# What causes the transition $J=0\to J^\prime=0$ absolutely forbidden?

For many-electron atoms, the dipole selection rules are $$\Delta M_J=0,\pm 1$$ and $$\Delta J=0,\pm 1$$ with the transition $$J=0\to J^\prime=0$$ absolutely forbidden. This remains true even for quadrupole transitions and magnetic dipole transitions.

Can this be proved or is it an empirical fact? Is $$j=0\to j^\prime=0$$ also forbidden for hydrogenic atoms?

If $$J_i$$ is the angular momentum of the initial state and $$L$$ is the angular momentum of the $$L$$-pole radiation, then the final state must have angular momentum $$J_f$$ that is in the coupling of $$J_i\otimes L$$. Thus, for dipole $$L=1$$ transitions, we have $$J_f=J_i+1, J_i$$ or $$J_i-1$$ provided $$J_i-1\ge 0$$. For quadrupole $$L=2$$ transitions we have $$J_i+2\le L_f \le \vert J_i-2\vert$$.
Using this with $$J_i=0$$, one simply notes that $$0\otimes L=L$$, and is thus $$J_f=0$$ is never allowed.
The less mathematical explanation is that radiation from and $$L$$-pole transition carries $$L$$ units of angular momentum, so that, if you start with $$J_i=0$$, returning to $$J_f=0$$ when the radiation carries some angular momentum would violate conservation of angular momentum.
• By $\vec{J}$, I meant $\vec{J}=\vec{L}+\vec{S}$ not only $\vec{L}$. – mithusengupta123 Dec 8 '18 at 4:35
• It makes no difference. Replace my $L_i$ and $L_f$ by $J_i$ and $J_f$: the angular momentum coupling rules are for the same for $J$ or $L$. – ZeroTheHero Dec 8 '18 at 4:37
• "when the radiation carries some angular momentum would violate conservation of angular momentum." But $\Delta J=0$, for example, $J_i=1$ to $J_f=1$ then also violate conservation of angular momentum? – mithusengupta123 Dec 8 '18 at 13:15
• @mithusengupta123 ... or slide 22 of this: slideshare.net/mobile/ibenk97/… which gives an example with $\ell=2$ and $\ell=1$, which would apply to dipole transitions from a $J_i=2$ state. – ZeroTheHero Dec 8 '18 at 13:29