The Schrödinger equation is generally formulated in position space
$$
i \hbar \frac{\partial}{\partial t}\psi(x,t) = \hat H_x \psi(x,t) = \left [ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\right ]\psi(x,t)
$$
or in momentum space
$$
i \hbar \frac{\partial}{\partial t}\psi(p,t) = \hat H_p \psi(p,t) = \left[ \frac{p^2}{2m} + V\left(\frac{\hbar}{i}\frac{\partial}{\partial p},t\right)\right ]\psi(p,t).
$$
But is it also possible to derive a Schrödinger equation for position $\times$ momentum $=$ phase space? I assume in that case it should be an equation acting on a wave function $\psi = \psi(x,p,t)$ . So I guess simply multiplying the two Schrödinger equations above is not enough. How could one then derive a Schrödinger equation for phase space, if it is possible at all?
Best Answer
There is no "wavefunction in phase space" because the wavefunctions $\psi(x)$ and $\psi(p)$ are obtained from the abstract state vector $\lvert \psi\rangle$ by $\langle x\vert \psi\rangle$ and $\langle p\vert \psi\rangle$, respectively. Since position and momentum don't commute, there are no $\lvert x,p\rangle$ to get a naive $\psi(x,p)$.
However, from any wavefunction we may obtain the Wigner quasiprobability distribution $W(x,p,t)$ on the classical phase space by the Wigner-Weyl transform. It obeys the equation $$ \partial_t W(x,p,t) = \{\{H(x,p,t),W(x,p,t)\}\}$$ with $H$ the classical Hamiltonian and the bracket on the r.h.s. as the Moyal bracket.