eigenstates of the main part.
A. The Hamiltonian
The Hamiltonian we wish to consider is one involving both atoms and (quantized)
electromagnetic fields. The pure electromagnetic field energy Hem will be given by Eq. (17.5),
which, after quantization, becomes Eq. (17.15). This Hamiltonian was the focus of the previous
chapter. In addition, there will be the interaction of all the electrons in all of the atoms, etc.
With the help of Eq. (9.20), we see that the full Hamiltonian will be
N
2
e
 1

H     Pj  eA  R j   B  R j   S j   V  R1 ,
m

j 1  2m
, R N   H em .
where V contains all of the interactions of the electrons with each other, or with the nuclei or
background fields, etc., and we have approximated g = 2 for the electron. Indeed, in general we
might want to also include lesser effects in V, such as the spin-orbit coupling within an atom, but
what we want to exclude is any interaction with external electromagnetic fields, which are
explicitly shown. Keep in mind that A and B are no longer mere functions (as they were
previously) but now are full operators. Recall that we are working in Coulomb gauge, and
therefore the electrostatic potential is determined by the instantaneous charge distribution.
We now wish to divide H up into “large” and “small” pieces, which we do as follows:
H  H atom  H em  W 1  W  2 ,
where
N
1 2
Pj  V  R1 ,
j 1 2m
H atom  
© 2016, Eric D. Carlson
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, RN  ,
XVIII. Photons and Atoms
and the perturbations are1
W
e N 
e2 N 2
 2

   A  R j   Pj  B  R j   S j  and W 
 A R j  .
m j 1
2m j 1
1
(18.2)
This distinction between the two perturbative terms is a natural one, because W(1) is first order in
the charge e, while W(2) is second order, and one way we will keep track of our perturbation
theory is by counting factors of e. Hence if we perform a computation to second order, we will
allow up to two factors of W(1), but only one factor of W(2). The explicit form of Eqs. (18.2) will
be needed later in terms of creation and annihilation operators, and using Eqs. (17.13) and
(17.14b), these can be written
W 1 
W
 2
e
m 2 0V
e2

4m 0V
N

j 1 k
N

 
j 1 k
1
k
e
e
ik R j
ik R j

 ik R j *
εk  Pj  i  S j  k  ak  e
εk  Pj  i  S j  k  ak† ,
εk ak  e
 ik R j *
k

ε ak†  e
ik R j
εk  ak   e
 ik R j *
†
k   k  
ε
kk 
k 
a
.
(18.3)
We now let our first two terms be our unperturbed Hamiltonian,
H0  Hatom  Hem .
Fortunately, these two terms are completely decoupled, and therefore to find eigenstates of H0,
we need only find eigenstates of these two parts separately. We will now assume that we have
somehow managed to find the exact eigenstates of
Hatom , which we call n
, where n describes
all the quantum numbers associated with the atom itself, such as l, j, mj, etc. This state will have
energy  n . In addition, there will be eigenstates of Hem, which we computed in the previous
chapter. The overall eigenstates will then be given by
n ; n1 , k1 , 1; n2 , k 2 ,  2 ; ; nM , k M ,  M .
and will have energy
E  n 
 n11  n22 
 nM M  .
We are now ready to start putting in the effect of our perturbations.
B. Absorption and Emission of Photons by Atoms
Consider first the effect of our interaction terms to leading order in e. To this order, we need
consider only W(1), given in Eqs. (18.3). We immediately note that this perturbation can only
create or destroy one photon. It follows that the final state must be identical with the initial state,
save for a single photon. The only matrix elements we will consider, therefore, will be of the
form
1
Note that
Pj  A  R j   A  R j   Pj , because we are working in Coulomb gauge where   A  r   0 .
XVIII. Photons and Atoms
304
© 2016, Eric D. Carlson
W 1 I ; n1 , k1 , 1;
F ; n1  1, k1 , 1;
The energy difference
EF  EI
will therefore be
EF  EI   F   n1  1   n22   nM M    I 
  F   I    FI    ,
where we have renamed
1  
.
n1  n22 
 nM M 
since this is the frequency that will interest us most. Since this
must vanish by Fermi’s Golden rule, we conclude that  F   I   , so we either emit a
photon and decrease the atom’s energy, or we absorb a photon and increase the atom’s energy.
Consider first the case where we absorb a photon, so that  F   I . Then only the annihilation
part of W(1) will contribute. Furthermore, in the sum, only the single term that matches the wave
number k1 and polarization 1 will contribute. We will rename these as k and , so we have
WFI1 
e N
F ; n  1, k,  ;

m j 1 2 0V 
ik R j
e
εk  Pj  i  S j  k  ak I ; n, k,  ;
.
The photon part of this expression is easy to find; it is just
n  1, k,  ;
ak n, k,  ;
 n n  1, k,  ;
n 1, k,  ;
 n,
so the whole matrix element is just
WFI1 
e
n N
ik R
F e j ε  Pj  i  S j  k  I .

m 2 0V  j 1
(18.4)
Comparing this, for example, with Eq. (15.25), we see that our computation now closely
follows what we found before. To leading order, we can approximate eikR  1 , and ignore the
spin term to obtain the electric dipole approximation
j
WFI1 
e
n
ε  F P I .
m 2 0V 
In a manner identical to before, we can use Eq. (15.27), relating F P I and F R I , and
rewrite this as
WFI1  ie
n 
n 
ε  F R I  ie
ε  rFI .
2 0V
2 0V
where we used the fact that we must have
I  F  
  FI .
Substituting this into Eq. (18.1b) then yields
 e2 n
2
ε  rFI   F   I     4 2
 0V
© 2016, Eric D. Carlson
305
1
2
 cn  

 ε  rFI  FI    ,
 V 
XVIII. Photons and Atoms
where we have replaced factors of e using the fine structure constant   e 4 0 c . A careful
comparison with Eq. (15.29) will convince you that they are identical formulas. The factor of
 cn  V  is simply the energy of the photons n  , divided by the volume (yielding energy
2
density) and multiplied by the speed of light, and hence is just the intensity .
Let’s now consider the reverse case, where the final energy is less than the initial. Then we
must increase the number of photons by one, which means we need the other half of our
perturbation, so
WFI1 
e N
F ; n  1, k,  ;

m j 1 2 0V 
ik R j
e
ε*  Pj  i  k  ε*   S j  ak† I ; n, k,  ;


This time the photon part of this matrix element is n  1, k,  ;
ak† n, k,  ;
.
 n  1 , which
yields
WFI1 
 n  1  eikR j ε*  P  i k  ε*  S   .
e N
  j I

F
j

m j 1 2 0V 
In the dipole approximation, this can be simplified in a manner very similar to before. Skipping
steps, the final answer is
2
 4 2   c  n  1   *
I  F   

 ε  rFI    IF  .
V



This time, the result is not what we got before. If you keep the term proportional to n, we
will get exactly the same result as before. The interesting thing is that we have a new term which
does not require the presence of photons in the initial state at all. This new process is called
spontaneous emission. The rate is given, in the electric dipole approximation, by
2
4 2 c *
I  F  
ε  rFI    IF  .
V
As it stands this formula is a bit difficult to interpret. It has a delta function, which makes it
look infinite, but it also has the reciprocal of the volume, which just means that in the limit of
infinite volume, the probability of it going to a particular wave number k is vanishingly small.
The way to avoid these double difficulties is to sum over all possible outgoing wave numbers,
and then take the limit of infinite volume, which gives us
I  F  

4 2 c *
ε  rFI
V
2
   IF   4 2 c 
 c
d k ε*  rFI
2 
k
2
2
 k dk  kc  IF  
d 3k
 2 
ε*  rFI    IF 
2
3
2
 k 2
d k ε*  rFI ,

2
2
d   I  F  IF3 *

ε  rFI .
2
d k
2 c
XVIII. Photons and Atoms
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© 2016, Eric D. Carlson
This rate is the angle-dependant polarized rate. If we do not measure the polarization of the
outgoing wave, then the effective rate is the sum of this expression over the two polarizations.
Without working through the details too much, the result is

d   I  F  IF3
2

rFI  kˆ  rFI
2
d k
2 c
2
.
where k̂ denotes a unit vector in the direction of k. For example, if
rFI
is a real vector, the last
factor is simply sin2, where  is the angle between k and rFI . If we don’t measure the direction
of the final photon, we can integrate over solid angle to get a final answer
I  F  
4 3
2
IF rFI .
2
3c
(18.5)
Eq. (18.5) assumes we have only a single final state F for decay. It is common that there will
be several final states, all of which are possible final states. In this case, the total spontaneous
emission rate would be given by
I     I  F  
F
4
2
3 r .
2  IF FI
3c F
(18.6)
It is then also helpful to define the branching ratio as the fraction of atoms that decay in a
particular way, given by
BR  I  F  
I  F 
.
I
(18.7)
Of course, we can go beyond the dipole approximation when needed. The atomic matrix
elements can be expanded in powers of k  R j to any order desired. The first three terms can be
written in the form
e
 ik R j
ε*  Pj  i  k  ε*   S j  I 
F e


m
ie
F  ieIF  ε*  R j   12 eIF  ε*  R j   k  R j    k  ε*    S j  12 L j    I .
m
The three terms correspond to the electric dipole, electric quadrupole, and magnetic dipole terms
respectively. Equations analogous to Eq. (18.5) can then be derived for each of these other two
cases.
C. The Self-energy of the Electron
We now see that to leading order in perturbation theory, W(1) causes atoms to absorb or emit
a single photon. What is the effect of W(2)? There are a variety of effects, but notice in particular
that it will shift the energy of an arbitrary atomic state  . To first order in time-independent
perturbation theory, the shift in energy will be. We therefore compute
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307
XVIII. Photons and Atoms
A R j    
k ,
e
2  V
ik R j
0
ak εk  e
ik R j

