Forbidden transition My question concerns something I just read on wikipedia whilst looking up forbidden transitions, here:
https://en.wikipedia.org/wiki/Forbidden_mechanism#Gamma_decay
Specifically, this sentence: 
"However, gamma emission is absolutely forbidden when the nucleus begins in a zero-spin state, as such an emission would not conserve angular momentum."
This seems to imply that gamma emission is not absolutely forbidden unless the nucleus begins in a zero-spin state.
I was under the impression (and other sources seem to suggest) that gamma emission was absolutely forbidden also in the case where the nucleus is not in an initial zero spin state, but where there is no change in spin between the initial and final states of the nucleus.
Could someone clarify that please?
 A: There's some missing background information: all even-even nuclei have ground state spin and parity $J^P = 0^+$.
Suppose a nuclear excited state with $J^P = 0^+$ were to decay to the ground state by emitting a single photon.  The lowest angular momentum that can be carried by a photon is $\hbar$, in dipole radiation.  (The parity of the EM radiation determines whether the field is "electric" or "magnetic" dipole, "E1" or "M1".)  So then you have an initial state with zero angular momentum in the excited nucleus, and a final state with zero angular momentum in the nucleus and nonzero angular momentum in the electromagnetic field.  This is forbidden.
Isomers with spin 2, 3, etc. may decay to a $0^+$ ground state by emitting "quadrupole" (E2/M2) or "octupole" (E3/M3) photons.  These states have less overlap with the nucleus than the dipole photons, and so are slower than E1/M1 transitions, but angular momentum is conserved in all reference frames.
You ask about transitions where the nuclear spin doesn't change, e.g. $2^+\to2^+$.  Here you can have dipole radiation emitted.  The initial and final spins are
\begin{align}
\vec J_\text{initial} &= \vec J_\text{isomer} \\
\vec J_\text{final} &= \vec J_\text{nucleus} + \vec J_\text{photon}
\end{align}
You may remember that there is some flexibility when you add quantum-mechanical angular momenta: the eigenvalue of $\vec J_\text{final}$ is constrained by
$$
\left|
J_\text{nucleus} - J_\text{photon}
\right|
\leq
J_\text{final}
\leq
\left|
J_\text{nucleus} + J_\text{photon}
\right|
$$
In a classical cartoon of angular momentum the range depends on the relative orientation of the contributions to the total: $\vec J_\text{nucleus}$ and $\vec J_\text{photon}$ may be parallel, antiparallel, or anywhere in between.  However if $J_\text{nucleus} = 0$ then there is no such freedom.
