# Why can an electric dipole transition allow $\Delta L=0,\pm2$ in hydrogen-like atoms?

I am studying the transition rule of electric dipole transition. I have some question about some rules in the Wikipedia: https://en.wikipedia.org/wiki/Selection_rule#Summary_table

In that table, they claim that in the LS coupling when $$\Delta s=0$$, $$\Delta l=0$$ is one of allowed transitions for hydorgen-like atoms, and even in the intermediate coupling $$\Delta l$$ can also be 0 or even $$\pm2.$$

However, how can I get this conclusion?

I tried to calculate this:

For the LS coupling cases, if I express the states in terms of $$|lm_l\rangle|s,m_s\rangle$$ and express $$\langle j',m'_j|er_q|j,m_j\rangle$$ :

$$|j,m_j\rangle=\sum_{m_s,m_l}\langle l,m_l;s,m_s|j,m_j\rangle|l,m_l;s,m_s\rangle$$

Then: ($$q=0,\pm1$$ $$\pm1$$ is for left/right circular polarization and 0 is for linear polarization.)

$$\begin{split} \langle j',m'_j|er_q|j,m_j\rangle&=\sum_{m_l,m_s,m_l',m_s'}\langle l',m'_l;s',m'_s|er_q|l,m_l;s,m_s\rangle \langle j',m'_j|l',m'_l;s',m'_s\rangle \langle l,m_l;s,m_s|j,m_j\rangle\\ &= \sum_{m_l,m_s,m_l',m_s'} \langle l',m'_l|er_q|l,m_l\rangle \langle s',m_s'|s,m_s\rangle \langle j',m'_j|l',m'_l;s',m'_s\rangle \langle l,m_l;s,m_s|j,m_j\rangle\\ \xrightarrow{\langle s',m_s'|s,m_s\rangle=\delta_{ss'}\delta_{m_s,m_s'}}&=\sum_{m_l,m_l',m_s} \langle l',m'_l|er_q|l,m_l\rangle \langle j',m'_j|l',m'_l;s,m_s\rangle \langle l,m_l;s,m_s|j,m_j\rangle\\ \end{split}$$

If the electric dipole moment does not vanish then $$\langle l',m'_l|er_q|l,m_l\rangle$$ can not vanish so $$\Delta l\neq0.$$

According to the Wigner-Eckart theorem,

$$\langle j',m'_j|er_q|j,m_j\rangle=\langle j',m'_j||er_q||j,m'_j\rangle \langle j,m_j;1,q|j',m'_j\rangle$$

The selection rule is:

$$m'=m+q,\\ j-1\leq j'\leq j+1$$

So according to this theorem the selection rule is: $$\Delta j=0,\pm1,\ \Delta m=q.$$ Even I pick $$\Delta j=0$$, I must still need to pick up two states whose $$\Delta l\neq0.$$ Also, how can I have two states whose $$\Delta s=1$$ in the intermediate coupling? Because as I know for hydrogen-like atoms there is only one valence electron so the spin angular momentum should always be $$1/2$$.

And how can two states with $$\Delta l=\pm2$$ be coupled in the intermediate coupling? And what does intermediate coupling means?

$$\Delta L = 2$$ refers to a quadrupole transition or a transition realized with two photons absorbed by your sample. In a typical single photon induced transition - dipole one, you have those "Rigorous rules" from the first column.

It does not mean that forbidden transitions do not happen at all, for example, a forbidden transition in Strontium: from $$^1S_0$$ to $$^3P_0$$ is used in atomic clocks.

Edit:

Sorry, I did not fully understand your question. Let me answer properly:

The mechanism of LS coupling was introduced to explain observed electric dipole transitions in multi-electron atoms. Hydrogen-like atoms are a special case for LS coupling because you are considering only one electron "jumping around". That significantly simplifies the structure of available atomic terms.