ak† ε*k  
k

2 V 
0
k,
1
e
k
ik R j *
k
ε
 ;1, k,  .
We therefore have
 
e2 N
e2
2

A
R





j
2m j 1
4m 0V
e2

4m 0V
N e2


2m 0
j 1 k , k
N
1
N
 
j 1 k ,
k , 
1
kk 
ε
k
 ε*k    ;1, k ,  e
ik R j
e
 ik R j
 ;1, k ,  
N e2 4  k 2 dk N 2 
  2 3   2m 0  2 3 0 ck   m 0 k dk   .
k
d 3k
Hence the atomic state will be shifted up in energy by an infinite amount! Note, however, that
this shift is completely independent of what state the electrons are in. So long as the number of
electrons does not change, it cannot contribute to differences in energy, and hence is irrelevant.
It is sometimes suggested that these infinities that seem to be appear have something inherent
to do with quantum mechanics. But the same problems plague classical mechanics. An electron
at the origin is surrounded by an electric field E  r   ke erˆ r 2 , which has an energy density
1
2
 0E2  r   kee2 8 r 4   c 8 r 4 . If we integrate the resulting energy density over all space,
we find there is an infinite energy associated with the electric field around an electron. Hence
this problem of infinities exists classically as well.
D. Photon Scattering
We turn our attention now to the subject of photon scattering. We will consider situations
where both the initial and final state includes a photon. We will assume only a single photon, so
the initial and final states are
I  I ; k I ,  I
and
F  F ; k F ,  F ,
where there is an implied “1” describing the number of photons in each case. The initial and
final state energy are EI   I  I and EF   F  F . We are interested only in the case
where the final photon differs from the initial one. We would like to calculate the probability
P  I  F  to second order in perturbation theory. The amplitude is
FI
 FW
 2
I  lim 
 0
m
F W 1 m m W 1 I
,
EI  Em  i
(18.8)
where we have changed our small variable to  to avoid confusion with the atomic energies  m .
We have left out W(1) in the first term because it can only create or annihilate one photon, and
1
therefore F W   I  0 . We have left out W(2) in the second term because this has more factors
of the coupling e, and therefore is higher order.
In the second term, what sort of intermediate states are possible? Since we have assumed the
final photon state is distinct from the initial, and each of the two perturbations can only
annihilate or create one photon, we conclude that one factor of W(1) must be eliminating the
XVIII. Photons and Atoms
308
© 2016, Eric D. Carlson
initial photon and the other creating the final photon. Depending on which is which, the
intermediate state will either have no photons in it, or it will have two photons, one each in the
initial and final state. So the only possibilities for the intermediate state are m  m or
m  m ; k I ,  I ; k F ,  F . These states have energies of Em   m or Em   m  I  F . Thus
the sum in Eq. (18.8) will become a double sum, one term for each of these cases, and we have
FI
  ; k ,  W 1 m m W 1 I ; k I ,  I
 F ; k F ,  F W  2 I ; k I ,  I  lim   F F F
 0 m
 I   m  I  i

F ; k F ,  F W 1 m ; k I ,  I ; k F ,  F m ; k I ,  I ; k F ,  F W 1 I ; k I ,  I 

(18.9)
.
 I   m  F  i

We now turn our attention to evaluating and using Eq. (18.9) in certain special cases, but before
we go on, it is helpful to come up with a more compact notation to keep track of what is going
on. When necessary, we will refer to the three terms in Eq. (18.9) as FI 1 , FI  2  , and
FI
 3
respectively.
E. A Diagrammatic Approach
When the expressions are getting as
complicated as Eq. (18.9) it is a good idea to
try to come up with an easier way to keep track
of what is going on. Fortunately, there is a
neat diagrammatic notation that helps us
understand what is happening. In this notation, Figure 18-1: The
an electron (or atom) will be denoted by a solid two diagrams
corresponding to the
line with an arrow on it, and a photon by a
interactions W(1).
wiggly line. Time will increase from left to
right, and interactions will be denoted by a dot.
Figure 18-2: The
For example, W(1) can create or annihilate one photon, and there are
three diagrams
correspondingly two corresponding diagrams, as illustrated in Fig. 18(2)
corresponding to
1. In contrast, W can either create two photons, annihilate two
the interaction W(2).
photons, or create and annihilate one of each. Hence there are three
corresponding diagrams, as illustrated in Fig. 18-2.
An equation like Eq. (18.9) could be written in the simple diagrammatic notation as:
I
F
I
+
kI
kF
m
kI
F
I
+
kF
F
m
kI
kF
We see that in the first diagram, the photons are emitted and reabsorbed at the same place and
time (W(2)), while in the other two diagrams, there are two factors of W(1). By looking at the
diagrams half way through we can see that the intermediate state contains only an atom in the
© 2016, Eric D. Carlson
309
XVIII. Photons and Atoms
middle diagram, but there is an atom and two photons in the intermediate state in the final
diagram.
F. Thomson Scattering
As our first computation of this process, consider a free electron, for which the “atomic”
Hamiltonian is simply given by H atom  P 2 2m . The eigenstates of this Hamiltonian will be
proportional to eiqr , and since we are working in finite volume, we can normalize them simply
as q  r   eiqr V . Hence our initial state will be labeled I  q I ; k I ,  I . For definiteness,
and to keep our computations simple, we will assume the initial electron is at rest, so qI  0 .1
Which of the three terms in Eq. (18.9) will be most important? The first one,
FI
1 , has a
matrix element of W  2  e2 A 2 2m . The second and third term each have factors that looks like
eA  P m and eB  S m . Because we chose the initial state to have momentum zero, it is easy to
see that P vanishes in the second factor of each of the second and third terms. Hence we must
look at eB  S m . The eigenvalues of S are of order , while B   A which will turn into a
factor of k  A , where k is the wave number of one of the photons. The first factor has a similar
term. Noting that the energy denominators contain terms like  , we therefore estimate the
order of magnitude of the second and third terms in Eq. (18.9) as
FI  2 
FI  3
e2 2 A 2 k 2
 m2
The relative size of these compared to the first term is therefore of order
 2
FI 1
FI
 3 k 2  
m mc2
FI 1
FI
Thus the second and third terms will be dominated by the first if the photons have an energy
small compared to the rest energy of the electron. Indeed, if this is not the case, then we need to
consider relativistic corrections for the electron, and our entire formalism is wrong. We
therefore ignore FI  2  , and FI  3 and have
FI

FI 1 
 
e2
qF ; k F , F A2  R  qI ; k I , I
2m
 ik R † *
 e ak εk    eik R ak  εk    eik R ak† ε*k    q I ; k I ,  I
q F ; k F ,  F W  2 q I ; k I ,  I 
e2 q F ; k F ,  F  eikR ak εk
k k  
 2m  2 0V kk 
.
This expression is not nearly as fearsome as it appears. The only non-vanishing terms in this
double sum occur when we annihilate the correct photon in the initial state, and create the correct
photon in the final state. There are exactly two terms in the double sum which survive, and they
are effectively identical.
1
This simplifies the computation, but does not affect the answer.
XVIII. Photons and Atoms
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FI



e2
q F ; k F ,  F eik F R ak†F F eik I R ak I  I ε*k F F  εk I  I q I ; k I ,  I ,
2m  0V kI kF
FI
  ε*F  ε I 
e2
q F eik I R ik F R q I .
2m 0V I F
(18.10)
This last matrix element is simple to work out:
q F eik I R ik F R q I 
1 iqF r ik I r ik F r iqI r 3
e
e
e d r   q F  k F ,q I  k I .
V
This matrix element does nothing more than ensure conservation of momentum. Substituting it
into Eq. (18.10) and then into Eq. (18.1b), and noting that squaring a Kronecker delta function
has no effect, we find
FI
  ε*F  ε I 
I  F  
2
e2
 q  k ,q  k ,
2m 0V I F F F I I
ε*F  ε I
2
(18.11a)
e4 2
 q k ,q k   EF  EI  .
4m2 02V 2I F F F I I
(18.11b)
Eq. (18.11b) is difficult to understand as it stands, because we are still in the infinite volume
limit, and because the probabilities of scattering into a particular state are small. To make sense
of it, sum over all possible final quantum states; that is, sum over the wave numbers for both the
electron and the photon, and sum over polarizations of the photon. Let’s also put in the explicit
form for the energy difference, where the energy is just the energy of the electron plus the
photon. We find
I  F  
  ε*F  ε I
k F  F qF
2
 e4
q
2m2 02V 2I F
F
 k F ,q I  k I
  EF  E I 
2 2


qF
 e4
  ε  εI
  F 
 I  ,
2 2 2
2m  0 V  I  F 
2m
k 

3
2 2
4
2


d kF *
q
1
e
I  F   
εF  εI
  F  F  I  .
3
2 2
V  F  2 
2m  0 I F 
2m

*
F
F
2
F
(18.12)
Now, if the initial wave number is kI, conservation of momentum will tend to make the final
photon and electron wave numbers comparable. We have, therefore,
2
qF2
2m
2
kI2
m
I2
  
 I  2I 
mc
 mc 
2
2
I .
Hence in Eq. (18.12) we can ignore the electron energy compared to the photon energies, so
  I  F , and we have
© 2016, Eric D. Carlson
311
XVIII. Photons and Atoms

2
1
e4
I  F  
k 2 dkF  d k ε*F  ε I   ck F  ck I 
2 2 2 2  F
V 16 m  0   F 0

2
2
e4 kF2
1
1 2 2
*
d

ε

ε

d k  ε*F  ε I ,
k F
I
2 2 2 2 
2

V 16 m  0  c
V mc
F
F
where in the last step we wrote our expression in terms of the fine structure constant
  e2 4 0 c .
One disturbing fact is that the rate is still inversely proportional to the total volume. This is
simply because we are trying to collide a single photon in the universe with a single electron.
Under such circumstances, the correct description is not in terms of rates, but in terms of cross
section, where the rate is the cross section times the density of targets (1/V in this case) times the
relative velocity (c in this case). Hence the cross section is
2
V
2 2
2 2
8 2 2
*
2
    I  F   2 2  d k  ε F  ε I  2 2  d k sin  
.
c
mc
mc
3m2c 2
F
(18.13)
If the polarized or unpolarized differential cross-section is desired instead, we can simply not
perform the final sums and integral. The angle  is the angle between the polarization of the
incoming wave and the direction of the outgoing wave. This cross-section can be calculated
classically, and is called the Thomson cross-section, but it is comforting to find the quantum
result agrees with it. With a bit more work, it can be shown that this cross-section is still valid if
the initial electron is moving, provided such motion is non-relativistic.
G. Scattering Away From a Resonance
Suppose that instead of a free electron, you have bound electrons in an atom, and that the
energy of the photon is insufficient to free the electron. We’ll also assume we are far from
resonance, so that the photon energy is not right, nor close to right, to excite the atom to an
intermediate state, so that  I  I   m for any intermediate state. We will furthermore assume
that the final state of the atom is identical to the initial state, so I  F   and F  I . We
need to calculate the matrix elements in Eq. (18.9), so we need
 I ; k F ,  F W 1 m m W 1 I ; k I ,  I
 I ; k F ,  F W I ; k I ,  I  lim  
 0 m
 I   m    i

 ; k ,  W 1 m ; k I ,  I ; k F ,  F m ; k I ,  I ; k F ,  F W 1 I ; k I ,  I 
 I F F
(18.14)
.
 I   m    i

 2
FI
We start with the first term, which in a manner similar to before, yields
*
FI 1   ε F  ε I 
XVIII. Photons and Atoms
N
e2
ik R ik R
I e I j F j I .

2m 0V  j 1
312
© 2016, Eric D. Carlson
Now, as we have argued before, the wavelength of light that causes transitions tends to be much
larger than the size of an atom, and we are by assumption working at energies of the same order
ik R ik R
or lower, so we can approximate e I j F j  1, and we have
*
FI 1   ε F  ε I 
N
e2
e2 N
*



ε

ε
 I I  F I  2m V  .
2m 0V  j 1
0
(18.15)
Now let’s look at the second term in Eq. (18.14), which, when expanded out a bit, is
N
I ; k F ,  F  A j  Pj  B j  S j  m m  Ai  Pi  Bi  Si  I ; k I , I
e2
lim
.


