9. Time-dependent Perturbation Theory
Background
Dirac delta and Definite integrals
Concepts of primary interest:
LS coupling vs. jj coupling:
Fermi’s Golden Rule is actually Dirac’s
Sample calculations:
SC1 - 
Helpful handouts:
Tools of the trade:
This handout is keyed to Griffiths Introduction to Quantum Mechanics, 2nd Ed. It is
not designed to be used independently.
Notation: The choice of letters originates from a now-obsolete system of categorizing
spectral lines as "sharp", "principal", "diffuse" and "fine", based on their observed fine
structure: their modern usage indicates orbitals with an azimuthal quantum number, l, of 0,
1, 2 or 3 respectively. After "f", the sequence continues alphabetically "g", "h", "i"… (l = 4, 5,
6…), although orbitals of these types are rarely required.
http://en.wikipedia.org/wiki/Electron_configuration
Thanks to Katherine DePooter (2009).
Electric Dipole Selection Rules:
The photon is a spin 1, odd parity particle and as such carries one unit of angular
momentum. In the case that the initial state of a system decays to a final state plus a
photon, the initial and final states of the system have opposite parity in the case that
the relative orbital angular momentum of the photon and the final-sate system is
zero.1
1
This approximation is appropriate for optical transition in atoms, not for gamma ray transitions in nuclei.
for use with Griffiths QM
Contact: tank@alumni.rice.edu
The atomic state has an angular momentum Ji and a final angular momentum of Jf.
Following the rules for the addition of angular momentum,
J f  J i  J  J f  J i ; J i  1 In the case J i  0; J f  1
Summary: J = 0,  1; except Ji = 0 requires J = 1 or Jf = 1
Orbital angular momentum: LS (SPIN-ORBIT) Coupling
n unrestricted
For the single electron making the transition,  = 1, m  0, 1 ; ms = 0
Overall orbital angular momentum: L = 0, 1; Li = 0  L = 1 or Lf = 1
S = 0
Orbital angular momentum: Intermediate Coupling (say f states and above) In this
regime, the coupling is often called j-j, and spins can be flipped due to the magnetic
coupling of the spins to the orbital angular momentum
L = 0, 1, 2 if S = 1
Relation of the Selection Rules to Matrix Elements
Y10 ( , ) 
1 3
cos 
2 
1 3
sin  e i
2 2
Y1,1 ( , ) 
pˆ  er
r
 
2
3
Y1,1 ( ,  ) 
2
3
 
Y1, 1 ( ,  ) iˆ  i
2
3
Y1,1 ( ,  ) 
2
3

Y1, 1 ( ,  ) ˆj 
4
ˆ
3 Y10 ( ,  ) k

Clearly the perturbation has an angular momentum one character. We factor the 
dependence out of the functions.


2
0
0
0
 n ' lm r  nlm   Rnl (r ) Rnl (r )r 2 dr  flm ( ) f lm ( )sin  d  e  im e i{0,1} eim d
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
2
0
e im e i{0,1} eim d generates the m  0, 1 rule
Using the spherical harmonics with their ,  dependences:

2
0
0
 n ' lm r  nlm   Rnl (r ) Rnl (r )r 2 dr 
2

0
0
 


0
Ylm ( , ) Y1,{0,1} ( , )Ylm ( , )sin  d d 
Ylm ( , ) Y1,{0,1} ( , )Ylm ( , ) sin  d d generates the  = 1 rule
It follows from the recurrence relation below and the orthogonality of the Ym.
m 1
m 1
 Y10 ( , )  m
m
Y
(

,

)

a
Y
(

,

)

a|m, ,1Y m1 ( , )  = 1 rule
 1



|m, ,1 1
m m 1
m m 1
Y1 ( , ) 
if  1
The same rules can be derived by formally evaluating commutators. This approach is
used by Griffiths and many others. Griffiths commutator approach:
[ Lˆz , x]  Lˆz x  xLˆ z  i y; [ Lˆ z , y ]  i x; [ Lˆ z , z ]  0
 n ' lm [ Lˆz , z ]  nlm   n ' lm Lˆz z  zLˆz  nlm  (m  m)  n ' lm z  nlm  0
 n ' lm z  nlm  0 unless m  0

