Confusion about selection rules

[AlonsoMcLaren]

While I am reviewing my undergrad physics for qualifying exam, I became confused about the numerous selection rules.

(1) We have the selection rules for spontaneous emission in a hydrogen atom: Δl=±1 ,Δml=0,±1. This rule is the easiest to understand by evaluating <n'l'm'|z|nlm>

(2) We have the selection rules for Zeeman effect.
Δj=0,±1 ,Δmj=0,±1. I don't understand the reasons for these rules.

(3) We also have the selections rules for a helium atom where one of its electron is excited: (Modern Physics by Randy Harris, Section 8.9): Δl_total=±1, Δs_total=0, Δj_total=0,±1 (0->0 forbidden). I am totally no clue on this one.

(4) What about a hydrogen atom, without external magnetic field, but in contrast to (1), we are now considering fine structure so the good quantum numbers are n,l,j,mj?

 

[DrClaude]

I'm not sure what kind of answer you are looking for. Electric dipole selection rules are always based on whether $\langle i | \hat{\mu} | f \rangle$, where $\hat{\mu}$ is the dipole operator, is zero or not. Or by "reason" do you mean the physical explanation behind the rule?

 

[Khashishi]

For 2, you need to know that a photon is a spin 1 particle and some basic rules of angular momentum addition. The total angular momentum of the atom and photon has to equal the total angular momentum of the excited atom.
An electric dipole only operates on the spatial (orbital) part of the wavefunction and not the spin part. When fine structure is involved, you have to consider that an eigenfunction is a superposition of up to two spatial wavefunctions (of n, l, m_l). That is,
$\psi_{nljm_j} = a \psi(x)_{nl,m_l=m_j-1/2} \psi_{m_s=+1/2} + b \psi(x)_{nl,m_l=m_j+1/2} \psi_{m_s=-1/2}$
So the problem including fine structure just reduces to the sum of some calculations with no fine structure (approximately).

 

[AlonsoMcLaren]

I'm not sure what kind of answer you are looking for. Electric dipole selection rules are always based on whether $\langle i | \hat{mu} | f \rangle$, where $\hat{mu}$ is the dipole operator, is zero or not. Or by "reason" do you mean the physical explanation behind the rule?

Yes I do mean the physical reasons behind these rules

 

[DrClaude]

The photon is a spin-1 particle, meaning that it has ħ angular momentum. Conservation of angular momentum is responsible for the selection rule Δl = ±1, and conservation of the projection of angular momentum for Δm_l = 0, ±1 (corresponding to π, σ⁺, and σ^- polarized light).

Δs = 0 because the EM field doesn't couple to spin.

The rules for Δj and Δm_j follow from the above rules for orbital angular momentum and spin. Δj = 0 is a bit more complicated to explain, but it comes from the fact that even though Δl = ±1, you can modify the orientations of l and s such that the resulting j vector is the same length, but this does not work if j = 0, so no 0 → 0 transition.