As you already know the LS coupling needs three good quantum numbers, $$L$$, $$S$$, and $$J$$ describing your whole system. Let me repeat the set of selection rules for LS coupling with a brief comment for each:

1. $$\Delta l_i= \pm 1$$ and $$\Delta l_{j \neq i} = 0$$, only one of the available electrons is changing its angular momentum (comment: one can find transitions where two electrons are changing their quantum numbers, but then the next rule applies);
2. An electric dipole transition has to change the parity of the electrons' wavefunction. The parity is given by: $$(-1)^{\sum l_i}$$
3. $$\Delta J = 0, \pm 1$$ and $$J'=0\not \leftrightarrow J=0$$; $$J$$ is resulting quantum number from coupling between $$L$$ and $$S$$ what is expressed as $$\mathbf J = \mathbf L + \mathbf S$$.
4. $$\Delta L = 0, \pm 1$$ and $$L' = 0 \not \leftrightarrow L = 0$$ where $$L=\sum l_i$$, i.e. $$L$$ is a sum over angular momenta of electrons for a given term.
5. $$\Delta S = 0$$, where $$S=\sum s_i$$ i.e. $$S$$ is a sum over oriented spins for a given term;

For the case of hydrogen-like atoms, rule number 1 prohibits the case you mentioned: $$\Delta L = 0$$, as $$L=l_1$$. However transitions, when that case can occur still exist - you just need more electrons. As an example we can take the transition in carbon:

$$2\text{p}^2 \space {^1}\text{D} \rightarrow 2\text{p}3\text{d} \space ^1 \text{D} ^\text{o}$$

at 148.8 nm. In this case, the electron configuration at the beginning creates available terms: $$^1\text{S}$$, $$^3\text{P}$$, and $$^1\text{D}$$. Your photon is exciting one electron from 2 to 3 and changes its momentum by 1 (from p to d). For the excited state configuration, you have another six terms: $$^1P^\text{o}$$, $$^1D^\text{o}$$, $$^1F^\text{o}$$, $$^3P^\text{o}$$, $$^3D^\text{o}$$, $$^3F^\text{o}$$, where superscript "o" indicates odd parity. You can see that this transition satisfies the parity rule, $$\Delta l_i = \pm 1$$, $$\Delta L = 0$$, and $$\Delta S = 0$$ (this example is taken from "Atomic and Laser Spectroscopy" by Alan Corney (link to publisher's site)).

LS coupling does not describe correctly all observed spectra, especially for heavier elements. For example, the mercury intercombination line is an example of an LS coupling model violation.

Regarding "intermediate coupling", Wikipedia cites "The Theory of Atomic Spectra" by E. U. Condon and G. H. Shortley (link). In that book, the intermediate coupling is defined as a coupling process that lies between LS coupling and jj coupling. Unfortunately, I don't have enough knowledge to comment on the selection rules presented in the table.

Edit 2:

I did some archeology. The table on Wikipedia in practically unchanged form dates back to 2005. Back then there was a comment as a part of the article: "Semi-forbidden transitions are electric dipole (E1) transitions for which the selection rule that the spin does not change is violated." This may suggest that the original author meant by "intermediate coupling" a coupling that allows for $$\Delta S \neq 0$$. That would explain why the table says "if $$\Delta S = \pm 1$$". In that case, the author probably meant singlet-triplet electric dipole transitions like the one in mercury (6s$$^2$$ $$^1$$S$$_0\rightarrow$$ 6s6p $$^3$$P$$_1$$) and did not refer to any specific model.

• Thank you so much for your reply. I still have some questions about the transitions in LS coupling of hydrogen-like atoms. How can I have a situation of $\Delta s=0,\ \Delta l=0?$ Because the parity of wave function should avoid it from happening. Another one is even in the two photons transition, how can a light 'add' angular momentum on hydrogen-like atoms and then cause $\Delta s=1?$ Commented Feb 17, 2023 at 17:49
• Hi @HsuBill, sorry for the late answer. I edited my post. Does that answer your questions?
– Noct
Commented Feb 19, 2023 at 0:12
• Thank you so much for your edit!! It helps a lot. I will check those links out. Commented Feb 19, 2023 at 1:53
• @HsuBill, I added also a comment about the Wikipedia article itself.
– Noct
Commented Feb 19, 2023 at 1:59