Appendix B: Hamiltonian formalism for spatially inhomogeneous weak turbulence.

Let us start with the GP equation written in the Hamiltonian form:

$$i\frac{\partial}{\partial t} \Psi_{\bf x} = \frac{\delta {\cal H}}{\delta \Psi_{\bf x}^*}.$$ (63)

The Hamiltonian for the GP equation () coincides with the total energy of the system:

$${\cal H} = \int d{\bf r}\left( \vert\nabla \Psi_{\bf x}\vert^2 +... ...{1}{2}\vert\Psi_{\bf x}\vert^4 + U({\bf x}) \vert\Psi_{\bf x}\vert^2 \right).$$ (64)

Let us first consider the case without a condensate. Applying the Gabor transformation to () we get

$$i\frac{\partial}{\partial t} \hat\Psi_{\bf x} = \widehat{\frac{\delta {\cal H}}{\delta \Psi_{\bf x}^*}}.$$ (65)

But if we notice that

$$\frac{\delta {\cal H}(\Psi) }{\delta \Psi_{\bf x}} = \frac{\delta {\cal H}(\hat\Psi) }{\delta \widehat{\Psi_{{\bf x},{\bf k}}} },$$

we obtain

$$i\frac{\partial}{\partial t} \hat\Psi_{\bf x} = \widehat{\frac{\delta {\cal H}}{\delta\widehat{\Psi_{{\bf x}, {\bf k}}^*}}}.$$ (66)

Thus, the time evolution of the Gabor transformed quantity is governed by the Gabor transformed Hamiltonian equation. However, we would like to obtain the equation of motion in Hamiltonian form without the Gabor transformation. Let us re-write () in terms of the slow amplitudes $a$ defined in ()

$$i \frac{\partial}{\partial t} a_{{\bf k},{\bf x}} = \int f({\bf ... ...\bf x'}) \frac{\partial {\cal H}}{\partial a_{ {\bf x'},{\bf k}}^*}d {\bf x'}.$$ (67)

Now, let us express the Hamiltonian () in terms of the slow variables $a$,

$${\cal H} = \int e^{i(k_1 - k_2)x}\, \left( -a_{\bf k_1, x}(-k_2^2 + 2 i k_2 ... ...f x}) a_{\bf k_1, x} a_{\bf k_2, x}^* \right)\, d{\bf r} d{\bf k_1} d {\bf k_2}$$

$$+ \int e^{i(k_1 + k_2-k_3 - k_4)x}\, a_{\bf k_1, x} a_{\bf k_2, x... ...^* a_{\bf k_4, x}^* \, d{\bf r} d{\bf k_1} d {\bf k_2} d {\bf k_3} d {\bf k_4}.$$

Here we have integrated by parts $\vert\nabla \Psi\vert^2$ and, while calculating the Laplacian of $\Psi$ in terms of slow variables, have kept only the first order gradients in $a_{k,x}$. Substituting () into () allows us to re-write this equation as

$$i \frac{\partial}{\partial t} a_{{\bf k},{\bf x}}= \frac{\delta H} {\delta a_{{\bf x},{\bf k}}^*},$$ (68)

where the filtered Hamiltonian $H$ can be represented as

$$H = \int f( x-x')\,e^{i(k_1 - k_2)x'}$$

$$\times\left( -a_{\bf k_1, x'} (k_2^2+ 2 i k_2 \nabla)a_{\bf k_2, ... ... k_1, x'} a_{\bf k_2, x}^* \right) \,d{\bf r} d{\bf r'} d{\bf k_1} d {\bf k_2}$$

$$+ \int F( {\bf k_1}+{\bf k_2 }-{\bf k_3 }-{\bf k_4 })\, a_{\bf k_1, x} a_{... ...^* a_{\bf k_4, x}^* \,d{\bf x} d{\bf k_1} d {\bf k_2} d {\bf k_3} d {\bf k_4},$$ (69)

and $F({\bf k})$ is the Fourier transform of the $f(\vec x)$.