FI  2  
m2  0 m j ,i 1
 I   m    i
As we have found before, the A j  Pj terms tend to dominate the B j  S j terms. Furthermore, in
the sum over all the photon operators, only the term that annihilates the correct photon and then
creates the correct photon contributes. So we find
N N  e
e2
I
2

lim



FI
2

m 2 0V   0 m j 1 i 1
As usual, we then approximate eik I Ri  1  e
We also replace  I   m   mI to yield
ik F R j
 ik F R j *
F
ε  Pj m m eik I Ri ε I  Pi I
 I   m    i
.
, and rewrite the sums as the total momentum.
I ε*F  P m m ε I  P I
e2
lim 
.
FI  2   
2 0Vm2  0 m
 mI    i 
(18.16)
Since we are assuming we are not near resonance, we can take the limit   0 . We rewrite
the denominator as
1
 mI   

1
mI

1
mI mI   
.
Substituting this into Eq. (18.16), we have
I ε*F  P m m ε I  P I  1
e2
1 
 
.
FI  2   
2 
2 0Vm m
mI
  mI   
We use our dipole trick Eq. (15.27), m P I  immI m R I , but for the first term we use
this only once, while for the second we use it twice, to yield
FI  2  
e2
2 0V

I
m
 i  ε  P I mI m ε I  R I 
ε*F  R m  m I

.
m
mI  


In the first term, we now note that we have a complete sum over intermediate states, and no other
factors depending on the intermediate state. On the final term we rewrite m R I  rmI so we
can rewrite this term in a nicer form to yield
© 2016, Eric D. Carlson
313
XVIII. Photons and Atoms
ie2
e2
*
I  ε F  R   ε I  P  I 
FI  2  
2m 0V 
2 0V

mI  ε*F  rmI*   ε I  rmI 
m
mI  
.
(18.17)
We have two terms done; we still have one to go. In a manner similar to before, we keep
only those terms that create the final state photon, and then annihilate the initial state photon.
We again ignore the exponential factors, and combine the momenta to obtain the total
momentum. We eventually arrive at an equation analogous to Eq. (18.16):
I ε I  P m m ε*F  P I
e2
lim 
.
FI  3  
2 0Vm2  0 m
 mI    i 
Notice the denominator is slightly different, since in this case we have an extra photon in the
intermediate state. We again expand out the denominator and then use the same trick
Eq. (15.27), so
 i I ε I  P m mI I ε I  R m 
*

 m ε F  R I
m
mI  
m 

2
2
mI  ε I  rmI*  ε*F  rmI 
ie
e
*

I  ε I  P   ε F  R  I 
.

2m 0V 
2 0V m
mI  
e2
FI  3 
2 0V

(18.18)
Now that we have the three pieces, all we need to do is put them together. To simplify
matters slightly, I’ll assume we are working with real polarizations or real dipole moments rmI .
Combining Eqs. (18.15), (18.17), and (18.18), we have
FI


FI
1 
FI
 2 
FI
 3
e2 N
ie2
ε*F  ε I  
I  ε*F  R   ε I  P    ε I  P  ε*F  R  I

2m 0V 
2m 0V 

e2
2 0V
 ε
m
F
 mI
mI 
*
 rmI   ε I  rmI  

.
 mI   mI   
(18.19)
Now, in the second term, we note that we have a commutator. Keeping in mind that we have
several electrons, this commutator can be worked out to yield
ε*F  R, ε I  P    ε*F  R j , ε I  Pi    i  ji  ε*F  ε I   Ni  ε*F  ε I  .
N
N
N
j 1 i 1
N
j 1 i 1
Substituting this into Eq. (18.16), we find the first two terms cancel exactly, so we have
FI
e2

 0V
mI
*
ε  rmI   ε I  rmI  .
2
2  F
mI  

m
(18.20)
We can then substitute this into Eq. (18.1b) to obtain a rate per unit volume, which in turn can be
converted into a cross-section.
Let us compare the size of the amplitude Eq. (18.20) we found for a bound electron to that
for a free electron, Eq. (18.11a). Ignoring matching factors, and making only order of magnitude
estimates, we find
XVIII. Photons and Atoms
314
© 2016, Eric D. Carlson
 bound 
FI  free 
FI
For a typical atom, we would expect mI
m 2mI rmI

2
mI
mc 2 2
2
 2 
.
and rmI ~
 mc  , so we have
 bound 
2
.
mI2   2
FI  free 
FI
Suppose that we have tightly bound atom, so that  mI
 mI 
2
1 . Then this ratio is of order
, and this gets squared to  mI  when you calculate the rate (which ultimately
4
becomes a cross-section). Hence a tightly bound atom has a much lower cross-section than a
free electron for photon scattering.
This has considerable significance in the early universe. The early universe was at very high
temperatures, hot enough that electrons were not generally bound into atoms, but rather free.
This meant that the early universe was opaque, and photons effectively were in near perfect
equilibrium. However, when the temperature of the universe dropped to about 3000 K, at an age
of about 380,000 years, the electrons became bound to the free protons to produce neutral
hydrogen atoms. Since the typical photon had an energy of about 3kBT  0.8 eV , and the first
excited state of hydrogen requires about 10.2 eV of energy, the cross-section dropped suddenly
by several orders of magnitude, and the universe became transparent. These photons have since
been red-shifted in the expanding universe, and now appear as a nearly uniform 2.73 K
background. This background gives us a snapshot of what the universe looked like at t =
380,000 y.
Our derivation of Eq. (18.20) assumed that we were working away from resonance.
However, as   mI , it is clear from Eq. (18.20) that something dramatic may occur, since
there will be a sudden increase in the amplitude. Under such resonance conditions, we need to
rethink how we are computing our amplitude. This complication is the subject of the next
section.
It should be understood that, as usual, we have included only the dipole approximation in our
scattering amplitude. This is generally a good approximation, since we are summing over all
intermediate states, and those which have non-zero dipole
I
I
moment will tend to dominate. But if we approach
m
resonance, so   mI for some state that does not have a
kF
kI
dipole moment, we will have to include “smaller” effects like
the electric quadrupole or magnetic dipole. These small
matrix elements will be greatly enhanced by the nearly vanishing denominators.
H. Scattering Near a Resonance
It is evident from Eq. (18.20) that our amplitudes may become large when   mI . The
large amplitude can be traced back to the contribution from
FI
 2 , corresponding to the
diagram sketched above. It is easy to see why this term dominates: the intermediate state has
© 2016, Eric D. Carlson
315
XVIII. Photons and Atoms
almost exactly the right energy to match the incoming energy. Indeed, we will focus exclusively
on this diagram, and only include one term in the sum, the one which has a small energy
denominator.1 The amplitude for this is given by Eq. (18.16), repeated here, but using
Eq. (15.27) to make the substitution m P I  immI rmI :
FI  2  
2
ε*F  rmI*  ε I  rmI  .
e2mI
lim
2 0V   0   mI  i
We are going to assume that we are very close to resonance, so we will approximate   mI
everywhere except the denominator of this equation. We simplify this expression to
 0
FI
ε F  rmI  ε I  rmI  .
e2mI

lim
2 0V  0   mI  i
*
*
(18.21)
The notation has changed slightly: we have dropped the (2), because when we are close to
resonance, this will effectively be the only term we need to worry about, and the superscript (0)
is added because in a moment we will be adding higher and higher order terms to try to figure
out what is going on. This diagram has zero loops in it.
If we were to continue with this expression, we would ultimately obtain a cross section that
diverges at   mI . What is the solution to this problem? The answer turns out, somewhat
surprisingly, to add more diagrams.
I
I
n
Consider now the one-loop diagram
m
m
sketched at right. In this diagram, the
kI
kF
k
ground state atom merges with a photon to
produce the resonant excited state, then it
splits into an intermediate photon of wave number k and the ground state again, which then
reemerges to produce the excited state before splitting back into a photon and the ground state.
Fortunately, nearly all the work has already been done, and we simply reuse it. The two new
vertices (in the middle) are virtually identical to the ones we had before, and we now have two
intermediate states that are purely the resonant state. The main new feature is an energy
denominator, which looks like
 I     n  k  i   I   m   I   n  k  i  mn  k  i .
So we have
*
ε*F  rmI
 εk  rnI  ε*k  rnI*  ε I  rmI 


 e2mI  e2mn 
k
,


lim
,
 

  0 
FI , n 
2
2

V
2

V




i





i





 0  0 
mI
mn
k
1
1
FI , n
1
 k,  
*
*
 mn  εk  rmn   εk  rmn 
.


 2 0 V    mI  i mn  k  i 
 0 
FI
e2
One complication is that there are often several intermediate states
m
with the same energy, usually related by
symmetry. However, since they are degenerate, it is always possible to change basis such that only one state has
non-zero matrix element
m P I
XVIII. Photons and Atoms
.
316
© 2016, Eric D. Carlson
This is the amplitude from this diagram for a particular intermediate momentum and
polarization, as well as a particular atomic state n . Of course, we don’t measure this
intermediate state, so we must actually sum over all such intermediate states. We then turn the
sum into an integral for the photon states in the infinite volume limit in the usual way, and have
1
FI

 0
FI
e2
  mI  i 2 0
1
d 3k mn εk  rmn

  2 3 mn  k  i .
n 
2
(18.22)
Note that there is an implied limit   0 in FI  , which applies to every factor in Eq. (18.22).
Now, we will be interested in the integral and sum appearing in Eq. (18.22). We will split it
into a real and imaginary part, which we write as
0
e2
 m  i m 
2 0
1
2
d 3k mn εk  rmn

  2 3 mn  k  i ,
n 
2
where m and m are real. Switching to spherical coordinates, and remembering that k  k c ,
and substituting the fine structure constant   e2 4 0 c , we have

k2 dk

2
m  im  2 2  nm   d k εk  rmn 
.
4 c n
mn  k  i

0
1
2
(18.23)
At the moment, I am not particularly interested in the real part. Indeed, if you continue the
computation, you will discover that it is divergent, but this divergence is only present because we
ik R
approximate the phase factors as e j  1 in all our matrix elements. The infinity is coming
from the high frequency modes, when this approximation breaks down, and one would
ultimately find some finite contribution to m .
Let’s concentrate on the imaginary part. In the limit   0 , the integrand in Eq. (18.23) is
zero almost everywhere, except for a small region right around k  mn , which can only happen
if mn  0 , or  m   n . We can therefore concentrate on this small region to find the imaginary
part. Let us take the limit   0 , but let’s distort the contour of integration in this region so
that we avoid the divergence. The i tells us how to do this: we want to keep the combination
mn  k  i positive imaginary, which we do by
Im(k)
letting k have a small negative imaginary part, as
mn
Re(k)
illustrated in Fig. 18-3. We can now let   0 in the
denominator, but for the little hop below the real axis,
Figure 18-3: The path for
we let k  mn   ei . Thus as  progresses from 0
integration for Eq. (18.23). The only
to  , the little half-loop will be followed. We will
part that contributes to the imaginary
treat  as so small that effectively it equals zero,
part is the little “hop” below that real
except in the critical denominator. Focusing on the
axis near
.
imaginary part, we substitute k  mn   ei into
Eq. (18.23), and find
© 2016, Eric D. Carlson
317
XVIII. Photons and Atoms
2


