Added Note 2: Conservation of parity in photon emission
In Section 18-1 of this chapter we considered the emission of light by an atom that goes from an excited state of spin $1$ to a ground state of spin $0$. If the excited state has its spin up ($m=+1$), it can emit a RHC photon along the $+z$-axis or a LHC photon along the $-z$-axis. Let’s call these two states of the photon $\ket{R_{\text{up}}}$ and $\ket{L_{\text{dn}}}$. Neither of these states has a definite parity. Letting $\Pop$ be the parity operator, $\Pop\,\ket{R_{\text{up}}}=\ket{L_{\text{dn}}}$ and $\Pop\,\ket{L_{\text{dn}}}=\ket{R_{\text{up}}}$.  What about our earlier proof that an atom in a state of definite energy must have a definite parity, and our statement that parity is conserved in atomic processes? Shouldn’t the final state in this problem (the state after the emission of a photon) have a definite parity? It does if we consider the complete final state which contains amplitudes for the emission of photons into all sorts of angles. In Section 18-1 we chose to consider only a part of the complete final state.  If we wish we can look only at final states that do have a definite parity. For example, consider a final state $\ket{\psi_{F}}$ which has some amplitude $\alpha$ to be a RHC photon going along $+z$ and some amplitude $\beta$ to be a LHC photon going along $-z$. We can write \begin{equation} \label{Eq:III:18:72} \ket{\psi_{F}}=\alpha\,\ket{R_{\text{up}}}+ \beta\,\ket{L_{\text{dn}}}. \end{equation} The parity operation on this state gives \begin{equation} \label{Eq:III:18:73} \Pop\,\ket{\psi_{F}}=\alpha\,\ket{L_{\text{dn}}}+ \beta\,\ket{R_{\text{up}}}. \end{equation} This state will be $\pm\,\ket{\psi_{F}}$ if $\beta=\alpha$ or if $\beta=-\alpha$. So a final state of even parity is \begin{equation} \label{Eq:III:18:74} \ket{\psi_{F}^+}=\alpha\{\ket{R_{\text{up}}}+ \ket{L_{\text{dn}}}\}, \end{equation} and a state of odd parity is \begin{equation} \label{Eq:III:18:75} \ket{\psi_{F}^-}=\alpha\{\ket{R_{\text{up}}}- \ket{L_{\text{dn}}}\}. \end{equation}  Next, we wish to consider the decay of an excited state of odd parity to a ground state of even parity. If parity is to be conserved, the final state of the photon must have odd parity. It must be the state in (18.75). If the amplitude to find $\ket{R_{\text{up}}}$ is $\alpha$, the amplitude to find $\ket{L_{\text{dn}}}$ is $-\alpha$.  Now notice what happens when we perform a rotation of $180^\circ$ about the $y$-axis. The initial excited state of the atom becomes an $m=-1$ state (with no change in sign, according to Table 17–2). And the rotation of the final state gives \begin{equation} \label{Eq:III:18:76} R_y(180^\circ)\,\ket{\psi_{F}^-}=\alpha\{\ket{R_{\text{dn}}}- \ket{L_{\text{up}}}\}. \end{equation} Comparing this equation with (18.75), you see that for the assumed parity of the final state, the amplitude to get a LHC photon along $+z$ from the $m=-1$ initial state is the negative of the amplitude to get a RHC photon from the $m=+1$ initial state. This agrees with the result we found in Section 18-1. 