8
$\begingroup$

We know that the Wigner function of a Gaussian quantum state is (up to a constant) a Gaussian distribution. The first moment and the covariance of this distribution uniquely specify a quantum state. Therefore a Wigner function uniquely determines a Gaussian state.

Are there any similar statements applying to non-Gaussian states?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

For any quantum state, we have a unique density matrix $\rho$.
For any $\rho$, we can do the Wigner transformation to get a unique Wigner function $P(x,p)$.
For any Wigner function $P(x,p)$, we can do the Weyl transformation to get back the unique $\rho$.
If the construction of the Wigner function from $\rho$ was not unique, then it would not be possible to define an inverse transformation (but we do have an inverse transformation, namely the Weyl transformation, so the Wigner transformation does generate a unique characterization of a quantum state).

It has also been pointed out on the Physics Stack Exchange, that the Wigner function contains all information about a quantum state, just like the density matrix.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Thanks! I also would like to share here that for any observable $H(x,p)$ of the form $H(x,p)=T(p)+U(x)$, its expectational value could be calculated by $\int W(x,p) H(x,p) dx dp$. (c.f. this stack exchange post) $\endgroup$
    – taper
    Aug 24, 2018 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.