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13. Applications of Approximation Methods
What Approximations Have We Made?
2
k
e
1
When describing hydrogen, we started with this Hamiltonian: H 
P2  e
2me
R
Some things that this doesn’t include:
• Recoil of nucleus
1
1
1
– Handled in chapter 7 by replacing electron


 me mN
mass me with reduced mass 
• Finite nuclear size
• Relativistic corrections
• Nuclear spin and magnetic field
Other, external effects
• Background electric or magnetic fields
• Van-der-Waals interactions
13A. Finite Nuclear Size
Electric Field from a Finite Nucleus
• We need the electric potential from a finite nucleus
• Imagine the nucleus is a sphere of uniform charge density
r
a
e
e
– Radius a
  4 3
– Charge density is
V
3a
• We will use Gauss’s Law to find electric field everywhere
– By spherical symmetry, the electric field points radially outwards E  Er rˆ
– The electric field depends only on the amount of charge closer than the
radius where you measure it
ke e
Er  2
• For r > a, it looks like a point charge at the origin of magnitude e
r
• But for r < a, we only see the field from the charge closer than r
• The charge contained within q  r   V   4  r 3  er 3 a 3
r
3
a sphere of radius r < a is
• So the electric field inside this is E  ke q  r  E  ke er
r
r
r2
a3
Electric Potential from Electric Field
• We now have the electric field
U
• We need the electric
E


E  U
r
potential, related by
r
• We therefore integrate:
U  r     Er dr
 ke e r 2 for r  a ,
Er  
3
k
er
a
for r  a .
 e
• Problem: Keep careful track of constant of integration!
• Solution: Potential is continuous, and vanishes at infinity
• Integrate in each region
C1  0
• Use fact that U() = 0
• Use fact that U is continuous at a
keer 1  C1  12 keer 2a 3  C2
r a
r a
ke ea 1   12 ke ea 1  C2
C2  32 ke ea 1
 ke er 1  C1
U r    1
2 3

k
er
 2 e a  C2
for r  a ,
for r  a .

ke er 1
for r  a ,
U r    1
2 3
1
3

k
er
a

k
ea
for r  a .
 2 e
2 e
The Perturbation
• We have Hamiltonian:

ke er 1
for r  a ,
U r    1
2 3
1
1 2
3

k
er
a

k
ea
for r  a .
H
P  eU  R 
 2 e
2 e
2m
2 1


k
e
for r  a ,
1 2
e r

P   1 2 2 3 3
1
2m
k
e
r
a

k
ea
for r  a .
2 e
2 e
2
2 1
We know how to find eigenstates of H0  P  2m  kee r
Our Hamiltonian differs from this only in the tiny region r < R
We therefore anticipate that we can use perturbation theory
We write H  H 0  W
0
for r  a ,

W   2 1 1 2 2 3 3
We therefore
1
k
e
r

k
e
r
a

k
ea
for r  a .
e
W

H

H

2 e
2 e
0
have
• Unperturbed states have wave function:  nlm  r   Rnl  r  Yl m  , 
• Unperturbed energies are
 c 2 2
n  
2n 2
•
•
•
•
•
First Order Correction Finite Nuclear Size
W  ke e2 r 1  12 ke e 2 r 2 a 3  32 keea 1
 nlm  r   Rnl  r  Yl  , 
m
for r  a
• First order correction to the energy is given by
*
   n, l , m W n, l , m   nlm
 r  nlm  r W  r  d 3r

  Yl  ,   d  R
2
m
0
2
nl
 r W  r  r
2
dr   Rnl2  r W  r  r 2dr
R
0
• Nucleus is much smaller than the atom
– Wave function hardly changes on the scale of the nucleus Rnl  r   Rnl  0
• We therefore approximate
   R  0  W  r  r dr  kee R  0   r  12 r 4 a 3  23 r 2 a 1  dr
2
nl
a
0
2
nl
 ke e 2 R
2
2
2
nl
 0   12 a 2  101 a5 a 1  12 a3a
• Comment: Rnl(0) = 0 for l  0
– Only non-vanishing for l = 0
a
0
1

 nl  101 kee2a2 Rnl2  0
 n 0  101 kee2 a2 Rn20  0
Magnitude of Nuclear Size Corrections
 n 0  101 kee2 a2 Rn20  0
• Wave function at origin is of order Rn20  0  a03
2
2
• Energy is of order
2
15
2



