I was wondering what happens to the spin of electrons after absorbing a photon.

As we all know, when a bound electron absorbs a photon of adequate wavelength, it jumps to an orbital higher in energy. In case the higher orbital is not already occupied by an electron, what happens to the spin of the electron that jumped to the higher orbital? Is the spin inversed or retained?

Following example from a lecture about organic chemistry:

enter image description here

Notice on the bottom right, an electron is getting promoted to a higher orbital without changing spin. This is in accord to what I learned in the lecture about inorganic chemistry (spin selection rule ΔS=0, no spin change).

However someone pointed out that the electron needs to change spin because it transfers a spin of +1. Is this correct or not.

Thanks in advance.

  • $\begingroup$ It is the whole atom that absorbs the photon, not the electron. Quantum number conservation will define the angular momentum of the electron in the orbital it is transferred by the atom absorbing the photon . $\endgroup$
    – anna v
    Commented Jul 16, 2022 at 19:40

2 Answers 2


The orbital angular momentum can change by unity: $\Delta m=1$ is allowed for electric dipole transistions.

  • $\begingroup$ So l changes, but the spin does not? $\endgroup$
    – 冰淇淋
    Commented Jul 16, 2022 at 12:18
  • $\begingroup$ You wuill have to read the section on "selection rules" in your quantum mechanics book. There is more than one rule: they depend on whether the transition is an electric dipole transistion or the rarer magnetic dipole transition, or even a "forbidden" transition. $\endgroup$
    – mike stone
    Commented Jul 16, 2022 at 12:39

As you correctly pointed out, a $\textit{bound}$ electron can absorb a photon (let's just focus on the H-atom for simplicity). To be more precise, one should actually say the system of electron and proton absorbs the photon. By conservation of angular momentum, the sum of photon and e-p angular momentum have to be equal to the angular momentum of e-p after absorption. There are three components to the total angular momentum of an atom (in its rest frame): spin of the nucleus (often denoted $I$), orbital angular momentum of the electron (denoted $L$) and spin of electron ($S$). The sum of orbital and spin angular momentum of the electron is mostly called $J = L+S$ and the total ang. momentum of the atom is called $F = J+I$.

If you look up selection rules, you will often see allowed changes in $L,m_L$ instead of $J$ or $F$, even though the total system electron-nucleus absorbs the photon. It just depends a bit on how exact the description of the atom has to be. If you only consider Coloumb attraction between nucleus and electron, then its angular momentum state is well described by $L$. If you add spin-orbit coupling, good quantum numbers for ang. momentum are $J$ and $m_J$ (called finestructure). If you also add spin-spin interaction between electron and nucleus, good quantum numbers are $F,m_F$ (called hyper-finestructure).

The photon has spin 1 (with only two projections $\pm 1$ instead of the usual triplet $+1, 0 -1$), so in the end, the total angular momentum of the e-p system has to be greater than one. So it is true that there are allowed transition where the electron spin ($S$) does not change, but then some other part of the bound system has to take the angular momentum.

Note that not all transitions where the angular momentum changes by the right amount are allowed, since the mixing of parity also has to be taken into account (and parity of a state is determined by $L$).

  • $\begingroup$ That makes sense, so if we assume that we are dealing with an electric dipole transition, m_s should be equal to zero and no spin flip occurs regarding the electron. Which part of the system takes the angular momentum then? $\endgroup$
    – 冰淇淋
    Commented Jul 16, 2022 at 12:59
  • $\begingroup$ en.wikipedia.org/wiki/Selection_rule#Summary_table Look at this summary table for dipole transitions, there are all the possibilities. If $\Delta S = 0$, then e.g. the angular momentum can increase (or decrease) by one. $\endgroup$
    – Samuel
    Commented Jul 16, 2022 at 13:09
  • $\begingroup$ Thank you, makes sense now $\endgroup$
    – 冰淇淋
    Commented Jul 16, 2022 at 13:12

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