# Conservation of angular momentum in rotational spectroscopy

In spectroscopy, transitions between rotational energy levels obey the selection rule $\Delta J = \pm1$. The rule follows from the fact that the photon absorbed has an angular momentum of $\pm \hbar$, so the molecule must change its angular momentum for the total angular momentum to be conserved.

However, the angular momentum of a molecule is given by $\hbar \sqrt{J(J+1)}$, so the change in molecular ang. mom. is $\hbar\sqrt{(J+1)(J+2)} - \hbar\sqrt{J(J+1)}$. As $\hbar\sqrt{(J+1)(J+2)} - \hbar\sqrt{J(J+1)} \neq \hbar$, it seems that the total angular momentum is not actually conserved.

Questions:

• Does the conservation of angular momentum apply to quantum systems?
• If yes, what is the source of the contradiction above?
• It is conserved, in the operator sense. That's what the Clebsch-Gordan formalism is all about - it normally takes a full chapter of late-undergrad QM textbooks. – Emilio Pisanty Oct 29 '17 at 23:44
• @EmilioPisanty Thanks for your comment! Could you please recommend a book / chapter? My study book is Atkins' Molecular Quantum Mechanics, and I couldn't find anything there – GingerBadger Oct 30 '17 at 0:10

If you want to think if $\sqrt{J(J+1)}$ as the length of the angular momentum, then be aware that adding two vectors does not result in a vector with a length that is the sum of the two original lengths. For instance, $\hat x$ is a vector of length $1$, as is $\hat y$, but their sum $\hat x+\hat y$ does not have length $2$.
$J$ should be understood at the largest possible projection (i.e. largest allowed value $J_z$) of a set of states that are eigenstates of $J^2$. $\Delta J=\pm 1$ simply means you go between two sets of states $J_{initial}$ and $J_{final}$ such that $\vert J_{initial}-J_{final}\vert=1$. Nothing more. There is no length of vectors involved.
• @Alex Angular momentum is a vector so you can't just compare magnitude. What you need to do is realize that the selection rules are compatible with angular momentum conservation since $J\otimes 1 = J+1\oplus J\oplus J-1$ as per angular momentum coupling theory. There are symmetry reasons (parity actually) why states with final angular momentum $J$ cannot appear. – ZeroTheHero Oct 30 '17 at 0:14
• Thanks! Two further questions / points: (1) I get your point about $\hat{x} + \hat{y}$ not being 2, and this only would explain everything if $\hbar\sqrt{(J+1)(J+2)} – \hbar\sqrt{J(J+1)}$ was smaller than $\hbar$, but it's in fact greater than that. (2) Could you please explain what you mean by the angular momentum coupling theory? I'm not familiar with it, but a simple "intuitive" explanation would suffice. Thanks again! – GingerBadger Nov 1 '17 at 21:07
• @Alex : if you want to think of angular momentum $J$ states as vectors of length $\sqrt{J(J+1)}$, then the angular momentum of the photon would have length $\sqrt{2}$ and $\sqrt{(J+1)(J+2)}-\sqrt{J(J+1)}\le \sqrt{2}$. – ZeroTheHero Nov 4 '17 at 12:34