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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 $T(f\star_M g)\neq f\star_Tg$.

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  • $\begingroup$ If we consider $T=e^{a\partial_p}$, where $a\in\Re$, then $\star_T=\star_M$, so $\lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T}=\{f,g\}$, but $\lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}=\{Tf,Tg\}\neq\{f,g\}$. Thinking about it further, we require $T$ to be expanded in a series of bidifferential operators. That would then mean that $T=e^{a\partial_p}$ not a valid operator? Is there a reason that such operators are not usually considered? $\endgroup$
    – user85503
    Commented Mar 9, 2017 at 17:45
  • $\begingroup$ If we use $T=e^{a\partial p}$, where $a$ is independent of $\hbar$ and apply the formula $\lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T}$, then the limit is just the Poisson bracket, rather than also incorporating $a$ (with $f,g$ also independent of $\hbar$) This makes me wonder if I am using the incorrect formula for the classical limit because it is as if we did not use $T$ in the first place. (As $T$ represents a shift in the momentum, would we not want the classical limit to also include a shift in the momentum?) $\endgroup$
    – user85503
    Commented Mar 9, 2017 at 21:36
  • $\begingroup$ And to work out the spectrum of the Husimi distribution, would we not just apply $T$ to $H\star_MW=E_nW$, such that we still get $\hbar(n+1/2)$ for the spectrum of $TH$? Looking at the spectrum of $H$ in the Husimi formulation, our spectrum of energies would be $E_n-1/2=\hbar n$ (assuming no silly calculation error). $\endgroup$
    – user85503
    Commented Mar 9, 2017 at 21:36
  • $\begingroup$ And looking at Husimi, no, you ignored that H is not TH, so H shifted: TH= H + ħ/2. That's the point: a mere change of language could not possibly net you a different spectrum!! $\endgroup$ Commented Mar 9, 2017 at 22:20

2 Answers 2

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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.

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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.)

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  • $\begingroup$ So, just to make sure that I understand one of the points - we have the formulas $[Tf,Tg]_{\star_T}=[f(x,p+a),g(x,p+a)]_{\star_M}$ and $[f,g]_{\star_T}=[f,g]_{\star_M}$. Both have slightly different limits as $\hbar\to0$ (remembering to also divide by $i\hbar$), but considering both as classical limits are valid? Would the physical difference be that in the first case, we are transforming $f,g$ to be in the same language as $T$ and $\star_T$, while in the second case, we want to see what happens in the limit for the original $f$ and $g$, but with $\star_T$ being the new binary operation? $\endgroup$
    – user85503
    Commented Mar 10, 2017 at 15:57
  • $\begingroup$ Though both may be "correct," it seems as if using $\lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}$ is better to use if we want to look at how $Tf$ and $Tg$ could be related in the limit of $\hbar\to0$ (giving rise to the Poisson bracket or similar quantity). Am I interpreting this right? I think my biggest concern is that if both $\lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}$ and $\lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T}$ are valid, I am not really sure how I should decide to use one or the other. $\endgroup$
    – user85503
    Commented Mar 10, 2017 at 15:59
  • $\begingroup$ I remain confused about the part of the question you are not discussing. There is no such thing as "the classical limit". You have the classical limit of several expressions, and you are free to do with them what you wish. If you start from operators, you get very different Wigner images f for M or T, but, by above, no change of language will result in a different classical limit! $\endgroup$ Commented Mar 10, 2017 at 16:06
  • $\begingroup$ Okay, so $\lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}$ is "a classical limit" and $\lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T}$ also "a classical limit." Starting from operators $\hat{f}$ and $\hat{g}$, we transform it into the Moyal formalism or the T formalism, apply $\lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}$, and we get one answer. If we apply $\lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T}$, we should get the same answer (if $T$ incorporates $\hbar$). If $T$ does not incorporate $\hbar$, then it is fine to either use either limit formula? $\endgroup$
    – user85503
    Commented Mar 10, 2017 at 16:25
  • $\begingroup$ I apologize for so many questions. Keeping the example of $T=e^{a\partial_p}$ ($a\neq a(\hbar)$) in mind, I hope to understand the physical rationale for the basis of $\lim_{\hbar\to0}\frac{1}{i\hbar}[Tf,Tg]_{\star_T}=\{f(x,p+a),g(x,p+a)\}$ and $\lim_{\hbar\to0}\frac{1}{i\hbar}[f,g]_{\star_T}=\{f(x,p),g(x,p)\}$. I guess that is why I am confused. If both formulas give a classical limit, but no change of language should result in a different classical limit, then both formulas should give the same classical limit. Yet, $\{f(x,p+a),g(x,p+a)\}\neq\{f(x,p),g(x,p)\}$. $\endgroup$
    – user85503
    Commented Mar 10, 2017 at 16:26

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