Wigner quasiprobability distribution

Remarks

$W_\rho$ is real-valued, and $\abs{W_\psi(x, p)} \le (\pi\hbar)^{-n}$ by the Cauchy–Schwarz inequality. For a mixed state $\rho = \sum_k \lambda_k \psi_k \ip{\psi_k}{\cdot}$ with orthonormal $\psi_k$, $\lambda_k \ge 0$ and $\sum_k \lambda_k = 1$, one has $W_\rho = \sum_k \lambda_k W_{\psi_k}$. For nice $\psi$ the marginals are $\int W_\psi(x, p)\, dp = \abs{\psi(x)}^2$ and $\int W_\psi(x, p)\, dx = \abs{\hat\psi(p)}^2$, where $\hat\psi$ is the unitary Fourier transform with the factor $\hbar$. For a symbol $a$ with Weyl quantization $A = \mathrm{Op}^w(a)$, under suitable integrability conditions, $\operatorname{tr}(\rho A) = \int_{\R^{2n}} a(x, p)\, W_\rho(x, p)\, dx\, dp$. Up to the factor $(2\pi\hbar)^{-n}$, $W_\rho$ is the Weyl symbol of $\rho$. Without the physical constants, $W_\psi$ is the Wigner distribution function of time–frequency analysis.

Example: for $n = 1$ and the oscillator ground state $\psi(x) = (m\omega/\pi\hbar)^{1/4} e^{-m\omega x^2/2\hbar}$, one has $\psi(x + y/2)\psi(x - y/2) = (m\omega/\pi\hbar)^{1/2} e^{-m\omega x^2/\hbar} e^{-m\omega y^2/4\hbar}$. The Gaussian integral in $y$ then gives $$W_\psi(x, p) = \frac{1}{\pi\hbar} \exp\!\left(-\frac{m\omega x^2}{\hbar} - \frac{p^2}{m\omega\hbar}\right),$$ a positive Gaussian of total integral $1$.

Non-example of positivity: for the first excited state of the same oscillator, $W(0, 0) = -1/(\pi\hbar) < 0$, so a Wigner function need not be a probability density.