are only known to circa $50\%$ accuracy so that the overall error is larger. Worse agreement for $\mu_N$ compared to $\mu_P$ is therefore no surprise.

With the standard choice $\varphi(u) \approx \varphi^{asy}(u)$ and the latest value $\chi(\mu=1\;\!{\rm{GeV}}^{2})=3.15\pm 0.3 \;\!{\rm{GeV}}^{-2}$ [@Ball:2002ps] [^3] we obtain the following results for the magnetic moments $$\begin{aligned}
\mu_P=& \phantom{(}  1+ (0.96 \pm 0.1)_{tw-2} + (0.72 \pm 0.18)_{tw-4} {\nonumber}\\
=&\phantom{(}2.68 \pm 0.28\\
\mu_N=&          (-1.93 \pm 0.2)_{tw-2} + (0.88 \pm 0.27)_{tw-4} {\nonumber}\\
=&\phantom{(}1.05 \pm 0.47 \mbox{.}\end{aligned}$$ Here we also included the different contributions of twist 2 and twist 4 to the magnetic moments and how the overall error is distributed.

In the following we will stick to the above standard choice for $\varphi(u)$.

The Nucleon-Delta-transition
============================

We will now expand the technique presented in the previous Section to the case of the nucleon-$\Delta$-transition. We will not go into the details of the calculation, except for the construction of a suitable Lorentz basis. This enables us to remove unwanted contributions of transitions including final states with isospin $3/2$ and spin $1/2$.

It is not necessary to treat the the $n\gamma \to \Delta^{0}$ transition separately from the $p\gamma \to \Delta^{+}$ transition, as the final formulae will only differ by the 