2

a
 10 m 
 a
n
10
a
k
e
2 2 3
~
e
~
~
10


~



 n 0 ~ ke e a a0 
10

n
 

a
10
m
n
0


a

a0  a0 
 0
• If we do hydrogen-like atom with nuclear charge Z, then
– Size of atom decreases by factor of Z
– Unperturbed energy increases by factor of Z2
– Nuclear size a increases by factor of A1/3 ~ Z1/3
 n 0 ~ ke e  Z a   a0 Z 
2
1/3
2
3
2
2
 a  5/3
ke e  a  11/3
10 5/3

Z


Z
~
10
Z
 

n
a0  a0 
 a0 
2
• Even with Z ~ 100, it is around a 10-7 size effect
• In atoms with multiple electrons, this effect is not that important
– 2s and 2p states are not degenerate anyway
Charge Radius of Proton Puzzle
• Another way to increase the effect is to replace the electron by a muon
• The size of muonic hydrogen is decreased by
2
a0  me
0.511
MeV/
c
1
~


a0 e m
105.7 MeV/c 2 206.8
• We can measure the “charge radius” of the proton three different ways:
– Electromagnetic scattering of electrons by a proton
– The split between the 2s and 2p levels of conventional hydrogen
– The split between the 2s and 2p levels of muonic hydrogen
• The first two methods are less exact, but lead to charge radius of
– This is not exactly the same as a from previous parts a p  877.5  5.1 am
• Muonic hydrogen gives a more precise value
– These disagree!
a p  840.87  0.39 am
• At present, an unsolved mystery
Sample Problem (1)
Suppose we model a proton as a hollow sphere of outer
radius a and inner radius a/2 with total charge +e. Find the
resulting shift in energy of all levels of hydrogen.
a
• The volume of this region is V   a    a   78  43  a 3
8 e
• The charge   e
 4 3
density is:
V
7 3a
Easiest way to handle this:
• Full sphere of radius a and charge +
8 e
– Charge of
qa   43  a 3  4 3 43  a 3   8 e
7
this sphere is
7 3a
• Anti-sphere of radius a/2 and charge –
8 e 4 1 3
3
– Charge of
4
1
qa /2    3   2 a    4 3 3   2 a    17 e
this sphere is
7 3a
• Note that total charge is +e
4
3
3
4
3
1
2
3
a/2
Sample Problem (2)
Suppose we model a proton as a hollow sphere of outer
radius R and inner radius R/2 with total charge +e. Find the
resulting shift in energy of all levels of hydrogen.
1
8
q


qa   7 e
a /2
7e
a
a/2
• The total potential will be the sum of the
V  r   Va  r   Va /2  r 
potentials from the sphere and the anti-sphere
• The perturbation is then
2 1
2 1
2 1
8
1




V
r

k
e
r

V
r

k
e
W  r   Va  r   Va /2  r   kee r
 a  7 e
  a /2   7 e r 
• First term is 8/7 of the result we found before for a sphere of radius a
• Second term is –1/7 of the result we found before, if we modify for radius a/2
• Our final answer, therefore is just
   nlm WR  WR/2  nlm   kee a R
8
7
1
  708  280
 kee2a2 Rn20  0
1
10
2
2
2
n0
 0
  ke e
1
7
1
10
31
 n0  280
kee2a2 Rn20  0
2
 a
1
2
2
Rn20  0 
13B. Relativistic Corrections
Types of Corrections
1
• The electron in hydrogen has speed of order v c ~   137
• This causes
2
2
2 1 2
2


