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

We now calculate the expectation value

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

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

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

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),

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

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),

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

We have therefore the final result

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

$$= N^{d} \left[ \frac{\alpha-\alpha_{0}+\lambda v_{0}^{2}}{2}\, (\ma... ...pha_{\mathfrak{N}}+3\lambda v_{0}^{2}}{2}\, \sigma_{\mathfrak{N}}^{2} + \right.$$

$$\hspace{2.2em} \left. + \frac{\lambda}{4}\, (\mathfrak{N}^{2}-1) ... ...da}{4}\, \sigma_{\mathfrak{N}}^{4} \right] g_{\mathfrak{N}}(n_{\mu}'-n_{\mu}'')$$ (A.7)

$$+ \left[ \alpha-\alpha_{\mathfrak{N}} + 3\lambda v_{0}^{2} + \lam... ...{\mu}(n_{\mu}'-n_{\mu}'')} } { [\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}]^{2} }.$$