Statistical averaging and graphs.

Let us consider the initial fields $a_k(0) = a^{(0)}_k$ are essentially RPA as defined above. We will perform averaging over the statistics of the initial fields in order to obtain an evolution equations, first for $Z$ and then for the multi-mode PDF. The ultimate goal of this exercise is to prove that the wavefield remains of the essentially RPA type over the nonlinear time.

Let us introduce a graphical classification of the above terms which will allow us to simplify the statistical averaging and to understand which terms are dominant. We will only consider here contributions from $J_1$ and $J_2$ which will allow us to understand the basic method. Calculation of the rest of the terms, $J_3$, $J_4$ and $J_5$, follows the same principles and can be found in Appendix 2. First, The linear in ${\epsilon}$ terms are represented by $J_1$ which, upon using (9), becomes

$$J_1 = \left<\prod_l \psi_l^{\mu_l} \sum_{j,m,n}(\lambda_j +\frac{\mu_j}... ..._{jn}^m a_m\bar a_n \bar\Delta_{jn}^m\delta_{j+n}^m\right)\bar a_j\right>_\psi.$$ (23)

Hereafter we omit, for brevity of notation, the super-script $\small (0)$ because no other super-scripts will appear from now on.

Let us introduce some graphical notations for a simple classification of different contributions to this and to other (more lengthy) formulae that will follow. Combination $V_{mn}^j \delta_{m+n}^j$ will be marked by a vertex joining three lines with in-coming $j$ and out-coming $m$ and $n$ directions. Complex conjugate $\bar V_{mn}^j \delta_{m+n}^j$ will be drawn by the same vertex but with the opposite in-coming and out-coming directions. Presence of $a_j$ and $\bar a_j$ will be indicated by dashed lines pointing away and toward the vertex respectively. ³ Thus, the two terms in formula (24) can be schematically represented as follows,

$$C_1= \parbox{40mm} { \begin{fmffile}{one} \begin{fmfgraph*}(110,6... ...mf{dashes_arrow}{v1,i2} \fmf{dashes_arrow}{o1,v1} \end{fmfgraph*}\end{fmffile}} \quad and \quad C_2= \parbox{40mm} { \begin{fmffile}{two} \begin{fmfgraph*}(110,6... ...mf{dashes_arrow}{i2,v1} \fmf{dashes_arrow}{o1,v1} \end{fmfgraph*}\end{fmffile}}$$

Let us average over all the independent phase factors in the set $\{\psi\}$. Such averaging takes into account the statistical independence and uniform distribution of variables $\psi$. In particular, $\langle\psi\rangle=0$, $\langle\psi_l \psi_m\rangle=0$ and $\langle\psi_l \bar \psi_m\rangle=\delta^m_l$. Further, the products that involve odd number of $\psi$'s are always zero, and among the even products only those can survive that have equal numbers of $\psi$'s and $\bar \psi$'s. These $\psi$'s and $\bar \psi$'s must cancel each other which is possible if their indices are matched in a pairwise way similarly to the Wick's theorem. The difference with the standard Wick, however, is that there exists possibility of not only internal (with respect to the sum) matchings but also external ones with $\psi$'s in the pre-factor $\Pi \psi_l^{\mu_l}$.

Obviously, non-zero contributions can only arise for terms in which all $\psi$'s cancel out either via internal mutual couplings within the sum or via their external couplings to the $\psi$'s in the $l$-product. The internal couplings will indicate by joining the dashed lines into loops whereas the external matching will be shown as a dashed line pinned by a blob at the end. The number of blobs in a particular graph will be called the valence of this graph.

Note that there will be no contribution from the internal couplings between the incoming and the out-coming lines of the same vertex because, due to the $\delta$-symbol, one of the wavenumbers is 0 in this case, which means ⁴that $V=0$. For $J_1$ we have

$$J_1 = \langle C_1 \rangle_\psi + \langle C_2 \rangle_\psi,$$

with

$$\langle C_1 \rangle_\psi = \parbox{40mm} { \begin{fmffile}{three}... ...arrow}{o1,v1} \fmfdot{i1} \fmfdot{i2} \fmfdot{o1} \end{fmfgraph*}\end{fmffile}} \quad + \quad \quad \parbox{40mm} { \begin{fmffile}{four} \begin{fmfgraph*}(110,62) \... ...\fmf{dashes_arrow}{o1,v1} \fmfdot{i1} \fmfdot{o1} \end{fmfgraph*}\end{fmffile}}$$

and

$$\langle C_2 \rangle_\psi = \parbox{40mm} { \begin{fmffile}{five} ... ...arrow}{o1,v1} \fmfdot{i1} \fmfdot{i2} \fmfdot{o1} \end{fmfgraph*}\end{fmffile}} \quad + \quad \quad \parbox{40mm} { \begin{fmffile}{six} \begin{fmfgraph*}(110,62) \f... ...\fmf{dashes_arrow}{v1,o1} \fmfdot{o1} \fmfdot{i1} \end{fmfgraph*}\end{fmffile}}$$

