Wigner function as average value of parity

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SUMMARY

The discussion centers on the equivalence of two definitions of the Wigner function in quantum mechanics. The first definition utilizes a Fourier transform of the density operator ##\rho (q+ y/2,q-y/2)##, while the second employs the Weyl transformation and the parity operator ##exp(i \pi \theta N)##, where N is the occupation number operator. The proof of their equivalence requires the Campbell identity, demonstrating that both definitions yield the same result through specific transformations and trace calculations.

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  • Familiarity with Fourier transforms in quantum contexts
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  • Proficiency in using the Campbell identity in operator algebra
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naima
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I found two definitions of wigner function on space time. the first uses a Fourier transform of ##\rho (q+ y/2,q-y/2)##
the second uses the Weyl transformation and parity operator ## exp (i \pi \theta N)##
where N is the occupation number operator.
Could you give me a link which shows the equivalence of the definition?
thanks
 
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the first definition is
##w_\rho (q, p) = (2π\hbar)^{−1} \int <q − y/2|ρ|q + y/2> e^ {i \pi y/\hbar} dy## (read Ballentine)
the other is
##w_\rho (q+i p) = 2 Tr(\rho D(p+iq) e^{i \pi a^\dagger a} D(-q -ip)##
(read Man'ko page 8)
I cannot prove that it is the same thing.
 
I found the answer in this paper.

The proof needs the Campbell identity where the commutatot = constant.
[tex]Exp( i \frac{p_0 \hat {x}}{ \hbar}) Exp(i \frac{-x_0 \hat {p}}{ \hbar}) = Exp (i \frac{p_0 \hat x -x_0 \hat p + x_0 p_0/2 }{ \hbar} )[/tex]
So
[tex]Exp( i \frac{p_0 \hat {x} - x_0 p_0/2}{ \hbar}) Exp(i \frac{-x_0 \hat {p}}{ \hbar}) = Exp (i \frac{p_0 \hat x -x_0 \hat p }{ \hbar} ) = D(\alpha][/tex]

[tex]D(\alpha) = Exp( i \frac{p_0 \hat {x} - x_0 p_0/2}{ \hbar}) Exp( x_0 \partial_x)[/tex]
with
[tex]<x|D(\alpha)|\Psi> =<br /> Exp( i \frac{p_0 x - x_0 p_0/2}{ \hbar}) \Psi(x+x_0)[/tex]

We have to compute
[tex]2 Tr(\rho D(x+ip) e^{i \pi a^\dagger a} D(-x -ip)[/tex]
[tex]= 2 <\Psi|D(\alpha) e^{i \pi a^\dagger a} D^\dagger(\alpha)|\Psi>[/tex]
It may be seen as a double sum of
[tex]= 2 <\Psi|D(\alpha)|y><y| e^{i \pi a^\dagger a}|x><x| D^\dagger(\alpha)|\Psi>[/tex]
[tex]\int \int Exp(- i \frac{p_0 y - x_0 p_0/2}{ \hbar}) \Psi^*(y+x_0)<y| e^{i \pi a^\dagger a}|x> Exp( i \frac{p_0 x - x_0 p_0/2}{ \hbar}) \Psi(x+x_0)[/tex]
It can be shown that ##<y| e^{i \pi a^\dagger a}|x> = \delta(y+x)## so after one integration we get (up to a normalizing constant) the formula of the first definition.
 

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