field $F_{\mu \nu}$. Expanding the correlator in Eq. in powers of the background field and taking only the terms linear in $F_{\mu \nu}$ corresponds to the single photon emission, also described by Eq.. We will refrain from a more detailed presentation of this expansion, but instead would like to refer to [@Novikov:1983gd] for a review on the background field method and to Chapter 2 in Ref. [@Ball:2002ps], which is dedicated to the expansion of correlation functions in an electromagnetic background field.

Expansion of the correlator
---------------------------

Applying the usual strategy of QCD sum rules, we have to calculate the correlation function Eq. in two different regimes. If the in- and out-going particle states are close to proton mass-shell, i.e. $p^2\approx m_P^2$ and $(p+q)^2 \approx m_P^2$, the hadronic representation of Eq. will be dominated by the process $p \to p \gamma$.

The contribution $T_P$ of the process $p \to p \gamma$ is then given by $$\begin{aligned}
T_P(p,q)=e^{(\lambda)}_{\mu} \frac{{\left\langle 0 \right|} \eta_P(0) {\left| P(p,s) \right\rangle} {\left\langle P(p,s) \right|} j^{\mu}(0) {\left| P(p+q,s) \right\rangle} {\left\langle P(p+q,s) \right|} \overline{\eta}(0) {\left| 0 \right\rangle}}{\left(m_P^2-p^2 \right) \left(m_P^2-(p+q)^2 \right)}\end{aligned}$$ Using Eqs., and the spin summation formula for Dirac spinors $$\begin{aligned}
\label{spinsum1}
\sum_{s} P^{(s)}(p) \overline{P}^{(s)}(p) = & \; \slashed{p}+m_{P}\end{aligned}$$ this 