  [ Lˆz , x]    n ' lm Lˆz x  xLˆz   (m  m)   x   i   y 
(m  m)   x   i   y 
  [ Lˆz , y]    n ' lm Lˆz y  yLˆz   (m  m)   y   i   x 
(m  m)   y   i   x 
(m  m) 2   x   1   x  ; (m  m) 2   y   1   y 
  x  and   y   0 unless (m  m) 2  (m) 2  1  m  1
Further (m  m)   x   i   y  means that one need only compute   x  or
  y  , not both!
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Hydrogen Atom Selection Rules
m 1
m 1
 Y10 ( , )  m
m
m
 1
Y ( , )   a|m, ,1Y 1 ( , )   a|m, ,1Y 1 ( , )
m m 1
m m 1
Y1 ( , ) 
if  1

Y ( , ) 
2
3 1,1
8
3
 
Y ( , ) iˆ  i
2
3 1,1
Y ( , ) 
2
3 1,1
Y1,1 ( , )  sin  e i

Y ( , ) ˆj 
2
3 1, 1
4
3 Y10 ( ,  )

1
3
2

Y ( , ) kˆ
4
3 10
cos 
m  0,1
l  1
n  no restrictio ns
Multi-electron Atom Selection Rules
L  1
S  0
J  0,1
(0 to 0 forbidden)
M J  0,1 (0 to 0 forbidden for J  0)
Relative Intensities for Zeeman components (Electric Dipole Radiation):
Following Shore and Menzel in Principles of Atomic Spectra
For transitions from and initial state i to a final state f the intensity of the radiation is
N ( J M ) ck2
4
 J Q (1)  J 
2
2
( J M ,1m|JM )2 ½(1  cos  ) m  1


2
2 J 1
m0
 sin 
The initial state i has quantum numbers J Mwhere  is all the quantum numbers
except for the angular momentum specific values J M
The final state f has quantum numbers JMwhere  is all the quantum numbers
except for the angular momentum specific values J M
J M) is the number of atoms in the initial state
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k is the wave-number
 J Q (1)  J  is the reduced matrix element, a value independent of M and M.
(J Mm|JM) is the Clebsch-Gordan coefficient for adding JMand m to get JM.
 is angle between the magnetic field and the line of sight. For the case of our Zeeman
experiment,  = 90o.
Relative Intensities in the Zeeman Effect:
This section demonstrates a computation method that reproduces the relative
intensities claimed in the manual for the Pacific Scientific experiment. A revised
version will be attempted in which the perturbation only operates on the orbital
portions of the wavefunction. The predictions of the two methods are to be
compared with experiment.
Consider the 5460.7 Å 3S  3P2 line in the spectrum of mercury. If the mercury is
1
placed in a strong uniform magnetic field, the 3S and 3P2 levels are each split into
1
several levels.
Into how many levels is the 3S level split? splits on mJ2J + 1 
1
Into how many levels is the 3P2 level? splits on mJ2J + 1= 5.
Compute the g factors for the 3S {J=1; S=1; L=0; g = 2}. 3P2 {J=2; S=1; L=1; g = 3/2}
1
The selection rule for allowed level to level transitions is: m = 0, 1
The magnetically split levels as corresponding to energies:
E0 (3 S1 )   k B Bext  E0 (3 S1 )  g B Bext mJ  E0 (3 S1 )  2 B Bext mJ
E0 (3 P2 )   j B Bext  E0 (3 P2 )  g B Bext mJ  E0 (3 P2 )  3 2 B Bext mJ
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3
|1 1ñ
S1
3

|1 -1ñ
P2
g=2
mJ = -1, 0, 1
E  E0 (3 S1 )  E0 (3 P2 )    B Bext
|1 0ñ
|1 +1ñ
3
E0 ( S1 )  2  B Bext
3
E0 ( S1 )
3
E0 ( S1 )  2  B Bext
|1 0ñ
|1 -1ñ
|2 2ñ
|2 1ñ
|2 0ñ
|2 -1ñ
|2 -2ñ
  2 to2 by ½
E0 (3 P2 )  3 B Bext
E0 (3 P2 )  3 2 B Bext
E0 (3 P2 )

