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

We now calculate the expectation value


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

Once again only the field-even part of the action will yield a non-zero result, so that we have


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

Using the form of $S_{V,{\rm even}}[\vec{\varphi}']$ shown in Equation (A.2), and if we already replace the expectation values of squared fields by $\sigma_{0}$ or $\sigma_{\mathfrak{N}}$ whenever possible, as well as replace $\left\langle\varphi_{\mathfrak{N}}'(n_{\mu}')\varphi_{\mathfrak{N}}'(n_{\mu}'')\right\rangle_{0}$ by $g_{\mathfrak{N}}(n_{\mu}'-n_{\mu}'')$, we get

\begin{eqnarray*}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')
\varphi_{\ma...
...\varphi_{\mathfrak{N}}'(n_{\mu}'')
\right\rangle_{0}
\right\}.
\end{eqnarray*}


We may now use the known value of the expectation value of the squared sum, found in Appendix B, Equation (B.15),


\begin{displaymath}
\left\langle
\left[
\sum_{i=1}^{\mathfrak{N}-1}
\varphi_...
...ngle_{0}
=
(\mathfrak{N}+1)(\mathfrak{N}-1)
\sigma_{0}^{4},
\end{displaymath}

as well as the fact that it can be shown that

\begin{eqnarray*}
\left\langle
\varphi_{\mathfrak{N}}'^{2}(n_{\mu})
\varphi_{...
...ak{N}}(n_{\mu}-n_{\mu}')\,
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}''),
\end{eqnarray*}


as one can also see in Appendix B, Equations (B.11) and (B.13), in order to write for our expectation value

\begin{eqnarray*}
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')
\varphi_{\ma...
...\mu}-n_{\mu}')\,
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}'')
\right\}.
\end{eqnarray*}


Next we group all terms containing $g_{\mathfrak{N}}(n_{\mu}'-n_{\mu}'')$ and simplify to get

\begin{eqnarray*}
\lefteqn
{
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')
...
...ak{N}}(n_{\mu}-n_{\mu}')\,
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}'').
\end{eqnarray*}


The sum over $n_{\mu}$ can now be done in all terms of the first group, yielding

\begin{eqnarray*}
\lefteqn
{
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')
...
...ak{N}}(n_{\mu}-n_{\mu}')\,
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}'').
\end{eqnarray*}


We must now perform the sum indicated. We get from Appendix B, Equation (B.5),


\begin{displaymath}
\sum_{n_{\mu}}^{N^{d}}
g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')\...
...}'')}
}
{
[\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}]^{2}
}.
\end{displaymath}

We have therefore the final result


$\displaystyle {
\left\langle
\varphi_{\mathfrak{N}}'(n_{\mu}')
\varphi_{\mathfrak{N}}'(n_{\mu}'')
S_{V}[\vec{\varphi}']
\right\rangle_{0}
}
$
  $\textstyle =$ $\displaystyle N^{d}
\left[
\frac{\alpha-\alpha_{0}+\lambda v_{0}^{2}}{2}\,
(\ma...
...pha_{\mathfrak{N}}+3\lambda v_{0}^{2}}{2}\,
\sigma_{\mathfrak{N}}^{2}
+
\right.$  
    $\displaystyle \hspace{2.2em}
\left.
+
\frac{\lambda}{4}\,
(\mathfrak{N}^{2}-1)
...
...da}{4}\,
\sigma_{\mathfrak{N}}^{4}
\right]
g_{\mathfrak{N}}(n_{\mu}'-n_{\mu}'')$ (A.7)
    $\displaystyle +
\left[
\alpha-\alpha_{\mathfrak{N}}
+
3\lambda
v_{0}^{2}
+
\lam...
...{\mu}(n_{\mu}'-n_{\mu}'')}
}
{
[\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}]^{2}
}.$