Conservation of angular momentum in rotational spectroscopy
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.