E0 (3 P2 )  3 2 B Bext
E0 (3 P2 )  3 B Bext

g = 3 /2 ;
mJ = -2, …. , 2
The relative intensities of the lines depend on the direction from which they are
viewed. For that reason, the form of the transition rate formulae prior to averaging
over directions must be used.
ba3
I ab 
 oc3
 fJfMf
p  eˆ  i J i M i
2
3
e2 ba

 oc3
 f J f M f r  eˆ  i J i M i
2
The overall angular momentum quantum numbers for the initial and final states are
displayed explicitly. The symbols i and f represent all the other quantum numbers
that remain fixed as the angular momentum options are investigated. For our
discussion, the only requirement is that i and f represent states of opposite parity.
The assumption is that the matrix element factors. This conjecture is supported by the
results for the n = 2 to n = 1 transition rates for atomic hydrogen that are computed in
the appendix.
I ab
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3
e2 ba

 oc3
 f r i
2
J f M f rˆ  eˆ J i M i
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2
Ch9-6
Note that ê is the polarization direction of the electric field.
The relative intensities follow as they are just the absolute squares of the ratios of the
angular momentum matrix elements. Let’s recall the angular momentum nature of the
position vector.
r  xiˆ  y ˆj  z kˆ  r (sin  cos  iˆ  sin  sin  ˆj  cos kˆ
r r

2
rˆ 
2
3 {Y1
3 {Y1
1
1
( , )  Y11 ( , )}iˆ  i
( , )  Y11 ( , )}iˆ  i
2
2
1
3 {Y1
1
3 {Y1
( , )  Y11 ( , )} ˆj 
( , )  Y11 ( , )} ˆj 
4
4
0
3 Y1
0
3 Y1
( , ) kˆ

( , ) kˆ
y
z
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Ch9-7
Consider the  lines which, when viewed from the transverse direction (in along the x
axis) are polarized parallel to the magnetic field. That is: they are polarized in the z
direction.
J f M f rˆ  kˆ J i M i
2
 JfMf
4
0
3 Y1 ( ,  ) J i M i
2
The perturbation z has a Y10 or |1 0ñ angular momentum character. The radial parts of
the wavefunctions are the same for the various Ym as long as  remains fixed. For the
chase above, Y10 or |1 0ñ is to be combined with |Ji Miñ to make |Jf Mfñ.
The three  lines have the following angular momentum characters:
| 1 1ñ  | 1 0ñ | 2 1ñ| 1 0ñ  | 1 0ñ | 2 0ñ and | 1 -1ñ  | 1 0ñ | 2 -1ñ.
Using the 1  1 Clebsch-Gordan table:
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| 1 1ñ  | 1 0ñ | 2 1ñ| 1 0ñ  | 1 0ñ | 2 0ñ and | 1 -1ñ  | 1 0ñ | 2 -1ñ.
11  10 
10  10 
2
3
21 
1
2
1
2
11
20  0 10 
1  1  10 
1
2
2 1 
1
2
1
3
00
11
Squaring the amplitudes, the intensities are predicted to be in the ratios:
1
/2 : 2/3 : 1/2 or 3:4:3.
For the + lines we only see the y projection of the polarization when viewed
transversely. Use i
2
1
3 {Y1
( , )  Y11 ( , )} . Note that the coefficients for the Y’s is
less by a factor of the square root of two (
2
3
vs.
4
3
) which is equivalent to a
factor of 2 in the predicted relative intensity.
| 1 1ñ  | 1 1ñ | 2 2ñ| 1 0ñ  | 1 1ñ | 2 1ñ and | 1 -1ñ  | 1 1ñ | 2 0ñ.
11  11  22
This 100% times 1/2 means the same as the 1/2 lines in the  line set.
10  11 
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1
2
21 
1
2
11
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Ch9-9
1  1  11 
1
6
20 
1
2
10 
1
3
00
Using the 1  1 Clebsch-Gordan table, the matrix elements are 1, 2-½, 6-½
corresponding to ratios of 1: 1/2 : 1/6 or, dividing by 2 to scale to the  lines,
1
/2: 1/4 : 1/12. For the collection of all the lines, we have:
1
/12: 1/4 : 1/2:1/2 : 2/3 : 1/2:1/2: 1/4 : 1/12
Scaling by 3/2 these become: 1/8: 3/8 : 3/4:3/4: 1 : 3/4:3/4: 3/8 : 1/8