E



1
mc

1

v
c

1
mc



kin
relativistic


corrections:
 v 2 3v 4  2 mv 2  3v 2 
  2  4  mc 
1 2 

2  4c 
 2c 8c 
• This implies corrections to the
2
 nl ~  n  v c    n 2 ~ 104  n
energy (eg. 2p vs. 2s) of order
• To fully understand these, need a relativistic theory of the electron
– The Dirac equation, chapter 16
• For hydrogen-like atoms, we will solve this exactly
• For other atoms, relativistic corrections must be approximated
• Since states 2s/2p are not degenerate for these atoms, corrections not important
• But corrections that depend on spin are important
Spin-Orbit Coupling
• There is a term in the Hamiltonian caused by
ge
e
W
B S  B S
the magnetic dipole moment of the electron
2m
m
• g2
But is there any magnetic field?
• In the rest frame of the nucleus, no magnetic field W  0
• But according to special relativity, particles moving
1
1
B   2 vE   2 pE
in an electric field experience a magnetic field
mc
c
• This suggests a perturbation
e
W   2 2 p  E  S
of the form:
mc
Which one is correct?
e
• Answer turns out to be the WSO   2 2  p  E   S
2m c
average of these two answers
• Note that we eE  eU  r   V r  rˆ dVc  r dVc
c 
r dr
dr
can write:
1 dVc
• So we have W   1 dVc  p  r   S
WSO 
L S
SO
2 2
2m 2 c 2 r dr
2m c r dr
Spin Orbit Coupling – Why It’s Hard
WSO
1 dVc

L S
2 2
2m c r dr
• We have spin in the perturbation, so must
n, l , s, m, ms
include spin in the unperturbed eigenstates
• These are eigenstates of
H , L2 , S 2 , L , S

0
z
z

• Energy for arbitrary atom depends on n, l, s
• The states with different m and ms are all degenerate
– Must use degenerate perturbation theory
• We need to calculate
n, l , m, s, ms WSO n, l , m, s, ms
the matrix elements
• Because LS doesn’t commute with Lz or Sz, there will be non-diagonal terms
– Means we will have to deal with matrices
– We need to find a better way
A Better Way
H , L , L , S , S 
2
0
2
z
z
WSO
1 dVc

L S
2 2
2m c r dr
n, l , m, s, ms WSO n, l , m, s, ms
• Instead of working with eigenstates of Lz and
2
2
2
H
,
L
,
S
,
J
, Jz
2
 0
Sz, we can work with eigenstates of J and Jz
• Eigenstates will now look like n, l , s, j , m j
n, l , s, j , mj WSO n, l , s, j , m j
• We therefore need matrix elements
• As proven in a homework problem: L  S  1  L  S 2  L2  S2   1 J 2  L2  S 2

2
2

• We therefore have
e
1 dVc 2
2
2




n, l , s, j , m j WSO n, l , s, j , m j 
n
,
l
,
s
,
j
,
m
J

L

S
n, l , s , j , m j


j
2 2
4m c
r dr
2
e
1 dVc
2
2
2

j  j  l  l  s  s  n, l , s, j , mj
n, l , s, j , m j
2 2 
4m c
r dr
• There is no angular
2
1 dVc
2
2
2
dependence in the matrix  n ,l , j , s 
j  j  l  l  s  s 

2 2 
4m c
r dr
element, so it is diagonal
Spin-Orbit Coupling Effects
 n,l , j , s 
2
1 dVc
j  j  l  l  s  s 

2 2 
4m c
r dr
2
2
2
• Expectation value of the Coulomb potential
 Vc  ~  nl
is of order the unperturbed energy
1 dVc
1
• We therefore have, approximately

 ~ 2  nl
r dr
a0
• So we have 2
2
2
p
v
 n,l , j ,s ~ 2 2 2  n,l ~ 2 2  n,l ~ 2  n,l ~  2 n,l
mc
m c a0
c
3p3/2
• Same size as other relativistic effects
• But other relativistic effects don’t split on the basis of spin
3p1/2
Consider, for example, Sodium
• Ground state, the outermost electron in 3s1/2
3s1/2
• First excited state has electron in 3p3/2 or 3p1/2 state
• Transition between 3p  3s causes wavelengths at 389.0 nm and 389.6 nm
Sample Problem
An electron (with spin ½) is trapped in a Coulomb potential VC(r) = m2r2/2.
Find the energy shift of the electron due to spin orbit coupling.
• The unperturbed Hamiltonian is a 3D harmonic oscillator n, l , m  n    n  32 
• Solved in homework problem, ignoring spin
n  l , l  2, l  4,
•
•
•
•
•
•
2
e
1 dVc
Spin orbit coupling
 n,l , j , s  2 2  j 2  j  l 2  l  s 2  s  

