Calculating the Wigner transform of operators

Question

Recently I started to study the formulation of quantum mechanics in the phase space. So I was introduced to the concept of Wigner function and Weyl transform. I learned that if F is an operator, then I can represent it by an integral as follows: \begin{equation} F = \int_{-\infty}^{+\infty}\frac{dpdq}{2\pi\hslash}f(p,q)\Delta(p,q) \end{equation} Where $f(p,q)$ is the Wigner transform given by: \begin{equation} f(p,q) = \int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}\langle p+\frac{u}{2}|F|p-\frac{u}{2}\rangle du \end{equation} and $\Delta(p,q)$: \begin{equation} \Delta(p,q) = \int_{-\infty}^{-\infty} e^{\frac{i}{\hslash}pv}|q+\frac{v}{2}\rangle \langle q-\frac{v}{2}|dv \end{equation} all of the above expressions were derived using the completeness relations as follows: \begin{equation} F = \int_{-\infty}^{+\infty}dp'dp''dq'dq''|q''\rangle\langle q''|p''\rangle \langle p''|F|p'\rangle\langle p'|q'\rangle \langle q'|\end{equation} and the following variable change was also taken \begin{equation} 2p =p'+p'', 2q = q'+q'', u = p''-p', v = q''-q' \end{equation}

Could someone show me what these calculations would look like for a numerical example of some observable operator. I tried to calculate for one of the pauli matrices, but I was stuck in the middle of the calculations. My learning becomes more consistent when I see practical examples, if anyone can help me with this problem, I will be very grateful.

Answer 1

Pauli matrices are mere constant matrices acting on 2d spinors, not functions of x or p, so you may be barking up the wrong tree.

I assume you or your text have evaluated the free particle hamiltonian, $$ h(p,q) = \frac{1}{2m}\int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}\langle p+\frac{u}{2}|\hat p^2|p-\frac{u}{2}\rangle du\\ \frac{1}{2m}\int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}(p-u/2)^2\langle p+\frac{u}{2}| p-\frac{u}{2}\rangle du\\ = \frac{1}{2m}\int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}(p-u/2)^2 ~\delta(u) ~ du =\frac{p^2}{2m}~~ . $$

The operators need not be observables. Try the hermitean parity operator, $$ P=\int \!\! dp ~~|-p\rangle\langle p| ~~\leadsto $$ $$ \Pi (q,p)= \int\!\!du dp' e^{iqu/\hbar} ~~\langle p+u/2 | -p'\rangle \langle p'|p-u/2\rangle \\ =\int\!\!du dp' ~ e^{iqu/\hbar} ~~ \delta( p+u/2 +p') \delta( p-u/2 -p') \\ =\int\!\!du ~ e^{iqu/\hbar} ~~ \delta(2 p ) =\frac{h}{2}\delta(q) \delta(p)~. $$