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Wigner function as average value of parity

  1. Jan 12, 2016 #1

    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
     
  2. jcsd
  3. Jan 12, 2016 #2

    naima

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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.
     
  4. Jan 20, 2016 #3

    naima

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    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> =
    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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