! PERFECT AGREEMENT WITH THE CHINESE MANUAL!
Relative Intensities of the lines in a Zeeman pattern
First, consider an atom with a single active electron. Following the lab setup, the
active perturbation is z for the  lines and y for the  lines. 
Consider a mythical atom with a fine structure that leads to a 2S1/2 and 2P1/2 states with
the same energy with no applied magnetic field. Lying at a higher level is a 2P3/2 state.
The level diagram is not drawn to scale. The energy separation between the 2P3/2 state
and the 2S1/2 and 2P1/2 states is E2 - E1 = 0.5 eV.
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Mythical Atom Fine Structure/Zeeman Level Diagram
mj= 3/2
mj= 1/2
2
E2
P3/2
g B B
mj= -1/2
mj= -3/2
n0
2
2
E1 2
S1/2
P1/2
g' B B
g'' B B
S1/2 mj= 1/2
2
P1/2 mj= 1/2
2
P1/2 mj= -1/2
2
S1/2 mj= -1/2
Exercise: a.) Compute the g factors for the levels. b.) What is the frequency of a 0.5
eV photon? c.) A magnetic field of strength 1 Tesla is applied. Label each Zeeman
level by its frequency shift per Tesla relative to the zero field level. For example, the
2P
2
3/2
, mj = - 3/2 state might be at – 28 GHz/T. d.) Suppose that the atom is initially in the
P3/2; mj = 3/2 level. What are the allowed transitions that the atom/electron can make?
e.) Repeat for the other three mj levels.
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Electric Dipole Selection Rules state that P to S transitions are allowed while P to P
transitions are forbidden. Remember that spins must not flip. Compute the angular part
of the matrix elements for the allowed transitions. The angular parts are associated
with the orbital angular momentum, not the total angular momentum. Find the
relative intensities of the allowed transitions.
The Battle Plan:
The intensities are computed for viewing along the x axis with the magnetic field in
the z direction. The  lines are horizontally polarized (z direction) while the  lines
appear vertically polarized (y direction).
1.) Decompose each level into its orbital angular momentum ket multiplying its spinor
representation. For example, 2P3/2; mj = ½ 
11 ½  ½ 
1
3
2
3
10 ½½ . In each
pair, the first ket is the orbital ket related to the spatial coordinates  and , and the
second is the spin ket related to the internal coordinates of the electron.
2.) The operators z and y are to be represented in terms of the spherical harmonics
2
revealing their ,  character. y 
Equivalently, y 
2
3 r{ 11
3 r{Y1
 1 1 }; z 
1
( , )  Y11 ( , )}; z 
4
3r
4
0
3 rY1
( , ) .
10 . These perturbations act in the
coordinate range of the orbital angular momentum and act as the identity operation,
multiplication by one, in spinor space. The Clebsch-Gordan tables can be used to
find the orbital kets equivalent to the operators acting on the orbital kets. For example,
z |11ñ
4
3r
10 11 
4
3r

