u^a(x) \mathcal{C} \gamma^{\lambda} u^{b}(x) \right)  \gamma_5 \gamma_{\lambda} d^{c}(x) \cdot \right.
\\ 
\left.\cdot\; {\overline{d}}^{c'}\left( 0\right)\gamma^{\lambda'} \gamma_5 \left({\overline{u}}^{a'}\left( 0 \right) \gamma_{\lambda'} \mathcal{C}{\overline{u}}^{b'}\left( 0 \right) \right)   \right \rbrace {\left| 0 \right\rangle}_F \varepsilon^{abc} \varepsilon^{a'b'c'}\end{gathered}$$ Using Wick’s theorem the calculation is straightforward. One has, however, to pay attention to the fact that we are working with massless quarks in a simultaneous electromagnetic and gluonic background field. The electromagnetic field $F_{\mu \nu}$ is just a plane wave, whereas the gluonic field $G_{\mu \nu}=G_{\mu \nu}^A t^A$ due to the presence of gluons in the hadron is unknown. The quark propagator then adopts the following form[^2] [@Balitsky:1987bk]: $$\begin{aligned}
\label{propagator}
\wick{1}{<1{q}(x) >1{\overline{q} }}(0)=&\frac{i\slashed{x}}{2 \pi^2 x^4}\left[x,0 \right]-\frac{ig}{16 \pi^2 x^2}\int_0^1 \! du \; \left[x,ux \right]  \left \lbrace {\overline{u}}\slashed{x} \sigma_{\alpha \beta} + u  \sigma_{\alpha \beta} \slashed{x} \right \rbrace G^{\alpha \beta}(ux) \left[ux,0 \right]  {\nonumber}\\
&-\frac{ie_q }{16 \pi^2 x^2} \int_0^1 \!du\; \left[x,ux \right] \left \lbrace {\overline{u}}\slashed{x} \sigma_{\alpha \beta}+u  \sigma_{\alpha \beta}\slashed{x} \right \rbrace F^{\alpha \beta}(ux)\left[ux,0 \right]+\ldots \mbox{  ,}\end{aligned}$$ here we used the abbreviation $$\label{pexp}
\left[ x,y \right]={\rm Pexp}\left \lbrace i \int_0^1 \! dt \;(x-y)_{\mu} \left[ e_q A^{\mu}(tx-\overline{t}y) + g B^{\mu}(tx-\overline{t}y) \right] \right \rbrace$$ for the path-ordered exponent ($A_{\mu}$ is the electromagnetic and $B_{\mu}$ the gluon field) and $\overline{a}=1-a$, $\forall a \in [0,1]$. The dots represent 