check for the numerical values of those non-perturbative parameters that are still under discussion, e.g. the magnetic susceptibility of the quark condensate $\chi$.

Definitions
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The transition matrix element $$\begin{aligned}
    \label{Protonvertex}
    {\left\langle P(p,s) \right|} j^{\mu}(0) {\left| P(p\hspace{-0.6pt}+\hspace{-0.6pt}q,s') \right\rangle} = P^{(s)}(p) \left[\gamma^{\mu} F_1(Q^2) -i \frac{1}{2m_p}\sigma^{\mu \nu} q_{\nu} F_2(Q^2) \right]\overline{P}^{(s')}(p\hspace{-0.6pt}+\hspace{-0.6pt}q) \mbox{ }\end{aligned}$$ can conveniently be parametrised in terms of Dirac and Pauli form factors, $F_1$ and $F_2$. Hereafter $P^{(s)}(p)$ is the proton spinor with momentum $p$ and spin $s$. The magnetic moment of a nucleon can then be defined as $$\begin{aligned}
\mu_N=F_1(0)+F_{2}(0) \mbox{ .}\end{aligned}$$ This allows us to take only real photons into account. As $F_{1}^{n}(0)=0$ and $F_{1}^{p}(0)=1$ are just the corresponding charges, it is only necessary to determine $F_{2}(0)$.

The process in Fig.\[PToP\] can be described by the correlation function $$\begin{aligned}
\label{ProtonKorrelator}
 \Pi^{\mu} \left(p,q \right) = i^2  \int\! d^4 x  \int \!d^4 y \;e^{i px + i qy} {\left\langle 0 \right|} \mathcal{T} \lbrace \eta_{p} \! \left( x\right) j^{\mu}\left( y \right)\overline{\eta}_{p}\! \left( 0 \right) \rbrace  {\left| 0 \right\rangle} \mbox{ .}\end{aligned}$$ Here $$\begin{aligned}
\label{EmCurrent}
j^{\mu}=e_d {\overline{d}}\gamma_{\mu} d + e_u {\overline{u}}\gamma_{\mu} u\end{aligned}$$ is the electromagnetic current, with $e_d=-1/3$ and $e_u=2/3$ being the quark charges. $e^{(\lambda)}_{\mu}$ is the four-polarisation vector of the emitted photon, $q 