εk  rmn 
2
i 2
i
m   2 2 Im  mn   d k  mn   e  d    e 

4 c
     ei 

 n
0

  i ei d 


2
2
3

d

ε

r
Im
3 r
sin 2 k d k ,


k k
mn
2 2  mn  
i
2  nm mn 
2 c n

 0  e  2 c n
4
2
3
(18.24)
m  2 mn
rmn .
3c n
We note that Eq. (18.24) is identical with Eq. (18.6). Thus m is the total decay rate of this
intermediate atomic state m .
Substituting Eq. (18.23) into Eq. (18.22) will then yield
1
FI
 0
FI

 m  12 i m
.
  mI  i
(18.25)
We don’t know what m exactly is, but we won’t need it in the subsequent discussion, but we do
know the explicit form of m . However, all we need to know for now is that they are both
finite.
Now, as we look at Eq. (18.25), it seems that our situation is worse than ever. Recall that our
0
original amplitude FI  given by Eq. (18.21) diverges at resonance. The new term Eq. (18.25)
has yet another denominator, and diverges even worse. Let’s try adding two loops and see what
we get, as sketched at
I
I
n'
n
right. This time we
m
m
m
can write down the
kF
kI
k
k'
answer instantly.
The effect of adding
another loop is simply to put yet another factor similar to the one in Eq. (18.25). The amplitude,
we immediately see, is given by
 2
FI

2
 0 
FI
m  12 im 

 .
   mI  i 
Now that we’ve established the pattern, we can continue indefinitely. We are most interested in
the sum of all these diagrams, which is a geometric series, which looks like
FI

 0
FI

1
FI

 2
FI


 0
FI
Substituting our explicit form Eq. (18.21) for
FI
n
 m  12 im 

 

n 0    mI  i 

 0
FI
 0 
FI
1
m  12 im 
1


 .
   mI  i 
, we have
*
*
ε*F  rmI
e2mI  ε*F  rmI
ε I  rmI 


 ε I  rmI  .
e2mI
  mI  i

lim

2 0V  0   mI  i   mI  i  m  12 im 2 0V   mI  m  12 im 
(18.26)
XVIII. Photons and Atoms
318
© 2016, Eric D. Carlson
Note that in the final setp, as if by miracle, we were able to take the limit   0 with impunity.
The decay rate m makes the denominator finite. We also now understand the meaning of m :
the energy of the state m has been shifted to  m  m .
We would like to go ahead and calculate the scattering cross-section from Eq. (18.26).
Substituting into Eq. (18.1b), we have
2
ε F  rmI ε I  rmI
2  e2mI 
I  F  
  EF  EI  .


2
2
 2 0V    mI   m   14  m
2
2
Let’s assume we’re working near resonance. We’ll neglect the small shift in the energy m . We
also want to sum over final state polarizations and outgoing final state directions. Then we have
2
ε F  rmI ε I  rmI
2 e4mI
I  F  
  EF  E I 
2 2 
4 0 V k ,   mI 2  14 m2
2
 e4mI2 ε I  rmI

2  02V  2
e  ε I  rmI

16 2  02V
4

2
mI
2
2
2
ε F  rmI
d 3k

  2     

3
mI




F2 dF
c3
0
 2mI2  2 ε I  rmI
2
2
Vc   mI   14 m2 


2
2
 14 m2
ε F  rmI
 d     
k
2
mI
  d k ε F  rmI 
2

  EF  EI 
2
 14 m2
  F   
4
8 2mI
ε I  rmI rmI
,
2
3Vc   mI   14 m2 


2
2
where at the last step, we once more used the approximation   mI in the numerator. Because
decay rates tend to be rather small, the denominator will have a dramatic increase right at
resonance. Note that the rate is still dependant on V, which is hardly surprising, since we are
trying to collide one photon with one atom. We need to convert this to a cross section, which we
do in the usual way by using the formula   n v , where n = 1/V is the density of target
atoms, and v  c is the relative velocity. We find
4
8 2mI
rmI ε I  rmI
   
.
2
3c 2   mI   14 m2 


2
2
(18.27)
A cross-section with the functional form of Eq. (18.27) is called a Lorentzian lineshape. Real
lineshapes will not be exactly Lorentzian because of a variety of approximations we have made
along the way.
Now, in all our computations, we have been assuming there is only one possible intermediate
state, and sometimes this is the case. More typically, however, there will be several degenerate
intermediate states, related to each other by rotational symmetry. To make our formulas work,
we can change basis such that for a single state m has a dipole moment in the direction of the
polarization, so that ε I  rmI  rmI , and the others all are perpendicular. Then our computations
are valid, and Eq. (18.27) can be simplified to
© 2016, Eric D. Carlson
319
XVIII. Photons and Atoms
   
4
8 2mI
rmI
.
2
3c 2   mI   14 m2 


4
(18.28)
If we perform the scattering right on resonance, at   mI , and note the similarity of the
numerator to Eq. (18.5), we find
 3c 2  m  I   6 c 2
2
    2 1 2 
  2  BR  m  I  
3c 4  m 
4mI
mI

2
8
(18.29)
Notice that all the factors of the coupling, matrix elements, and so on, have disappeared.
Typically, branching ratios are of order 1; indeed, if m is the first excited state, it must be
one. The cross-section Eq. (18.29) is enormous; it is of the order of the wavelength squared,
which will be tens of thousands of times greater than scattering off of a free electron. This is the
power of resonant scattering.
One way to think of this process is that the atom is actually absorbing the photon, staying in
an excited state for a brief time, and then re-emitting the photon. Over what range of energies
will this absorption occur? We can see from Eq. (18.28) that the cross section will be of order its
maximum value if   mI  12  . This means that the energy is “wrong” by an approximate
quantity E
 . The atom will “hang onto” the energy of the photon for a typical time
t  1 . Multiplying these two quantities yields the time-energy uncertainty relationship,
Et 12 .
1
2
Problems for Chapter 18
1. An electron is trapped in a 3D harmonic oscillator potential, H  P 2 2m  12 m02 R 2 . It is in
the quantum state nx , ny , nz  2,1,0
(a) Calculate every non-vanishing matrix element of the form nx , ny , nz R 2,1,0 where the
final state is lower in energy than the initial state.
(b) Calculate the decay rate  210  nx , ny , nz for this decay in the dipole approximation


for every possible final state, and find the corresponding branching ratios.
2. A hydrogen atom is initially in a 3d state, specifically, n, l , m  3, 2, 2 .
(a) Find all non-zero matrix elements of the form n, l , m R 3, 2, 2 , where n  n . Which
state(s) will it decay into?
(b) Calculate the decay rate in s-1.
XVIII. Photons and Atoms
320
© 2016, Eric D. Carlson
3. An electron is trapped in a 3D harmonic oscillator potential, H  P 2 2m  12 m02 R 2 . It is in
the quantum state nx , ny , nz  0,0, 2 . It is going to decay directly into the ground state
0,0,0
(a) Convince yourself that it cannot go there via the electric dipole transition. It can,
however, go there via the electric quadrupole transition.
(b) Calculate every non-vanishing matrix element of the form 0,0,0 Ri R j 0,0, 2 .
(c) Calculate the polarized differential decay rate d pol  002  000  d  for this decay.
This will require, among many other things, converting a sum to an integral in the infinite
volume limit.
(d) Sum it over polarizations and integrate it over angles to determine the total decay rate
  002  000  .
4. The present day density of electrons is about 0.21/m3; however, these electrons are not all
free.
(a) Calculate the cross-section for free photon-electron scattering in m2. If all the electrons
were free, what would be the rate at which they scatter, today? How many scatterings
would have occurred to an average photon in the age of the universe, 13.7 Gyr?
(b) The last time the electrons were free was when the universe was 1092 times smaller in all
three directions, and it was only 380,000 years old. The number of electrons was about
the same then (though the number density was much higher, since the universe was
smaller). Redo part (a) at this time.
5. An electron is trapped in a 3D harmonic oscillator potential, H  P 2 2m  12 m02 R 2 . It is in
the ground state nx , ny , nz  0,0,0 . Photons are fired at the electron with frequency
  0 , going in the z-direction, and polarized in the x-direction, ε  xˆ . Calculate the
differential cross section d d  (summed over final polarizations, not integrated), and the
total cross section  .
6. A Hydrogen atom in the 1s-state is hit by photons polarized in the z-direction very close to
3
   2 mc 2 2
4 Ryd ; that is, quite close to the 2p-resonance, where Ryd
is the binding frequency of the ground state of Hydrogen.
(a) Find all non-zero components of the matrix elements 210 R 100 . Find the decay rate
210 for this state.
(b) Calculate a formula for the cross-section as a function of . Assume that only one
diagram (the one calculated in class) is relevant, the one that has a resonance at the 2p
state. Treat the energy shift 210 as zero.
(c) Sketch the cross-section in units of a02 as a function of  in the range
   0.75  105  Ryd . I recommend making the vertical scale logarithmic.
© 2016, Eric D. Carlson
321
XVIII. Photons and Atoms
Appendix A. Some Mathematical Tools
A. Vectors in Three Dimensions
Vectors are denoted in this book by bold face (a) and scalars in italics (s). They can be
specified by giving their components either as an ordered list, or in terms of the unit vectors in
Cartesian coordinates:
a   ax , ay , az   ax xˆ  ay yˆ  az zˆ
The magnitude of a vector is simply
a  ax2  ay2  az2 .
Addition of vectors is commutative and associative:
a  b  b  a and
a  b   c  a  b  c  .
Multiplication of a vector by a scalar can be on the right or the left. Such multiplication is
commutative, distributive and associative:
sa  as ,
 st  a  s ta  ,  s  t  a  sa  ta ,
and s  a  b   sa  sb .
Multiplication of two vectors can take two forms: the dot or scalar product, and the cross or
vector product. The dot product is the product of the lengths of the two vectors times the cosine
of the angle between them:
a  b  a b cos    ai bi .
(A.1)
i
The cross-product a  b is a vector whose length is the product of the lengths times the sine of
the angle between them, and whose direction is perpendicular to both vectors in a right-hand
sense:
a  b  a b sin  .
In components, it can be written as
a  b   aybz  az by  xˆ   az bx  axbz  yˆ   axby  a ybx  zˆ .
Another way of writing this is in terms of the Levi-Civita symbol  ijk . The Levi-Civita
symbol is completely anti-symmetric, so that interchange of any two indices changes the sign:
 ijk   jik   ikj   kji ,
(A.2)
and  xyz  1 . This is sufficient to determine all of its components:
 xyz   yzx   zxy  1,  xzy   zyx   yxz , all others vanish.
Appendix A. Some Mathematical Tools
322
(A.3)
© 2016, Eric D. Carlson
Any two Levi-Civita symbols that are multiplied can be combined into combinations of
Kronecker delta-functions. In particular,


