in quantum electrodynamics the Fadeev-Popov ghost field does not couple to any other fields. Its only effect is to subtract the contribution of the unphysical longitudinal photon polarization to the free energy, and one has to keep it only if one is interested in the absolute value of the latter. Neglecting all of these terms, we thus consider the following action: S\[[|]{},;A\_\] &=& dx [|]{}\_(x) \_(x)\
&&-60pt + dxA\_(x)A\_(x)\
&&-60pt - ie dx (x)n(x) + dx [B]{}(x)\_(x)\
&&-60pt + dx\_[\_1]{}(x)\_[\_1 \_2]{}((x)) \_[\_2]{}(x).\
\[eq:3.1\] Here $x \equiv ({\bm x},\tau)$ comprises real space position ${\bm x}$ and imaginary time $\tau$, and $\int dx \equiv \int_V d{\bm x} \int_0^{1/T} d\tau$. $\bar\psi_{\sigma}$ and $\psi_{\sigma}$ are Grassmann-valued fields for electrons with spin projection $\sigma$, and the first term in Eq.(\[eq:3.1\]) describes free electrons with chemical potential $\mu$.[@Negele_Orland_1988] $A_{\mu} \equiv (\phi,-A_i)$ ($\mu = 0,1,2,3$; $i=1,2,3$) denotes the electromagnetic potential, with $\phi$ the scalar potential and ${\bm A} = (A_1,A_2,A_3)$ the vector potential, and the second term in in Eq. (\[eq:3.1\]) describes the free electromagnetic field in Feynman gauge (i.e., $\rho=1$ in Eq. (\[eq:A.4c\])).[@Kaputsa_1989] Note that both 4-vector potential fields in the second term carry covariant indices; that is, the $A$-action is euclidian.[@Euclidian_footnote] $n(x) =
{\bar\psi}_{\sigma}(x)\,\psi_{\sigma}(x)$ and 