1
2
21 
1
2

11 .
3.) Compute all the initial to final matrix elements. The ket above is multiplied by its
spinor ket as are the final state orbital bras. The inner products are evaluated are the
product of the orbital bra-kets and the spinor bra-kets. When the operator y is used,
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either the plus or minus term alone is considered depending on whether the intensity
of the + or the - line is in question.
Extended Battle Plan: The Zeeman components in the 546 nm transitions of mercury.
Mercury has two valence electrons in the spin symmetric S =1 configuration for this
line. Match this with an anti-symmetric two particle spatial functions. The dipole
operator must include both electrons, p  er1  er2 . Proceed using the components of
the previous battle plan with the two particle complications.
??? What is the two particle coordinate perturbation
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

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LS coupling vs. jj coupling:
Consider atoms with two valence electrons. We believe that the overall angular
momentum should be a conserved quantity and hence we consider paths to finding J
by combining the orbital angular momenta and spin angular momenta. If we first
combine the orbital angular momenta and separately the spin angular momenta and
then combine the total orbital with the total spin, we have LS coupling. If we combine
the orbital angular momentum and spin angular momentum of each electron separately
and then combine the resultant j’s, we have jj coupling. The LS coupling
representation has states that most nearly represent atoms for low Z atoms. The jj
coupling representation has states that most nearly represent atoms for higher Z atoms.
Atomic states have designations 2S+1LJ where S is the total spin. In low Z atoms, LS
coupling is the in play and S is a semi-conserved quantity. That is: to a good
approximation one sees optical transition between singlet states (S = 0) and between
triplet states (S = 1), but rarely between a singlet state and a triplet state. As Z
increases, transitions between a single become more likely state and a triplet state
become more likely and the jj coupling more representative.
Appendix I: Hydrogen Dipole Matrix Elements: Compute the dipole matrix
elements connecting the n = 1 state to each of the n = 2 states. As s = 0, we will
assume that all the electrons are spin up.
The definite integrals handout and hydrogen atom math handouts provide the
information:
The spatial parts of the hydrogen atom wave functions have the form:
un m (r , , )  Rn (r ) Y m ( , )
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Ch9-15
The full time-dependent eigenfunction is (r , t )  un m (r, , ) ein t . The Y m ( , ) are
the spherical harmonics, the QM eigenfunctions of angular momentum. They satisfy
the ortho-normality relation:

 
0
2
0
Y / m/ ( ,  )  Y m ( , )sin  d d  
*
/
 mm
/
1 if m  n
0 if m  n
 mn  

and: (n  1)  0 xn e x dx  n!
The radial wavefunctions and the spherical harmonics are discussed in the H-Atom
Math handout. They can also be found in Griffiths in pages 139 and 154. For our case
the atomic number Z = 1.
R10(r) =
a 
R20(r) =
 
R21(r) =
 
3/ 2
Z
2
0
3/ 2
Z
2 a0
3/ 2
Z
2 a0
  Z r
e

a0 
Y00 ( ,  ) 
 Zr

 2  Z r  e 

a0 

2 a0 


Y10 ( , ) 
1  Z r    Z r 2 a0 
e

a0 
3
1
4
1 3
cos 
2 
Y1,1 ( , ) 
1 3
sin  e i
2 2
We will push the calculations into the form:
*
 Rn (r ) r Rn  (r ) r dr  Y m ( , ) f ( , ) Y m ( , ) sin  d d
2
k
Toward this end, the electric dipole operator is expressed using the spherical
harmonics.
r  xiˆ  y ˆj  z kˆ  r (sin  cos  iˆ  sin  sin  ˆj  cos kˆ
r r

2
3 {Y1
1
( , )  Y11 ( , )}iˆ  i
2
1
3 {Y1
( , )  Y11 ( , )} ˆj 
4
0
3 Y1
( , ) kˆ

The dipole operator is e r . Compute with just r . Consider 200| r |100ñ

200| x|100ñ =  R20 (r ) r R10 (r ) r 2dr  Y00* ( , )
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2
3 {Y1
1
( ,  )  Y11 ( ,  )}Y00 ( , ) sin  d d
SP425 Notes –Time-Dependent Perturbation Theory
Ch9-16
=  R20 (r ) r R10 (r ) r 2dr
1
4
0
 Y0 ( , )

