From my knowledge, selection rules consists of (correct me if I am wrong):

  • |m-m'|=<1
  • |j-j'|=<1
  • |l-l'|=1
  • s=s'

where prime denotes the final state and unprimed denotes the initial state.

From my knowledge, selection rules provides a set of necessary conditions for a transition to be an allowed transition.

However, is the opposite case true in the sense that: can I safely assume that if a transition that obeys the set of selection rules then it is an allowed transition? In other words, is obeying all selection rules a sufficient condition for a state to be an allowed transition?


2 Answers 2


It is possible to have specific transitions which have $0$ probability but are still allowed by the selection rules. Suppose $\hat A^\ell_k$ is a tensor operator. The transition probability is usually computed through the matrix element $$ \langle j’m’\vert \hat A^\ell_k \vert j m\rangle = \langle j’\Vert \hat A^\ell \Vert j\rangle \frac{\langle j’m’\vert \ell k; j m\rangle}{\sqrt{2j’+1}} $$ using the Wigner-Eckart theorem, with $\langle j’\Vert \hat A^\ell\Vert j\rangle$ the reduced matrix element and $\langle j’m;\vert \ell k;j m\rangle$ a Clebsch-Gordan coefficient.

The selection rules are extracted from the reduced matrix element, i.e. the initial an final states must be so that $j’\in \ell\otimes j$ as angular momentum coupling. In your case $\ell=1$ so that $j’=j-1,j,j+1$ in accordance with your stated selection rule.

It is possible, however, for the Clebsch-Gordan coefficient to have an accidental $0$ even if the reduced matrix element is $\ne 0$. That would make this specific transition probability nil, although it would be non-zero for some other component $\hat A^\ell_{k’}$ with $k’\ne k$. For instance, $\langle j’ 0\vert 1 0;j 0\rangle$ is $0$ unless the sum $j+j’$ is odd. This means that dipole-type transitions in the $+z$ direction cannot connect to states with $j’=j$.


There is a popular rule of the thumb in this case: Anything that's not forbidden is mandatory. If a potential transition is compatible with the selection rules, and doesn't violate any other conservation law either, it has to be an allowed transition. It is a sufficient condition.

However, even though it may be allowed, how frequently these transitions occur, is a different subject altogether. There are allowed processes that proceed at a very slow rate, so much so that with a bad choice of a sampling interval, they may not be detected. That might come across as the process being forbidden, while it is not.

One has to look at the transition amplitudes and probabilities, to be sure.

  • $\begingroup$ Is there a formal proof that it is a sufficient condition, or is it just a rule of thumb because it is true for most cases (in fact, is it true for ALL cases)? $\endgroup$
    – user148792
    May 13, 2018 at 3:25
  • $\begingroup$ See, formal theoretical proofs are just going to be compatibility with the conservation laws and the selection rules, and the fact there is a non-vanishing transition probability. But even with these theoretical proofs under the belt, the ultimate test of a conjecture is in a lab - the only way to be sure is to actually observe whether it happens or not. The rule of the thumb originates from years of experience of particle physicists, analyzing collision debris in their experiments, and decay products of various species, to construct all possible reaction/decay modes of a species. (contd.) $\endgroup$
    – 299792458
    May 13, 2018 at 3:34
  • $\begingroup$ (contd.) But of course, the line of distinction between a forbidden transition, and an allowed transition that proceeds at a very slow rate, if difficult to analyze experimentally. The choice of sampling interval becomes crucial. But weird examples, such as the proton decaying with an estimated half life of $10^{34}$ years will always blur the bounary of distinction between these two. $\endgroup$
    – 299792458
    May 13, 2018 at 3:40

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