Expanding $U(x')$ as $U(x)+(x-x')\nabla U(x)$ and taking into account that $(x-x')$ can be interpreted as $-i\partial_{k_2} e^{i k_2 (x-x')}$, we have

$$H = \int\Big( ( k^2 + U(x) ) \vert a_{k, x}\vert^2 - \frac{i}{2} (\nabla U(x))\hat\Psi_{k, x}\partial_{k}\hat\Psi_{k, x}^* + i k a_{k, x}\nabla a_{k, x}^*$$

$$+ \frac{i}{2} (\nabla U(x))\hat\Psi_{k, x}^*\partial_{k}\hat\Psi_{k, x} - i k a_{k, x}^*\nabla a_{k, x}\Big)\, d {\bf k} d {\bf x}$$

$$+ \int F( {\bf k_1}+{\bf k_2 }-{\bf k_3 }-{\bf k_4 })\, a_{\bf k_1, x} a_{... ...^* a_{\bf k_4, x}^* \,d{\bf x} d{\bf k_1} d {\bf k_2} d {\bf k_3} d {\bf k_4},$$

Since $\omega_{k,x}= k^2 + U(x)$ we can represent the above formula as

$$H = \int \Big( (\omega_{k, x} - x \nabla \omega_{k, x}) \vert a_{k, x}\vert^2 + \frac{i}{2}(\n... ... \left( a_{k, x}^*\nabla_{k}a_{k, x} - a_{k, x}\nabla_{k}a^*_{k,x} \right)$$

$$+ \frac{i}{2}(\nabla_k \omega_{k, x}) \left( a_{k, x}\nabla_{x}a_{k, x}^* - a_{k, x}\nabla_{k}a^*_{k,x} \right)\Big) \,d {\bf k} d {\bf x}$$

$$+ \int F({\bf k_1} +{\bf k_2 }-{\bf k_3 }-{\bf k_4 })\, a_{\bf k_1 x} a_{\bf k_2 x}... ...}^* a_{\bf k_4 x}^*\, d{\bf x} d{\bf k_1} d {\bf k_2} d {\bf k_3} d {\bf k_4},$$ (70)

Now, we will show that if a condensate is present then the quadratic part of the Hamiltonian can also be written in the same canonical form as in (). Let us start from the equation () for $\Lambda$

$$\partial_{t}\Lambda = \partial_{k}\omega\cdot\nabla\Lambda + \fr... ...k^{2}}\vec{k}\cdot\nabla\varrho - \nabla\omega\cdot\partial_{k}\Lambda + i J ,$$ (71)

with

$$J = \frac{t\Lambda\omega}{k^{2}}\vec{k}\cdot\nabla\omega + \frac... ...bla\varrho - \frac{tk^{2}\Lambda}{\omega}\nabla\varrho\cdot\partial_{k}\omega,$$

Expression () for the waveaction in this case allows us to guess the form of the normal variable,

$$a_{\vec k,\vec x} = \frac{\sqrt{\varrho\omega_{\vec k,\vec x}}}{k} \Lambda_{\vec k,\vec x}\,e^{i\omega_{\vec k,\vec x}}.$$

Note that this expression is consistent with the waveaction considered above for the case with no condensate. This can be checked by taking the limit $\rho\to 0$. In terms of normal variable $a_{\vec k,\vec x}$ equations () and () acquire the following form:

$$\dot a_{\vec k,\vec x} = i \omega_{\vec k,\vec x} a_{\vec k,\vec x} + \partial_{\vec k}\omega_{\vec k,\vec x}\nabla a_{\vec k,\vec x} + \nabla \omega_{\vec k,\vec x} \partial_{\vec k} a_{\vec k,\vec x}$$

$$- 2 i a_{\vec k,\vec x}\vec k\cdot{\vec v} - i a_{\vec k,\vec x}\vec x\cdot\nabla\omega_{\vec k,\vec x}.$$ (72)

This equation can be represented in the form of a Hamiltonian equation of motion with a quadratic Hamiltonian as in () when the frequency is replaced by its Doppler shifted value,

$$\omega \to \omega + 2 \vec k\cdot{\vec v}.$$

Note that the Doppler shift does not enter into the equation for the waveaction because it leads to terms that are of second order in $\epsilon$ and therefore should be neglected.