Does a Wigner function uniquely determine a quantum state?

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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?

taper

Posted 2018-08-23T22:53:44.517

Reputation: 183

Answers

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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.

user1271772

Posted 2018-08-23T22:53:44.517

Reputation: 8 162

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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)

– taper – 2018-08-24T13:55:38.467