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I know that I can calculate the probability distributions of $x$ and $p$ from the Wigner quasiprobability distribution, and I can calculate the probability distributions of other operators by calculating their eigenstates, Wigner transforming them, and projecting them onto the Wigner distribution of interest.

Is there some way to use the Wigner distribution to calculate probability distributions of operators without finding the eigenstates in the operator representation of QM? For example, how would I go about finding the probability that the energy is within a certain range?

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If I understood you correctly, this is a very basic application of the Wigner function, see, for example, eq.11 of This summary by W. Case –  bechira May 15 '14 at 14:41
@bechira: That summary doesn't include getting the full probability distributions for the Weyl transform of an operator, just expectation values. –  Dan May 21 '14 at 2:34
gotcha, good question! –  bechira May 21 '14 at 2:57

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The question might well be too broad. One may, of course, reconfigure the variables x and p into different ones, and integrate w.r.t. the "irrelevant" one , e.g. the angular variable on phase space, to produce a marginal quasi-probability distribution in the other, e.g. the angular variable squared, $(x^2+p^2)/2$ which happens to be the (rescaled variables') Weyl symbol of the energy operator for the oscillator, cf. Ref. 1. where a multiple of this variable is termed z, and behold the relevant figures.

But recall that a WF for each pure state will have a unique *-genvalue E (identical to the eigenvalue of the Weyl-ordered correspondent operator on that state) so a δ-spike in your notional distribution w.r.t. energy: a sharp spectral line, not a probability distribution. The quasi-distribution in z that has resulted is not a probability distribution (Laguere polynomials for the oscillator take negative values!) and does not represent the distribution you are envisioning--- even though quantum tomography applications of WFs utilize such constructs.

If you used it to find the expectation of z you'd recover the above E, of course: one value, not a distribution.

A different WF with a different characteristic E would replicate the above with a spike centered on a different value. Suppose you have a superposition of 2 orthogonal energy states, a and b. The corresponding pure state WF will, of course, yield a unique linear combination of the two eigen-energies in its expectation value of the energy observable, just as in Hilbert space.

Suppose, now, you have a mixed state of these 2 states, corresponding to a diagonal 2x2 density matrix for them. This is a bona-fide distribution of the 2 states, with the positive diagonal coefficients A and B representing the respective probabilities of the two states with the two energies. I understand you are plotting A and B on your notional y-axis and $E_a$, $E_b$ on your x-axis. It is trivial to transcribe this to phase space quantization, through a WF of the form $Af_{aa}+Bf_{bb}$. You might now generalize to an arbitrary number of states, including a continuum of them for a continuous spectrum. But the mixed state WF will never give you something more than its corresponding Weyl image density matrix. By itself, the phase-space formulation will not solve the spectral problem of the operator!

Can you specify over what states you are ranging to produce the probability distribution sought? I fear the phase space variations defocus the problem rather than sharpen it. The simplest way to reformulate your question might be to reformulate it in Hilbert space with density matrices ρ and operators, and then it would be child's play to recast it in phase-space language.


  1. Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014. The PDF file is available here.
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I was trying to derive an "ionized population" operator. That is, I was trying to see if I could derive an operator that would give the fraction of the wavefunction in energy eigenstates with eigenvalues greater than 0 without calculating all of the eigenstates of the hamiltonian. Failing that, a way to calculate this value from the Wigner distribution itself would also be helpful. –  Dan Sep 12 at 20:26
Yes, knowledge of the star-genstates is necessary to resolve the components of a mixture. The overlap integral of the mixture WF with the star-gen-WFs, mutually orthogonal, will produce the component coefficients, A, B, etc.... –  Cosmas Zachos Sep 12 at 20:34
So, then, to make it more explicit, for $f=Af_{aa} +Bf_{bb}$, star-orthogonality retrieves $A=h\int dx dp f f_{aa}$, etc... –  Cosmas Zachos Sep 12 at 21:32

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