# Time Evolution of Wigner Function

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?

• I found one paper which starts with the von Neumann equation of evolution and then takes the Weyl Transform of the commutator. Could that be a start? Aug 20, 2020 at 9:03
• Do you understand Moyal brackets? Aug 20, 2020 at 10:32
• Yes there is actually an exercise on Moyal brackets previously to this one. I guess the way to solve it is by starting from the von Neumann equation and taking the Weyl Transform. The LHS then is just the time derivative of the wigner function because the transform and derivative commute. The RHS then I guess becomes the Moyal bracket? Aug 20, 2020 at 10:37
• Yes. See WP. Aug 20, 2020 at 10:41
• There are half a dozen textbooks on the subject, like this one by us. They all get you there... Aug 20, 2020 at 13:23

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:

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

• If I recall correctly, the series on the right will truncate for the h.o. Aug 21, 2020 at 12:55
• I thought about it again. Intuitively higher order terms I think must vanish, because the Weyl transformed Hamiltonian is quadratic in $x$ and $p$, however the next order involves derivatives of third order, so those should vanish I guess, but I'd have to do an exact calculation. Aug 22, 2020 at 12:06
• I think I got it now. I will update the solution tomorrow. Sep 29, 2020 at 17:13
• @MrProof Do you mean how the commutator gets mapped to the moyal bracket? Feb 2 at 8:39
• @MrProof I am not an expert in QM but pretty much any book on QM should be able to explain it. As far as I know the commutator is something like a Poisson bracket in classical mechanics. The commutator is also important for deriving the Heisenberg's uncertainty principle. Hope that helps. Feb 2 at 8:45