Classical limit in deformation quantization

Question

In deformation quantization, when dealing with the Moyal product, the classical limit is

\begin{align} \lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_M}=\{f,g\}\, \end{align}

which is just the Poisson bracket. Let's say that we have an operator, $T$, such that $$T(f\star_Mg)=Tf\star_TTg,$$ where $$\star_T=\star_MT^{-1}[\overleftarrow{\partial}]T[\overleftarrow{\partial}+\overrightarrow{\partial}]T^{-1}[\overrightarrow{\partial}].$$ Here, $T^{-1}[\overleftarrow{\partial}]$ means to replace all derivatives of $\partial_p$ in $T^{-1}$ with $\overleftarrow{\partial}_p$ and similarly for $\partial_q$. If we want to calculate the classical limit, do we use

\begin{align} \lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T} \end{align}

or do we use

\begin{align} \lim_{\hbar\to0}\frac{1}{i\hbar}T[f,g]_{\star_M}= \lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}? \end{align}

I'm thinking that we use this second equation because usually $Tf\neq f$ and $Tg\neq g$. Further, if we have a system described by Hamiltonian $H$, then we apply $T$ to it, in this new formulation, this system is no longer described by $H$, but by $TH$. If we don't use $Tf$ and $Tg$, it doesn't seem as if we are applying cohomological equivalence correctly because $T(f\star_M g)\neq f\star_Tg$.

Answer 1

The following comments seem relevant to OP's post:

1. In deformation quantization an associative star product $$\star:~~C^{\infty}(M)[[\hbar]]\times C^{\infty}(M)[[\hbar]]\to C^{\infty}(M)[[\hbar]] \tag{0} $$ on a Poisson manifold $(M,\{\cdot,\cdot\}_{PB})$ should satisfy the correspondence principle $$\lim_{\hbar\to 0} f\star g~=~\left(\lim_{\hbar\to 0}f\right) \left(\lim_{\hbar\to 0}g\right),\tag{1} $$ $$\lim_{\hbar\to 0} \frac{[f\stackrel{\star}{,} g]}{i\hbar}~=~\left\{\lim_{\hbar\to 0}f,~\lim_{\hbar\to 0}g\right\}_{PB}. \tag{2} $$ Here we have defined the star commutator $$[f\stackrel{\star}{,} g]~:=~f\star g-g\star f. \tag{3} $$

2. One may show that the correspondence principle (1) & (2) also holds for another associative star product $$f\star^{\prime}g ~:=~T^{-1}((Tf)\star (Tg)),\tag{4} $$ if the operator $$T:~~C^{\infty}(M)[[\hbar]]~\longrightarrow C^{\infty}(M)[[\hbar]] \tag{5} $$ is of the form $$T~=~{\bf 1} + i\hbar\Delta+{\cal O}(\hbar^2)\tag{6}$$ where $$\Delta:~~C^{\infty}(M)~\longrightarrow C^{\infty}(M)\tag{7}$$ is (at most) a second-order differential operator.

Answer 2

It might be worth reviewing formulas such as (122,34,5; 131) of our book, but, frankly, I am not sure I understand what underlies and follows the question, and especially the classical limit discussion.

To start with, in QM, f,g and T normally involve ħ, since they are Wigner transforms of quantum operators (Lemma 0.12): this creates subtleties in the ħ ↝ 0 limit, which need not commute with T; but I'll try your comment dictate that none depend on ħ, which simplifies everything. Let's save space by calling $\star_M=\star$.

Thus, since T is linear, $$T([f, g]_\star)= [Tf,Tg]_{\star_T} \Longrightarrow [f, g]_\star = T^{-1} ( [Tf,Tg]_{\star_T}), $$ so the "classical limit", is something like the ħ ↝ 0 limit of this.

This is a simple transcription of the "Moyal" expression yielding the PB you started your question with: it is a mere change of language.

Applying your test $T=e^{a \partial_p}$, a translation in p, gives you the same expression, so, then, since here $\star_T=\star$, exceptionally and felicitously, the two expressions above amount to $$ [f(x,p+a),g(x,p+a)]_\star \mapsto [f(x,p),g(x,p)]_\star . $$ The respective classical limits are then $$ \{f(x,p+a),g(x,p+a)\} \mapsto \{f(x,p),g(x,p)\} . $$ It is up to you to choose which expression you wish to consider the classical behavior of. The PB algebra is the same and it is up to you to specify the functions in the argument; you may even change variables to make them look the same. (That's why your "correct" is so puzzling. All expressions have classical limits, and it might be interesting to know all such, but we saw how they are related by equivalence. Also remember, several different quantum systems have the same classical limit.)

If your a-shift were linear in ħ, the classical limit of the two expressions would be the same.

---

In the Husimi prescription, $$ f_{_H} (x,p) =T[f]= \exp \left({\hbar\over 4}(\partial_x^2 +\partial_p^2 ) \right) f \\ =\frac{1}{\pi\hbar}\int dx' dp' \exp \left( -{(x'-x)^2+(p'-p)^2 )\over \hbar} \right) f(x',p') , $$ and likewise for the observables.

(So, for instance, the oscillator hamiltonian now becomes $\quad H_H= (p^2+x^2+\hbar)/2$.)
The representation-changing map T is a Weierstrass transform that collapses to the identity in the classical limit.

With the distinctly different $$\star_T= \star_H= \exp\left( \frac{\hbar}{2} (\overleftarrow{\partial}_x \overrightarrow{\partial}_x + \overleftarrow{\partial}_p \overrightarrow{\partial}_p ) \right) ~~\star~= \exp\left( \frac{\hbar}{2} (\overleftarrow{\partial}_x -i \overleftarrow{\partial}_p) (\overrightarrow{\partial}_x + i\overrightarrow{\partial}_p ) \right) , $$ it is, in fact, much easier to solve the *-genvalue equation for this hamiltonian, (Exercise 0.21), up to some care at the origin, than in the Moyal picture: the resulting ODE is only first order!

Both pictures, naturally, produce the same spectrum of stargenvalues, as they should. (But, no room for complacency, here: the ground state stargenfunction in the Husimi picture is the square root of that in the Wigner-Moyal picture. Can you check that? They both become δs at the origin in the classical limit.) An arbitrary nonstatic Husimi distribution, just like a Wigner function, or a classical object, rotate rigidly in phase space for an oscillator.

---

The takeaway is that operators, $[ \mathfrak {F}, \mathfrak{G}]$, Moyal brackets, $[f(x,p),g(x,p)]_\star$, any other prescription brackets, $[Tf,Tg]_{\star_T}$, etc... are, of course, unequal, but equivalent: any operation in any language will produce answers equivalent under the invertible maps (functors) taking you from one to another. In particular, since this is quantum mechanics, all expectation values should be identical in the end. As a result, the classical limit thereof will be strictly identical, even if the steps leading up to it appear different: think of a calculation in polar versus Cartesian coordinates: the Laplacian looks very different in each system, but the answers should be the same, after transcribed into any language. (Of course, the Moyal picture is the "Cartesian" system here, as described in that book.)