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My preconception is that when you fill an empty orbital by exciting an electron (with a photon), the empty orbital should be filled by an electron with the same spin. However, I've seen examples of that happening and not happening. For example, on this University of Calgary web page there is an image of the electron spin being flipped:

The lowest energy transition is that between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) in the ground state. The absorption of the EM radiation excites an electron to the LUMO and creates an excited state. The more highly conjugated the system, the smaller the HOMO-LUMO gap, DE, and therefore the lower the frequency and longer the wavelength.

Excitation that doesn't conserve spin

Does it just not matter that much energetically?

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  • $\begingroup$ @orthocresol Does that mean the spin angular momentum of a photon at fotoexcitation generally leads to change of an electron orbital angular momentum but not the spin one, correct ? $\endgroup$
    – Poutnik
    Jan 18 at 10:09
  • $\begingroup$ @Poutnik I think so, that’s a typical selection rule for spectroscopy. $\endgroup$
    – orthocresol
    Jan 18 at 10:29
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The picture shows a triplet excited state returning to the ground state by emitting a photon, i.e. phosphorescence. This can only happen if there is also an interaction that couples angular momentum change with the transition, such as spin orbit coupling.

The electron has two quantum numbers; the spin $S=1/2$ but is not the spin of the electron that is changing but the other quantum number 'z' , 'azimuthal' or 'magnetic' spin quantum number $m_z=\pm 1/2$.

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    $\begingroup$ I know it looks as if the photon is being emitted (that was my first thought as well), but especially placed in context of the webpage, it’s depicting an excitation process. I think maybe I shouldn’t have removed the contextual text, sorry. $\endgroup$
    – orthocresol
    Jan 18 at 10:29
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    $\begingroup$ No apology needed. Well the arrow is the wrong way round if it is excitation, and if it were absorption it would predict a singlet to triplet process which would still need the coupling I mention to break the $\Delta S =0$ rule. $\endgroup$
    – porphyrin
    Jan 18 at 18:34
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TL;DR Excitation of an electron typically conserves spin; that is to say, the spin must be left unchanged by the process. If the initial state has no net spin (one spin-up and one spin-down electron), then the final state should also have no net spin (one spin-up and one spin-down electron). This rule is reliable for small-ish atoms, but often breaks down with heavier atoms.

Note that this doesn't prescribe which electron gets excited (spin-up or spin-down); in general we can't say anything about that.


The reason is to do with Fermi's golden rule. The transition probability is governed by a quantum mechanical integral (the "transition dipole moment") which looks like

$$\langle \psi_\text{final} | \hat{H} | \psi_\text{initial} \rangle.$$

Here, $\psi_\text{initial}$ and $\psi_\text{final}$ are the initial and final states respectively, and $\hat{H}$ is the perturbation Hamiltonian. For an electromagnetic wave process this is the dipole moment operator $\hat{\mu}$, which can be expressed in terms of the position operators:

$$\hat{\mu} = \mu_x\hat{x} + \mu_y\hat{y} + \mu_z\hat{z}$$

The key things to note are that:

  1. The electron wavefunctions $\psi$ can typically be factorised into a spatial part and a spin part, which I denote as $\phi$ and $\chi$ respectively:

$$\begin{align} |\psi_\text{final}\rangle &= |\phi_\text{final}\rangle \cdot |\chi_\text{final}\rangle, \\ |\psi_\text{initial}\rangle &= |\phi_\text{initial}\rangle \cdot |\chi_\text{initial}\rangle. \\ \end{align}$$

  1. The perturbation Hamiltonian $\hat{H} = \hat{\mu}$ only depends on spatial degrees of freedom, and not spin.

Therefore, the transition dipole moment can be factorised into a spatial part and a spin part:

$$\begin{align} \langle \psi_\text{final} | \hat{H} | \psi_\text{initial} \rangle &= \langle \phi_\text{final}\chi_\text{final} | \hat{\mu} | \phi_\text{initial}\chi_\text{initial} \rangle \\ &= \langle \phi_\text{final} | \hat{\mu} | \phi_\text{initial} \rangle \langle \chi_\text{final} | \chi_\text{initial} \rangle \end{align}$$

In particular, the spin part $\langle \chi_\text{final} | \chi_\text{initial} \rangle$ is zero, and hence the transition probability is zero, unless the initial and final states have the same spin. Processes which involve spin flips are often called "spin-forbidden".

Under certain conditions, the assumption (1) breaks down: that is to say, the wavefunction $\psi$ cannot be cleanly factorised into spatial and spin parts $(\phi, \chi)$. This happens particularly when spin–orbit coupling is present: heavier elements (i.e. elements with larger atomic number) typically exhibit greater spin–orbit coupling.

Thus, for light atoms like hydrogen, spin-forbidden processes are genuinely "forbidden", and therefore almost never happen. However, so-called "spin-forbidden" processes can actually happen for heavier atoms. Some examples include:

  • the pale pink colour of $\ce{Mn^2+}$ salts results from a spin-forbidden excitation process. The fact that it is spin-forbidden just means that the colour is a lot weaker than that of a typical transition metal. There is a lot more discussion of this particular case at Why is manganese(II) coloured although the transition should be spin-forbidden?.

  • the purple colour of iodine $\ce{I2}$ results from a singlet to triplet excitation. The fact that this colour is pretty intense (as opposed to the pale pink of $\ce{Mn^2+}$ simply reflects the fact that iodine has a large atomic number.

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