4th and 5th-order terms of the Hamiltonian

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4th and 5th-order terms of the Hamiltonian

First part of $\mbox{$\cal H$}_4$ is

$\displaystyle {\mbox{$\cal H$}_4}^{1st}$

$\textstyle =$

$\displaystyle -{1\over 2} \int 2\mbox{$\cal \Psi$}^{(3)} H{\mbox{$\cal \Psi$}^{(1)}}^\prime du =\cr$

Second part of $\mbox{$\cal H$}_4$ is

$\displaystyle {\mbox{$\cal H$}_4}^{2nd}$

$\textstyle =$

$\displaystyle -{1\over 2} \int \mbox{$\cal \Psi$}^{(2)} H{\mbox{$\cal \Psi$}^{(2)}}^\prime du =\cr$

In the

-space it acquires the form:

$\displaystyle {\mbox{$\cal H$}_4}^{2nd} =$

$\textstyle \frac{1}{4\pi}$

$\displaystyle \int \{(\vert k_1 k_2\vert-k_1 k_2) \vert k_2+k_4\vert + (\vert k_1\vert k_2 - k_1 \vert k_2\vert) (k_2+k_4)\}\times\cr$

(1.1)

First part of $\mbox{$\cal H$}_5$ is

$\displaystyle {\mbox{$\cal H$}_5}^{1st} = -$

$\textstyle \int$

$\displaystyle \mbox{$\cal \Psi$}^{(4)} H{\mbox{$\cal \Psi$}^{(1)}}^\prime du =\cr \frac{-1}{(2\pi)^{3\over 2}}$

Second part of $\mbox{$\cal H$}_5$ is

$\displaystyle {\mbox{$\cal H$}_5}^{2nd} = -$

$\textstyle \int$

$\displaystyle \mbox{$\cal \Psi$}^{(3)} H{\mbox{$\cal \Psi$}^{(2)}}^\prime du =\cr$

Hamiltonian can be written in symmetrical form:

$\displaystyle \mbox{$\cal H$}= \frac{1}{2}$

$\textstyle \int$

$\displaystyle (g \vert z_k\vert^2 + \vert k\vert\vert\mbox{$\cal P$}_k\vert^2) dk +\cr +\frac{1}{2\sqrt{2\pi}}$

where $S_{k_1 k_2 k_3}$ , $L_{k_1 k_2}$ and $F^{k_1}_{k_2 k_3}$ are defined in (3.2), and

$\displaystyle M^{k_1 k_2}_{k_3 k_4}$

$\textstyle =$

$\displaystyle (\vert k_3\vert+\vert k_4\vert)(k_1 k_2 +\vert k_1 k_2\vert) +\cr$

$\displaystyle N^{k_1 k_2 k_3}_{k_4 k_5}$

$\textstyle =$

$\displaystyle \cr$

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Dr Yuri V Lvov 2007-01-17