4m c
r dr
adds a shift
The matrix element we need is
1 dVc
1 d 1
2 2
1

 
m

r



  m 2 r   m 2    m 2
2
r dr
r dr
r
Spin is ½ for a single electron, so
e 2 m 2 2
2
3



j

j

l

l


n ,l , j
Total angular momentum ranges from
4
2 2
4m c
j = |l – s| to l + s
If l > 0, this means j = l – ½ or j = l + ½.
Simply substitute these two expressions for j into this expression
Beyond Spin-Orbit Coupling?
n, l , s, j , m j
• Atomic states look like
Energy is non-degenerate based on:
• n: governs overall electronic configuration
• l: for multiple electrons, energy is different due to screening
• s: spin state of multiple electrons affects symmetry and hence energy
• j: due to spin-orbit coupling
Can anything break the remaining degeneracy?
• Hamiltonian rotationally invariant
• As we perform a rotation of the atom, mj changes
• Therefore energy independent of mj
Unless:
• The nucleus is not rotationally invariant (spin)
• There are external forces breaking the degeneracy
– Such as a magnetic field
13C. The Hyperfine Splitting
Nuclear Magnetic Dipole Moment
• The proton is itself a rotating electric charge
g pe
• We would expect it would have a magnetic
μp 
I
2m p
dipole moment in the direction it is spinning:
– Where I is the spin of the proton
• If the proton were elementary, we
would predict gp 2, but actually g p  5.585
a
• We will approximate the proton as a uniformly
magnetized sphere of radius a:
3
3
4

M

μ

a
μ

r
a
r  a,

• Magnetization:


p 3
0  p
A

• Vector potential, Jackson eq. (5.111):
  μ p  r  r 3 r  a .

– 0 is magnetic permeability of free space
• Magnetic field:
3

2
μ
a
r  a,
p
0 
B   A 

  3rˆ  rˆ  μ p   μ p  r 3 r  a .

Shrinking the Nucleus
A
•
•
•
3
3

2
μ
a
r  a,

p
0 
0  μ p  r  a r  a,
B


  3rˆ  rˆ  μ p   μ p  r 3 r  a .
  μ p  r  r 3 r  a .


Again, the nucleus is small, so we’d like to take the limit a  0
– Can we just always use the lower formula?
How large is the magnetic field integrated just over the volume of the nucleus?
3
2μ p 3
3
2
μ

2

4

a
p
0
0
Bin   Bd r 
d
r

0μ p

3
r a

3
r

a
4
a
3
4 a
3
In the limit a  0, must add a contribution of this magnitude right at the origin
0 
2
3

ˆ
ˆ
B
3
r
r

μ

μ


μ

r 


p
p
0 p
3 
 r
3
• No comparable contribution from A, since it is smaller by factor of r near origin
0
A
μ r
3  p
 r
Shrinking the Nucleus
A
•
•
•
3
3

2
μ
a
r  a,

p
0 
0  μ p  r  a r  a,
B


  3rˆ  rˆ  μ p   μ p  r 3 r  a .
  μ p  r  r 3 r  a .


