Selection rules in electric dipole appoximation

[blue_leaf77]

Some literatures say that the selection rule in electric dipole approx. for angular momentum $\Delta j = 0,-1,1$ some other say $\Delta l = -1,1$. I follow the notation used in my references, despite the difference I think since j and l are both angular momenta which fulfill angular momentum commutation relations, I can regard them to be the same. But why are they different?

 

[DrClaude]

$\Delta l = \pm 1$ is due to conservation of angular momentum, as a photon has spin 1. But since $\hat{\jmath} = \hat{l} + \hat{s}$, you can find combinations of $l$ and $s$ (or $m_l$ and $m_s$) such that $l$ changes by 1, but $m_l$ changes also such that $j$ doesn't change. Note that this is not possible when $j = 0$, such that $j = 0 \rightarrow j'=0$ transitions are forbidden (i.e., only for $j \geq 1$ is $\Delta j = 0$ allowed).

 

[blue_leaf77]

but $m_l$ changes

Did you mean $s$ in place of $m_l$?

 

[DrClaude]

Did you mean $s$ in place of $m_l$?

No. The electromagnetic field doesn't couple to spin, so you have $\Delta S = 0$ for many-electron atoms.

 

[blue_leaf77]

If I take an example of transition from $1s_{1/2}$ to $2p_{1/2}$, would that be a good example?

 

[DrClaude]

Starting from $l=0, m_l = 0, s= 1/2, m_s = 1/2, j= 1/2, m_j = 1/2$, the atom can absorb a $\sigma^-$ photon ($\Delta l = 1$, $\Delta m_l = -1$), to end up in the state with $l=1, m_l = -1, s= 1/2, m_s = 1/2, j= 1/2, m_j = -1/2$.

Note that the same photon can also lead to a transition to $l=1, m_l = -1, s= 1/2, m_s = 1/2, j= 3/2, m_j = -1/2$.