# Selection rule $\Delta S=0$: Why does a photon not interact with an electrons spin?

When talking about selection rules in atomic physics, many books state that the photon interacts with the electrons angular momentum such that that $\Delta l=\pm 1$. Absorbed/emitted photons exchange angular momentum with the electrons of the atom. For example: When an incoming circular polarised photon with spin $1\hbar$ is absorbed by the atom, an electron has to change its angular momentum $\vec{l}$ by $1\hbar$.

But why is there no such interaction between the electrons spin $\vec{s}$ and photons? What is the origin of the selection rule $\Delta S=0$?

I'm interested in both mathematical and especially intuitive explanations :-)

The selection rule $\Delta S=0$ is an approximation, nothing more, and in suitable circumstances it can easily break. One prominent example of this is the hydrogen 21cm line.

Electromagnetic atomic and molecular transitions are arranged into a series in order of multipolarity, which describes the atomic operators that enact the interaction hamiltonian, with higher multipolarities scaling down in coupling strength as powers of $a/\lambda$, i.e. the ratio between the system's size and the radiation's wavelength, which is generally small. Thus, you get

• as the leading-order term, electric dipole (E1) transitions;
• weaker by a factor of ${\sim}a/\lambda$,
• magnetic dipole (M1), and
• still weaker by another factor of ${\sim}a/\lambda$,
• electric octupole (E3) transitions;
• and so on.

Generally, the no-spin-flips rule holds only for electric dipole transitions, for which the interaction hamiltonian is the electric dipole operator $\hat{\mathbf d}$, which does not couple sectors with different spin (unless you have strong spin-orbit coupling).

However, it is perfectly possible to have magnetic dipole transitions between states of spin that differ by $\Delta S=1$. Here the coupling is weaker (so you will need a higher intensity, or longer pulse times, to excite them), and thus typically the linewidth is smaller (so you will require a sharper laser), also giving a longer decay lifetime, but those are things that make the transition harder to observe, not impossible.

When you have only one electron then $\Delta S=0$ makes intuitive sense: you can change the angular momentum $l$ of the atom by changing it's internal structure (by pushing the electron in "another orbit" if you will), while you certainly can't change the internal structure of the electron to change $s$.

Would it be possible to change $s$ then you could change the internal structure of the electron, but since, as far as we know it, spin is intrinsic you can't do this and $s$ stays fixed.

The selection rule $\Delta S=0$ is also valid if you have two (unpaired) electrons in the atom. You can see this if you write the total wavefunction of both electrons in the dipole matrix element as a product of position function and spin function:

\begin{align}M_{ik} & =\int \Psi^*_i(\vec r_1,\vec r_2)(\vec r_1+\vec r_2)\Psi_k(\vec r_1,\vec r_2)d\tau_1\tau_2\\ &=\int \psi^*_i(\vec r_1,\vec r_2)\chi(s_1,s_2)(\vec r_1+\vec r_2)\psi_k(\vec r_1,\vec r_2)\chi(s_1,s_2)d\tau_1\tau_2 \end{align}

where $d\tau_i$ indicates that we only integrate over the electrons positions.

But first we have to come back to the Pauli Exclusion principle: another way to state the Pauli exculsion principle is that the total wavefunction must be antisymmetric with respect to the exchange of the two electrons. This in turn means that either the position function $\psi$ or the spin function $\chi$ must be antisymmetric. When the position function is symmetric, then it doesn't change sign if you change one electron for the other:

$$\psi_s(1,2)\to+\psi_s(2,1)$$

while the antisymmetric position function changes sign:

$$\psi_a(1,2)\to-\psi_a(2,1)$$

Next, there are four ways to combine two electrons: three with total spin one (triplett state) and the singulett state with spin zero. The triplett state have a antisymmetric position function, while the singlet state has a symmetric position function.

Now, if you want the matrix element to be unchanged if you exchange the electrons then both states $\Psi_i$ and $\Psi_k$ must be either a triplett state or a singulett state. Because when both are in a trilett state (S=1) and you exchange the electrons then the sign stays the same. When both are in a singulett state then you get a minus from both states if you exchange the electrons, the minus then cancels.

On the other hand, if one is a singulett and the other a triplett state the one state gets a minus from the exchange and the other a plus. Since electrons are indistinguishable, $M_{ik}$ must not change if you exchange them. Because $M_{ik}$ changes sign when you go from S=1 to S=0 this transition is forbidden, or $\Delta S=0$. See Para- and ortho-helium as an example.

But why is there no such interaction between the electrons spin $\vec{s}$ and photons?

Actually, maybe there is. See Compton scattering. The incident photon is partially absorbed and decelerated in the vector sense, whilst the free electron moves. IMHO you can visualize this by drawing repeated circles on a piece of paper without lifting your pencil. Now repeat while somebody pulls the paper down and to the left. The electron spin is no longer spherically symmetric. What is the origin of the selection rule ΔS=0 ?

Actually, I don't know, but I would hazard a guess that it's something to do with no spin-flip occurs. Or perhaps at some deeper level it's the electron's intrinsic spin makes it what it is. You don't change an electron into something else by throwing a photon at it. You only do that to a photon, in gamma-gamma pair production.