Again, the nucleus is small, so we’d like to take the limit a  0
– Can we just always use the lower formula?
How large is the magnetic field integrated just over the volume of the nucleus?
3
2μ p 3
3
2
μ

2

4

a
p
0
0
Bin   Bd r 
d
r

0μ p

3
r a

3
r

a
4
a
3
4 a
3
In the limit a  0, must add a contribution of this magnitude right at the origin
0 
2
3

ˆ
ˆ
B
3
r
r

μ

μ


μ

r 


p
p
0 p
3 
 r
3
• No comparable contribution from A, since it is smaller by factor of r near origin
0
A
μ r
3  p
 r
The Hyperfine Perturbation for Hydrogen
0
A
μ r
3  p
 r
μp 
g pe
2m p
I
B
0 
2
3

ˆ
ˆ
3
r
r

μ

μ


μ

r 


p
p
0 p
3 
 r
3
ke e2 ge
1
2
H

BS
 P  eA  
2m
R
2m
• Recall the Hamiltonian for
electromagnetic interactions:
• Drop A2, and approximate g = 2:
2
k
e
1
e
2
e

P  eP  A  eA  P  
 BS

2m
R
m
• Compare to unperturbed Hamiltonian
P 2 ke e 2
H0 

• The perturbation is therefore:
2m R
e
WHF   A  P  B  S 
m
0 e 1 
2 0 e
3
ˆ
ˆ


μ

R

P

3
R

S
R

μ

μ

S

μ

S

R 

p
p
p
3  p

4 m R
3m
2
0 e 2 g p 1
g

e
ˆ S R
ˆ  I  I  S   p 0 I  S 3  R 
 R  P   I  3 R

 3mmp
8 mmp R3 






Hyperfine Perturbation for s-Waves
eg 1
g e
ˆ
ˆ
 R  P   I  3  R  S  R  I   I  S  
W 
 3mm
8 mm R 
2
0
p
HF
p
2
0
3
p
2
p
I  S 3  R 
2
0 e g p 1
g

e
• Replace
ˆ S R
ˆ  I  I  S   p 0 I  S 3  R 
I  L  3 R
WHF 
R  P = L:
 3mmp
8 mmp R3 
• We first need unperturbed states
– Have to include nuclear spin now!
n, l , m, ms , mI
– mI is eigenvalue associated with Iz
• Because of spin-orbit coupling, it is better to add L and S to get total electron
angular momentum J = L + S
• So better choice of basis states would be n, l , j, m j , mI
• Degenerate perturbation theory again
n, l , j , mj , mI WHF n, l , j , m j , mI
• So we will need matrix elements
• For l = 0 (s-waves), the angular momentum operator always vanishes
• Less obvious: for l = 0, the next two terms also vanish
– Next slide from now, plus proof by homework problem



Sample Problem
Prove that for s-wave states, only the final term contributes to hyperfine splitting
WHF
2
0 e 2 g p 1
g

e
ˆ S R
ˆ  I  I  S   p 0 I  S 3  R 
I  L  3 R

 3mmp
8 mmp R3 



L n, l  0, m  0  0
• All s-waves (l = 0) have no angular dependence, so
• So contributions to perturbation theory from this term will be n,0,0 L n,0,0  0
• For the other two terms, write it out as
3
3
1  ˆ
1  ˆ
ˆ   I S 
ˆ

 WHF    3 3 R  S R  I  I  S     3 3 R j R
k
jk
a b



R
R
j 1 k 1
• Because the wave function has no angular
 d   3rˆ j rˆk   jk 
dependence, the angular integral is
• This can be shown to always vanish (nine expressions, six independent)
• For example, let j = k = z, then

 d   3rˆ rˆ
z z

  

  zz   0 d 1 3cos   1 d  cos    2  cos   cos   1  0
2
1

2

3
1
Total Atomic Angular Momentum
n, l , j , mj , mI WHF n, l , j , m j , mI
WHF 

g p 0 e 2
3mmp
I  S 3  R 
• We are still working in basis states of J2 and Jz, where J = L + S
• But as before, we could then combine the electron angular momentum with the
proton’s spin to get the total internal
F  J I  LSI
angular momentum of the atom:
• Eigenstates of H0 will now look like: n, l , j, f , m f
2
2
2
F

f
  f  , Fz  mF
– These are eigenstates of
• Since we have l = 0, ignore L and so F = S + I
• Since electron has s = ½ and proton has i = ½, total spin is f = 0 or f = 1
2
• Use addition of angular
1 
I  S  2  I  S   I 2  S2   12  F 2  I 2  S 2 


momentum trick to write
• Therefore,
g p 0 e 2 1
 f 2  f  2  34 2  34 2   3  R  n, l  0, j , f , m f
WHF n, l  0, j, f , m f 

