Example: Nonlinear Schrödinger Equation -- waves on condensate.

Bose-Einstein Condensate (BEC) is a state of matter that arises in dilute gases with large number of particles at very low temperatures [19,20,21,22,23,24]. BEC can be described by the Nonlinear Schrödinger equation (also known as Gross-Pitaevskii equation [25]). Here, we apply the theorem to this well studied model. The evolution of the state function $\psi$ is described by the following equation

$$i\frac{\partial \psi}{\partial t}+\triangle\psi-\vert\psi\vert^2\psi+\kappa(t)\psi=0,$$

with a corresponding Hamiltonian

$$H=\int(\vert\nabla\psi\vert^2+\frac{1}{2}\vert\psi\vert^4-\kappa(t)\vert\psi\vert^2)d\textbf{x}.$$

The term $\kappa(t)\psi$ is introduced for convenience as will become clear later. Following [7], let us consider the amplitude-phase representation of the order parameter $\psi$ :

$$\psi=Ae^{i\varphi}.$$

Now, we introduce Hamiltonian momentum

$$p=2A\varphi,$$ (77)

and rewrite Eq. (87) in terms of new canonical variables $A$ and $p$ as

$$A_t = \frac{\delta H}{\delta p},$$ (78)

$$p_t = -\frac{\delta H}{\delta A},$$

where

$$H=\int\left((\nabla A)^2+\frac{1}{2}A^4-\kappa(t)A^2+\frac{1}{4}\left(\nabla p-\frac{p\nabla A}{A}\right)^2\right)d\textbf{x}.$$ (79)

Let us consider weak perturbations on background of a strong condensate,

$$A=A^{(0)}+A^{(1)},\ \ p=p^{(0)}+p^{(1)},\ \ \vert A^{(1)}\vert\ll \vert A^{(0)}\vert.$$ (80)

We now choose $\kappa(t)=(A^{(0)})^2$ which gives us $p^{(0)}\sim\varepsilon$ . Substituting Eq. (90) into Eq. (89) we have

$$H=H_0+H_2+H_3,$$

where the subscripts denote the order of the term with respect to perturbation amplitudes. Since in this paper we study the linear dynamics, we only consider the quadratic part of the Hamiltonian

$$H_2=\int\left(\left(\nabla A^{(1)}\right)^2+(A^{(0)})^2(A^{(1)})^... ...(1)}+\frac{1}{4}p\left(\nabla\ln A^{(0)}\right)\cdot\nabla p\right)d\textbf{x}.$$

Here, we used the fact that the spatial derivative adds one order in $\varepsilon$ and we neglected the terms of the order two and higher.

In order to apply the theorem to $H_2$ we first transform to Fourier space and then switch to normal variables. Let us denote $R=\left(A^{(0)}\right)^2$ , $S=p^{(0)}/A^{(0)}$ and $T=\nabla_x\ln A^{(0)}$ . We have $R=O(1)$ and $S,T=O(\varepsilon)$ . Transforming $H_2$ into Fourier space we obtain

$$H_2=\int\left((\textbf{k}_1\cdot\textbf{k}_2\delta_2^1+R_{2-1})A_... ...2^1-T_{2-1}\textbf{k}_1\cdot(\textbf{k}_2-\textbf{k}_1)\Big)p_1p^*_2\right)d12,$$

where we used the following simplified notations: $1\equiv \textbf{k}_1$ , $2\equiv \textbf{k}_2$ and subscript $2-1\equiv \textbf{k}_2-\textbf{k}_1$ . Next, we switch to normal variables using the transformation

$$A_\textbf{k} = \frac{1}{\sqrt{2}}(a_\textbf{k}+a^*_{-\textbf{k}}),$$ (81)

$$p_\textbf{k} = -\frac{i}{\sqrt{2}}(a_\textbf{k}-a^*_{-\textbf{k}}).$$

In normal variables, $H_2$ reads

$$H_2 = \int\Bigg(\left(\frac{5}{4}\textbf{k}_1\cdot\textbf{k}_2\delta_2^1+R_{2-1}+\frac{1}{8}T_{2-1}(\textbf{k}_2-\textbf{k}_1)^2\right)a_1a^*_2+$$

$$+ \frac{1}{2}\left(\frac{3}{4}\textbf{k}_1\cdot\textbf{k}_2\delta_2... ...2-1}\textbf{k}_1\cdot(\textbf{k}_2-\textbf{k}_1)\right)a_1a_{-2}+c.c.\Bigg)d12.$$

The only part of the coefficient in the second parenthesis we are interested in is the one that satisfies Eq. (65):

$$\frac{3}{4}\textbf{k}_1\cdot\textbf{k}_2\delta_2^1+R_{2-1}-\frac{... ...}\textbf{k}_1\cdot\textbf{k}_2+\frac{1}{8}T_{2-1}(\textbf{k}_2-\textbf{k}_1)^2.$$

