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Calculating the Wigner transform of operators

Physics Asked by Wagner Coelho on June 9, 2021

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}{2pihslash}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”ranglelangle q”|p”rangle langle p”|F|p’ranglelangle 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.

One Answer

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, $$ h(p,q) = frac{1}{2m}int_{-infty}^{+infty}e^{frac{i}{hslash}qu}langle p+frac{u}{2}|hat p^2|p-frac{u}{2}rangle du frac{1}{2m}int_{-infty}^{+infty}e^{frac{i}{hslash}qu}(p-u/2)^2langle p+frac{u}{2}| p-frac{u}{2}rangle du = frac{1}{2m}int_{-infty}^{+infty}e^{frac{i}{hslash}qu}(p-u/2)^2 ~delta(u) ~ du =frac{p^2}{2m}~~ . $$

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

Correct answer by Cosmas Zachos on June 9, 2021

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