ijk ilm
  jl km   jm kl ,
(A.4)
i
where  ij  1 if i  j , and  ij  0 if i  j . The components of a cross product are then
 a  b i    ijk a j bk .
(A.5)
j ,k
It is trivial to see from Eqs. (A.1), (A.5), and (A.2) that the dot product is symmetric and the
cross product anti-symmetric:
a  b  b  a and a  b  b  a .
(A.6)
Products of more than two vectors can often be simplified with the help of Eqs. (A.1) through
(A.6). Some of the more useful ones are listed here:
 a  b   c   b  c   a  c  a   b    ijk aib j ck
(A.7a)
a   b  c   c  b   a  b a  c   c a  b 
(A.7b)
 a  b    c  d   a  b c  d   a  d b  c 
(A.7c)
i , j ,k
It should be noted that though all of the equations in this section are valid for vector and scalar
quantities, they are not necessarily true if the vectors or scalars are themselves operators which
might not commute. For example, using Eqs. (A.6) we see that a  a  0 for any vector, and yet
for any angular momentum-like operators, Eq. (7.7) tells us J  J  i J .
B. Calculus with One Variable
Quantum Mechanics takes extensive advantage of calculus, and you should familiarize
yourself with techniques of calculus. The two basic functions of calculus are the derivative and
the integral. The derivative of a function of one variable f  x  can be written as df  x  dx or
f   x  . Finding derivatives of most functions is not difficult. The following table gives
derivatives of the most common simple functions
d
c 0,
dx
d n
x  nx n1 ,
dx
d x
e  ex ,
dx
d
1
sin 1  x  
,
dx
1  x2
d
d
1
cos x   sin x ,
cos 1  x  
dx
dx
1  x2
d
1
ln x  ,
dx
x
d
tan x  sec2 x ,
dx
© 2016, Eric D. Carlson
d
sin x  cos x ,
dx
323
d
1
tan 1  x  
.
dx
1  x2
Appendix A. Some Mathematical Tools
Notice that many other functions can be rewritten in terms of these functions; for example,
x  x1/2 and 1 x  x1 . In addition, there are rules for taking the derivatives of sums,
differences, products, quotients, and composition of functions:
d
 f  x   g  x  
dx 
d
 f  x  g  x  
dx 
d  f  x 


dx  g  x  
f   x  g  x  ,
f  x g  x  f  x g x  ,
f  x g  x  f  x g x
,
g 2  x
d
f  g  x   f   g  x  g  x  .
dx
As a special case, note that when a function has a constant added to it, the constant has no effect;
when a function has a constant multiplied by it, the constant simply multiplies the derivative as
well.
We will sometimes also use hyperbolic functions, defined as
cosh x
1
2
ex
e
x
1
2
, sinh x
ex
x
e
sinh x
.
cosh x
, tanh x
Obviously, the derivatives of these functions are easily worked out:
d
cosh x
dx
sinh x ,
d
sinh x
dx
d
tanh x
dx
cosh x ,
1
.
cosh 2 x
A very useful formula from the calculus of one variable is the Taylor expansion, which says
that a function in the neighborhood of any point a can be expanded in terms of the value of the
function and its derivatives at the point a. The Taylor expansion is
f  x  f a   x  a f a 
1
1
 x  a 2 f   a    x  a 3 f   a  
2!
3!
,
where n! 1 2 3 n . This formula will strictly speaking be true only if the function f is
sufficiently smooth between a and x, and if the sum converges. If the sum does not converge, a
finite sum often will still serve as an excellent approximation for the function, though it will not
be perfect.
The Taylor expansion comes up in lots of situations, but some of the more useful cases are
1  x 
n
 1  nx  2!1 n  n  1 x2  3!1 n  n  1 n  2  x3 
e x  exp  x   1  x  2!1 x 2  3!1 x3  4!1 x 4 
cos x
sin x
,
,
(A.8a)
(A.8b)
1
1
2!
x2
1
4!
x4
1
6!
x6
,
(A.8c)
x
1
3!
3
1
5!
5
1
7!
7
,
(A.8d)
x
x
x
The notation exp  x  is an alternate notation for e x , useful because we will often make the
argument of this function quite complicated, and it is easier to read, for example, exp   12 Ax 2 
Appendix A. Some Mathematical Tools
324
© 2016, Eric D. Carlson
1 Ax 2
2
1 ; the other three
converge for all complex x. Another nice feature about Eqs. (A.8) is that if we replace x with
any square matrix, the formulas still make sense, so we can now take exponentials of matrices,
for example. Euler’s theorem, which we will use extensively, can be derived from Eqs. (A.8b),
(A.8c), and (A.8d):
then it is to read e
. Eq. (A.8a) is convergent for any complex number x
eix
cos x i sin x .
(A.9)
This formula also allows one to write cosine and sine in terms of exponentials:
cos x  12  eix  eix   cosh  ix  ,
sin x  21i  eix  eix   1i sinh  ix  .
(A.10a)
(A.10b)
The other thing you do in the calculus of one variable is to integrate. There are two types of
integrals, definite and indefinite. A definite integral has limits of integration and is defined as
the area under a specific curve between two limits. It looks like this:
b
 f  x  dx
a
The other type of integral is an indefinite integral. It uses the same symbol, except that no limits
are specified, and it is defined as the anti-derivative; that is, the function whose derivative is
f  x  . In other words, if we let F  x  be the indefinite integral of f  x  , then
F  x    f  x  dx  F   x   f  x  .
As we already mentioned, if you add a constant to a function F, its derivative is unchanged, so
that the indefinite integral is ambiguous up to an additive constant. Hence, in general, the
indefinite integrals should always end with C , an unknown constant. The fundamental
theorem of calculus tells us that these two types of integrals are related. Specifically, if the
indefinite integral of f  x  is F  x  , then
b

a
b
f  x  dx   F   x  dx  F  x  a  F  b   F  a  .
b
(A.11)
a
Because of the subtraction, the constant of integration cancels out, and therefore is unnecessary
and irrelevant when calculating an indefinite integral. Note that if you ever have to exchange the
two limits of an integral, the resulting integral changes sign, as is clear from Eq. (A.11). Also
note that in any definite integral, the variable being integrated (x in this case) disappears after the
substitution of the limits, and hence can be replaced by any other unused variable with impunity.
Unlike derivatives, there are no simple rules for doing integrals. Generally you use a few
steps to try to convert your integral into smaller, more manageable pieces, and then either look
up the integral in an integral table, or use some tool like Maple to do the integral. Two of the
simple rules that allow you to convert complicated integrals into simpler ones are
© 2016, Eric D. Carlson
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Appendix A. Some Mathematical Tools
 cf  x  dx  c cf  x  dx ,
  f  x   g  x  dx   f  x  dx   g  x  dx .
Also useful is integration by parts, a technique one can use when you are integrating the product
of two functions f  x  g  x  , one of which you know the integral of, F  x  . The rule states
 f  x  g  x  dx  F  x  g  x    F  x  g   x  dx .
In addition, it is possible to change variables. Suppose we have an integral and we wish to
change variables x  g  y  . Then we can substitute this integral in to find
 f  x  dx   f  g  y  d  g  y    f  g  y  g   y  dy .
Note that if you are dealing with an indefinite integral, you will have to substitute back in to
convert to the original variable x using y  g 1  x  . If you are doing a definite integral, this may
not be necessary, but you will have to redo the limits of integration appropriately.
Once the equation has been massaged into a relatively well-controlled form, you will often
be left with some integrals that you need to complete. I recommend looking them up in an
integral table (see section C) or using a program like Maple. For example, suppose you have the
surprisingly difficult integral
e
Ax 2
dx . This integral is defined only if A is positive, or at
least has a positive real part; otherwise, the integral is undefined. To determine the integral, you
must tell Maple that this constant is positive. To find this integral, enter the commands1
> assume(A>0);integrate(exp(-A*x^2),x=-infinity..infinity);
and it will come back with the answer:
e
x2
.
C. Tables of Integrals
For the following indefinite integrals, it is to be understood that there is a common constant
of integration + C to be added to all integrals. It is also to be understood that and are nonzero.
1
 ax  b n1 , n  1
a  n  1
x
1
n
n 1
n2
 x  ax  b  dx  a  n  1  ax  b   a2  n  1 n  2  ax  b  , n  1, 2
  ax  b 
2
 x  ax  b  dx 
n
n
dx 
x2
2x
n 1
n2
 ax  b   2
 ax  b 
a  n  1
a  n  1 n  2 
2
 3
 ax  b n3 , n  1, 2, 3
a  n  1 n  2  n  3
1
(A.12a)
(A.12b)
(A.12c)
To make the input look like this, you have to open Maple worksheet, rather than the conventional Maple program.
Appendix A. Some Mathematical Tools
326
© 2016, Eric D. Carlson
dx
1
 ax  b  a ln  ax  b 

dx
 ax2  b
1


xn
b
b  ax2 dx 
(A.12e)

(A.12f)

1
sin 1 x b a  x b  ax 2
2
2 a
1
x
x
 e dx   e
1 n  x n n1  x
n x
 x e dx   x e    x e dx
n n1 n n 1 n 2
n!
x
x
e x , n an integer, n
2
3
n 1

xn e x dx
2

dx
1

tan 1 x a b
2
b
ab
1
x

tan 1 x a b 
2b  ax2  b 
2b ab
 ax

(A.12d)
(A.12g)
(A.12h)
(A.12i)
0.
1
 sin  x  dx    cos  x 
1
 cos  x  dx   sin  x 
x
1
 x sin  x  dx    cos  x   
2
(A.12j)
(A.12k)
(A.12l)
sin  x 
(A.12m)
x 1
x 1
 sin  x  dx  2  4 sin  2 x   2  2 sin  x  cos  x 
(A.12n)
1
1
 sin  x  cos  x  dx   4 cos  x   12 cos 3 x 
(A.12o)
1
1
 sin  x  cos  x  dx  8 x  32 sin  4 x 
(A.12p)
2
2
2
2
 sin  x  sin   x  dx 
sin     x  sin     x 

,  
2    
2    
(A.12q)
 sin     x  sin     x   cos     x  cos     x 
x
sin

x
sin

x
dx

x








2
2

2



2







2



2    




 
(A.12r)
Also useful are the following definite integrals. Note that for the first one, it is the definition
of the gamma function   n  . In all the formulas below, it is assumed that Re    0 ..