which correspond to the following expressions,

$$\langle C_1 \rangle_\psi = \sum_{j \ne m \ne n}(\lambda_j +\frac{\mu_j}{2A_j^2}) V_{mn}^j A_... ...a(\mu_m +1) \delta(\mu_n +1) \delta(\mu_j-1) \prod_{l \ne j,m,n} \delta (\mu_l)$$

$$+ \sum_{m}(\lambda_{2m} +\frac{\mu_{2m}}{2A_{2m}^2}) V_{mm}^{2m} A_... ...{mm}^{2m} \delta(\mu_m +2) \delta(\mu_{2m}-1) \prod_{l \ne m,2m} \delta (\mu_l)$$ (24)

and

$$\langle C_2 \rangle_\psi = = 2\sum_{j \ne m \ne n}(\lambda_j +\frac{\mu_j}{2A_j^2}) \bar V_{jn... ...a(\mu_m +1) \delta(\mu_n -1) \delta(\mu_j-1) \prod_{l \ne j,m,n} \delta (\mu_l)$$

$$+ 2\sum_{n}(\lambda_n +\frac{\mu_n}{2A_n^2}) \bar V_{nn}^{2n} A_{2n... ...n}^{2n} \delta(\mu_{2n} +1) \delta(\mu_n -2) \prod_{l \ne n,2n} \delta (\mu_l).$$ (25)

Because of the $\delta$-symbols involving $\mu$'s, it takes very special combinations of the arguments $\mu$ in $Z\{ \mu \}$ for the terms in the above expressions to be non-zero. For example, a particular term in the first sum of (25) may be non-zero if two $\mu$'s in the set $\{ \mu \}$ are equal to 1 whereas the rest of them are 0. But in this case there is only one other term in this sum (corresponding to the exchange of values of $n$ and $j$) that may be non-zero too. In fact, only utmost two terms in the both (25) and (26) can be non-zero simultaneously. In the other words, each external pinning of the dashed line removes summation in one index and, since all the indices are pinned in the above diagrams, we are left with no summation at all in $J_1$ i.e. the number of terms in $J_1$ is $O(1)$ with respect to large $N$. We will see later that the dominant contributions have $O(N^2)$ terms. Although these terms come in the ${\epsilon}^2$ order, they will be much greater that the ${\epsilon}^1$ terms because the limit $N \to \infty$ must always be taken before ${\epsilon}\to 0$.

Let us consider the first of the ${\epsilon}^2$-terms, $J_2$. Substituting (9) into (20), we have

$$J_2 = \frac{1}{2}\langle\prod_l\psi_l^{(0)\mu_l} \sum_{j,m,n,\kappa,\nu}(\lambda_j+\lambda_j^2A_j^2-\frac{\mu_j^2}{2A_j^2})$$

$$\hspace{1cm} \times (V_{mn}^ja_ma_n\Delta_{mn}^j\delta_{m+n}^j+2\... ...r a_{\kappa}a_{\nu}\Delta_{j\nu}^{\kappa}\delta_{j+\nu}^{\kappa})\rangle_{\psi}$$

$$= \langle B_1 + B_2 + \bar B_2 + B_3 \rangle_{\psi},$$ (26)

where

$$B_1 = \hspace{1cm} \parbox{35mm} { \begin{fmffile}{n7} \begin{fmf... ..._arrow}{o2,v2} \fmf{dots_arrow, label=$j$}{v2,v1} \end{fmfgraph*}\end{fmffile}} B_2 = \hspace{1cm} \parbox{35mm} { \begin{fmffile}{n8} \begin{fmfgra... ...ow}{v2,o2} \fmf{dots_arrow, label=$j$}{v2,v1} \end{fmfgraph*}\end{fmffile}} and B_3 = \hspace{1cm} \parbox{35mm} { \begin{fmffile}{n9} \begin{fmf... ..._arrow}{v2,o2} \fmf{dots_arrow, label=$j$}{v2,v1} \end{fmfgraph*}\end{fmffile}}$$ (27)

Here the graphical notation for the interaction coefficients $V$ and the amplitude $a$ is the same as introduced in the previous section and the dotted line with index $j$ indicates that there is a summation over $j$ but there is no amplitude $a_j$ in the corresponding expression.

