\cdot e^{(\lambda)}=0$. The current $$\begin{aligned}
\label{IoffeCurrent}
\eta(x)=& \left(u^a(x)\mathcal{C}\gamma^{\lambda} u^{b}(x) \right) \gamma_5 \gamma_{\lambda }d^{c}(x)\varepsilon^{abc}\end{aligned}$$ is the usual Ioffe current [@Ioffe:1981kw] for the proton. $\mathcal{C}$ is the charge conjugation matrix, $a,\;\!b,\;\!c$ are colour indices and $\varepsilon^{abc}$ is the three dimensional Levi-Civita symbol. Note that the current has fixed isospin $1/2$. The isospin relation between proton and neutron then assures that the formulae for the neutron magnetic moments can be obtained from those of the proton by exchanging $e_u \leftrightarrow e_d$. Therefore we will only consider the proton. The coupling $\lambda_P$ of the Ioffe current Eq. to the proton is defined by $$\begin{aligned}
\label{ProtonCoupling}
{\left\langle 0 \right|} \eta(0){\left| P(p,s) \right\rangle}=\frac{\lambda_{P}}{(2\pi)^2} P^{(s)}(p) \mbox{ .}\end{aligned}$$

By introducing an electromagnetic background field of a plane wave $$\begin{aligned}
\label{Fmunu}
F_{\mu \nu} = i\left(e^{(\lambda)}_{\nu} q_{\mu}-e^{(\lambda)}_{\mu} q_{\nu} \right) e^{iqx}=f_{\mu \nu} e^{iqx} \mbox{ ,}\end{aligned}$$ it is possible to absorb the emitted photon into the background field. We consider the following object $$\begin{aligned}
 \label{Protoncorrelator}
 \Pi_{P}^{\mu} \left(p,q \right) e^{\left( \lambda \right)}_ {\mu}= i \int\! d^4 x   \;e^{i px} {\left\langle 0 \right|} \mathcal{T} \lbrace \eta_{p} \! \left( x\right) \overline{\eta}_{p}\! \left( 0 \right) \rbrace  {\left| 0 \right\rangle}_F \mbox{ .}\end{aligned}$$ Here the subscript $_{F}$ indicates that the v.e.v. has to be evaluated in the background 