From this question I understood that the uncertainty principle is causing a problem because two points $x,p$ and $x',p'$ in phase space can be confused. Why exactly is this a problem? I don't grasp the connection to Kolmogorov's third axiom yet.


1 Answer 1


I'm not sure "problem" is the right word. Feature? Quasiprobability distributions simply deal with points in phase space which are not mutually exclusive, so you cannot add their probabilities as you would via Kolmogorov's 3rd, namely σ-additivity. Big words indicating you don't just add the probability of being at (x,p), to that of being at (x',p'), to that of being at (x'',p''), etc, because you are either here or there or further down, to get arbitrary phase-space location probabilities. Typically, a simple real probability distribution could be two narrow spikes near each other, the sum of the probabilities of two disjoint alternatives.

Not so for quasiprobability distributions, however. They can never be the above nearby spikes! If you are $\hbar$-close to another point, you don't really know you are at that point or where you thought you were. That is, your "location" distribution is constrained by the uncertainty principle.

For the Wigner distribution, $W(x,p)$ (take it centered at the origin so $\langle \mathfrak{x}\rangle=\langle \mathfrak{p}\rangle=0 $), this means your location probability at (x,p) also knows about that at (x',p') and influences (constrains) its value, since the uncertainty principle dictates $$ \langle \mathfrak{x}^2\rangle \langle \mathfrak{p}^2\rangle= \left(\int dx dp~ x^2 W\right )\left(\int dx'dp' ~ p'^2 W\right) \geq \hbar^2/4~, $$ no matter what. For the direct proof of the HUP in this peculiar setting, see, e.g. this paper.

Physically, this is just fine, a feature!, since you may not measure your location with more than $\hbar$ accuracy, and so, of course, W "knows" how to constrain its collective values at the $\hbar$-microscale. (You can thus see how the above two spikes, e.g., on the p-axis, cannot be too close in x, since the small p spread constrains the x spread from below.)

In practice, W may go negative at such small scales, all the while ensuring by such constraints that the negative regions are smaller than this ambit and never yield negative or even too small values for $\langle \mathfrak{x}^2\rangle \langle \mathfrak{p}^2\rangle$.

Having said all that, at the end of the day you do with W what you do with real probability distributions, namely you take expectation values integrating the Wigner transform of your operator in phase space with W as a measure, and they are provably correct! $$\langle { {\mathfrak G}} \rangle =\int\! dx dp~ W~ g(x,p). $$

The Husimi distribution Q, $$ Q(x,p)=\frac{1}{\pi\hbar}\int dx' dp' \exp \left( -{(x'-x)^2+(p'-p)^2 \over \hbar} \right) W(x',p') $$ is the invertible linear Weierstrass transform of the W. It can be show to never go negative, which looks significantl, but is largely irrelevant. (You may convince yourself that, a Gaussian $W=\exp\bigl(-(ax^2 +bp^2)/\hbar\bigr)$ Weirstrass-transforms to a broader one, $Q=\exp\bigl(-(ax^2/(1+a) +bp^2/(1+b))/\hbar\bigr) /\sqrt{(1+a)(1+b)} $ .)

It yields the same expectation values as above; but you must fold in the Weirstrass transform on the Wigner transform of the operator, to turn it into an operator Husimi transform, $g_H$, $$ \langle {{\mathfrak G}}\rangle = \int\! dx dp~ g(x,p)~ \exp\!\left(-{\hbar\over 4}(\partial_x^2+\partial_p^2 ) \right) ~Q = \int\! dx dp~ g_{_H} ~ e^{\hbar (\overleftarrow {\partial }_x \overrightarrow{\partial }_x + \overleftarrow {\partial }_p \overrightarrow{\partial }_p )/2 } ~Q. $$

That is, as emphasized in the penultimate paragraph of your linked question, there is an extra operator intercalated in the integral, preventing naive probability interpretation of the positive-semidefinite Q.

But, beyond that, Q too is highly constrained by the uncertainty principle. Using the above, find $$ \langle \mathfrak{x}^2\rangle \langle \mathfrak{p}^2\rangle \\ = \left(\int dx dp~ x^2\exp\!\left(-{\hbar\over 4}(\partial_x^2+\partial_p^2 ) \right) Q\right )\left(\int dx'dp' ~ p'^2\exp\!\left(-{\hbar\over 4}(\partial_x'^2+\partial_p'^2 ) \right) Q\right) \\ \geq \hbar^2/4~. $$ Admittedly, this is far less transparent than the above W constraint case, but it is an equally powerful a-priori constraint on the form of Q as well, mooting any possible interpretation as a bona-fide additive probability distribution.

Positive semidefiniteness is the least of the issues of quasiprobability distributions.


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.