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In quantization, one studies maps from functions on the phase space to operators acting on the Hilbert space. Let's fix one such map and call it $Q$.

Deformation quantization is based on the idea that $Q$ can be studied indirectly, by endowing the linear vector space of functions over the phase space with a non-commutative star product:

$$ f \star g = Q^{-1} \left( Q(f) \,Q(g) \right). $$

Kontsevich gives an explicit formula for the star product that can be applied to any compact phase space and gives an associative algebra with correct behavior in the $\hbar \rightarrow 0$ limit. It is therefore often claimed that Kontsevich formula solves the long-standing problem of proving that any compact symplectic manifold admits a quantization.

However, the other important ingredient of Quantum Mechanics is the trace of an operator. Traces are essential for making physical prediction, i.e. expectation values of observables are traces of the corresponding operators multiplied by the density matrix.

The Kontsevich formula does not give me a quantization map, only the star product. So how do I compute $\text{tr} Q(f)$ by only knowing $f$?

One possible answer that I see is that the classical formula holds: $$ \text{tr} Q(f) = \int \omega^{\wedge n} f. $$

Here $\omega^{\wedge n} = \omega \wedge \omega \wedge \dots \wedge \omega$ is the volume form associated to the symplectic form $\omega$, and the integral is over the phase space.

But I have never heard anyone say definitively that indeed this phase space integral is the counterpart of the operator trace in deformation quantization, and I can't come up with a good argument to show that $\mathcal{O}(\hbar)$ corrections don't appear.

My questions are:

  1. Do $\mathcal{O}(\hbar)$ corrections to the phase space integral appear in general?
  2. If they do, is there an explicit formula for the trace?
  3. If they do not, how do I convince myself of that?
Qmechanic
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Prof. Legolasov
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1 Answers1

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Wikipedia says the following properties to uniquely determine the trace operation (up to scalar multiples):

  1. $\mathrm{tr}(cA) = c\mathrm{tr}(A)$
  2. $\mathrm{tr}(A + B) = \mathrm{tr}(A) + \mathrm{tr}(B)$
  3. $\mathrm{tr}(AB) = \mathrm{tr}(BA)$

For any linear $Q$, $\mathrm{tr} Q(f)$ will satisfy all three properties. $\int f d\Omega $ clearly satisfies (1) and (2). For (3), we want to show that $\int f \star g d\Omega = \int g \star f d\Omega $. It's easy to show that the $O(1)$ and $O(\hbar)$ terms vanish for sufficiently nice $f,g$ (using integration by parts and the equivalence of mixed partials). However, I don't understand Kontsevich graphs well enough to confidently extend this argument to higher orders in $\hbar$. If you can find a reference or an explanation, let me know. Assuming the argument does extend, we find that $\mathrm{tr} Q(f)$ and $\int f d\Omega $ are equivalent up to a scalar multiple.

Expectation values are given by $\mathrm{tr}(Q(f)\rho)$, so we can choose to normalize our trace operation such that $$\mathrm{tr}(\rho) = \int Q^{-1}(\rho)d\Omega = 1$$ This should be enough to uniquely determine all the physics. You could define in an $O(\hbar)$ term in the original integral formula, but once you normalize your density matrix it has no physical effect.

Daniel
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