3mm 2 
p
Hyperfine Splitting for Hydrogen s-waves
g p 0 e 2 1
 f 2  f  2  34 2  34 2   3  R  n, l  0, j , f , m f
WHF n, l  0, j, f , m f 

3mmp 2 
• We need matrix elements n, 0, j , f , mf WHF n, 0, j , f , m f

g p 0 e 2
6mmp
g p 0 e 2
2
 f 2  f  32  n, 0, j, f , mf  3  R  n, 0, j, f , m f
2
2 3
• Energy is
2
3
3
 f  f  2   ff  m f mf    r   n 00  r  d r

now diagonal
6mmp
• Most important effect is difference between f = 0 and f = 1 energy
g p 0 e 2 2 2
2 3
2
3
3
3 

E f 1  E f 0 
1  1  2    0  0  2     r   n 00  r  d r


6mm
p

g p 0 e
2 2
3mmp
 n 00  0 
2
E f 1  E f 0 
g p 0 e 2
2
12 mmp
Rn20  0 
Comments on Hyperfine Splitting
4 ke
g p 0 e 2 2 2
• Permittivity of free space is
0  2 E f 1  E f 0 
Rn 0  0 
related to Coulomb’s constant by
c
12 mmp
• Therefore
3
4 g p m c
 ke e m 

 
2
3 
2
2
3 
2
3mc mp a0 3mc m p a0 
3mc m p
3m p

• By comparison, spin-orbit coupling is of order 2E1s
– So hyperfine is suppressed by m/mp
• But this is the only contribution that distinguishes these spin states
E1s 
g p ke e 2
2
R102  0  
4 g p ke e 2
2
4 g p ke e 2
2
2
4
2 2
8g p 2 m
• Transition from f = 1 to f = 0 generates electromagnetic radiation with a
wavelength of 21 cm.
• The 21 cm line is used to track atomic hydrogen throughout the galaxy
• The comparable transition in Cesium is used for atomic clocks
3mp
E1s
13D. The Zeeman Effect
Weak Magnetic Fields
• For an isolated atom in vacuum, including the hyperfine
n, l , s , j , f , m f
interaction, general state now looks something like
• Since rotating the atom changes mf different mf values must be truly degenerate
• Consider a uniform magnetic field acting on an atom
B  Bzˆ
• Ignore hyperfine splitting, since it’s so small
n, l , s, j , m j
• Unperturbed states are therefore
• Assume a weak field, so that spin-orbit coupling dominates magnetic effects
– So energy in absence of magnetic field depends on n, l, s, and j
• Presence of magnetic field adds to the Hamiltonian a perturbation:
– See chapter 9
eB
W
 Lz  gS z 
– Lz and Sz are sums over all electrons
2m
Zeeman Effect Computation
eB
• We need
n, l , s, j , mj WB n, l , s, j , m j
W
 Lz  gS z 
2m
• Fortunately, Lz and Sz both commute with Jz = Lz + Sz
– Therefore, Jz eigenvalues can’t change
   m j   n, l , s, j , m j WB n, l , s, j , m j
• Insert complete basis |n, l, s, ml, ms
eB
 mj  
n, l , s, j, m j  Lz  gS z  n, l , s, ml , ms n, l , s, ml , ms n, l , s, j, m j

2m ml ,ms
eB

 ml  g ms  n, l , s, j, m j n, l , s, ml , ms n, l , s, ml , ms n, l , s, j, m j

2m ml ,ms
2
eB
• These matrix elements
 mj  
 ml  gms  l , s; ml , ms j, m j

2m ml ,ms
are just CG coefficients
• Though it looks like a double sum, it really isn’t
– Recall, CG coefficients vanish unless mj = ml + ms
• Can show in general that shift is proportional to mj
eB

