The Groenewold-Moyal (phase space) picture of quantum mechanics describes the evolution of a probability density corresponding to a wavefunction that evolves as described by Schrödinger's equation. The central equation for Groenewold-Moyal quantum dynamics is the Moyal equation;

$$\frac{\partial \rho}{\partial t} = -\{\{ \rho,H\}\} = \frac{2}{\hbar}\rho\sin\left(\frac{\hbar}{2}\left(\overleftarrow{\partial_x}\overrightarrow{\partial_p}- \overleftarrow{\partial_p}\overrightarrow{\partial_x}\right)\right)H = -\left[\rho,H\right] + O(\hbar^2)$$

where $\{\{.,.\}\}$ is the Moyal bracket, $\left[.,.\right]$ is the Poisson bracket, $H$ is the Hamiltonian and $\rho$ is the phase space probability density.

The classical analogue, Liouville's equation, is simply $$\frac{\partial \rho}{\partial t} = -\left[\rho,H\right].$$ This paints a very clear mathematical picture of the correspondence between classical and quantum dynamics, namely that for $\hbar\rightarrow0$ we recover classical dynamics from the quantum picture.

This, however, opens up a few questions about the physical interpretation of what quantum mechanics "is" relative to classical mechanics in this picture. For example,

  1. Is there any physical reason why the "deformation" of the Poisson bracket into the Moyal bracket is specifically sinusoidal? Does it directly come from some fundamental assumption in the derivation of the Moyal equation?

  2. What is the physical role of $\hbar$ in this formulation of quantum mechanics? How would changing $\hbar$ change the evolution of the phase space probability density in a semi-intuitive sense?


2 Answers 2


Cosmas Zachos has already given a nice answer. He correctly points out that the sine function in the $\star$-commutator originates from the exponential function in the $\star$-product.

Question: But why the exponential function, then?

Answer: Consider the following ansatz$^1$ for the $\star$-product:

$$\star~=~f\left( \stackrel{\leftarrow}{\partial} \wedge\stackrel{\rightarrow}{\partial} \right),\tag{1}$$ where $f:\mathbb{C}\to \mathbb{C}$ is a general (sufficiently nice) function. We want to prove that $f$ must be an exponential function.

In eq. (1) we used the notation $\partial_I\equiv \frac{\partial}{\partial z^I}$, where the coordinates $z^I$ are the phase space variables. If $$\alpha~=~\sum_I\alpha_I\mathrm{d}z^I\qquad\text{and}\qquad \beta~=~\sum_I\beta_J\mathrm{d}z^J\tag{2}$$ are one-forms, then the constant (=$z$-independent) Poisson structure is $$\alpha \wedge\beta~=~\sum_{IJ}\alpha_I\omega^{IJ}\beta_J ~\in~\mathbb{C}.\tag{3}$$

Exercise: Prove that $$e^{\alpha\cdot z} \star e^{\beta\cdot z} ~=~f\left(\alpha \wedge\beta\right)e^{(\alpha+\beta)\cdot z}. \tag{4}$$

We next use that the $\star$-product should be associative. In particular, it should hold that $$ \left(e^{\alpha\cdot z} \star e^{\beta\cdot z}\right)\star e^{\gamma\cdot z}~=~e^{\alpha\cdot z} \star \left(e^{\beta\cdot z}\star e^{\gamma\cdot z}\right). \tag{5}$$ Eqs. (4) & (5) imply that $$ f\left(\alpha \wedge\beta\right) f\left((\alpha+\beta) \wedge\gamma\right) ~=~f\left((\alpha\wedge(\beta+\gamma)\right) f\left(\beta\wedge\gamma\right), \tag{6}$$ or equivalently $$f(t)f(r+s)~=~f(t+s)f(r), \qquad r,s,t\in\mathbb{C}.\tag{7}$$ Now put $t=0$: $$f(0)f(r+s)~=~f(s)f(r), \qquad r,s\in\mathbb{C}.\tag{8}$$ It is well-known that the solutions to the functional equation (8) are the exponential functions! (Conversely, one may show that every $\star$-product on exponential form are associative.) If we furthermore impose the correspondence principle $$\star~=~1 ~+~ \frac{i\hbar}{2}\stackrel{\leftarrow}{\partial} \wedge\stackrel{\rightarrow}{\partial} ~+~ {\cal O}(\hbar^2) \tag{9} $$ between classical & quantum mechanics, then the $\star$-product (1) is uniquely given by the Groenewold-Moyal formula $$\star~=~\exp\left(\frac{i\hbar}{2}\stackrel{\leftarrow}{\partial} \wedge\stackrel{\rightarrow}{\partial}\right).\tag{10} $$ $\Box$


$^1$ In this answer we consider only constant (=$z$-independent) (possibly degenerate) Poisson structures. We should stress that even within the class of constant Poisson structures, there exist many $\star$-products that are not on exponential form and that don't satisfy the ansatz (1). Let us also emphasize that for non-constant Poisson structures, the exponential form of the $\star$-product does generically not apply. For these more general Poisson structures, one should use Fedosov or Kontsevich quantization, cf. e.g. this Phys.SE post.

  • $\begingroup$ Thank you for the elucidation! This provides very helpful context to Cosmas's answer regarding how the sine function pops out of the way we define the $\star$-product. I will be sure to look further into it. $\endgroup$ Commented Jun 20, 2018 at 17:41

Frankly, I am confused as to why you suspect you have to revisit QM interpretations in this picture; but your two circumscribed questions are answerable.

  1. No physical reason--mere mathematical convenience. See Refs 1 & 2. The sine comes from the fact the relevant Groenewold star product is the exponential of the PB, a matter of coincidence, built into the Wigner map taking you from Hilbert space operators to phase-space c-number variables. Under this map, the quantum commutator antisymmetrizes these exponentials into a sine function of PBs. But other strictly mathematically equivalent star products, such as the one in Husimi's picture (cf. eqn. (124) of Ref. 1) will correspond to very different, in general messier, brackets. They all expand past the PB to $O(\hbar^2)$ terms including the strictly quantum deformation features, which are only neat powers of the sine expansion for the Groenewold-Moyal product... and awful messes in (most) other of the half-dozen pictures I know. It is these terms that specify the dynamics' departure from classical flows, as you may notice on the entertaining movies of Ref. 2, and give QM its distinctive diffusive and compressible flavor.
  2. $~\hbar$ is a dimensionfull parameter, and, as such bimodal. Qualitatively, there are two cases: zero (classical), and nonzero (QM). Typically, it normalizes the phase space cell area, or the PB, or... into dimensionless entities. That is, you observe its (QM) effects for variables $xp/\hbar$ which are not enormous-ginormous... microscopic effects where $\hbar \partial_x \partial_p$ are not infinitesimal; when such action variables are enormous, as in a speeding locomotive or a mosquito, the QM deformation correction effects are normally invisible. The actual size of $\hbar$ only matters in the scale of objects falling into this microscopic quantum ambit. You might be invited to visualize a world of larger $\hbar$, so larger Bohr radius, so larger atoms, etc... The WP movies would look pretty much the same. Liouville's theorem is violated.


  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.
  2. WP article.
  • $\begingroup$ Thank you for the comprehensive answer, Cosmas! I suspect that most of my confusion is coming from not properly understanding the Wigner-Weyl transform, which seems to be the "source" of the differences between the classical and quantum formulations in phase space. I will be sure to read your book and obtain insight on it! $\endgroup$ Commented Jun 20, 2018 at 17:38
  • $\begingroup$ Indeed, you got it. The Wigner-Weyl transform underlies all... $\endgroup$ Commented Jun 20, 2018 at 18:53

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