2
3 {Y1
1
( ,  )  Y11 ( ,  )} sin  d d = 0
Used Y00 = [4]-½. The angular integrals vanish by the orthogonality of the spherical
harmonics.
Clearly, the same fate awaits 200| y|100ñ and 200| z|100ñ

210| x|100ñ =  R21 (r ) r R10 (r ) r 2dr  Y10 ( , )

2
3 {Y1
1
( ,  )  Y11 ( ,  )} ( 1 ) sin  d d
4
=0
Used Y00 = [4]-½. The angular integrals vanish by the orthogonality of the spherical
harmonics.
Clearly, the same fate awaits 210| y|100ñ.
210| z|100ñ =  R21 (r ) r R10 (r ) r 2 dr  Y10 ( , )
=  R21 (r ) r R10 (r ) r 2 dr
1
3

4
0
3 Y1
( ,  ) ( 1 ) sin  d d
4

0
0
 Y1 ( , ) Y1 ( ,  ) sin  d d 
1
3
 R21(r ) r R10 (r ) r dr
2
Used Y00 = [4]-½. The angular integral is one by the ortho-normality of the spherical
harmonics.

Let's attack

0
R21 (r ) r R10 (r ) r dr  
2
 
= 16
1/ 2
1
ao3


0
 r e
 a 
 0

0
 
1
3/ 2
2 a0


  3r
2 a0 

r
 r  e 


3  a0 
1
   23  a 
r 3 dr  16
1/ 2
5
o

2 a0 
a 
1
0

0
3/ 2
2
e u u 4 du 
e
  r 
 a0 
r 3 dr

25 (4)(3)(2)
35 6
ao



Used u = 3r 2a0   0 R21 (r ) r R10 (r ) r 2 dr  235 6 ao 
7

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210| z|100ñ 
1
 R21 (r ) r R10 (r ) r dr
21,1| x|100ñ =  R21 (r ) r R10 (r ) r 2 dr  Y11 ( , )
=
7
= 235 2 ao
2
3
 R21 (r ) r R10 (r ) r dr 
2
1
6

1
6
2
3 {Y1
1
( ,  )  Y11 ( ,  )} ( 1 ) sin  d d
4
27 a
35 o
21,1| y|100ñ =  R21 (r ) r R10 (r ) r 2dr  Y11 ( , )
=  R21 (r ) r R10 (r ) r 2 dr  i

i

 = i 23
7
5
2
3
{Y ( , )  Y
1
1
1
1
( ,  )} ( 1 ) sin  d d
4
ao
Summary:
200| x|100ñ =200| y|100ñ =200| z|100ñ = 0
210| x|100ñ210| y|100ñ
7
210| z|100ñ = 235 2 ao
21,1| x|100ñ =
27 a
35 o
21,1| y|100ñ = i 235 ao
7
21,1| z|100ñ = 0
Appendix II: The canonical interaction: q p  A 
One learns that the interaction energy of an electric dipole in an Electric field is: Udipole
=  pE  Dip  E . (The symbol
p
is adopted for the electric dipole moment to avoid confusing
it with the momentum.) This result is not directly applicable to the evaluation of the
interaction energy of an electromagnetic wave and an atom. The result was derived by
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SP425 Notes –Time-Dependent Perturbation Theory
Ch9-18
considering two equal, but opposite charges in an electrostatic field. E-M radiation is
scarcely electrostatic. As this is a deep mystery, the resolution is rarely presented prior to
graduate school. Here the path to the answer is to be sketched, but not presented in detail.
Just follow the flow of the ideas. By the time the dust settles, the conclusion is that the
equation for the basic interaction energy does not change although the way that one
regards it may change.
Matter-EM wave interaction: H1 =  pE  Dip  E
The canonical method to identify a quantum operator for a quantity is to begin with its
representation in classical Hamiltonian mechanics in terms of coordinates and
momenta. Unfortunately, we need to develop the lagrangian as a step in identifying
the canonical momenta and the hamiltonian itself. We could start with that T – U
prescription.
 L  ½ m xi xi  q  ( x, y, z )
Summation notation is invoked so repeated indices are summed from 1 to 3. Relativity
could guide me to the incorporation of magnetism. The electrostatic potential  is the
zero element of a four vector and magnetic interactions involve the velocity. The fourpotential and the four-velocity are:
A  (c1, Ax , Ay , Az )  and  v   ( c,  vx ,  vy ,  vz ) 
The lagrangian is a scalar so we try the inner product of these vectors as the potential
term. A v   (  v  A) This term obviously includes relativistic corrections that
we are ignoring in this course and in the expression for the kinetic energy. In the  
1 limit, L  ½ m xi xi  q (  v  A) .
The previously development seems suspect to validate the proposed function. The
lagrangian is whatever it takes to reproduce the classical equations of motion. What do
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Ch9-19
we expect for the magnetic part of the Lorentz force? The equation for v  B is to be
developed with Bi replaced by  ijk