Let us now perform the phase averaging which corresponds to the internal and external couplings of the dashed lines. For $\langle B_1\rangle_{\psi}$ we have

$$\langle B_1\rangle_{\psi} = \hspace{1cm} \parbox{35mm} { \begin{f... ...\fmfdot{i1} \fmfdot{i2} \fmfdot{o1} \fmfdot{o2} \end{fmfgraph*}\end{fmffile}} + \parbox{35mm} { \begin{fmffile}{n11} \begin{fmfgraph*}(70,50) \fmfke... ...nu$}{v2,v1} \fmfdot{i1} \fmfdot{i2} \fmfdot{o1} \end{fmfgraph*}\end{fmffile}} + \parbox{35mm} { \begin{fmffile}{n12} \begin{fmfgraph*}(70,50) \fm... ...$2 m$}{v1,v2} \fmfdot{i1} \fmfdot{i2} \fmfdot{o1} \end{fmfgraph*}\end{fmffile}}$$

$$+2 \hspace{.5cm} \parbox{35mm} { \begin{fmffile}{n13} \begin{fmfg... ...\fmf{dashes_arrow, right=.7, label= $n$}{v1,v2} \end{fmfgraph*}\end{fmffile}} ,$$ (28)

where

$$\parbox{25mm} {\fmfreuse{x}} = \frac{1}{2}\sum_{j\neq m\neq n\neq\kappa\neq\nu}(\lambda_j+\lambd... ...r\Delta_{\kappa\nu}^j\delta_{m+n}^j\delta_{\kappa+\nu}^jA_mA_nA_{\kappa}A_{\nu}$$

$$\hspace{2.2cm} \times \delta(\mu_m+1)\delta(\mu_n+1)\delta(\mu_{\kappa}-1)\delta(\mu_{\nu}-1) \prod_{l\neq m,n,\kappa,\nu}\delta(\mu_l)$$

$$\parbox{35mm} {\fmfreuse{fish}} = \frac{1}{2}\sum_{m\neq n\neq\nu}(\lambda_{2\nu}+\lambda_{2\nu}^2A... ...\nu}^2}{2A_{2\nu}^2}) V_{mn}^{2\nu}\delta_{m+n}^{2\nu}\bar V_{\kappa\nu}^{2\nu}$$

$$\times \Delta_{mn}^{2\nu}\bar\Delta_{\kappa\nu}^{2\nu}A_mA_nA_{\n... ...a(\mu_m+1)\delta(\mu_n+1)\delta(\mu_{\nu}-2) \prod_{l\neq m,n,\nu}\delta(\mu_l)$$

$$2 \hspace{.5cm} \parbox{35mm} {\fmfreuse{theta}} =\prod_{l}\delta... ... \vert V_{mn}^{j}\vert^2 \vert\Delta_{mn}^{j}\vert^2 \delta_{m+n}^{j}A_m^2A_n^2$$

We have not written out the third term in (29) because it is just a complex conjugate of the second one. Observe that all the diagrams in the first line of (29) are $O(1)$ with respect to large $N$ because all of the summations are lost due to the external couplings(compare with the previous section). On the other hand, the diagram in the second line contains two purely-internal couplings and is therefore $O(N^2)$. This is because the number of indices over which the summation survives is equal to the number of purely internal couplings. Thus, the zero-valent graphs are dominant and we can write

$$\langle B_1\rangle_\psi = \prod_l\delta(\mu_l)\sum_{j,m,n}(\lambd... ...}^{j}\vert^2 \vert\Delta_{mn}^{j}\vert^2 \delta_{m+n}^{j}A_m^2A_n^2[1+O(1/N^2)]$$ (29)

For $\langle B_2\rangle_\psi$ we have

$$\langle B_2\rangle_{\psi} = \hspace{1cm} \parbox{35mm} { \begin{f... ...arrow}{v2,v3} \fmfdot{i1} \fmfdot{o1} \fmfdot{v3} \end{fmfgraph*}\end{fmffile}}$$

$$+ \hspace{1cm} \parbox{35mm} { \begin{fmffile}{n17} \begin{fmfgra... ...ht=1., label= $m$}{o1,v1} \fmfdot{i1} \fmfdot{o1} \end{fmfgraph*}\end{fmffile}}$$ (30)

where

$$\parbox{25mm} {\fmfreuse{x1}} = \sum_{j\neq m\neq n\neq\kappa\neq\nu}(\lambda_j+\lambda_j^2A_j^2-... ...lta_{j\nu}^{\kappa}\delta_{m+n}^j\delta_{j+\nu}^{\kappa}A_mA_nA_{\kappa}A_{\nu}$$

$$\hspace{2.2cm} \times \delta(\mu_m+1)\delta(\mu_n+1)\delta(\mu_{\kappa}-1)\delta(\mu_{\nu}+1) \prod_{l\neq m,n,\kappa,\nu}\delta(\mu_l)$$

$$\parbox{35mm} {\fmfreuse{hammock}} = \sum_{j,m,n}(\lambda_j+\lambda_j^2A_j^2-\frac{\mu_j^2}{2A_j^2}) V_{mn}^j V_{j-m}^{n}\Delta_{mn}^j$$

