# Dequantizing Dirac's quantization rule

In this blog post, Lubos Motl claims that any

commutator may be shown to reduce to the classical Poisson brackets: $$\lim \limits_{\hbar \to 0} \frac{1}{i\hbar} \left[ \hat{F}, \hat{G} \right] = \{F, G\},$$

where $\hat{F}$ and $\hat{G}$ are the Hermitian operators corresponding to the classical observables $F$ and $G$. How is this done?

Edit: As ACuriousMind points out, the proof is trivial if you start with a classical Hamiltonian and then quantize it via a reasonable quantization procedure. But what I have in mind is starting with a quantum Hamiltonian (and the canonical commutation relation $[\hat{x}_i, \hat{p}_j] = i\, \delta_{ij}$), then taking some limit $\hbar \to 0$ and showing that the resultant emergent classical theory has Poisson brackets that agree with the quantum commutators. Under these assumptions, you can't use any facts about your quantization procedure, because you never quantize a classical Hamiltonian at all.

• – AccidentalFourierTransform Dec 7 '16 at 23:57
• Can't you use Ehrenfest's theorem to derive Hamilton's equations from a quantum system? – tparker Dec 8 '16 at 1:03
• More generally, the real world is not described by a classical Hamiltonian that has been quantized; it's described by a quantum system which, in certain regimes, appears classical. The quantum mechanics comes first, but just from those few postulates one could, in principle, derive all of classical mechanics in regimes in which $\hbar$ becomes irrelevant. In particular, one could derive the value of Poisson brackets between arbitrary classical observables; my question is whether there's a simple proof that the result is always proportional to the quantum commutators. – tparker Dec 8 '16 at 1:05
• @CountIblis Could you be more specific? I understand why the classical solution dominates the path integral for actions much larger than $\hbar$, but I don't see how to get Poisson brackets out of commutators. – tparker Dec 8 '16 at 2:28
• Ah! so much fuss! You cannot prove that equation because it simply makes no sense: the l.h.s. is an operator and the r.h.s. is a $c$ number. You won't be able to prove it because it simply cannot hold. Also, you cannot take $\hbar\to 0$. This doesn't make sense either. You can only take limits of dimensionless numbers, but $\hbar$ is dimensionfull. Something that could make sense is $\lim_{n\to\infty}\langle n|[F,G]|n\rangle=\{f,g\}$, where $n$ is a quantum number for some basis $|n\rangle$. I don't think that you can prove this either. At least, not in general. The classical limit of (1/2) – AccidentalFourierTransform Dec 8 '16 at 15:12

I do not know about deep questions. And people seem to give pretty deep answers here. My contribution is to show

$$\lim_{\hbar \to \infty} \frac{1}{i\hbar}[ F(p,x) , G(p,x)] = \{F(p,x), G(p,x)\}_{P.B.}$$

where $[ F, G] = F G - G F$ and

$$\{ F(p,x), G(p,x) \} = \frac{\partial F}{\partial x} \frac{\partial G}{\partial p} - \frac{\partial G}{\partial x} \frac{\partial F}{\partial p}.$$

Preliminars.

With $[x, p] = i \hbar$, you can show the following two equalities:

$$[x, f(p) ] = i \hbar \frac{\partial f}{\partial p}$$

and

$$[p , g(x) ] = - i \hbar \frac{\partial g}{\partial x}.$$

I think this is almost mandatory for every QM course, so I will skip this derivation. In any case, the standard route is to consider the commutator of x with increasing powers of p; then use induction when developing $f(p)$ as a Taylor series.

A more illustrative example is the following:

$$[x^{2} , f(p) ] = [x ,f(p) ] x + x [x, f(p)] \\ = i \hbar f'(p)\, x + i \hbar x \, f'(p) = 2 i\hbar x f'(p) - i\hbar [x , f'(p)]\\ = 2 i \hbar x f'(p) - (i \hbar)^{2} f''(p)$$

where I have introduced the pretty useful notation $f'(p) = d f /dp$.

By now you can see the fun is in arbitrary powers of $x$. You should more or less be able to guess the result and prove it by induction.

Lemma.

$$[x^{n} , f(p) ] = \sum_{j=1}^{n} (-)^{j+1} \binom{n}{k} \, (i \hbar)^{j} x^{n-j} \, f^{(j)}(p)$$

Proof: You do it. Use induction. It should be more or less straightforward. By the way, $\binom{n}{k}$ denotes the binomial coefficient.

Moment of truth.

