Calculation of $J_4$

$$J_4 = \left<\prod_l \psi_l^{(0)\mu_l} \sum_j \left[\frac{\lambda_j^2}{2... ...\frac{\lambda_j \mu_j}{2A_j^2} \right] (a_j^{(1)}\bar a_j^{(0)})^2 \right>_\psi$$

$$= \langle \prod_l \psi_l^{\mu_l} \sum_{j,m,n, \kappa, \nu} \left[\f... ...j}{4A_j^4}\left(\frac{\mu_j}{2}-1\right)+\frac{\lambda_j \mu_j}{2A_j^2} \right]$$

$$\left(V_{mn}^j a_m a_n\Delta_{mn}^j\delta_{m+n}^j +2\bar V_{jn}^m... ...kappa \bar a_\nu \bar\Delta_{jn}^m\delta_{j+n}^m\right) \bar a_j^2\rangle_\psi.$$ (79)

Graphically, the 4 terms to be averaged in this expression are

$$\parbox{40mm} { \begin{fmffile}{n20} \begin{fmfgraph*}(80,40) \fm... ...mf{dashes_arrow}{v3,v1} \fmf{dashes_arrow}{v3,v2} \end{fmfgraph*}\end{fmffile}}$$

$$%YL + \quad \parbox{40mm} { \begin{fmffile}{n22} \begin{fmfgraph*... ...mf{dashes_arrow}{v3,v1} \fmf{dashes_arrow}{v3,v2} \end{fmfgraph*}\end{fmffile}}$$

Note that there is no dotted lines in these graphs because for each summation index there is a corresponding wave amplitude present. As a consequence, the rule for the number of surviving summations is somewhat different from what we had so far. Namely, the number of the summation indices after the phase averaging is one less than the number of the purely internal couplings. Easy to see that the phase averaging of the above terms always leads to an external coupling of the dashed lines $j$ which removes the $j$ summation. Moreover, no more than one purely internal coupling of the dashed lines is possible in any of these graphs. ⁶ Thus, $J_4$ contains no summation at all and is only a $O(1/N^2)$ correction to the main terms in $J_2$ and $J_3$.