Time Evolution of Wigner Function

Question

The Wigner Function is defined as: $$W(x,p,t)=\frac{1}{2\pi\hbar}\int dy \rho(x+y/2, x-y/2, t)e^{-ipy/\hbar}\tag{1}$$ Where $\rho(x, y, t)=\langle x|\hat{\rho}|y\rangle$. I am supposed to find the time evolution of the Wigner function for the Harmonic Oscillator starting from the von Neumann evolution equation given by: $$i\hbar\frac{\partial \rho}{\partial t}=\left[H,\rho\right].\tag{2}$$ I am not sure how to start, because the von Neumann evolution equation involves the commutator of the Hamiltonian and the operator of interest. However the Wigner function is a function, how can I evaluate the commutator?

Answer 1

Starting from the von Neumann equation: $$i\hbar\partial \hat{\rho} / \partial t=[\hat{H}, \hat{\rho}]$$ We now take the Weyl Transform on both sides and noting that the partial derivative commutes with the transform and the commutator gets mapped to the Moyal bracket: $$i\hbar\partial \tilde{\rho} / \partial t=-2i\tilde{H} sin(\hbar \Lambda/2) \tilde{\rho}$$ where the tilde implies they Weyl transform of the operator and $\Lambda = \frac{\partial}{\partial p}\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\frac{\partial}{\partial p}$ Where the first partial derivative acts to the left and the second to the right. Now the Weyl transform of the Hamiltonian of the harmonic oscillator can be shown to be just $\tilde{H}=p^2/2m+m\omega^2x^2$ Now expanding the sine function in a Taylor Series we get: $$i\hbar\partial \tilde{\rho}=-2i\left((p^2/2m + m\omega^2 x^2)\left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{\hbar}{2}\right)^{2n+1}\left(\frac{\partial}{\partial p}\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\frac{\partial}{\partial p}\right)^{2n+1}\right)\tilde{\rho}\right)$$ Now we express the first term of the sum seperately and we get: $$i\hbar\partial \tilde{\rho}=-2i\left((p^2/2m + m\omega^2 x^2)\left(\left(\frac{\hbar}{2}\right)\left(\frac{\partial}{\partial p}\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\frac{\partial}{\partial p}\right)+\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{\hbar}{2}\right)^{2n+1}\left(\frac{\partial}{\partial p}\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\frac{\partial}{\partial p}\right)^{2n+1}\right)\tilde{\rho}\right)$$

Now applying the first term of the sum we get: $$i\hbar\partial \tilde{\rho}=-i\hbar\left((p/m\frac{\partial}{\partial x} - 2 m\omega^2 x\frac{\partial}{\partial p})\tilde{\rho}+\tilde{H}\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{\hbar}{2}\right)^{2n+1}\left(\frac{\partial}{\partial p}\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\frac{\partial}{\partial p}\right)^{2n+1}\right)\tilde{\rho}\right)$$

The term on the left and the first two terms on the right outside the sum resemble precisely Lioville's equation. Since the harmonic oscillator Hamiltonian is quadratic in $x$ and $p$ and doesn't have any higher order terms the terms of higher order vanish, leaving us with:

$$\partial \tilde{\rho}+(p/m\frac{\partial}{\partial x} + 2 m\omega^2 x\frac{\partial}{\partial p})\tilde{\rho}=0$$