x
e dx    n  , n  0
n 1  x
(A.13a)
0
  n  1  n  n  ,  1  1,   12    ,   n  1  n! for n an integer
© 2016, Eric D. Carlson
327
(A.13b)
Appendix A. Some Mathematical Tools

x e
n  x
dx 
  n  1

x e
n  x
dx 
0

x e
n  x2
n!
 n1
dx 
0

 x
e
2
 x
, n an integer
1
2
dx 


 n 1
2
n  x2   x



xe x
2
 
e

2
dx 

(A.13e)
 4 
 d   
dx 
e
  d   
 x
(A.13f)
  
e
2 
2
2
 4  

 4 
2  x2   x

(A.13g)
(A.13h)
 1
 2   2
dx  
 2
e
 2 4  

sin 2  x 
 x2 dx  

xe
(A.13d)
  n21 
n
xe

(A.13c)
 n1
0
 4 
(A.13i)
(A.13j)
Finally, there are a lot of integrals for the Dirac delta function,   x  , which should be
thought of as a function which vanishes everywhere except x
the origin. It is defined as
0 , but has a spike with area 1 at
b
 f  x    x  dx  f  0
a
for any function f  x  , assuming the range of integration includes the point x = 0. This leads to
the following more general relations for the delta function:
 f  x    x  a  dx  f  a 
 f  x    ax  b  dx 
 f  x    g  x    
xi  g 1 0
1
f b a 
a
1
f  xi 
g   xi 
1
  x
k
1
  g  x   
  x  xi 
xi  g 1  0 g   xi 
  kx  
(A.14a)
(A.14b)
(A.14c)
(A.14d)
(A.14e)
The expression xi  g 1  0  means that the sums should be performed over all points xi such that
g  xi   0 .
Appendix A. Some Mathematical Tools
328
© 2016, Eric D. Carlson
D. Calculus with Several Variables
In math we often work in one dimension; in physics, at least when we are doing realistic
calculations, there are at least three. In Cartesian coordinates, a point would be denoted by
r   x, y, z    rx , ry , rz   xxˆ  yyˆ  zzˆ ,
where x̂ , ŷ , and ẑ are unit vectors in the directions of increasing x, y, and z respectively. A
scalar function in three dimensions would be a single ordinary function like f  r   f  x, y, z  .
A vector function in three dimensions would actually be a triplet of functions A  r  , with
A  r   xˆ Ax  r   yˆ Ay  r   zˆ Az  r  .
We can then take the derivatives of scalar functions or the components of vector functions
using partial derivative, which we write as
x,
y , and
z .1 It is helpful to define a sort
of vector derivative, which we denote as
xˆ
yˆ
x
zˆ
y
z
.
This vector derivative, in three dimensions, can be used to produce a vector function from a
scalar function using the gradient, defined by
xˆ
f
f
x
f
y
yˆ
f
,
z
zˆ
or in components,
 f i 
f
.
ri
From a vector function A  r  , we can produce a scalar function by using the divergence, defined
as
A
Ax
x
Ay
Az
z
y
Ai
.
ri
i
We also define the curl, a vector quantity, as
A
xˆ
Az
y
Ay
z
yˆ
Ax
z
Az
x
zˆ
Ay
x
Ax
.
y
This can be written in components more conveniently in terms of the Levi-Civita symbol,
Eq. (A.3):
1
Note that in this notation, expressions like
f   r  make no sense. We must either write out the partial derivative
explicitly, or develop a new, more sophisticated shorthand.
© 2016, Eric D. Carlson
329
Appendix A. Some Mathematical Tools
  A i    ijk
j ,k

Aj
k
(A.15)
Finally, it is useful to define the Laplacian, a second derivative of a scalar function, given by
2
f
f
2
f
x
2
2
f
y
2
2
f
z
2
2
i
f
ri 2
.
The gradient, divergence, and Laplacian generalize in an obvious way to higher dimensions; the
curl, in contrast, does not so simply generalize. Some other combinations of two derivatives
automatically vanish, thanks to the anti-symmetry of the Levi-Civita symbol:
  f   0,
(A.16a)
    A   0.
(A.16b)
Related to these expressions, one can show that any vector function that has no curl can be
written as a divergence, and any vector function that has no gradient can be written as a curl, so
A 0
B 0
A
(A.17a)
(A.17b)
f
A
B
  fg  , can be generalized in a variety of
The product rule for single variables,  fg   f g
ways to vector functions. Three useful examples are
  fg    f  g  f g
(A.18a)
   fA    f   A  f   A
(A.18b)
  fA    f   A  f  A
(A.18c)
   A  B     A   B  A    B 
(A.18d)
  A  B   A    B    B   A  B    A    A   B
(A.18e)
Just as you can take derivatives for functions of multiple variables, you can also do integrals.
We will focus almost exclusively on volume integrals, which are a type of definite integral where
we add up the value of some function f  r  over a volume V. A volume integral would be
written something like
 f r  dV   f r  d r
3
V
V
Such an integral is really three one dimensional integrals, one over each of the three Cartesian
coordinates x, y, and z. Writing out more explicitly, you can think of the integral as of the form
 f r  d r   dx dy  dz f  x, y, z  .
3
V
In other words, we have three nested integrals, one inside the other (they can generally be
performed in any order). As set up here, one would first perform the inner z integral, then the
intermediate y integral, and finally the outermost x integral. The limits can be quite tricky; for
Appendix A. Some Mathematical Tools
330
© 2016, Eric D. Carlson
example, the innermost integral (z) may have limits that depend on the outer two variable (x and
y), while the intermediate integral may have limits that depend on the outermost variable (x), but
not the inner variable (z), and the outermost integral (x) may not depend on the other two
variables. Furthermore, when performing the integral, the variables x and y should be treated as
constants inside the inner-most integral, while z is a variable; when you perform the intermediate
integral, y is a variable, x is a constant, and z should have disappeared, and in the outermost
integral, only the variable x should appear.
For example, suppose that I were given a region V defined by the four conditions
V   x, y, z  : x  0 , y  0, z  0 , x  y  z  1 .
Now, suppose I was asked to find the volume of this region, which is
V
1d 3r
dx dy dz .
What limits should I use? For the innermost integral, the simultaneous conditions z 0 and
x y z 1 set the interval of integration as  0,1  x  y  . For the intermediate integration, we
don’t know what z is (other than positive), but clearly y 0 and x
integration, we know only that 0 x 1 . So the correct limits are
1
1 x
1 x  y
0
0
0
1d r   dx  dy 
3
V
1
1 x
1
0
0
0
y
1, and for the outermost
y 1 x
dz   dx  1  x  y  dy   dx 1  x  y  12 y 2  y 0
1
1
2
2
3
  dx 1  x   12 1  x     12  13 1  x   16 .


0
0
As a notational matter, it is worth noting that if the integral is over all space, it is common to
leave the limit symbol V out of it entirely.
Is there any multi-dimensional analog of the fundamental theorem of calculus, Eq. (A.11),
that allows you to find the integral of a derivative? There are several, in fact, depending on what
type of integral you are dealing with. The one we will need most often is Gauss’s Law, which
relates an integral over a volume V to a surface integral over the boundary of V when you are
integrating a divergence of a vector function.
   A r  dV   nˆ  A r  dS ,
(A.19)
S
where S is the boundary of the volume V and n̂ is a unit normal sticking out of the integration
region. The funky integral symbol on the right side of Eq. (A.19) merely indicates that the
integral is being taken over a closed surface. Gauss’s Theorem reduces the triple integral to a
double integral in three dimensions.
Stokes’ Theorem will also prove useful. It relates the integral of the curl of a vector field
over a surface to an integral around the boundary of that same surface:
 nˆ   A r  dS   A r   dl .
S
(A.20)
C
Once again, the number of integrals is reduced by one.
Finally, there is the divergence theorem, which assists you in performing a path integral of
the divergence of a function from one point to another along a path:
© 2016, Eric D. Carlson
331
Appendix A. Some Mathematical Tools
B
 f r   dl  f  B   f  A .
(A.21)
A
E. Coordinates in Two and Three Dimensions
In dimensions higher than one, it is common to use alternate coordinates besides the
conventional Cartesian coordinates. When you change coordinates from a set of coordinates
x   x1 , x2 , , xN  to a new set of coordinates y   y1 , y2 , , yN  , any integration over the Ndimensional volume must be transformed as well. The rule is

 x
f  x  d N x   f  x  y   det  i
 y j

 N
 d y .

(A.22)
Furthermore, all of the derivative operators must be rewritten in the new basis as well.
Let’s start with two dimensions. In addition to standard Cartesian coordinates  x, y  , the
most common set of coordinates are polar coordinates   ,   . The relations between these two
coordinate systems are
 x   cos  
   x 2  y 2 
and



.
1
 y   sin  
  tan  y x 
(A.23)
For any pair  x, y  there are multiple values of   ,   that satisfy these equations. To make the
relationships Eq. (A.23) unambiguous, we therefore restrict the coordinates   ,   by
0
, 0
2 .
(A.24)
The inverse tangent must be considered carefully, since there will be two values of  in the range
0
2 that satisfy   tan 1  y x  . The ambiguity can be removed with the help of
if y is positive and
y
sin : we choose 0
2 if y is negative. If y vanishes,
then we pick
if x is negative. If x and y both vanish, then is
0 if x is positive and
ambiguous; it is a bad spot in the coordinate system.
Scalar functions in one coordinate system are unchanged in the new coordinate system, but a
vector quantity will now take the form
A
ρˆ A
φˆ A .
These components can be related to the ordinary Cartesian components by
Ax
Ay
cos
sin
sin
cos
A
,
A
A
A
cos
sin
sin
cos
Ax
.
Ay
(A.25)
We present, without proof, formulas for the gradient, divergence, Laplacian, and integral in
these new coordinates:
Appendix A. Some Mathematical Tools
332
© 2016, Eric D. Carlson
f
f
A 
2
f
ρˆ
1 f
φˆ ,
(A.26a)
1 
1 A
 A  
,

 
 
1
f
1 2f
2
2
(A.26b)
,
(A.26c)
 f r  d r    d   d f   ,  .
2
(A.26d)
If one integrates over all space, the limits of integration in Eq. (A.26d) are implied by Eq. (A.24).
In three dimensions, there are two coordinate systems that are commonly used in addition to
Cartesian. The first are cylindrical, where  x, y, z  are replaced with   ,  , z  , defined by the
relations
   x2  y 2 
 x   cos  




1
 y   sin   and   tan  y x  ,
 zz 


zz




(A.27)
where the last expression in each triplet means simply that the coordinate z is unchanged. These
coordinates are obviously very similar to polar coordinates in two dimensions. The restrictions
Eqs. (A.24) apply in this case as well, and the inverse tangent must be interpreted carefully in a
manner similar to the comments after Eqs. (A.24). Vector functions in the two coordinates are
related in a manner almost identical to Eq. (A.25):
Ax
Ay
Az
cos
sin
0
sin
cos
0
0 A
0 A ,
1 Az
A
A
Az
cos
sin
0
sin
cos
0
0 Ax
0 Ay .
1 Az
The various differential and integral relations look nearly identical to Eqs. (A.26), except we
include the curl:
f
f
zˆ ,
z
1 
1 A Az
A 
 A  

