the sum of all perturbative terms that will contribute the electromagnetic current () we will denote by $V$:

$$V \equiv W_1 + W_2 + W_3 \ . \eqno(\VDEF)$$

In addition, the part of $V$ containing (linear) $\omega$-terms will be called $\omega \, V_1 $ and the $\omega$-independent part will be denoted by $V_0$. Hence, the full perturbative term reads:

$$V = V_0 \ \omega^0 + V_1 \ \omega^1  \ .
\eqno(\VDEFa)$$

To obtain () we have computed the trace over the Euclidean time in the $\omega$-space, the Fourier conjugate to $x_4$. To this end we have substituted in ():

$$\int dx_4 \ f(\partial_{x_4}) \rightarrow
\int {d\omega\over 2\pi} \ f(+i\omega)
\ . \eqno(\TIMESP)$$

Also, the formulae (–) from the Appendix have been used.

Applying () and () we should have in mind that there are, in general, two types of terms in (): linear and bilinear in ${\cal B}$. Hence, the products of the terms (-) can contribute as well. Also, those products should be linear in $\Omega$ and $A_{\mu}$. This implies that the only $W$-term contributing to the second order FSD expansion is of the form: $W_1 \times W_3$ — the first one linear in the potential $A_{\mu}$ and the 