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The Wigner transform is the bridge between Hilbert space operators to phase-space quantities (c-numbers). Use for issues relating to the Weyl correspondence (the inverse of the Wigner transform), the Wigner function (the Wigner-transform of the density matrix) and, in general, Quantum Mechanics in phase space issues, such as the *-product, the Wigner transform of the operator multiplication operation. May also use for distributions such as the Husimi.

2 votes

What is the most general wave function of a minimum uncertainty (Gaussian) state in quantum ...

You don't get anything "new" by rotating in the $xp$ plane. Simply define \begin{align} X&=x\cos\theta -p \sin\theta \, ,\\ P&= x\sin\theta +x \sin\theta \, . \end{align} Since $[X,P]=[x,p]$, the ac …
ZeroTheHero's user avatar
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6 votes

What is wrong with Weyl-Wigner representation?

The $\star$-product is highly non-trivial except for HW. Even for $SU(2)$ this is quite complicated. And without a $\star$-product you are reduced to pictures or to the TWA, which only does so much. …
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2 votes

Negative probabilities with Wigner quasi-probability distributions

The negativity of the Wigner function is certainly one of the most interesting feature of this approach. One can easily show that, if two states are orthogonal, $\langle \phi\vert\psi\rangle=0$, then …
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2 votes

Can the Wigner function be described using coherent states?

Coherent states exist for a large number of algebraic systems so it's not hard to define a $Q$-function. The difficulty is that, for other values of $s$, the displacement operator takes on much more …
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1 vote

Is the Moyal-Liouville equation $\frac{\partial \rho}{\partial t}= \frac{1}{i\hbar} [H\stack...

In this approach the dynamics is expressed as a differential equation in the phase-space coordinates. The number of phase space coordinates is fixed and independent of the number of particles, wherea …
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2 votes
Accepted

How is the complex integration done for the Wigner function in coherent state representation?

Since $\lambda=x+ip$, $d\lambda^* \,d\lambda$ is basically $dx \, dp$ (as expected). Moreover, the factor $e^{-\lambda \lambda^*/2}$ is a Gaussian tail, which usually means integration by parts will c …
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5 votes
Accepted

How can the Wigner function of squeezed states be non-negative?

There is no contradiction because positivity of the Wigner is not enough to guarantee classicality: squeezed states are precisely examples of this. What is known for pure states (Hudson’s theorem) is …
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