The previous argument can be used to include an analytic function of $x$. Consider

$$[ g(x) , f(p)] = \Biggl[ \, \sum_{k=0}^{\infty} \frac{1}{k!} g^{(k)} (0) x^{k}, \, f(p) \Biggr] = \sum_{k=0}^{\infty} \frac{1}{k!} g^{(k)} (0) \Biggl[ x^{k}, f(p) \Biggr] \\ = \sum_{k=0}^{\infty} \frac{1}{k!} g^{(k)} (0) \sum_{j=1}^{k} (-)^{j+1} C^{k}_{j} \, (i \hbar)^{j} x^{k-j} \, f^{(j)}(p) \\ = \sum_{j=1}^{\infty} (-)^{j+1} \, (i \hbar)^{j} g^{(j)}(x) \, f^{(j)}(p).$$

The trick in the fourth equality is to switch the sums (and then expand $C^{k}_{j}$... everything fits nicely).

It is interesting to notice that the double summations collapsed into one. This somehow makes sense by dimensional analysis, powers of x and p decrease together so that $\hbar$ appears.

The final part is the most subtle point. A general $f(x,p)$ is tricky because $x$ and $p$ do not commute. So you would have problems with "hermiticity" and ordering. I will choose every $p$ to be the left of every $x$. Once this is agreed, a general $F(p,x)$ can be written as

$$F(p, x) = \sum_{n=0}^{\infty} \alpha_{n} (p) \,\, f_{n} (x).$$

Now, we can compute

$$\Biggl[ F(p,x) , G(p,x) \Biggr] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \Biggl[ \alpha_{n} (p) \,\,f_{n} (x), \, \beta_{m} (p) \, \, g_{m} (x) \Biggr] \\ = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \alpha_{n} (p) \biggl[ \, f_{n} (x) , \beta_{m} ( p) \biggr] g_{m} (x) + \beta_{m} (p) \, \biggl[ \, \alpha_{n} ( p) , g_{m} (x) \biggr] \, f_{n} (x) \\ = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \, \alpha_{n} (p) \, \biggl( \sum_{j=1}^{\infty} (-)^{j+1} \, (i \hbar)^{j} f^{(j)}_{n} (x) \, \beta^{(j)}_{m} (p) \biggr) \, g_{m} (x) + \beta_{m} (p) \biggl( \sum_{j=1}^{\infty} (-)^{j} \, (i \hbar)^{j} g^{(j)}_{m}(x) \, \alpha^{(j)}_{n}(p) \biggr) \, f_{n} (x)$$

specially using

$$\sum_{n=0}^{\infty} \alpha_{n} (p) \, f_{n}^{(j)} (x) = \frac{\partial^{j}}{\partial x^{j}} \biggl( \sum_{n=0}^{\infty} \alpha_{n} (p) \, f_{n} (x) \biggr) = \frac{\partial^{j}}{\partial x^{j}} F(p,x)$$

you see that you get the desired result (after changing the summations):

$$\Biggl[ F(p,x), G(p,x) \Biggr] = \sum_{j=1}^{\infty} (-)^{j} \frac{(i \hbar)^{j}}{j!} \Biggl( \frac{\partial^{j} G}{\partial x^{j}} \frac{\partial^{j} F}{\partial p^{j}} - \frac{\partial^{j} F}{\partial x^{j}} \frac{\partial^{j} G}{\partial p^{j}} \Biggr)$$

because you see the only term that survives after dividing by (i \hbar) is the first one. This gives you the Poisson bracket. I didn't do any involved computations because they are long. It is more or less convincing.

• What you've shown in the language of the existing answers is that the deformed bracket/star product belonging to deformation quantization for anti-canonical ordering ($p$s before $x$s, classical phase space function $x^n p^m$ goes to $\hat{p}^m \hat{x}^n$) is given by the r.h.s. of your last equation. – ACuriousMind Dec 9 '16 at 2:51
• You assumed that all the $p$'s come before all the $x$'s, but it's very easy to extend this proof to the case of arbitrary ordering. Commuting an $x$ through a $p$ spins off of commutator proportional to $\hbar$. So if you start with an arbitrary ordering of $p$'s and $x$'s and rearrange it so that the $p$'s come first, then the commutator terms will all be higher order in $\hbar$, and so those terms will vanish in the limit $\hbar \to 0$. – tparker Dec 9 '16 at 6:01
• Also, to write "choose" in Latex, you can include the amsmath package and use \binom{k}{j} to get $\binom{k}{j}$. – tparker Dec 9 '16 at 6:04

Let me rearrange the logic of the Moyal Bracket that @ACuriousMind discussed neatly, by visiting a notional planet where people somehow discovered classical mechanics and quantum mechanics independently; but suffered a terrible mental block that prevented them from appreciating there was a connection between the two, at first.

