Calculation of $J_3$

$$J_3 = \left<\prod_l \psi_l^{(0)\mu_l} \sum_j (\lambda_j + \frac{\mu_j}{2A_j^2})a_j^{(2)}\bar a_j^{(0)} \right>_\psi$$

$$= \langle \prod_l \psi_l^{(0)\mu_l} \sum_j (\lambda_j + \frac{\mu_j}{2A_j^2})$$

$$\sum_{j,m,n, \kappa, \nu} \left[ 2 V^j_{mn} \left( -V^m_{\kappa \... ..._{n \kappa},\omega^j_{mn}]\delta^\kappa_{m + \nu}\right) \delta^j_{m+n} \right.$$

$$\left. + 2 \bar V^m_{jn} \left(-V^m_{\kappa \nu}\bar a_n a_\kappa... ...\nu j},-\omega^m_{j n}] \delta^\kappa_{m + \nu} \right) \delta^m_{j+ n} \right.$$

$$\left. + 2 \bar V^m_{jn} \left( \bar V^n_{\kappa \nu}a_m \bar a_\... ...jn}]\delta^\kappa_{n + \nu}\right)\delta^m_{j+n} \right] \bar a_j \rangle_\psi.$$ (76)

The terms to be averaged here can be drawn as

$$\parbox{35mm} { \begin{fmffile}{n30} \begin{fmfgraph*}(60,45) \fm... ..._arrow}{v2,o2} \fmf{dots_arrow, label=$m$}{v1,v2} \end{fmfgraph*}\end{fmffile}}$$

$$+ \quad \parbox{35mm} { \begin{fmffile}{n33} \begin{fmfgraph*}(60... ..._arrow}{v2,o2} \fmf{dots_arrow, label=$n$}{v2,v1} \end{fmfgraph*}\end{fmffile}}$$ (77)

Let us now average over the random phases. Again, the leading order terms will be given by the diagrams with the largest number of internal couplings. They will arise from the $V\bar V$ terms (the 2nd, 3rd and the 6th graphs) because they allow 2 internal couplings in each of them. There are also possibilities to have one internal and two external couplings of the dashed lines, - such terms will give a $O(1/N)$ correction to the leading order. Therefore,

$$J_3 = \left( \parbox{35mm} { \begin{fmffile}{n28} \begin{fmfgraph*}(70,... ...w, right=.7, label= $j$}{v2,v1} \end{fmfgraph*}\end{fmffile}} \right)[1+O(1/N)]$$

$$= 4 \prod_l\delta(\mu_l) \sum_{j,m,n} \lambda_j \left[ - \vert V_{m... ... V_{jn}^m\vert^2 E(0,-\omega_{jn}^m)\delta_{j+n}^m (A_m^2 -A_n^2) \right] A_j^2$$

$$\hspace{7cm} \times [1+O(1/N)]$$ (78)