这是一份ucl伦敦大学学院 PHAS0073作业代写的成功案
$$
j_{\mu}^{e m} \stackrel{U_{c}}{\rightarrow}-j_{\mu}^{e m}
$$
But the electromagnetic field $A_{\mu}$ satisfies the equation
$$
\square^{2} A_{\mu}=j_{\mu}^{e m} .
$$
Thus from Eq. (75), it follows that
$$
A_{\mu} \stackrel{U_{c}}{\rightarrow}-A_{\mu}
$$
Since a photon is a quantum of electromagnetic field, it follows that the $C$-parity of photon is $-1$ viz.
$$
\eta_{c}(\gamma)=-1
$$
The decays $\pi^{0} \rightarrow 2 \gamma$ and $\eta^{0} \rightarrow 2 \gamma$ have been observed. Hence if these reactions proceed via electromagnetic interaction, it then follows from $C$-conservation that
$$
\begin{aligned}
&\eta_{c}\left(\pi^{0}\right)=+1 \
&\eta_{c}\left(\eta^{0}\right)=+1
\end{aligned}
$$
PHAS0073 COURSE NOTES ：
which changes sign under charge conjugation. Hence
$$
\Gamma\left(\phi \rightarrow K^{+} K^{-}\right)=\cos ^{2} \theta \Gamma\left(\omega_{8} \rightarrow K \bar{K}\right) .
$$
Therefore,
$$
\Gamma\left(\phi \rightarrow K^{+} K^{-}\right)=\cos ^{2} \theta \frac{\gamma^{2}}{4 \pi}\left(3 \times \frac{2}{3}\right)\left(\frac{p_{K K}^{3}}{m_{\phi}^{2}}\right)
$$