Expectation values as a phase space average of Wigner functions

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Gabriel Maia
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Hi. I'm trying to prove that

[tex][\Omega] = \int dq \int dp \, \rho_{w}(q,p)\,\Omega_{w}(q,p)[/tex]


where

[tex]\rho_{w}(q,p) = \frac{1}{2\pi\hbar} \int dy \, \langle q-\frac{y}{2}|\rho|q+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})[/tex]
is the Wigner function, being \rho a density matrix. On the other hand
[tex]\Omega_{w}(q,p) = \frac{1}{2\pi\hbar} \int dy \, \langle q-\frac{y}{2}|\Omega|q+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})[/tex]
is the Wigner representation of the operator I'm interested in. The expectation value of an operator can be calculated from

[tex][\Omega] = Tr(\rho\Omega) = \int dp \, \langle p|\rho\Omega|p\rangle[/tex]
[tex]= \int dp^{\prime} \int dp \, \langle p|\rho|p^{\prime}\rangle\langle p^{\prime}|\Omega|p\rangle[/tex]


Now, the matrix elements [tex]\langle p | \rho | p^{\prime} \rangle \,\,\, \mathrm{and} \,\,\, \langle p | \Omega| p^{\prime} \rangle[/tex]

Should lead to the Wigner function and the Wigner representation of the operator but they only do so if
[tex]p=p^{\prime}[/tex]

Do you know how can I solve this?

Thank you.
 
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The solution to this problem is to use a Fourier transform of the matrix elements. That is, we can write\langle p | \rho | p^{\prime} \rangle = \frac{1}{2\pi\hbar}\int dy \, \langle p-\frac{y}{2}|\rho|p+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})and \langle p | \Omega | p^{\prime} \rangle = \frac{1}{2\pi\hbar}\int dy \, \langle p-\frac{y}{2}|\Omega|p+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar}).Substituting these expressions into the expectation value equation yields[\Omega] = \int dq \int dp \, \rho_{w}(q,p)\,\Omega_{w}(q,p),as desired.