Fine structure of spectral lines in the hydrogen atom

Question

I'm studying atomic physics from the book "Physics of atoms and molecules" by Bransden. On chapter 5 it discusses the effects of relativistic corrections on the hydrogen spectral lines. It basically says that since, in the dipole approximation, the matrix element $\langle \psi_b | \overline{r} | \psi_a \rangle$ does not depend on the spin the selection rule $\Delta l = \pm 1$ is still valid. Then it states the selection rule $\Delta j = 0, \pm 1$ which the book says follows from the previous selection rule. According to the former selection rule the transition $3d_{5/2} \to 2p_{1/2}$ is fine, but it is forbidden according to the latter. So, what is the correct selection rule and why?

Answer 1

Relativistic corrections determine splitting of degenerate energy levels and are different from selection rules which determine whether specific transitions are allowed or forbidden. So, whether the transition $3d_{5/2}\rightarrow 2p_{1/2}$ will occur or not is given by the selection rules, but they do not say what the energy of the emitted radiation will be.

Also, for atomic transitions, both selection rules apply. This means the transition is forbidden.

Answer 2

Well, this is the exact question ihv been stuck in for quite a time, still i dont have the very precise answer to how the later (∆j) follows but this is what i guess may be we can discuss too. First of all, in that very book the selection rules ∆l=+-1,∆s=0 are derived for 'ATOMIC TRANSITIONS' under 'E1-APPROXIMATION' in chapter 4. But in chapter 5 it directly says to be followed as ∆j=0,+-1. Here, i believe he uses J=L+S and hence ∆J=∆L+∆S then ∆J=0(in absence of L) and ∆J=+-1(in absence of S) but can be any of 0,+1,-1 when both L and S are present. Well, now which is correct? So here, i guess both are correct in their own rights, own domains in the sense, the previous one (l,s) were for atomic transitions where we didnt consider any need of LS coupling/fine structures i.e. the resultant J doesnt come into our picture of study at all. But when considered, j indeed plays a vital role so we need a new set of rule for j. So the ∆J we just saw was an additional rule to the previously found ones for L and S.

Well, any opinions are appreciated this is just my opinion.

Answer 3

The selection rules follow from the Wigner-Eckart theorem and the fact that the dipole operator involved in E1 transitions for hydrogen is proportional to $\hat{r}$.

The conditions for the Wigner-Eckart theorem to apply depend on the commutation relations.

Since $\hat{J} = \hat{L} + \hat{S}$ and $\hat{r}$ commutes with $\hat{S}$, it follows $\hat{r}$ has the same commutation relations with both $\hat{J}$ and $\hat{L}$ so the Wigner-Eckart theorem applies to both.

However, there is an additional restriction that $\Delta l \neq 0$ in E1 transitions because $<\psi_1(\textbf{r})|\hat{r}|\psi_2(\textbf{r})>$ is zero if the two wavefunctions have the same parity:

$<\psi_1(\textbf{r})|\hat{r}|\psi_2(\textbf{r})> = <\psi_1(-\textbf{r})|-\hat{r}|\psi_2(-\textbf{r})> = -P_1P_2<\psi_1(\textbf{r})|\hat{r}|\psi_2(\textbf{r})>$

$P = (-1)^l$ for the hydrogen wavefunctions.