$$\hskip 5cm \times %YL \Delta_{j-m}^{n}\delta_{m+n}^jA_n^2A_mA_{-m} \delta(\mu_m+1)\delta(\mu_{-m}+1) \prod_{l\neq m,-m}\delta(\mu_l)$$

$$\parbox{25mm} {\fmfreuse{A}} = \sum_{j\neq m\neq n\neq\kappa}(\lambda_j+\lambda_j^2A_j^2-\frac{\mu_j^2}{2A_j^2}) V_{mn}^j V_{jn}^{\kappa}$$

$$\hskip 5cm \times %YL \Delta_{mn}^j \Delta_{jn}^{\kappa}\delta_{m+n}^j\delt... ...m+1)\delta(\mu_n+2)\delta(\mu_{\kappa}-1) \prod_{l\neq m,n,\kappa}\delta(\mu_l)$$

$$\parbox{35mm} {\fmfreuse{fish1}} = \sum_{ m\neq \kappa\neq\nu}(\lambda_{2m}+\lambda_{2m}^2A_{2m}^2-\frac{\mu_{2m}^2}{2A_{2m}^2}) V_{mm}^{2m} V_{2m\nu}^{\kappa}$$

$$\hskip 5cm \times %YL \Delta_{mm}^{2m} \Delta_{2m\nu}^{\kappa}\delta_{2m+\nu}^{\kappa}A_m^2A_{\kappa}A_{\nu} \hspace{1.2cm}$$

$$\hspace{2.2cm} \times \delta(\mu_m+2)\delta(\mu_{\kappa}-1)\delta(\mu_{\nu}+1) \prod_{l\neq m,\kappa,\nu}\delta(\mu_l)$$

$$\parbox{35mm} {\fmfreuse{scorpion1}} = \sum_{ m}(\lambda_{2m}+\lambda_{2m}^2A_{2m}^2-\frac{\mu_{2m}^2}{2A_{2m}^2}) V_{mm}^{2m} V_{2mm}^{3m}$$

$$\hskip 5cm \times %YL \Delta_{mm}^{2m} \Delta_{2mm}^{3m} A_m^3 A_{3m} \delta(\mu_m+3)\delta(\mu_{3m}-1) \prod_{l\neq m, 3m}\delta(\mu_l)$$

$$\parbox{35mm} {\fmfreuse{scorpion2}} = \sum_{ m}(\lambda_{2m}+\lambda_{2m}^2A_{2m}^2-\frac{\mu_{2m}^2}{2A_{2m}^2}) V_{mm}^{2m} V_{2m -m}^{m}\Delta_{mm}^{2m}$$

$$\hskip 5cm \times %YL \Delta_{2m \, -m}^{m} A_m^3 A_{-m} \delta(\mu_m+1) \delta(\mu_{-m}+1) \prod_{l\neq m, -m}\delta(\mu_l)$$

The second term in (31) contains one summation because its graph has one purely internal coupling. This term is $N$ times smaller than the largest terms in $\langle B_1 \rangle_\psi$ (which have 2 surviving summation indices). All the other terms in (31) contain no summation at all because all their dashed lines are coupled externally.

Similarly, the leading contribution to $\langle B_3 \rangle_\psi$ will be given by the zero-valent graph with the maximum possible number of internal couplings (which is equal to 2 in this case). Because of the $\delta$'s, there are no graphs with just one internal coupling, but there are graphs with all the dashed lines coupled externally. Thus,

$$\langle B_3 \rangle_\psi = \parbox{25mm} { \begin{fmffile}{n19p} \begin{fmfgraph*}(70,50) \f... ...s_arrow, left=.7, label= $n$}{v2,v1} \end{fmfgraph*}\end{fmffile}} [1+O(1/N^2)]$$

$$%YL =2 \prod_{l}\delta(\mu_l)\sum_{j,m,n}(\lambda_{j}+\lambda_{j}... ...vert^2 \vert\Delta_{jn}^{m}\vert^2 \delta_{j+n}^{m}A_m^2A_n^2 \; [1+O(1/N^2)] ,$$ (31)

Summarising the results of this section we can write for $J_2$:

$$J_2 = \prod_{l}\delta(\mu_l)\sum_{j,m,n}(\lambda_{j}+\lambda_{j}^... ... \vert\Delta_{jn}^{m}\vert^2 \delta_{j+n}^{m} \right] A_m^2A_n^2 \; [1+O(1/N)].$$ (32)

Thus, we considered in detail the different terms involved in $J_2$ and we found that the dominant contributions come from the zero-valent graphs because the have more summation indices involved. This turns out to be the general rule that allows one to simplify calculation by discarding a significant number of graphs with non-zero valence. After this observation finding the rest of the terms, $J_3$ to $J_5$, becomes a routine task and we therefore move it to the Appendix 2.