• For example, if s = 0, then ms = 0 and CG coefficients are 1   m j  s 0  2m m j
– And mj = ml
13E. Van Der Waals Interaction
The Hamiltonian
• Consider two atoms that are close, but not too close to each other
– For definiteness, two hydrogen atoms
• Treat the nuclei as fixed point positive charges separated by a
• Assume they are far enough apart that the electron wave functions are not
overlapping
– Then we don’t have to worry about anti-symmetrizing the wave function
• Treat the electrons quantum mechanically,
R1
R
2
with positions R1 and – R2 relative to
their nuclei
a
• Then the Hamiltonian is:
2
2
2
2
2
2
1
k
e
k
e
k
e
2
2
k
e
k
e
k
e
e
e
e
e
e
e
H
P1  P2  






2m
R1
R2
a
a  R1 a  R 2
aR R
1
2
The Perturbation
2
2
2
2
2
2
k
e
k
e
k
e
k
e
k
e
k
e
1
2
2
e
e
H
P

P
 e  e  e
 e


1
2 
2m
R1
R2
a
a  R1 a  R 2 a  R1  R 2
• The unperturbed Hamiltonian is:
• This has energy eigenstates:
2
2
k
e
k
e
1
H0 
P12  P22   e  e

2m
R1
R2
ke e 2
 nn   n   n ,  n   2
nlm, nl m
2n a0
2
2
2
2
k
e
k
e
k
e
k
e
e
• The perturbation is given by:
W e  e
 e

a
a  R1 a  R 2 a  R 1  R 2
• For definiteness, choose: a  azˆ
• We want to calculate this in the limit that a >> R
aR 
1
1/2
1/2
 a 1 1  2Z a  R 2 a 2 
  a  Z   X  Y 


2
2 2

2
2
 2Z R  1 1
 2Z R 
1
1
Z
3
Z

R
3
1
 1    2  
 2   2  2  2 
 2    2
3
a
a
a
a
a
a
a
2
a




 

2
2
2
The Perturbation Simplified
ke e 2
ke e 2
ke e 2
ke e 2
W



a
a  R1 a  R 2 a  R 1  R 2
aR
1
1 Z 3Z 2  R 2
  2
a a
2a 3
• Substitute and simplify:
a 1   a 1  a 2 Z1  32 a 3 Z12  12 a 3R12   a 1  a 2 Z 2  23 a 3Z 22  12 a 3R 22  

 

2
W  ke e 

2
2
1
2
3 3
1 3


 a  a  Z1  Z 2   2 a  Z1  Z 2   2 a  R1  R 2 



 


 kee2 a 3  32 Z12  12 R12  32 Z22  12 R 22  32  Z1  Z2   12  R1  R2 
2
 kee2a3  3Z1Z2  R1  R2 
2

W  kee2a3  2Z1Z2  X1 X 2  YY
1 2
• Let’s do first order perturbation theory on the ground state |100;100:
 g  100,100 W 100,100  kee2a3 100,100  2Z1Z2  X1 X 2  YY
1 2  100,100
 kee a
2
3
 2 100 Z 100
2
 100 X 100  100 Y 100
2
2

0
Second Order Perturbation Theory
W  kee2a3  2Z1Z2  X1 X 2  YY
1 2
• Let’s try second order perturbation
1

g  
nlm, nl m W 100,100
theory on ground state:
nlm , nl m  11   nn
2 4
e
6
k e
 g 
a
2 nlm Z 100 nl m Z 100  nlm X 100 nl m X 100
1

 nlm Y 100 nl m Y 100
nlm , nl m 11   nn
• This sextuple sum (!) should exclude the state |100,100
• Technically, it also contains unbound states
• As previously shown, matrix elements like nlm|R|n'l'm' only non-vanishing if l
and l' differ by exactly 1
• Therefore, only l, l' = 1 contributes (which means n, n'  2)
• Note that energy denominator is always negative, times a positive number
2
2
A Lower Limit on the Energy Shift
2 4
e
6
k e
  
a
2 nlm Z 100 nl m Z 100  nlm X 100 nl m X 100
1

 nlm Y 100 nl m Y 100
nlm , nl m 11   nn
2
• The largest matrix elements come from n = 2, or n' = 2
• All terms negative, so dropping any terms leads to an overestimate of the energy
• Let’s include only n = n' = 2 (and recall that only l = l' =1 contributes):
2 21m Z 100 21m Z 100  21m X 100 21m X 100
2
k e
1

a m,m 11   22
 21m Y 100 21m Y 100
For example, if m = 0, then only 210|Z|100  0, and then forced to pick m' = 0
Previous homework problem: 210 Z 100  215 235 a0
2
Substitute this in
ke e 2 ke e 2
3
k
e
e
11   22  21  2 2  



