Quantum corrections in the phase space formulation

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 $$\left(\beta, \hbar)$$ 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

(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 $${\cal O}(\hbar^2)$$.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 5 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.