Example: an advection-type system

Let us consider an advection-type system. For simplicity of calculations let us restrict our attention to a one dimensional case, although a general dimensional case can also be considered. An advection-type system has a Hamiltonian of the form

$$H=i\int U(x)\big[\psi(x)\nabla_x\psi^*(x)-\psi^*(x)\nabla_x\psi(x)\big]dx.$$ (49)

with the corresponding equation of motion

$$i\dot{\psi}(x) = -i\nabla_x(U(x)\psi(x))-iU(x)\nabla_x{\psi(x)}$$

$$= -i(2U(x)\nabla_x\psi(x)+\nabla_x U(x)\psi(x)).$$ (50)

In the Fourier space, this system is described by the Hamiltonian

$$H=\int\Omega(k,k_1)\hat{\psi}_k\hat{\psi}_{k_1}^*dkdq_{k_1}.$$

with the kernel

$$\Omega(k,k_1)=(k+k_1)\hat{U}(k_1-k),$$ (51)

After applying Lemma to Hamiltonian (53), we obtain the following canonical form

$$H_f=\int \check{\psi}_{kx} [\omega_{kx}-x\nabla_x \omega_{kx}+i\{\omega_{kx},\cdot\}]\check{\psi}_{kx}^*dk dx,$$ (52)

where $\check{\psi}$ are the new variables.

$$\omega _{kx}=2kU(x).$$ (53)

is a position dependent frequency. Note that in this case we have $\nabla_k\nabla_x\omega _{kx}\neq 0$ and the near-canonical change of variables given by Eq. (41) had to be performed.

We can also obtain the same result by directly applying the Gabor transform to Eq. (54). Using the slow dependence of $U(x)$ on $x$ (disregarding the second derivative and higher), we obtain

$$\Gamma[U(x)\nabla_x\psi(x)]\approx$$

$$\big(U(x)-x\nabla_xU(x)+i\nabla_xU(x)\nabla_k\big)(\nabla_x+ik)\tilde{\psi}_{kx}.$$ (54)

Similarly, we have

$$\Gamma[\nabla_xU(x)\psi(x)]\approx\nabla_xU(x)\tilde{\psi}_{kx}.$$ (55)

Substituting Eqs. (58) and (59) into Eq. (54), we obtain

$$i\frac{\partial}{\partial t}{\tilde{\psi}}_{kx}=\big(-i2U(x)\nabla_x+2kU(x)+2ix\nabla_x U(x)\nabla_x-2xk\nabla_xU(x)+2\nabla_xU(x)\nabla_{kx}+$$

$$2i\nabla_xU(x)+2i\nabla_xU(x)k\nabla_k -i\nabla_x U(x)\big)\tilde{\psi}_{kx}.$$

Using Eq. (57), we rewrite Eq. (60) as

$$i\frac{\partial}{\partial t}{\tilde{\psi}}_{kx}=\left(\omega _{kx... ...nabla_x}+i(\nabla_x\omega _{kx}\nabla_k- \nabla_k\omega _{kx}\nabla_x)+ \right.$$

$$\left. \underline{ix\nabla_k\nabla_x\omega _{kx}\nabla_x}+ \frac{1}{2}i\nabla_k\nabla_x\omega \right)\tilde{\psi}_{kx}$$ (56)

As in the Lemma, we neglect the higher order terms with the two derivatives over $x$ (underlined in Eq. (61)). In order to obtain the canonical form of the equation of motion, we need to make a near-canonical transformation

$$\hat{\psi}_{kx}=f \check{\psi}_{kx},$$

where $f$ satisfies

$$\nabla_k\omega _{kx}\nabla_xf-\nabla_x\omega _{kx}\nabla_kf= \frac{1}{2}f\nabla_k\nabla_x\omega _{kx}.$$ (57)

For this special case, we obtain

$$U(x)\nabla_xf=\nabla_xU(x)\left(\frac{1}{2}f+k\nabla_kf\right)$$ (58)

We have to find the solution for Eq. (63) such that $f\rightarrow 1$ when $\nabla_xU(x)\rightarrow 0$ . Therefore, we need to find a solution in the form $f=1+g$ , where $g$ satisfies $\vert g(k,x)\vert\ll 1$ . Let us try to find a solution in the form $f=f(x)$ , i.e., independent of $k$ . Then we have

$$f(x)=C\sqrt{U(x)},$$

where $C$ is an arbitrary constant. We expand $U(x)$ around some point of reference $x_0$ as

$$U(x)\approx U(x_0)+(x-x_0)\nabla_xU(x_0).$$

Let us choose the constant to be $C=1/\sqrt{U(x_0)}$ then we obtain

$$f\approx 1+\frac{1}{2}(x-x_0)\frac{\nabla_xU(x_0)}{U(x_0)}$$

If $\nabla_xU(x_0)\sim\varepsilon$ and $\vert x-x_0\vert\ll 1/\varepsilon$ then $f\approx 1$ and the transformation is near-canonical.