Then, one day, their Groenewold observed that, starting from QM, where capitals denote QM operators, $[P,Q]=PQ-QP=-i\hbar,\,$, etc..., and lower case denote classical phase-space entities, he could take any operator function of P and Q, Φ, and package all of its matrix elements into the following c-number generating function, $$f(q,p)= 2 \int_{-\infty}^\infty \text{d}y~e^{-2ipy/\hbar}~ \langle q+y| \Phi (Q,P) |q-y \rangle,$$ (what we'd recognize here as our Wigner map to phase space on our planet), that is, to say, completely specified by the totality of its matrix elements, $$\langle x| \Phi |y \rangle = \int_{-\infty}^\infty {\text{d}p\over h} ~e^{ip(x-y)/\hbar}~ f\left({x+y\over2},p\right) .$$ He thus discovered that the operator Φ could actually be extracted out of inverting the above, so it is an operator functional of the quantum c-number function f(q,p), which of course also depends on $\hbar$, in general, $$\Phi [f] = \frac{1}{(2\pi)^2}\iint\!\!\! \iint f(q,p) ~e^{i(a(Q-q) +b(P-p))}~ \text{d}q\, \text{d}p\, \text{d}a\, \text{d}b.$$

Observe how this form expresses Φ(Q,P), with its complicated and capricious ordering of strings of Qs and Ps, now in a form where Qs and P are completely symmetric (the exponential being the formal infinite power series development thereof).

(On our planet, this inverse map is called the Weyl map, and was discovered first, in a misguided effort to start with classical quantities f(q,p) and somehow, magically!, be led to their quantum correspondents, which know about $\hbar$, so with more information appearing out of thin air, but no matter. Still Kubo was the one to appreciate this procedure automatically Weyl-orders arbitrary operators, i.e. yields equal operators in this special ordering, in general looking different.)

Moreover, this Wigner map maps Hilbert space operator commutators $[\Phi,\Gamma]/(i\hbar)$ to what we call the Moyal Bracket, $$\frac{2}{\hbar} ~ f(x,p)\ \sin \left ( {{\tfrac{\hbar }{2}}(\overleftarrow{\partial}_x \overrightarrow{\partial}_{p}-\overleftarrow{\partial}_{p}\overrightarrow{\partial }_{x}} )\right ) \ g(x,p),$$ where you note the leading term in the Taylor series w.r.t. $\hbar$ is just $\{f,g\}$, the Poisson Bracket. Hilbert space traces map to phase-space integrals.

(Full disclosure: an expansion of these moves can be found in our booklet A Concise Treatise on Quantum Mechanics in Phase Space by Curtright, Fairlie, and Zachos, WS 2014, cf. online update, or most other popular texts on the subject.) So far, absolutely no physics, or insight: through a technical change of language, plain QM was simply re-expressed in c-number phase space.

Now, however, our Tralfmadorean Groenewold must have been very pleased indeed, since he also knew this was within the scope of classical mechanics, so he could discuss both QM and classical mechanics in the same breath. He could then observe that most "large", macroscopic, systems and entities involving large quantum numbers, and large actions compared to $\hbar$, behave as classical c-number functions of phase space familiar from classical mechanics (corrected by $\hbar$-fuzz, ignorable for very small $\hbar$), the Moyal Bracket for slowly varying functions (on the scale of $\hbar$ again, where waviness and interference rule), devolved to the Poisson brackets, etc... He must have been beside himself with the emergent classical mechanics limit he found.

So, even though f, g, etc, depend on $\hbar$, as full quantum objects, those that have a nonsingular limit as $\hbar\to 0$ reduce to neat engineering physics (freshman lab) quantities free from the frustrating complications of quantum mechanics. Oh, dear: variables are effectively commutative, when you sacrifice (quantum) information... Suddenly, talking about trajectories, in general, could make sense! (But then chaos and entropy reared their ugly heads. But we are digressing.)

OK, this is the outline of emergent classical behavior. Several subtleties are swept under the rug, including macroscopic quantum systems, etc..., but ginger treading conquers the fog of $\hbar$, and decoherence is a friend.

The invertible maps above, nevertheless, have nothing to do with quantization--they are mere changes of variables. But they help you monitor it, if you wished to go the Dirac way, and hence the misnomer "deformation quantization": you pretend you start with $\hbar$-independent fs and the PB and "cleverly deform it" to the MB by guessing the $\hbar$-corrections on intuitive beauty principles. But you'll never get the correct square of the angular momentum this way. Quantization is an art, a mystery.

Convenience Edit to connect to antistandard ordering: @OkThen replicates the antistandard ordering prescription, that Kirkwood 1933 utilized, in eqn (121) of the book cited above; I couldn't resist the teachable moment. It is, of course, equivalent to the Wigner-Weyl map discussed here, as @ACuriousMind and @tparker point out. All of these Hilbert-space to phase-space maps are, where agreement to the classical entities at $O(\hbar^0)$ is essentially enforced as a boundary condition, so failure of the Dirac correspondence would be evidence of an error, as emphasized by @ACuriousMind.

