Consider a two-level atom of with energy levels $|g\rangle$ and $|e\rangle$, such that dipole transitions are allowed between these two levels. Let $l_{e}=l_{g}+1$ where $l_{e},l_{g}$ are the total angular momentum to the states and let $m_{e},m_{g}$ be the angular momentum along the $z$ axis. Suppose the atom is placed in an optical cavity oriented along the $x$ axis, such that the cavity can support two linear polarisations of light, $\hat{\textbf{e}}_{y}$ and $\hat{\textbf{e}}_{z}$ and such that the cavity is resonant with the transition frequency.
If $m_{e}$=$m_{g}$, the transition would be a so called $\pi$ transiton, whereby a photon of polarisation $\hat{\textbf{e}}_{z}$ is absorbed to excite from $|g\rangle$ and $|e\rangle$. This make sense to me, as such a photon has $m_{\gamma}=0$, so $m_{g}+m_{\gamma}=m_{e}$, hence angular momentum is conserved.
Now suppose $m_{e}$=$m_{g}+1$. The dipole matrix element of such a transition would have a unit vector of $\sigma^{+}=\frac{1}{\sqrt{2}}(\hat{\textbf{e}}_{x}+i\hat{\textbf{e}}_{y})$, indicating that light of this polarisation drives the transition with maximum strength. This also makes sense to me as now the photon is circularly polarised, and $m_{\gamma}=+1$ $m_{g}+m_{\gamma}=m_{e}$ is still true.
What I can't understand is the idea of single photons being absorbed from/emitted into the cavity. Because $\hat{\textbf{e}}_{y}$ is not perpendicular to the dipole matrix element for this transition, mathematically, it seems to me the you can excite from $|g\rangle$ to $|e\rangle$ by absorption of a photon that is linearly polarised in $\hat{\textbf{e}}_{y}$ (indicated by the non-zero value for the atom cavity coupling, $g$). However when I think about this from the perspective of angular momentum conservation, it seems to me that $m_{g}+m_{\gamma}\neq m_{e}$, since $m_{\gamma}=0$.
Am i incorrect in assuming $m_{\gamma}=0$ in this case? Or am i thinking about conservation of angular momentum in the quantum regime incorrectly?
