I'm trying to reconcile the following two statements:

  1. Quantum Mechanics gives physical predictions which are different than the predictions that are obtained in the $\hbar \rightarrow 0$ limit, that is, the predictions of quantum mechanics generally depend on $\hbar$.
  2. In the phase space formulation of QM, the Moyal product and the quantization map are related by the following formulae: $$ Q(a) Q(b) = Q(a * b),$$ $$ \text{tr} Q(a) = \frac{1}{2 \pi \hbar} \int dx \int dp a(x, p). $$

Here $Q(a)$ is a quantization of the phase space function $a(x, p)$, and $a * b$ is the Moyal product given by $$ a * b = a e^{\frac{i \hbar}{2} \omega^{ij} \overleftarrow{\partial_i} \overrightarrow{\partial_j}} b, $$ the arrow notation means that $\overleftarrow{\partial}$ acts on $a$ only and $\overrightarrow{\partial}$ acts on $b$ only.

Now consider a fairly generic problem from classical mechanics. Let's say I have an observable $E(x, p)$, and a probability distribution $\rho(x, p)$ (it can be for example the energy of the system and the Gibbs distribution for some value of $\beta$). The question is: what is the expectation value of $\left< E \right>$? According to classical physics,

$$ \left< E \right> = \int dx \int dp E(x, p) \rho(x, p). $$

Now if we try to compute the quantum corrections to this formula, we end up with

$$ \left< E \right>_{\hbar} = 2 \pi \hbar \text{tr} (Q(E) Q(\rho)) = \int dx \int dp E(x, p) * \rho(x, p). $$

It seems as if the corrections are encoded in the Moyal product. However, that can't be true: it is easy to show that for any $a, b$,

$$ \int dx \int dp \; a * b = \int dx \int dp \; a b. $$

(Demonstrate this by integrating by parts and observing that each non-zero-order term contains a product of antisymmetric $\omega^{i j}$ and symmetric $\partial_i \partial_j b$ and therefore vanishes).

So there are actually no quantum corrections in $\left< E \right>_{\hbar}$ defined above.

On the other hand, it is well known that there are quantum corrections to classical expectation values. Take the expectation value of energy a quantum oscillator for example. Computing it in the Hamiltonian eigenstate basis yields a value $\left<E \right>(\beta, \hbar)$ that depends on $\hbar$ explicitly.

Therefore we have a contradiction, so I must conclude that some of the assumptions I've made before must be incorrect. In particular, I see two possibilities:

  1. The trace formula isn't correct: $$ \text{tr} ( Q(a) Q(b) ) \neq \frac{1}{2 \pi \hbar} \int dx \int dp \; a * b. $$
  2. The quantization of the probability distribution isn't $Q(\rho)$ but somehow acquires extra corrections.

Both possibilities seem strange to me. What's going on here?


2 Answers 2


(I've done this in "reverse" notation as you in the sense that you have (in my notation) $W_A(x,p)\mapsto \hat A$ so my $W_A(x,p)$ is your $a(x,p)$ and your $Q(a)$ is my $\hat A$. I suppose it's just "semantics" in a way but I'm also more comfortable in starting with operators and finding the symbols rather than quantizing symbols.)

It is true that, if $\hat A\mapsto W_A(x,p)$, then (up to a constant factor) $$ \hbox{Tr}(\hat A\hat B)\propto \int dx dp W_A(x,p)\star W_B(x,p) = \int dxdp W_A(x,p)W_B(x,p). \tag{1} $$ If you already have the symbols, there is no point is using the $\star$-product if you want to do a phase space integration as in (1).

In general, finding the symbols for polynomials is where the action is, although for $xp$ space the $\star$ product has a reasonably manageable form in terms of the exponential as you suggest. You can then use $$ W_{AB}(x,p)=W_A(x,p)\star W_B(x,p) $$ to build up symbols for complicated observables from the symbols of the simpler ones. Of course $W_A(x,p)\star W_B(x,p)\ne W_B(x,p)\star W_A(x,p)$.

The corrections tend to be more visible in the evolution of an observable or a state since $$ i\partial_t W_{A}=\{W_H,W_B\}_M:= W_H\star W_A - W_B\star W_H\, . \tag{2} $$ For this type of computation, there is no phase space integration and the $\star$-product is unavoidable, even if you already have the symbols.

If $W$ is the Wigner symbol, then the leading term of the Moyal bracket in (2) should basically $i\hbar$ times the Poisson bracket of the symbols $W_H$ and $W_a$, plus the corrections you're looking for, of size ${\cal O}(\hbar^3)$. If you use different ordering (say $P$- or $Q$-functions) the corrections are of size ${\cal O}(\hbar^2)$.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    Sep 5, 2021 at 19:48

I figured it out.

Indeed, the quantization of $Q(\rho)$ acquires extra corrections, in a very peculiar way. For example, take the Gibbs distribution for a 1d harmonic oscillator:

$$ \rho(x, p) = e^{- \beta H(x, p)} = e^{- \frac{1}{2} \beta (x^2 + p^2)} $$

(I took $m = \omega = 1$ to simplify the equations).

The crucial observation is that $$ Q(\rho) = Q(e^{- \beta H}) \neq e^{- \beta Q(H)}. $$

If you take $Q(\rho)$ as your dennsity matrix, like I originally did, you will arrive at the value of $\left< H \right>(\beta)$ (the energy expectation) that doesn't depend on $\hbar$.

The correct density matrix is, however, the operator $$ e^{- \beta Q(H)} $$ and not $$ Q(e^{- \beta H}). $$

This is explicitly stated at the start of Wigner's paper.

I still don't understand which physical considerations went into choosing the former expression rather than the latter, however, I now understand where the corrections are coming from.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.