we obtain, for the interaction part of the action, $$S_I = -\frac{ie}{2}\ \int d^4q\, \{[A_\mu(q),\varphi(q)],\partial^\mu\varphi(q)\} = -ie  \int d^4q\, A_\mu(q)[\varphi(q),\partial^\mu\varphi(q)],$$ where the cyclicity of $\int d^4q$ was used. We obtain therefore the interaction of $A_\mu$ with a current $j^\mu(q) = -ie[\varphi(q),\partial^\mu\varphi(q)]$, and the classical Euler-Lagrange equations for $A_\mu$ take the form $$\partial_\mu F^{\mu \nu} -ie[A_\mu, F^{\mu \nu}] = - j^\nu,$$ with the field strength defined by $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu -ie[A_\mu,A_\nu]$ and transforming as $F_{\mu\nu} \to UF_{\mu\nu}U^*$.

In order to understand the physical meaning of this interaction, we now assume $A_0(q) = 0$, $A_h(q) = \frac12\ep_{hjk} B^jq^k$, the potential corresponding to an external constant magnetic field $\boldsymbol{B}$ in the classical spacetime limit ($\lambda_P \to 0$) and we again neglect in  the quadratic term in $A_\mu$, thus obtaining $$\label{eq:Binteraction}\begin{split}
\cL_I &= -\frac{ie}{4}\ep_{hjk}B^j\big\{\big[q^k,\ph(q)\big],\partial^h\ph(q)\big\} =\frac{e}{2}\ep_{jkh}B^jQ^{k\mu}\{\partial^\mu \ph(q), \partial^h  \ph(q)\},
\end{split}$$ where in the second equation we used the identity $[q^\nu, f(q)] = iQ^{\nu\mu} \partial_\mu f(q)$.

The above term corresponds then to the energy of a total magnetic moment $\boldsymbol{M}$ with components, in generic units, $$\label{eq:magnetic}
M_j =  (e/2) \lambda _P ^2 \int_{q^0=t}d^3q\,\left[\frac12(\{\partial_l\ph,\partial^l\ph\}\de_{jk}-\{\partial_j\ph,\partial^k\ph\})m_k-\ep_{jkh}\{\partial_0\ph,\partial^h\ph\}e_k\right], \qquad j=1,2,3,$$ where $e_k := Q^{0k}$ and $m_k := \frac 1 2 \ep_{khl} Q^{hl}$ are respectively the electric 