,

 
  z
A 
 1 Az A   A Az 
1 
 A  


zˆ ,
 ρˆ  
 φˆ     A  
  
 
   z   z  
2
1
f
1 2f
f
2
f
,
2
2
2
z
f
ρˆ
1 f
φˆ
 f r  d r    d   d  dz f   , , z  .
3
(A.28a)
(A.28b)
(A.28c)
(A.28d)
(A.28e)
Spherical coordinates  r , ,   are related to Cartesian coordinates by the relations
© 2016, Eric D. Carlson
333
Appendix A. Some Mathematical Tools
 r  x2  y 2  z 2 
 x  r sin  cos  






1
2
2
x  y z ,
 y  r sin  sin   and   tan
 z  r cos q 


1


   tan  y x 



(A.29)
with coordinate restrictions
0
r, 0
, and 0
2 .
The ambiguity in the variable is resolved in a manner identical to that in polar coordinates in
two dimensions or cylindrical in three. There is no corresponding ambiguity for . They are
related to cylindrical coordinates by
 r   2  z2 
   r sin  




1
     and   tan   z  .
 z  r cos  


 




(A.30)
Vector functions in spherical coordinates are related to Cartesian by
Ax
Ay
Az
sin cos
sin sin
cos
cos cos
cos sin
sin
sin
cos
0
Ar
A ,
A
Ar
A
A
sin cos
cos cos
sin
sin sin
cos sin
cos
cos
sin
0
Ax
Ay .
Az
and they are related to cylindrical coordinates by
A
A
Az
sin
0
cos
cos
0
sin
0 Ar
1 A ,
0 A
Ar
A
A
sin
cos
0
0
0
1
cos
sin
0
A
A .
Az
The various differential and integral identities in this coordinate system look like
1 f ˆ
1
f
(A.31a)
θ
φˆ ,
r
r sin
1 
1 
1 
(A.31b)
 A  2  r 2 Ar  
A
sin  A  
r r
r sin  
r sin  
A 

A 
1 
1  1 Ar 
1 
 A 
sin  A     rˆ  
  rA  θˆ    rA   r  φˆ ,


r sin   
 
r  sin   r
r  r
 

(A.31c)
2
2
1 
1
 
f 
1
 f
2 f 
rf   2
,
(A.31d)
 sin 
 2 2
2 
r r
r sin   
  r sin   2
1   f 
1
 
f 
1
2 f
2 f  2  r 2   2
sin


.
(A.31e)


r r  r  r sin   
  r 2 sin 2   2
f
f
rˆ
r
There is also a Dirac delta functioin in three dimensions, defined as
Appendix A. Some Mathematical Tools
334
© 2016, Eric D. Carlson
 3 r     x   y   z  .
The relations corresponding to the one dimensional Eqs. (A.14) are then
 f r   r  a  d r  f a 
3


3
f  r   3  r  a  d 3r  det 
f  r   3  g  r  d 3r 
 3  g  r  

rk g 1  0

det  gi
det  gi rj r

rk g 1  0
1
(A.32a)

r 
f
a
1
1
j r
k
(A.32b)
f  rk 
(A.32c)
1
 3  r  rk 
k
(A.32d)
F. Fourier Transforms
We start with the following “identity”:

e
ikx
dx  2  k  .
(A.33)

This identity cannot really be proven, simply because it isn’t quite true, but it can be justified by
considering the following integral, which can be evaluated with the help of Eq. (A.13f):

ikx
 e dx  lim

 0

ikx  x
 e dx  lim
2
 0

 ik 
e

2
4
 lim
 0
 k
e

2
4
.
(A.34)
Now, as  gets small, Eq. (A.34) describes a slender and tall function, which is essentially a
Dirac delta function. To confirm this, we simply have to make sure that it has area 1, which can
be checked by using Eq. (A.13f) again:

 dk

 k
e

2
4


4  2 .

This shows that the right side of Eq. (A.34) is 2  k  , confirming Eq. (A.33).
Let f  x  be any reasonably smooth function. We will normally also demand that f vanish
not too slowly, say, faster than 1/x at large x. Then define the Fourier transform f  k  as
f k  



dx
f  x  eikx .
2
Consider now the Fourier transform of the Fourier transform, which we will evaluate with the
help of Eq. (A.33):
© 2016, Eric D. Carlson
335
Appendix A. Some Mathematical Tools
f  x 





 ikx
1
iky
  f  y  e dy  e dk  2

dk
1
f  k  eikx 
2
2
  ik  x  y  
   e dk  f  y  dy
  


   x  y  f  y  dy  f   x  .

Up to sign, the Fourier transform of a Fourier transform is the function back again. We therefore
summarize this information as two relations, here presented together.
f k   


 dk
dx
f  x  eikx , f  x   
f  k  eikx .

2
2
(A.35)
One other relation that is not hard to prove is that the magnitude squared integral of a function
and its Fourier transform are the same. The proof is given below.


f  k  dk 
2



f  k  f  k  dk 
*


1
2




  dx
   dy

 ikx
f
x
e
dx
f  y  e iky dy  dk



  2
   2








f  x  dx  f *  y  dy  eikx eiky dk 


 f  x  f  x  dx   f  x 
*

2


f  x  dx  f *  y  dy  y  x 

dx .
(A.36)

All of these formulas can be generalized to three (or more) dimensions easily, so we give the
resulting equations without proof.
e
ik r
f k   
d 3r
 2 
3/2

