I have a really basic qustion about spin and electronic transitions.

In one-electron systems (like H or $\ce{He+}$), will the electron change its spin during an electronic transition? Let's say the electron de-excites from 5p UP to 3s, will the spin be UP or DOWN in the 3s orbital?

From crystal-field theory I remember that a d-d transition where the electron had to switch to opposite spin was spin-forbidden (although it may still happen due to certain vibrational states). Is this still true for one-electron systems?


1 Answer 1


In general, transitions in which the "direction" of spin changes are forbidden. More formally, the spin conservation rule (also referred to as the Wigner rule) states that

For both radiative and radiationless transitions, transitions between terms of the same multiplicity are spin-allowed, while transitions between terms of different multiplicity are spin-forbidden.

But spin-orbit coupling relaxes this rule, so in general, lines corresponding to spin-"forbidden" transition are just much weaker and not totally absent from the spectrum. And the bigger the atomic number, the stronger is the spin-orbit coupling, and consequently, the bigger are the intensities of spin-"forbidden" transitions.

$ \newcommand{\linop}[1]{\hat{#1}} \newcommand{\ket}[1]{\left|\,{#1}\right\rangle} \newcommand{\abs}[1]{\left|\,{#1}\right|} $ With respect to your question on addition of spin angular momentum you raised in comment section: you formulated it incorrectly which confused me, so my comments (now deleted) were wrong. Thus, I decided to write a short answer here.

So, for angular momentum $J$ the simultaneous eigenstates of $\linop{J}^{2}$ and $\linop{J}_{z}$ operators are given as follows, $$ \begin{align} \linop{J}^{2} \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\ \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, , \end{align} $$ where $j$ could take non-negative integer and half-integer values, $$ j = 0, 1/2, 1, 3/2, \cdots \, , $$ and $m$ ranges from $-j$ to $j$ in integer steps $$ m = -j, -j+1, \cdots, j-1, j \cdots \, . $$

When there are two sources of angular momentum (like two electrons with their spins) we first we have to make a choice when specifying the state of the system: either we use the uncoupled picture $\ket{j_{1}, m_{1}, j_{2}, m_{2}}$, which leaves the total angular momentum unspecified, or we use the coupled picture $\ket{j_{1}, j_{2}, j, m}$, which leaves the individual components unspecified.

The very language OP uses (i.e. talking about multiplicities of states) suggest that we choose the second picture for further discussion. In the coupled picture the question arises as to the permissible values of total angular momentum quantum numbers $j$ and $m$. It is known from quantum theory of angular momentum that the allowed values of $m$ follow immediately from the relation $m = m_{1} + m_{2}$, while the situation with the allowed values of $j$ is more complicated: its permited values are given by $$ j = j_{1} + j_{2}, j_{1} + j_{2} - 1, \dotsc, \abs{j_{1} - j_{2}} \, . $$

All this is true for any kind of angular momentum (orbital or spin) with the only one exception: with respect to half-integer values of $j$ we note that for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible.

Now for each and every electron we have a source of spin angular momentum characterised by quantum numbers $j = 1/2$ and $m = \pm 1/2$. Thus, in accordance with what is written above for a system of two electrons we have 4 possible state which in the coupled picture are written as follows: $\ket{1/2, 1/2, 1, -1}$, $\ket{1/2, 1/2, 1, 0}$, $\ket{1/2, 1/2, 1, 1}$ and $\ket{1/2, 1/2, 0, 0}$. The first three states are three components of a triplet, the last one is the only component of a singlet.

Note, however, that projection quantum numbers for individual components ($m_1$ and $m_2$) are not specified in the coupled picture, so while it might be tempting to think that for $\ket{1/2, 1/2, 1, -1}$ state they are $m_1 = m_2 = -1/2$ this is wrong. And that what was essentially wrong with the question raised by OP in the comments section: he used the coupled picture terminology ("multiplicities") and at the same time specified the values of projection quantum numbers for individual components ("two electrons both have spin of -1/2"). This could not be dome simultaneously.

  • $\begingroup$ Thank you. Follow-up question: What happens when two valence electrons in different orbitals both have spin of -1/2? According to the Clebsch-Gordon series we should get a total spin angular momentum of -1 and 0, leading to multiplicities of -1 and 1. Is that correct? Or should I take the absolute value of the sum of the two electron spins? $\endgroup$
    – Yoda
    Commented Sep 25, 2015 at 19:20
  • $\begingroup$ As a minor point, if you are looking for more information, phys. chem. and spectroscopy textbooks usually call the two coupling schemes described here as $j j$ and Russell -Saunders coupling $\endgroup$
    – porphyrin
    Commented Jul 16, 2016 at 16:12

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