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what do we mean by the selection rule $\Delta S=0$?

Can you give me some example for hydrogen atom?

For example if I want to go from $1s$ to $2p$ how can I calculate $S$ for $1s$ or for $2p$?

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This question has already been asked in different forms. [Refer : 1) Selection rule ΔS=0: Why does a photon not interact with an electrons spin? 2) Does a photon interact with the spin of an electron?].

I will try to give you a summary of the very good points brought up. $\Delta S = 0$ actually is not a binding rule. Basically, in the semi-classical model of light where you consider oscillating electric and magnetic dipoles, only the magnetic dipoles can change the spin of the electron. This process is called 'spin-flip'. It is the same mechanism for the famous Hydrogen 21 cm astronomical limit. This is a well known case of spin flip and where the selection rule actually breaks down. In fact, spin-flip is very possible and observed in PL spectra of many materials - although, you would need a high intensity pulse (of a very short duration) because magnetic dipole interactions are very weak compared to electric dipole interactions (because in the former, the velocity is restricted by the speed of light - Read 'Fine structure constant in Hydrogen').

Therefore, in optical processes, you typically consider only electric dipole interactions which are stronger in strength, and in this case, as you can see, the spin selection rule $\Delta S = 0$ is very much true.

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  • $\begingroup$ My lecturer has mentioned this by explaining as saying that the electric dipole transition does not couple to spin, could you explain this a little? $\endgroup$
    – Thormund
    Commented May 11, 2020 at 18:15
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    $\begingroup$ @Thormund It's very easy to reconcile if you're familiar with magnetism in condensed matter, which is inherently quantum in nature. You can imagine electron spins behaving like tiny magnets that can be aligned parallel or anti parallel to an external magnetic field. The electric field has no effect in 'orientation', it can only accelerate the electron. $\endgroup$
    – Xivi76
    Commented May 11, 2020 at 18:31
  • $\begingroup$ Understood, thanks!! $\endgroup$
    – Thormund
    Commented May 11, 2020 at 18:32
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$\newcommand\bs\boldsymbol$ We note that the spin $S$ of an atom is the spin angular momentum sum of all the electrons (and nuclei) in the atom, as given by $$\bs{S} = \sum_i \bs{S}^{(i)} .$$

Most often, selection rules help us specify what happens to the atom upon interaction with a photon, which has spin angular momentum of norm $\sqrt{2}\hbar$, and a projection of $\pm\hbar$. (The orbital angular momentum of a photon is beyond the scope necessary for atomic-physics.)

When the atom absorbs/releases a photon, the $orbital$ angular momentum $\bs{L}$ changes, but spin orbital angular momentum $\bs{S}$ does not change, and total angular momentum is conserved. This is what we mean by $\Delta \bs{S} = 0, \Delta \bs{L} = \pm 1$.

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  • $\begingroup$ Ok thank you, but why for example in a triplet state $S=1$ and in a singlet $S=0$? $\endgroup$
    – Salmon
    Commented May 12, 2020 at 14:00
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    $\begingroup$ @Salmone For 2 electron systems, there are 4 ways to arrange these 2 electrons (up to normalisation). Three of them will will add to $S^2| electrons \rangle = \hbar\sqrt{2} | electrons \rangle$, which we call them as triplet states, and 1 way for which we get 0. You may want to read into Clebsh-Gordan coeffecients and angular momentum addition. $\endgroup$
    – Thormund
    Commented May 12, 2020 at 15:46

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