# 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?
• To the first question: Yes Oct 29 '17 at 12:41
• @user1420303 thanks! What about the second one! Oct 29 '17 at 12:43
• Angular momentum is a vector and the addition of the angular momentum of the photon to that of the molecule has to follow the rules for vector addition. That means that you first have to add these two vectors together before you can take the operator of the total angular momentum.
– Paul
Oct 30 '17 at 8:11
• @Paul, I see why $\hbar\sqrt{(J+1)(J+2)} – \hbar\sqrt{J(J+1)}$ can be smaller than $\hbar$ because of vector addition, but it's in fact greater than that, so I don't see how this helps -- could you please clarify? Also, what do you mean by "add vectors together before your can take the operator of the total angular momentum" -- do you mean add vectors, then get the magnitude of the sum? If so, this still doesn't explain why the increase is greater than what one would expect even if vectors were aligned. And thanks for your help! Nov 1 '17 at 21:11
• Before the absorption takes place the total angular momentum in the system is given by $\vec{J}=\vec{j}+\vec{j}_\text{ph}$ where $\vec{j}$ and $\vec{j}_\text{ph}$ are the angular momenta of the molecule and photon respectively. The total angular momentum operator is thus $\hat{J}^2=(\hat{j}+\hat{j}_\text{ph})^2=\hat{j}^2+\hat{j}_\text{ph}^2+2\hat{j}\hat{j}_\text{ph}$. In your treatment you have ignored the last term (you only add the angular momentum of the individual molecule and photon and ignore their interaction). A proper evaluation involves quite some angular momentum algebra.
– Paul
Nov 2 '17 at 11:28

Sorry for the late answer, but I could not find the time to write something earlier. A concept that you encounter often in quantum mechanics is the so-called coupled and uncoupled basis. Your question is a variation on that theme: before the absorption the uncoupled representation is the more natural where the angular momentum of the photon and the angular momentum of the molecule (and their projections) are good quantum numbers, whereas after the absorption the coupled representation is more suitable and the total angular momentum of the system (and it's projection) is a good quantum number. In Dirac's braket notation we have

$$|jm_j\rangle |j_\text{ph}m_{j_{\text{ph}}}\rangle$$ for the uncoupled representation and $$|JM_J\rangle$$ for the coupled representation.

Your question concerns the conservation of the total angular momentum of the system $\hat{J}^2$ and in the coupled representation the expectation values are easily obtained because $J$ is a good quantum number $$\hat{J}^2|JM_J\rangle=\hbar^2 J(J+1)|JM_J\rangle$$

In the uncoupled representation the total angular momentum is given by $\hat{J}^2=\left ( \hat{j}+\hat{j}_\text{ph} \right)^2=\hat{j}^2+\hat{j}_\text{ph}^2+2\hat{j}\cdot \hat{j}_\text{ph}$

The first two terms of this expression are again easy, but the thrid term is a bit tricky. Using the ladder operators you can write it as $$2\hat{j}\cdot \hat{j}_\text{ph}=2(\hat{j}_x\hat{j}_{\text{ph},x}+\hat{j}_y\hat{j}_{\text{ph},y}+\hat{j}_z\hat{j}_{\text{ph},z})=\hat{j}_+\hat{j}_{\text{ph},-}+\hat{j}_-\hat{j}_{\text{ph},+}+2\hat{j}_z\hat{j}_{\text{ph},z}$$ where $$\hat{j}_\pm|jm_j\rangle=\hbar\sqrt{j(j+1)-m(m\pm1)}|jm_j\pm1\rangle$$ and it thus couples states of different $m_j$. Let us, for example, couple $|jm_j\rangle=|1m_j\rangle$ and $|j_\text{ph}m_{j_\text{ph}}\rangle=|1m_{j_\text{ph}}\rangle$ so that we have $(2j+1)(2j_\text{ph}+1)=9$ basis states: $$|1,-1\rangle|1,-1\rangle,\\|1,-1\rangle|1,0\rangle,\\|1,-1\rangle|1,1\rangle,\\ |1,0\rangle|1,-1\rangle,\\|1,0\rangle|1,0\rangle,\\|1,0\rangle|1,1\rangle,\\ |1,1\rangle|1,-1\rangle,\\|1,1\rangle|1,0\rangle,\\|1,1\rangle|1,1\rangle$$

We determine all expectation values of the total angular momentum operator using the 9 basis states which gives us a total of 81 values that we can order in a 9x9 matrix (in units of $\hbar^2$)

6   0   0   0   0   0   0   0   0
0   4   0   2   0   0   0   0   0
0   0   2   0   2   0   0   0   0
0   2   0   4   0   0   0   0   0
0   0   2   0   4   0   2   0   0
0   0   0   0   0   4   0   2   0
0   0   0   0   2   0   2   0   0
0   0   0   0   0   2   0   4   0
0   0   0   0   0   0   0   0   6


We can diagonalize this matrix and find the eigenvalues 0, 2, 2, 2, 6, 6, 6, 6, 6. These are the same eigenvalues (including the $2J+1$ degeneracy) of $\hat{J}^2$ in the coupled representation!

So the expectation value of the total angular momentum is the same in both cases, the difference is that in the uncoupled case the matrix has off-diagonal elements and it is a bit more work to find its eigenvalues. In the coupled case, the matrix for $\hat{J}^2$ is diagonal and you get the expectation values directly. You might ask if there is an easier way to convert from the coupled to the uncoupled case and back. As you might have guessed there is such a way and it involves the so-called Clebsch-Gordan coefficients.