quantum field theory - Optical theorem for the given diagram

Assume we have the abelian gauge theory with single fermion. Suppose the following diagram:

Here the initial "photon" $\gamma$ is in the same state as the final one, so this is the diagram of self-energy. I want to check whether the optical theorem holds for it.

The optical theorem for the total amplitude reads $$ \tag 1 2\text{Im}M_{\gamma \to \gamma}= (2\pi)^{4}\sum_{n}|M_{\gamma \to n}|^{2}\delta (p_{n} - p_{\gamma}) \simeq \Gamma_{\gamma} $$

The question: have I also take into account the diagram which is also 6th order of perturbation theory?

An edit

It seems that the formal answer is indeed yes. Really, the rhs of $(1)$ contains the half of the squared diagram, and the diagram above is needed in the lhs.