In this paper, the following is stated:

It is well known that the uncertainty principle makes the concept of phase space in quantum mechanics problematic. Because a particle cannot simultaneously have a well defined position and momentum, one cannot define a probability that a particle has a position $q$ and momentum $p$, i.e. one cannot define a true phase space probability distribution for a quantum mechanical particle.

I'm confused by this. Aren't the square modulus of the momentum and position representations of a particle's state precisely probability distributions in $p$ and $q$ respectively? (i.e. wouldn't the function $\psi^*(q)\phi^*(p)\psi (q) \phi (p)$ meet the criteria?) Of course such a definition won't be so nice for expectation value calculations (not like, for example, the Wigner distribution function), but still - I can't see how this function wouldn't contradict the statement I cited.

Of course, I'm fairly certain I'm under a misapprehension. I'd just like to know where and how this misapprehension lies.

  • $\begingroup$ Have you actually tried to see whether your $\psi^\ast \phi^\ast \psi \phi$ gives the correct expectation values? That's the first thing you should check when claiming it somehow encodes the information about the quantum state: That it actually reproduces the predictions of standard QM. $\endgroup$
    – ACuriousMind
    Mar 24, 2016 at 21:48

1 Answer 1


The classic review you quote is absolutely right. It means position and momentum at the same time! Well, if you write down something that replicates QM, it must somehow comport with the uncertainty principle.

The expression (bilinear in wave functions, not the present squared version!) you wrote down is actually quite close to the standard ordering prescription quasi probability distribution, Exercise 0.19 in this book, and, in fact, can be transformed to the Wigner function--it is a nonlocal change of variables of it; but it has extreme problems. (These can be addressed systematically, but by the time they have, the victory is pyrrhic... you have to stick in some unwieldy star product in that prescription to take expectation values of composite operators of x and p multiplying each other, such as angular momenta.)

If you do not square the wavefunctions, they are not positive semidefinite, and might even integrate to zero! A spike at a given x and p might have a negative expectation value--negative probability for being at that point in phase space.

If your cc-square your expression (as per your edit, now) you do get something positive semidefinite, but you lose non-commutativity: try to get the expectation value of $[ \hat{x}, \hat{p} ]= i\hbar$. One presumes you'll use commutative variables x and p, not differential operators, as one does when one works either in coordinate, or momentum space. Of course, the uncertainty principle dictates that ψ(x) and φ(p), being the Fourier transform of each other, cannot be both peaked arbitrarily on a single point in phase space, as a probability in classical mechanics could.

A bona-fide probability density must be positive semi-definite and have each point of its independent variables' domain represent distinct disjoint alternatives, which QM cannot allow by the uncertainty principle. Wigner addressed all possibilities in the 30s and settled on his eponymous function, which bypasses, magically, many of these problems, and contains the uncertainty principle automatically, in a surprising way.

If you complete and massage your expression, then you have written an allowable quasi-probability distribution, but 9 times out of 10 you will have to go through major *-product stunts to get correct expectation values of operators that are not simple sums of x dependent and p dependent operators, but, instead, arbitrary products in fancy orderings.

  • $\begingroup$ Thank you very much! I corrected a few typos that this answer addressed (e.g. I meant $\psi^*(q)\phi^*(p)\psi(q)\phi(p)$, not $\psi(q)\phi(p)$). Of course, you addressed this also in the third paragraph (loss of non-commutativity). $\endgroup$ Mar 24, 2016 at 21:30
  • 1
    $\begingroup$ Oh, OK, in that case your expression is quartic in wave functions, and not related to the Wigner function. But then taking the expectation of the Heisenberg commutation relation spells inconsistency. $\endgroup$ Mar 24, 2016 at 21:34

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