Deformation quantization procedure is a well-known way to quantize a classical phase space (at least formally for Poisson manifolds which is known as formal deformation quantization). Although it is a beautiful and an elegant subject, I can not understand uncertainty principle in this framework. The basic equation is Moyal equation which is as follows (suppose that the phase space is just $R^{2n}$)

$$ \frac{\partial f}{\partial t}=\{\{f,H\}\}, $$

where the right is the Moyal bracket. One can find Wigner function $f$ and compute expectation values of $x\star x=x^2, p\star p=p^2 $ and establish uncertainty principle in terms of $(\triangle x)^2=\langle (x-\langle x\rangle)^2\rangle$ and similarly for $p$ as follows

$$ (\triangle x)^2.(\triangle p)^2\geq\frac{\hbar^2}{4} $$

It might be a good argument for having uncertainty principle in deformation quantization, but consider a particle with oscillator Hamiltonian $H=\frac{1}{2}(p^2+x^2)$, it is clearly depends on $x$ and $p$ and in order to be good and a well defined function on our phase space, we should know $x$ and $p$ of the particle simultaneously which is not acceptable from uncertainty principle.

Does it violate uncertainty principle ontologically?

(I searched the links in site, and there was not a related topic).

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    $\begingroup$ " in order to be good and a well defined function on our phase space, we should know 𝑥 and 𝑝 of the particle simultaneously, which is not acceptable from the uncertainty principle". Where did this absurdity come from? The c-number function is well specified, and so is the Wigner distribution. It is only expectation values for "knowing" quantities that are restricted by the uncertainty principle in this formulation. In fact, If you read on, you'd see Wigner functions are very restricted, precisely so as to satisfy this uncertainty principle. Ontology? $\endgroup$ Apr 1, 2022 at 22:56
  • $\begingroup$ Can we know momentum and position simultaneously according to uncertainty principle? if not, how we talk about c-number functions on phase space? and why do we need to restrict ourselves to expectation values from theoretical point of view ? $\endgroup$
    – Arian
    Apr 2, 2022 at 6:12

1 Answer 1


Ontology is not yet a thing in this SE—perhaps the philosophy SE could be more forgiving to it. In any case, it is assumed that you are in complete and unequivocal peace with the uncertainty principle (UP), the way it has been settled in the last 95 years or so, and you merely wish to understand how it presents itself in deformation quantization, without expecting new handles or insights into it, somehow unavailable to the operators-in-Hilbert-space traditional formulation.

The knowables in QM are mathematical (=logical) facts, and the expectation values of measurements, and, significantly, nothing else. There is no answerable "why" there. QM is a gimmick producing reliable expectation values.

Can we know momentum and position simultaneously according to uncertainty principle? if not, how we talk about c-number functions on phase space? and why do we need to restrict ourselves to expectation values from theoretical point of view ?

In that sense, the answer to your first question here is: of course not, whenever "know" dares go beyond expectations of measurable quantities. The third was already summarily dispatched above: it is a cornerstone of the theory; those not at peace with it have been walking in hapless circles for a century, mumbling. So, I will try to explain the second one.

Deformation quantization works in phase space, where you may "think" about points (x,p) in it, without the naive, fallacious association with feasible measurements. One may think about them the way thinks about wavefunctions, that is formal ingredients leading to the answers.

The UP is baked into the theory from its foundations, perhaps counterintuitively, which lots of physicists (including Dirac) struggled to understand in the late forties, and Moyal triumphantly succeeded in. (His free online biography, by his wife details how). For starters,

  • The Wigner function f(x,p) characterizing the state cannot be a δ function, since $|f(x,p)|\leq 2/h$, a trivial consequence of its definition. In fact, since it is normalized to 1 in phase space, you may see that its boxy-skyscraper height L=1/A where A is its footprint area; so the particle is condemned to spread over a phase-space area larger than h/2. We can talk about arbitrary xs and ps, as long as we don't picture a particle as something localized in an area smaller than h/2. Moreover,

  • The Wigner function f is not an arbitrary function in phase space. For a pure state, it satisfies an almost "mystical" property (in the sense it used to mystify novices in the area). For an arbitrary function g, $$ \langle g^* \star g\rangle = \int\!\! dx dp ~ f (g^* \star g) \geq 0. $$ This is the intrinsic constraint serving to prove the UP in this formulation, tactically equivalent to the positivity of the norm in Hilbert space. Finally,

  • It is inappropriate to think of the Wigner function as a probability versus mutually exclusive contingencies. Two nearby points in phase space necessarily participate in describing the state (particle) and cannot represent independent locations. Dramatically, the Wigner function can go negative in such small areas, $O(h)$ (only). The uncertainty principle prevents you from visualizing a "particle galvanometer" in such small areas, to "locate" ("know the phase-space location") of a particle. The UP is the invisible cloak that prevents direct observation of negative f regions and localizing particles.

The upshot is that, contrary to the intuition you developed from classical statistical mechanics, the peculiar rules of the nonlocal star product endow phase-space variables, observables, and distributions with completely different properties than one's naive imagination would suggest.

You saw how different. Any phase-space function, such as your oscillator hamiltonian, is poised to be star-multiplied, so, then, perform the function of an operator. In that sense, it means something quite different than in classical mechanics. (Even though, even there, it is but an input to the Poisson-Bracket evolution operator...)

  • $\begingroup$ Thanks for your answer. I am familiar with the properties of Wigner function. ontologically? I mean the formulation allows us to talk about $x$ and $p$ simultaneously in c-number functions which are observables. You said it is just formal ingredients which leads to answers, I think the phase space is the same but the algebra of the functions over it are different in classical mechanic and phase space formulation of quantum mechanics. do you say that $H=p^2+x^2$ has a different meaning over $R^{2n}$ which is a valid phase space for classical mechanic and quantum mechanic? $\endgroup$
    – Arian
    Apr 2, 2022 at 17:40
  • $\begingroup$ Indeed, in general, functions such as H have a different meaning in deformation quantization. For instance, $H^2\neq H\star H$, the analog of the QM $\hat H ^2$. Thinking of Weyl symbols as anything different than "operators in a peculiar language", a prefix to a star product, is a recipe for (well-deserved) conceptual disaster... In the Hilbert space formulation, avoidance of association between operators and their classical limit is precisely so one won't get confused this way... $\endgroup$ Apr 2, 2022 at 17:53
  • $\begingroup$ Thank you so much dear Cosmos Zachos, I need to think more about it. A minor question, do you work on a version of classical mechanic with Moyal bracket? for example, the basic equations would be $\frac{\partial x}{\partial t}=\{\{x,H\}\}$ and $\frac{\partial p}{\partial t}=\{\{p,H\}\}$ as a generalization of Poisson bracket . A comment would be enough for me, if there are related ideas. $\endgroup$
    – Arian
    Apr 2, 2022 at 18:37
  • $\begingroup$ For classical mechanics, only PBs work, never MBs, since the latter include ℏ corrections. But there is a formulation of classical mechanics in Hilbert space, the Koopman-Von Neumann one... $\endgroup$ Apr 4, 2022 at 21:59
  • $\begingroup$ Thanks a lot dear Cosmos Zachos, I mean a classical mechanics which is defined by MBs, It is not a problem, if there was a new parameter $\hbar$ and possible corrections. The equations are deterministic but with a new additional parameter. However, if we consider $x$ and $p$, the MB and PB is the same {{x,H}}={x,H} and there is no interesting $\hbar$ corrections. $\endgroup$
    – Arian
    Apr 5, 2022 at 9:13

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