Since $A^{(0)}$ and $p^{(0)}$ are slowly varying functions of $\textbf{x}$ , so are $R(\textbf{x})$ , $S(\textbf{x})$ and $T(\textbf{x})$ . Therefore, their Fourier transforms are peaked around zero making the terms proportional to $T_{2-1}(\textbf{k}_2-\textbf{k}_1)^2$ of the second order in $\varepsilon$ , which can be neglected. Finally, we can write down the Hamiltonian in the form given in Eq. (4)

$$H_2=\int \left(A(\textbf{k}_1,\textbf{k}_2)a_1a^*_2+\frac{1}{2}(B(\textbf{k}_1,\textbf{k}_2)a_1a^*_2+c.c.)\right)d12,$$ (82)

where

$$A(\textbf{k}_1,\textbf{k}_2) = \frac{5}{4}\textbf{k}_1\cdot\textbf{k}_2\delta_2^1+R_{2-1}$$ (83)

$$B(\textbf{k}_1,\textbf{k}_2) = \frac{3}{4}\textbf{k}_1\cdot\textbf{k}_2\delta_2^1+R_{2-1}-\frac{i}{2}S_{2-1}\textbf{k}_1\cdot\textbf{k}_2$$

In terms of window transformations, which we denote here as $a$ the Hamiltonian reads

$$H_f=\int a (\mu-\textbf{x}\cdot\nabla\mu+i\{\mu,\cdot\})a^*d\text... ...xtbf{x}\cdot\nabla\lambda+i\{\lambda,\cdot\}]a_{-}d\textbf{k}d\textbf{x}+c.c.],$$

where

$$\mu = \int e^{i\textbf{m}\cdot\textbf{x}}A(\textbf{k}-{\textbf{m}}/2,\textbf{k}+{\textbf{m}}/2)d\textbf{m}=\frac{5}{4}\textbf{k}^2+R(\textbf{x}),$$ (84)

$$\lambda = \frac{1}{2}\int e^{i\textbf{m}\cdot\textbf{x}}B(\textbf{k}-{\text... ...{i}{2}\textbf{k}^2S(\textbf{x})+\frac{1}{2}\textbf{k}\cdot\nabla S(\textbf{x}).$$

Up to the first order in $\varepsilon$ , we have

$$\nu = \frac{3}{4}\textbf{k}^2+R(\textbf{x}),$$ (85)

$$\tilde{\nu} = -\frac{1}{2}\textbf{k}^2S(\textbf{x}).$$

Here, $\mu$ is an even function of $\textbf{k}$ which means that $\mu_{ev}=\mu$ and $\mu_{od}=0$ . Then, the position dependent frequency of the small perturbations in the presence of the condensate becomes

$$\omega=\sqrt{\mu^2-\nu^2}=\vert\textbf{k}\vert\sqrt{R(\textbf{x})+\textbf{k}^2}.$$

Bogolyubov's transformation, $a=ub+vb^*_-$ , is given by the following coefficients

$$u = \frac{\mu}{\sqrt{\mu^2-\nu^2}}=\frac{5\textbf{k}^2+4R(\textbf{x})}{4\vert\textbf{k}\vert\sqrt{\textbf{k}^2+R(\textbf{x})}},$$

$$v = -\frac{\nu}{\sqrt{\mu^2-\nu^2}}=-\frac{3\textbf{k}^2+4R(\textbf{x})}{4\vert\textbf{k}\vert\sqrt{\textbf{k}^2+R(\textbf{x})}}.$$

In terms of variables $b$ the Hamiltonian takes the following form

$$H_f = \int b(\omega-\textbf{x}\cdot\nabla_{\textbf{x}}\omega+i\{\omega,... ...ev}^2}{\nu}\left\{\varphi,b_-\right\}\right]d\textbf{k}d\textbf{x}+c.c.\right),$$ (86)

where

$$\sigma = \frac{\mu^2}{2\nu}\textbf{x}\cdot\nabla\sqrt{1-\frac{\nu^2}{\mu^2... ...{\vert\textbf{k}\vert\sqrt{\textbf{k}^2+R(\textbf{x})}}+ iS(\textbf{x})\right),$$

$$\varphi = \sqrt{1-\frac{\nu^2}{\mu^2}}=\frac{4\vert\textbf{k}\vert\sqrt{\textbf{k}^2+R(\textbf{x})}}{5\textbf{k}^2+4R(\textbf{x})}.$$

Finally, we perform the near-identity transformation $b=c+\alpha c^*_-+\beta\{\gamma,c^*_-\}$ where

$$\alpha = \frac{\textbf{x}\cdot\nabla R(\textbf{x})}{4(\textbf{k}^2+R(\text... ...\frac{i\vert\textbf{k}\vert S(\textbf{x})}{4\sqrt{\textbf{k}^2+R(\textbf{x})}},$$

$$\beta = \frac{i(5\textbf{k}^2+4R(\textbf{x}))^2}{8\vert\textbf{k}\vert\sqrt{\textbf{k}^2+R(\textbf{x})}(3\textbf{k}^2+4R(\textbf{x}))},$$

$$\gamma = \frac{4\vert\textbf{k}\vert\sqrt{\textbf{k}^2+R(\textbf{x})}}{5\textbf{k}^2+4R(\textbf{x})}.$$

The resulting Hamiltonian attains the canonical form (5).