x j
Ak 
Fmag  qv  B  q  ijk x j kmn xm An  q  kij kmn x j xm An

A
Fmag, i  q  im jn   in jm  x j xm An  q xn Axn  xm x i
m
i

It is time to test our candidate for the lagrangian.
L  ½ m xk xk  q (  xm Am )
L
xi
L
xi
 m xi  q Ai  pi ;
 
Am
 q ( 
x  xm x )
i
i
 

d L  L  m x  q    Ai  q x Am
i
m x
xi
xi
t
dt xi
i
x
k
Ai
xk

Realizing that we are free to re-label dummy indices and noting that the electric field
is E    t A , it is confirmed. The proposed lagrangian works. The hamiltonian
follows as H  pk xk  L  m xk xk  q xk Ak  ½ m xk xk  q (  xm Am )  .
H  pk xk  L  ½ m xk xk  q
The answer appears pretty simple, but we have yet to eliminate the coordinate
velocities in favor of the momenta as required. xi  m 1  pi  q Ai  .
H  21m  pk  q Ak  pk  q Ak   q 
p p
2m
q


q2


 m p  A  2m A  A  q
Electromagnetic processes By Robert Joseph Gould Google Books
Electromagnetic interaction hamiltonian non-relativistic.
Choosing the gauge   A  0 and  = 0,



q
q2
Hˆ    2m p  A  A  p  2m A  A

The interaction hamiltonian above is correct for a point charge in an external field. For
particles with intrinsic structure that leads to a magnetic moment, the magnetic dipolefield interaction must be added.
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Ch9-20
Hˆ      B     (  A)
The H term is active in the interaction electromagnetic transitions of an atom or
molecule, and the H term is active in the interaction of the field with an intrinsic (say
electron) spin. Terms describing the interaction of extended charge (and current)
distributions such as orbiting electrons can be found inside H. With a great deal of
luck, I may find a discussion of them.
Problems
1.) The matrix element
2

0
0
 
Ylm ( , ) Y1,{0,1} ( , )Ylm ( , ) sin  d d factor coupled
with the orthogonality of the spherical harmonics yields the  = 1 rule.
a.) Express Y10(,)Y11(,) as a sum of the form
a
b.) Express Y1,-1(,)Y11(,) as a sum of the form
Y m ( , ) .
m
a
Y m ( , ) .
m
c.) Discuss you results in the context of the  = 1 rule.
References:
1. David J. Griffiths, Introduction to Quantum Mechanics, 2nd Edition, Pearson
Prentice Hall (2005).
2. Richard Fitzpatrick, Quantum Mechanics Note Set, University of Texas.
3.) Robert J. Gould, Electromagnetic Processes, Princeton University Press (2006).
4.) Bruce W. Shore and Donald H. Menzel, Principles of Atomic Spectra, John Wiley
and Sons (1968).
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Ch9-21