Wigner function as average value of parity

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The discussion centers on two definitions of the Wigner function in spacetime, one derived from a Fourier transform and the other from the Weyl transformation involving the parity operator. The user seeks a link demonstrating the equivalence of these definitions, referencing specific equations and sources for each. A proof is mentioned that utilizes the Campbell identity, highlighting the relationship between the operators involved. The conversation includes detailed mathematical expressions and integrals to illustrate the connection between the two definitions. The thread emphasizes the complexity of proving the equivalence and the necessary mathematical tools required for such a demonstration.
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.
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} )
So
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]

D(\alpha) = Exp( i \frac{p_0 \hat {x} - x_0 p_0/2}{ \hbar}) Exp( x_0 \partial_x)
with
&lt;x|D(\alpha)|\Psi&gt; =<br /> Exp( i \frac{p_0 x - x_0 p_0/2}{ \hbar}) \Psi(x+x_0)

We have to compute
2 Tr(\rho D(x+ip) e^{i \pi a^\dagger a} D(-x -ip)
= 2 &lt;\Psi|D(\alpha) e^{i \pi a^\dagger a} D^\dagger(\alpha)|\Psi&gt;
It may be seen as a double sum of
= 2 &lt;\Psi|D(\alpha)|y&gt;&lt;y| e^{i \pi a^\dagger a}|x&gt;&lt;x| D^\dagger(\alpha)|\Psi&gt;
\int \int Exp(- i \frac{p_0 y - x_0 p_0/2}{ \hbar}) \Psi^*(y+x_0)&lt;y| e^{i \pi a^\dagger a}|x&gt; Exp( i \frac{p_0 x - x_0 p_0/2}{ \hbar}) \Psi(x+x_0)
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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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