Calculation of $J_5$

Expression for $J_5$ seemingly involves a great number terms. However, this number can be dramatically reduced by the following speculation. Just as in $J_4$ there is no dotted lines in the graphs involved in $J_5$ because for each summation index there is a corresponding wave amplitude present. Thus, we have the same rule for the number of summations surviving the phase averaging (i.e. one less than the number of internal couplings). In order to be of the same order as the leading terms in $J_2$ and $J_3$, we must have 3 purely internal couplings and, therefore, no external couplings. This is only possible when the number of dotted lines directed to the vertices is equal to the number of them pointing away which is true for the $V\bar V$ terms but not true for the $V V$ and $\bar V \bar V$ terms. Thus we will only consider the $V\bar V$ terms. Further, the fact that there is no external couplings means that such terms are only non-zero when all $\mu$'s are zero. Thus, there will be no contribution from the second part of $J_5$ which has a $\mu_k$ pre-factor.

$$J_5 = {1 \over 2} \left<\prod_l \psi_l^{(0)\mu_l} \sum_{j \ne k}\lambda... ...{(0)}+\bar a_j^{(1)} a_j^{(0)})a_k^{(1)}\bar a_k^{(0)} \right>_\psi [1+O(1/N)],$$ (80)

where the $V\bar V$ terms to be averaged here are

$$C_1 = \hspace{1cm} \parbox{50mm} { \begin{fmffile}{n24} \begin{fmfgraph... ...mf{dashes_arrow}{o2,v2} \fmf{dashes_arrow}{v4,v2} \end{fmfgraph*}\end{fmffile}}$$

$$\hspace{5cm} = - \prod_l\psi_l^{(0)\mu_l}\sum_{j\neq k,m,n,\mu,\n... ...u}^\kappa a_\kappa \bar a_\nu \bar a_k \Delta_{mn}^j \bar\Delta_{k\nu}^\kappa ,$$

$$C_2 = \hspace{1cm} \parbox{50mm} { \begin{fmffile}{n25} \begin{fmfgraph... ...mf{dashes_arrow}{v2,o2} \fmf{dashes_arrow}{v4,v2} \end{fmfgraph*}\end{fmffile}}$$

$$\hspace{5cm} = - \prod_l\psi_l^{(0)\mu_l}\sum_{j\neq k,m,n,\mu,\n... ...\kappa+\nu}^k a_\kappa a_\nu \bar a_k \bar \Delta_{jn}^m \Delta_{\kappa \nu}^k,$$

$$C_3 = \hspace{1cm} \parbox{50mm} { \begin{fmffile}{n26} \begin{fmfgraph... ...mf{dashes_arrow}{o2,v2} \fmf{dashes_arrow}{v4,v2} \end{fmfgraph*}\end{fmffile}}$$

$$\hspace{5cm} = {2}\prod_l\psi_l^{(0)\mu_l}\sum_{j\neq k,m,n,\mu,\... ...u}^\kappa a_\kappa \bar a_\nu \bar a_k \Delta_{jn}^m \bar \Delta_{k\nu}^\kappa,$$

$$C_4 = \hspace{1cm} \parbox{50mm} { \begin{fmffile}{n27} \begin{fmfgraph... ...mf{dashes_arrow}{v2,o2} \fmf{dashes_arrow}{v4,v2} \end{fmfgraph*}\end{fmffile}}$$

$$\hspace{5cm} = \frac{1}{2}\prod_l\psi_l^{(0)\mu_l}\sum_{j\neq k,m... ..._{\kappa+\nu}^k a_\kappa a_\nu\bar a_k\bar \Delta_{mn}^j \Delta_{\kappa \nu}^k.$$

By coupling the dashed lines we have in the leading order

$$\langle C_1 \rangle_\psi = 2 \hspace{.1cm} \parbox{35mm} { \begin{fmffile}{n40} \begin{fmfgraph... ...2} \fmf{dashes_arrow, left=.7, label= $k$}{v1,v2} \end{fmfgraph*}\end{fmffile}} = - 2\prod_l\delta(\mu_l) \sum_{j\neq k,n}\lambda_j\lambda_k \vert V_{kn}^j\vert^2 \delta_{k+n}^j A_j^2 A_n^2 A_k^2 \vert\Delta_{kn}^j\vert^2 ,$$

$$\langle C_2 \rangle_\psi = 2 \hspace{.1cm} \parbox{35mm} { \begin{fmffile}{n41} \begin{fmfgraph... ...1} \fmf{dashes_arrow, left=.7, label= $k$}{v1,v2} \end{fmfgraph*}\end{fmffile}} = - 2\prod_l\delta(\mu_l) \sum_{j\neq k,n}\lambda_j\lambda_k \ver... ...{j+n}^k A_j^2 A_n^2 A_k^2 \vert\Delta_{jn}^k\vert^2 = \langle C_1 \rangle_\psi,$$

$$\langle C_3 \rangle_\psi = \hspace{.1cm} \parbox{35mm} { \begin{fmffile}{n42} \begin{fmfgraph*}... ...} \fmf{dashes_arrow, right=.7, label= $m$}{v2,v1} \end{fmfgraph*}\end{fmffile}} = 2\prod_l\delta(\mu_l) \sum_{j\neq k,m}\lambda_j\lambda_k \vert V_{jk}^m\vert^2 \delta_{j+k}^m A_j^2 A_m^2 A_k^2 \vert\Delta_{jk}^m\vert^2 .$$

Term $C_4$ does not survive the averaging because $j \ne k$ and a triple internal coupling is not possible. Summarising,

$$J_5= 2 \prod_l\delta(\mu_l) \sum_{j\neq k,n}\lambda_j\lambda... ...elta_{jk}^n\vert^2 \right] A_j^2 A_n^2 A_k^2 \;\; [1+O(1/N)].$$ (81)