The Expectation Value of $\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\varphi}']$

Let us now calculate the expectation value


\begin{displaymath}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\varphi}']
\right\rangle_{0}.
\end{displaymath}

Since $S_{0}[\vec{\varphi}']$ is field-even, all expectation values of field-odd observables are zero when calculated in its measure. Therefore it is necessary that the observables be field-even if their expectation values are to be non-zero. Since in this case we have an explicit factor of $\varphi_{\mathfrak{N}}'(n_{\mu}')$, it follows that only the field-odd part of $S_{V}[\vec{\varphi}']$ will contribute to this expectation value,


\begin{displaymath}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\...
...'(n_{\mu}')S_{V,{\rm odd}}[\vec{\varphi}']
\right\rangle_{0}.
\end{displaymath}

If we write the expectation value out, using the form of the action $S_{V,{\rm odd}}[\vec{\varphi}']$ given in Equation (A.1), we get

\begin{eqnarray*}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\va...
...)\varphi_{\mathfrak{N}}'(n_{\mu}')
\right\rangle_{0}
\right\}.
\end{eqnarray*}


The expectation values in the first three terms turn out to be just the position-space propagator for the $\varphi_{\mathfrak{N}}'(n_{\mu})$ field component. From Appendix B, Equation (B.2), we get


\begin{displaymath}
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')
=
\frac{1}{N^{d}}
\su...
..._{\mu}-n_{\mu}')}}
{\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}},
\end{displaymath}

which is just the statement that $g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')$ is the inverse Fourier transform of the momentum-space propagator. We also have the corresponding result for the other field $\mathfrak{N}-1$ components, with $i\neq\mathfrak{N}$,

\begin{eqnarray*}
g_{0}(n_{\mu}-n_{\mu}')
& = &
\left\langle
\varphi_{i}'(n_...
...^{d}k_{\mu}(n_{\mu}-n_{\mu}')}}
{\rho^{2}(k_{\mu})+\alpha_{0}}.
\end{eqnarray*}


The second expectation value that we must calculate, with $i\neq\mathfrak{N}$, is simply

\begin{eqnarray*}
g_{0}(0)
& = &
\left\langle
\varphi_{i}'^{2}(n_{\mu})
\ri...
... \sum_{k_{\mu}}^{N^{d}}
\frac{1}{\rho^{2}(k_{\mu})+\alpha_{0}}.
\end{eqnarray*}


The third expectation value that we must calculate can be found in Appendix B, Equation (B.9), and can be shown to be given in terms of the first one by


\begin{displaymath}
\left\langle
\varphi_{\mathfrak{N}}'^{3}(n_{\mu})\varphi_{...
...\sigma_{\mathfrak{N}}^{2}\,g_{\mathfrak{N}}(n_{\mu}-n_{\mu}').
\end{displaymath}

We are thus left with a simpler form for the expectation value,

\begin{eqnarray*}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\va...
...m}{4ex}
-
j_{0}
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')
\right\}.
\end{eqnarray*}


In all terms the only quantity still depending on $n_{\mu}$ is $g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')$, so that we can write this equation as


\begin{displaymath}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\...
...}
\sum_{n_{\mu}}^{N^{d}}
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}').
\end{displaymath}

Using Equation (B.3), which gives this final sum, we may finally write


\begin{displaymath}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')S_{V}[\vec{\...
...thfrak{N}}^{2}
\right]
-
j_{0}
}
{\alpha_{\mathfrak{N}}}.
\end{displaymath} (A.4)