336
d 3r   2   3  k  ,
3
f  r  eikr ,
f r   
(A.37a)
d 3k
 2 
f k  d 3k   f r  d 3r .
2
2
© 2016, Eric D. Carlson
3/2
f  k  eikr ,
(A.37b)
(A.37c)
Index
accidental degeneracy, 118
addition of angular momentum, 128
dimensionality check, 130
resulting quantum numbers, 130
adjoint operator, 42
Aharonov, Akir, 155
Aharonov-Bohm effect, 155
and magnetic monopoles, 156
derivation, 153
angular momentum
addition of. See addition of angular
momentum
commutator, 105
commutator with scalar operator, 133
commutator with spherical tensor, 134
commutator with tensor operator, 134
commutator with vector operator, 133
eigenvalues, 106, 107
generalized, 106
identities, 106
internal. See spin
j = ½, 108
j = 1, 108
j = 2, 122
ladder operators, 106
effect on eigenstates, 107
spherical coordinats, 110
orbital, 128
spherical coordinates, 109
squared, 106
total, 128
z-component
eigenvalue, 99
in spherical coordinates, 100
anti-electron. See positron
anti-muon, 120
anti-particle, 283
anti-proton, 120
anti-symmetric, 161
anti-symmetrization operator, 162
basis, 37
changing, 46, 52
matrix notation, 46
postulates, 58
© 2016, Eric D. Carlson
complete. See complete basis
complete, orthonormal. See complete,
orthonormal basis
independent, 37
orthogonal. See orthogonal basis
orthonormal. See orthonormal basis
tensor product space, 158
with spin, 126
Bell, John Stewart, 195
Bell's inequality, 195
Bell's Inequality, 195
Bessel equation, 101
Bessel functions, 101
block diagonal, 49, 50
Bohm, David, 10, 155
Bohr radius, 116
Bose, Setyndra Nath, 165
Bose-Einstein Condensate, 165
boson, 161
complete, orthonormal basis, 163
boundary conditions
Dirac delta function, 28, 94
finite potential, 22, 25
infinite potential, 26, 27
bra, 34
bra-ket notation, 34
center of mass operators, 114
CG coefficients. See Clebsch-Gordan
coefficients
characteristic equation, 50
Clebsch-Gordan coefficients, 130
adding ½ to ½, 133
adding j = ½, 132
identities, 131
non-zero constraints, 130
normalization, 131
recursion relations, 131
symmetry, 132
coherent state, 75
collapse of state operator, 187, 192, 193
collapse of state vector, 56, 60, 176, 190,
191, 194, 197, 198, 201
and normalization, 58
commutator, 41
337
Index
angular momentum, 105
identities, 42
ladder operators, 72, 77, 80
position and momentum, 41
spin, 125
complete basis, 37
complete set of commuting observables, 49
complete, orthonormal basis, 38, 39, 41, 44,
48, 49, 55, 67, 186
boson, 163
fermion, 163
completeness relation, 41
spherical harmonics, 113
complex numbers
classical fields, 6
for classical waves, 1
wave functions, 6
composite particles, 165
conservation of probability
in a volume, 21
local, 20
Copenhagen interpretation, 10, 54, 202
covector, 34
cross-product
and spherical tensors, 136
crystal momentum, 93
CSCO, 49
Davisson, Clinton, 5
de Broglie hypothesis, 2, 6
de Broglie, Louis, 5, 10
Debye, Peter, 6
degeneracy
hydrogen, 118
delayed choice experiments, 196
delta function. See Dirac delta function
density matrix. See state operator
Dirac delta function
boundary conditions, 28, 94
Dirac equation, 149
Dirac notation, 33
Dirac quantization condition, 156
dot product
and spherical tensors, 135
Ehrenfest theorem, 65
with electromagnetic fields, 148, 149
eigenstate, 19, 47
Index
harmonic oscillator. See harmonic
oscillator, eigenstate
uncertainty, 69
eigenvalue, 47, 55
characteristic equation, 50
Hermitian operator, 47
matrix notation, 49
unitary operator, 47
eigenvector, 46, See also eigenstate
Hermitian operator, 48
matrix notation, 49
unitary operator, 47
Einstein Podolsky Rosen Paradox. See EPR
paradox
Einstein separability, 194
Einstein, Albert, 1, 165
electron
magnetic dipole moment, 152
spin, 126
energy shift irrelevance, 68
entangled state, 190
entropy, quantum mechanical, 186, 188, 193
EPR paradox, 190
expectation value, 12, 62
evolution, 64
from state operator, 187
momentum, 12
position, 12
faster than light, 190, 192, 194
Fermi degeneracy pressure, 167
Fermi energy, 166, 167
Fermi momentum, 166, 167
Fermi, Enrico, 166, 275
fermion, 161
complete, orthonormal basis, 163
fermions
example particles, 162
Feynman path integral, 178, 179
connection to classical mechanics, 180
fine structure constant, 118
finite square well, 89
Fourier transform
solving free Schrödinger equation, 17
wave function, 11
free will, 196
gauge choice, 147
338
© 2016, Eric D. Carlson
gauge function, 143, 145
gauge invariant, 147, 148, 150, 153
gauge transformation, 144, 146, 155
generator
for rotation, 97
for translation, 97
Gerlach, Walthier, 150
Germer, Lester, 5
gyromagnetic ratio, 149, See also magnetic
dipole moment
Hamiltonian, 40, 55
energy shift, 68
harmonic oscillator. See harmonic
oscillator, Hamiltonian
multiple particles, 159
with electromagnetic fields, 147
with electromagnetic fields and spin, 149
harmonic oscillator, 65
coherent state. See coherent state
complex, 80
complex correspondences, 81
complex Hamiltonian, 80
coupled, 78, 82
eigenstate, 73, 74, 75
eigenvalues, 73, 78
expectation values of operators, 74
from general potential, 71
Hamiltonian, 71, 72
ladder operators. See ladder operators
spherical, 123
uncoupled, 77
Heisenberg formulation, 181
Schrödinger equation, 182
Hermitian conjugation
bras and kets, 35
complex numbers, 36
general procedure, 43
matrix notation, 46
operator, 42
Hermitian operator, 43, 54
eigenvalue, 47
eigenvector, 48
matrix notation, 46
Hertz, Heinrich, 1
hidden variables, 194
Hilbert space, 33
© 2016, Eric D. Carlson
hydrogen
degeneracy, 118, 127
division of the Hamiltonian, 114
energy eigenvalues, 118
Hamiltonian, 114
-like. See hydrogen-like
principal quantum number, 118
quantum numbers, 118, 127
radial equation, 115
radial wave function, 117
explicit, 121
recursion formula, 117
spin-orbit coupling. See spin-orbit
coupling
uncertainty and stability, 13
wave function, 117
hydrogen-like
energy eigenvalues, 119
size, 119
independent basis, 37
infinite square well, 25
1D, 26, 67
3D, 27
circular, 101
square, 91
wave function, 26
inner product, 32
interference, 3, 5, 154, 155
internal angular momentum. See spin
kernel. See propagator
ket, 34
kinematic momentum, 146
Kronig Penney model, 94
energy eigenvalues, 95
ladder operators, 72
acting on eigenstates, 73, 78
angular momentum. See angular
momentum, ladder operators
commutator, 72, 77, 80
momentum and position operators, 72
Landau levels, 151
Landau, Lev, 151
Lennard-Jones 6-12 potential, 81
Levi-Civita symbol, 105, 133
Lorentz force, 145, 149
lowering operator. See ladder operators
339
Index
magnetic dipole moment, 149
electron, 152
neutron, 149
proton, 149, 232
magnetic field, 145
and vector potential, 145
effects on an atom, 152
free particle in, 150
Lorentz force, 149
measuring spin, 150
magnetic flux, 154
magnetic monopole, 156
many worlds interpretation, 10
Many Worlds Interpretation, 201
matrix notation, 44
basis change, 46
bra, 45
eigenvalues and eigenvectors, 49
Hermitian conjugation, 46
Hermitian operator, 46
ket, 45
operator, 45
Maxwell’s equations, 6, 143
measurement, 4, 10, 55, 196
measuring device, 197, 199
mixed state, 185
momentum
kinematic, 146
operator, 8, 40
in terms of ladder operators, 72
photon, 1
probability density, 11
wave-number relation, 2, 6
multiple particles
tensor product space, 159
muon, 120
neutron
magnetic dipole moment, 149
spin, 126
neutron star, 168, 172
normalization
1D, 9
3D, 9
and Schrödinger equation, 58
collapse of state vector, 58
radial wave function, 111
Index
nuclear size, 119
observable, 48, 54, 55
commuting, 62
simultaneous, 62
Ockham, William of, 203
Ockham's razor, 203
operator, 39
acting on bras, 41
diagonalizing, 49
Hermitian. See Hermitian operator
linear, 39
product, 39
self-adjoint. See Hermitian operator
unitary. See unitary operator
orbital angular momentum, 128
orthogonal basis, 37
deriving, 37
orthogonal matrix, 79, 85
orthonormal basis, 37
finding, 38
outer product, 40
parity, 40, 44, 88
effect on operators, 88
spherical harmonics, 113
parity invariant, 89
Paschen-Back effect, 153
Pauli Exclusion Principle, 163
Pauli matrices, 108
multiplication rules, 108
permutation operator, 160
photoelectric effect, 1
photon, 1, 3
momentum, 1
pilot wave, 10
Planck constant, 1
reduced, 1
Planck, Max, 1
position bra, 34
position operator, 40
in terms of ladder operators, 72
positive definite. See vector space, positive
definite
positron, 120, 159, 283
positronium, 120, 159
ortho- and para-, 160
postulates
340
© 2016, Eric D. Carlson
basis choice, 58
consistency, 58
continuous variables, 56
list, 55
potential, 7
principal quantum number, 118
probability amplitude. See wave function
probability current, 19, 20, 23, 156
probability density
1D, 9
3D, 9
momentum space, 11
propagator, 176
free particle, 177
how to compute, 177
Schrödinger equation, 177
proton, 13
magnetic dipole moment, 149, 232
quarks, 149
spin, 126, 233
pseudovector operator, 88
pure state, 185
quantum electrodynamics, 152
quantum tunneling, 24
transmission probability, 25
quantum Zeno effect, 69
radial wave function
hydrogen, 121
normalization, 111
polar equation, 100
spherical equation, 110
raising operator. See ladder operators
reduced mass, 114
reduced matrix element, 137
relative motion operators, 114
relativity, 1, 118, 119, 149, 151
roots of unity, 92
rotation, 86, 124
acting on operators, 86, 87
from angular momentum, 98
identities, 86
on wave function, 86
rotation invariance
continuous, 99
discrete, 91
rotation matrix, 85
© 2016, Eric D. Carlson
explicit form 2D, 86
explicit form 3D, 86
orthogonal, 85
proper and improper, 85
scalar operator, 133
commutator with angular momentum, 133
scalar potential, 143, 145
Schrodinger equation
for time evolution operator, 174
Schrödinger equation
1D, 8, 16
3D, 8, 16
for propagator, 177
for state operator, 186
free, 1D, 7
free, solution of, 16
general, 40, 55
Heisenberg formulation, 182
linearity, 16
time-independent, 18, 66
solution of, 19, 66
with electromagnetic fields, 146
Schrödinger, Erwin, 6
Schwartz inequality, 32, 52
self-adjoint operator. See Hermitian operator
separable system, 190
SI units, 13, 114, 344
Slater Determinant, 164
solenoid, 154
spectrum, 190
spherical coordinates, 109
spherical harmonics, 111
and coordinate polynomials, 122
azimuth dependence, 111
completeness relation, 113
explicit, 121
general formula, 112
orthonormality, 113
parity, 113
products, 138, 139
quantum numbers, 111
spherical symmetry
Hamiltonian, 108
radial wave function, 110
spherical tensor, 134
commutator with angular momentum, 134
341
Index
multiplying, 135
spin, 125
and symmetry, 162
commutators, 125, 127
effect on states, 126
electron, 126
neutron, 126
proton, 126
quantum numbers, 126
s = ½, 126
unpolarized, 188
spin matrix, 124
from spin, 125
identities, 124
spin-orbit coupling, 127
spin-statistics theorem, 162, 165
state operator, 10, 185
expectation value, 187
postulates, 187
Schrödinger equation, 186
state vector, 10, 55
collapse. See collapse of state vector
reduction. See collapse of state vector
stationary phase approximation, 180
stationary state, 67, 75
step boundary, 21
reflection and transmission, 23, 24
Stern, Otto, 150
Stern-Gerlach experiment, 150, 194, 196
Stokes’ Theorem, 154
symmetric, 161
symmetrization operator, 162
symmetry
complex eigenvalues and degeneracy, 92
continuous, 98
generator. See generator
mixed, 161
parity. See parity
rotation. See rotation
translation. See translation
tensor
commutator with angular momentum, 134
rank, 134
spherical. See spherical tensor
tensor operator, 134
tensor product, 158
Index
tensor product space, 158
basis, 158
multiple particles, 159
time evolution operator, 174
Schrödinger equation, 174
time-independent Schrödinger equation. See
Schrödinger equation, time-independent
total mass, 114
trace, 183
partial, 184, 191
translation, 83
effect on operators, 84, 85
from momentum, 97
identities, 83
on eigenstates, 85
on wave function, 83
translation invariance
continuous, 99
discrete, 93
true vector, 88
uncertainty, 2, 63
classical, 2
coherent state, 81
definition, 12
eigenstate, 69
harmonic oscillator, 75, 76
momentum, 13
position, 13
uncertainty relation
general, 64
position and momentum, 2, 64
unitary operator, 44
eigenvalue, 47
eigenvector, 47
unpolarized spin, 188
vector operator, 87, 133
commutator with angular momentum, 133
pseudovector, 88
true vector, 88
vector potential, 145
vector space
3D Euclidean, 32, 33, 45
complex, 55
positive definite, 32, 55
subspace, 35
superspace, 35
342
© 2016, Eric D. Carlson
wave function, 9
meaning, 8
reduction. See collapse of state vector
wave number
covector, 34
© 2016, Eric D. Carlson
eigenstate, 35
wave packet, 2, 23, 148
white dwarf, 168, 172
Wigner-Eckart Theorem, 136, 137
reduced matrix element, 137
343
Index
Units and Constants
Fundamental SI Units
Measures
Name Abbr.
Length
meter
m
Time
second
s
Mass
kilogram kg
Temperature Kelvin
K
Charge
Coulomb
C
Physical Constants
Name
Planck’s Constant
Planck’s Reduced Constant 
Metric Abbreviations
Name Abbr. Multiple
tera
T
1012
giga
G
109
mega M
106
kilo
k
103
centi
c
10-2
milli
m
10-3
micro

10-6
nano
n
10-9
pico
p
10-12
femto
f
10-15
Derived SI Units
Measures
Name Abbr.
Force
Newton N
Energy
Joule
J
Power
Watt
W
Frequency Hertz
Hz
Pressure
Pascal
Pa
Current
Ampere A
El. Potential Volt
V
Mag. Field Tesla
T
Symbol
h
Speed of light
Fundamental Charge
Coulomb’s Constant
Permittivity of free space
Permeability of free space
Avogadro’s number
Electron mass
c
e
ke
0
µ0
NA
me
Proton mass
mp
Neutron mass
mn
Boltzmann’s Constant
kB
Equiv.
kgm/s2
Nm
J/s
s-1
N/m2
C/s
J/C
Ns/m/C
Value
86,400 s
3.1558107 s
1.602210-19 J
1.660510-27 kg
= 931.494 MeV/c2
= 6.582110-16 eVs
2.9979108 m/s
1.602210-19 C
8.9876109 Nm2/C2
8.854210-12 C2/N/m2
410-7 Ns2/C2
6.02211023 /mol
9.109410-31 kg
= 511.00 keV/c2
1.672610-27 kg
= 938.27 MeV/c2
1.674910-27 kg
= 939.56 MeV/c2
1.380710-23 J/K
= 8.617310-5 eV/K
Useful Combinations
Combination
Value
hc
1.986410-25 Jm
= 1239.8 eVnm
c
3.161510-26 Jm
= 197.33 eVnm
a0 

Non SI Units
Name
Abbr.
Day
d
Year
y
Electron volt
eV
Unified mass unit u
Value
6.626110-34 Js
= 4.135710-15 eVs
1.054610-34 Js
2
ke mee2
kee2
c
MeV/c
MeV/c2
0.052918 nm
1
137.04
= 0.0072974
5.34410-22 kgm/s
1.7826610-30 kg