Explicitly, sticking a extra factor $\exp(i\hbar ab/2)$ to the exponential of the above Φ converts the above operator kernel to $e^{ib(P-p)} e^{ia(Q-q)}$ yielding a slightly different Φ', mappable invertibly to Φ, of course. The corresponding image of the Moyal bracket is, as given, a bit less symmetric, $~f(1-\exp(i\hbar(\overleftarrow{\partial}_x \overrightarrow{\partial}_{p}-\overleftarrow{\partial}_{p}\overrightarrow{\partial }_{x} ))g/i\hbar$, but of course mappable to the MB invertibly, by the same map. This was actually Dirac's original thesis observation, that correspondence of q with Q and p with P automatically yields the boundary condition discussed, so it could not fail. It was only subsequent cookie-cutter quantization scheme seekers who unwisely insisted on applying such maps to quantization, now safely excluded by Groenewold.

Note added on Bracken's emergence : In a remarkable 2003 paper, Bracken observes that the obverse side of the standard quantization relation $MB=\frac{2}{\hbar}\sin (\hbar ~PB /2)=PB + O(\hbar^2)$ is $PB=\frac{2}{\hbar}\arcsin (\hbar ~MB /2)=MB + O(\hbar^2)$, so emergent classical mechanics is an infinite asymptotic series of $\hbar$ quantum corrections to the quantum result: the magic here is the complete cancellation of all $\hbar$ dependence, analogous to the destructive interference of quantum phases in the functional integral yielding the classical extremizing result. It's good to know as a formal wisecrack, but I have never seen a brass-tacks utilization of it in a cogent nontrivial calculation.

The statement is true by the very definition of quantization, i.e. there is nothing to show. So let's talk about the definition of quantization, which is a map from classical observables to quantum observables.

There is no quantization map $f\mapsto \hat{f}$ that sends classical observables (functions on phase space) to quantum observables (self-adjoint operators on Hilbert space) that fulfills

1. $$\widehat{\{f,g\}} = \frac{1}{\mathrm{i}\hbar}[\hat{f},\hat{g}]\tag{1}$$ for all classical observables $f,g$.
2. For all polynomials $p$, $\widehat{p(f)} = p(\widehat{f})$ for all classical observables $f$.
3. The representation of the algebra of observables is irreducible.

This is known as the Groenewold-van Howe theorem. The precise technical assumptions about the quantization map vary, but these are the main points it should naively, in "canonical quantization", fulfill, but cannot.

In order to allow for a quantization map one has to weaken an assumption. One option is deformation quantization where $(1)$ is only supposed to hold up to quantum corrections of order $\hbar^2$, and the usual deformation of the Poisson bracket is then the Moyal bracket $\{\{-,-\}\}$, which agrees with the naive canonical quantization recipe for the brackets of the coordinates $x_i,p_j$ as $$\frac{1}{\mathrm{i}\hbar}[\hat{x}_i,\hat{p}_j] = \widehat{\{x_i,p_j\}} = \widehat{\{\{x_i,p_j\}\}}$$ but deviates for higher polynomials in $x,p$ from the Poisson bracket at order $\hbar^2$ and higher. So, by definition of deformation quantization, we have $$\frac{1}{\mathrm{i}\hbar}[f,g] = \{\{f,g\}\} = \{f,g\} + \mathcal{O}(\hbar^2)$$ where taking $\hbar\to 0$ on both sides clearly yields $$\lim_{\hbar\to 0} \frac{1}{\mathrm{i}\hbar}[f,g] = \{f,g\}.$$

If you want start with a quantum system with canonical commutation relations $[x_i, p_j] = \mathrm{i}\hbar\delta_{ij}$ "without having obtained it by quantization", then that's just impossible - you may not have "obtained" it my quantization, but it is nevertheless the same as the result of standard quantization:

By the Stone-von Neumann theorem, all representations of this commutation relation are unitarily equivalent. So we can always obtain the part of the quantum algebra of observables generated by $x_i,p_j$ as the deformation quantization of the corresponding classical system, and the equality between the commutator and the Poisson bracket in the classical limit is again immediate from the definition of the quantization procedure.

• This isn't quite what I meant; I have edited my question to clarify. – tparker Dec 8 '16 at 0:35
• @tparker I believe I answer your question in every technically meaningful way. I have added an edit that explains why you cannot have quantum systems with the CCR that are not in the image of such a deformation quantization, i.e. why the answer explains the equality of commutators and Poisson brackets for all systems where Poisson brackets make sense. – ACuriousMind Dec 9 '16 at 3:02