1. 13. Applications of Approximation Methods
What Approximations Have We Made?
When describing hydrogen, we started with this Hamiltonian:
Some things that this doesn’t include:
Recoil of nucleus
Handled in chapter 7 by replacing electron 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

2. Electric Field from a Finite Nucleus
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
Radius a
Charge density is
We will use Gauss’s Law to find electric field everywhere
By spherical symmetry, the electric field points radially outwards
The electric field depends only on the amount of charge closer than the radius where you measure it
For r > a, it looks like a point charge at the origin of magnitude e
But for r < a, we only see the field from the charge closer than r
The charge contained within a sphere of radius r < a is
So the electric field inside this is

3. Electric Potential from Electric Field
We now have the electric field
We need the electric potential, related by
We therefore integrate:
Problem: Keep careful track of constant of integration!
Solution: Potential is continuous, and vanishes at infinity
Integrate in each region
Use fact that U() = 0
Use fact that U is continuous at a

4. The Perturbation We have Hamiltonian:
We know how to find eigenstates of
Our Hamiltonian differs from this only in the tiny region r < R
We therefore anticipate that we can use perturbation theory
We write
We therefore have
Unperturbed states have wave function:
Unperturbed energies are

5. First Order Correction Finite Nuclear Size
First order correction to the energy is given by
Nucleus is much smaller than the atom
Wave function hardly changes on the scale of the nucleus
We therefore approximate
Comment: Rnl(0) = 0 for l  0
Only non-vanishing for l = 0

6. Magnitude of Nuclear Size Corrections
Wave function at origin is of order
Energy is of order
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
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

7. 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
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
Muonic hydrogen gives a more precise value
These disagree!
At present, an unsolved mystery

8. 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
a/2
The volume of this region is
The charge density is:
Easiest way to handle this:
Full sphere of radius a and charge +
Charge of this sphere is
Anti-sphere of radius a/2 and charge –
Note that total charge is +e

9. 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. a
a/2
The total potential will be the sum of the potentials from the sphere and the anti-sphere
The perturbation is then
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

10. 13B. Relativistic Corrections
Types of Corrections
The electron in hydrogen has speed of order
This causes relativistic corrections:
This implies corrections to the 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

11. Spin-Orbit Coupling
There is a term in the Hamiltonian caused by the magnetic dipole moment of the electron
g  2
But is there any magnetic field?
In the rest frame of the nucleus, no magnetic field
But according to special relativity, particles moving in an electric field experience a magnetic field
This suggests a perturbation of the form:
Which one is correct?
Answer turns out to be the average of these two answers
Note that we can write:
So we have

12. Spin Orbit Coupling – Why It’s Hard
We have spin in the perturbation, so must include spin in the unperturbed eigenstates
These are eigenstates of
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 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

13. A Better Way
Instead of working with eigenstates of Lz and Sz, we can work with eigenstates of J2 and Jz
Eigenstates will now look like
We therefore need matrix elements
As proven in a homework problem:
We therefore have
There is no angular dependence in the matrix element, so it is diagonal

14. Spin-Orbit Coupling Effects
Expectation value of the Coulomb potential is of order the unperturbed energy
We therefore have, approximately
So we have
Same size as other relativistic effects
But other relativistic effects don’t split on the basis of spin
Consider, for example, Sodium
Ground state, the outermost electron in 3s1/2
First excited state has electron in 3p3/2 or 3p1/2 state
Transition between 3p  3s causes wavelengths at nm and nm
3p1/2
3p3/2
3s1/2

15. 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
Solved in homework problem, ignoring spin
Spin orbit coupling adds a shift
The matrix element we need is
Spin is ½ for a single electron, so
Total angular momentum ranges from 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

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17. Beyond Spin-Orbit Coupling?
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

18. 13C. The Hyperfine Splitting Nuclear Magnetic Dipole Moment
The proton is itself a rotating electric charge
We would expect it would have a magnetic 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
We will approximate the proton as a uniformly magnetized sphere of radius a:
Magnetization:
Vector potential, Jackson eq. (5.111):
0 is magnetic permeability of free space
Magnetic field: a

19. Shrinking the Nucleus
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?
In the limit a  0, must add a contribution of this magnitude right at the origin
No comparable contribution from A, since it is smaller by factor of r near origin

20. Shrinking the Nucleus
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?
In the limit a  0, must add a contribution of this magnitude right at the origin
No comparable contribution from A, since it is smaller by factor of r near origin

21. The Hyperfine Perturbation for Hydrogen
Recall the Hamiltonian for electromagnetic interactions:
Drop A2, and approximate g = 2:
Compare to unperturbed Hamiltonian
The perturbation is therefore:

22. Hyperfine Perturbation for s-Waves
Replace R  P = L:
We first need unperturbed states
Have to include nuclear spin now!
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
Degenerate perturbation theory again
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

23. Sample Problem
Prove that for s-wave states, only the final term contributes to hyperfine splitting
All s-waves (l = 0) have no angular dependence, so
So contributions to perturbation theory from this term will be
For the other two terms, write it out as
Because the wave function has no angular dependence, the angular integral is
This can be shown to always vanish (nine expressions, six independent)
For example, let j = k = z, then

24. Total Atomic Angular Momentum
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 angular momentum of the atom:
Eigenstates of H0 will now look like:
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
Use addition of angular momentum trick to write
Therefore,

25. Hyperfine Splitting for Hydrogen s-waves
We need matrix elements
Energy is now diagonal
Most important effect is difference between f = 0 and f = 1 energy

26. Comments on Hyperfine Splitting
Permittivity of free space is related to Coulomb’s constant by
Therefore
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
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

27. 13D. The Zeeman Effect Weak Magnetic Fields
For an isolated atom in vacuum, including the hyperfine 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
Ignore hyperfine splitting, since it’s so small
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
Lz and Sz are sums over all electrons

28. Zeeman Effect Computation
We need
Fortunately, Lz and Sz both commute with Jz = Lz + Sz
Therefore, Jz eigenvalues can’t change
Insert complete basis |n, l, s, ml, ms
These matrix elements 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
For example, if s = 0, then ms = 0 and CG coefficients are 1
And mj = ml

29. 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, with positions R1 and – R2 relative to their nuclei
Then the Hamiltonian is: R2 R1 a

30. The Perturbation The unperturbed Hamiltonian is:
This has energy eigenstates:
The perturbation is given by:
For definiteness, choose:
We want to calculate this in the limit that a >> R

31. The Perturbation Simplified
Substitute and simplify:
Let’s do first order perturbation theory on the ground state |100;100:

32. Second Order Perturbation Theory
Let’s try second order perturbation theory on ground state:
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

33. A Lower Limit on the Energy Shift
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):
For example, if m = 0, then only 210|Z|100  0, and then forced to pick m' = 0
Previous homework problem:
Substitute this in
Find energy denominator:
Include m, m' = 1
Put it all together

34. A Lower Limit on the Energy Shift
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):
Non-vanishing matrix elements:
Energy denominator:
Put it all together

35. An Upper Limit on the Energy Shift (1)
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
Technically, sum excludes ground state, but this matrix element vanishes
Use completeness

36. An Upper Limit on the Energy Shift (2)
Substitute in:
Because of spherical symmetry, cross terms do not contribute

37. Combining the Limits Combining the limits, we have
Sophisticated analysis yields  = 6.50
Attractive potential that goes like a–6
This is for hydrogen-hydrogen
Will it apply in general?
For neutral atoms, since they are in states of definite l, we will have
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