Find energy denominator:
a0
4a0
4a0
Include m, m' = 1
2 4
30
30
30
33
2 5




k
e

4
a
2
2
2
2
k
e
4
4
4
0
e a0
Put it all together    e 

4 20 a0  20 a0  20 a0     20 6
6
2 
a  3kee   3
3
3
3 a

  
•
•
•
•
•
•
2 4
e
6
A Lower Limit on the Energy Shift
2 4
e
6
k e
  
a
2 nlm Z 100 nl m Z 100  nlm X 100 nl m X 100
1

 nlm Y 100 nl m Y 100
nlm , nl m 11   nn
2
• The largest matrix elements come from n = 2, or n' = 2
• All terms negative, so dropping any terms leads to an overestimate of the energy
• Let’s include only n = n' = 2 (and recall that only l = l' =1 contributes):
2 4
e
6
2 21m Z 100 21m Z 100  21m X 100 21m X 100
2
k e
1

a m,m 11   22
 21m Y 100 21m Y 100
• Non-vanishing matrix elements:
210 Z 100  215 235 a0 , 21  1 X 100  2735 a0 , 21 1 Y 100  2735 ia0
• Energy
3ke e2
ke e 2 ke e 2


denominator: 11   22  21  2 2  
4a0
a0
4a0
• Put it all
33
2 5
2 4
30
30
30




2
k
e
k
e

4
a
2
2
2
4
4
4
e a0
0
together    e 
    20 6
4 20 a0  20 a0  20 a0 
6
2 
3 a
a  3kee   3
3
3

  
An Upper Limit on the Energy Shift (1)
1
   
nlm, nl m W 100,100
nlm , nl m  11   nn
2
• The smallest magnitude energy denominator is n = n' = 2
• Small energy denominator causes large (negative) contribution to the energy shift
• If we replace energy denominator by 11 – 22 on every term, we are
overestimating the (negative) energy shift, so
1
   
100,100 W nlm, nl m nlm, nl m W 100,100
nlm , nl m  11   22
• Technically, sum excludes ground state, but this matrix element vanishes
• Use completeness
100,100 W 2 100,100
  
11   22
An Upper Limit on the Energy Shift (2)
W  kee2a3  2Z1Z2  X1 X 2  YY
1 2
100,100 W 2 100,100
  
11   22
11   22   3kee2  4a0 
• Substitute in:
4a0 ke2e4
2

 
100,100  2Z1Z 2  X 1 X 2  YY
1 2  100,100
2
6
3ke e a
4a0 ke e2
2 2
2
2
2 2



100,100
4
Z
Z

X
X

Y
1 2
1
2
1 Y2   cross terms   100,100
6

3a
• Because of spherical symmetry, cross terms do not contribute
4a0 ke e2
  
3a 6
4a0 ke e2

3a 6
4 100 Z
4  a

2
100  100 X 100  100 Y 100
2
2
   a    a  
2 2
0
2 2
0
2 2
0
2
2
2
8a05 ke e2
   
a6

Combining the Limits
233 ke e2 a05
    20 6
3 a
•
•
•
•
Combining the limits, we have
Sophisticated analysis yields  = 6.50
Attractive potential that goes like a–6
This is for hydrogen-hydrogen
8a05 ke e2
   
a6
   
 ke e2 a02
2.64    8
a6
Will it apply in general?
• For neutral atoms, since they are in states of definite l, we will have g W g  0
• So always have to do second order perturbation theory
• So we generally get attractive a–6 potential
• Only the factor of  changes
– Larger if electron easily excited to state with different l value