# Constructing quantum operators

1. Oct 15, 2012

### HJ Farnsworth

Greetings,

Regarding operators, my understanding until today was that given the operators $\hat{x}$ for x and $\hat{p}$ for p, you could construct the operator corresponding to any classical quantity Q by expressing Q in terms of x and p, and then swapping x and p for $\hat{x}$ and $\hat{p}$, ie., Q = Q(x,p) $\Rightarrow$ $\hat{Q}$ = $\hat{Q}$($\hat{x}$,$\hat{p}$). I got this from Griffiths QM - in the 1st edition, he states this outright on page 16 (or at least, he says that to find <Q>, you do this direct substitution method and take the standard expectation value integral).

Today, however, my mind was blown - I was told that this is only true for expressions where x and p don't multiply each other. If they do, the fact that they don't commute results in not being able to simply substitute their operators, because there is an uncertainty term to worry about. The person who told me about this also told me to look up the terms "Weyl quantization" and "quantization ambiguity" for a more complete explanation.

Since Griffiths has pretty much been my QM bible up to this point, I tried to find a direct confirmation of what I was told today, but have so far failed. The stuff I've found on Weyl quantizations and quantization ambiguity is sparse, and also very abstract (some heavy abstract algebra, which I've only just started studying). I get the impression that Poisson brackets play into this subject a lot, but I have yet to find a simple intuitive explanation of what's going on.

So what I'm looking for is...

1. Confirmation or denial that the direct substitution method I learned in Griffiths is, in general, wrong.

2. A more general method to construct quantum operators from the equations for classical quantities.

3. For whatever general way operators are constructed, an intuitive explanation of why we are "allowed" to construct them from classical equations that way. If there is simply no way to explain this without delving into abstract algebra and Poisson brackets, please let me know.

-HJ Farnsworth

2. Oct 16, 2012

### strangerep

In general, there is indeed an ambiguity. If you do a Google search for

Groenwald "van Hove"

you'll find lots of stuff on the relevant theorem.

To quantize a classical theory involving nontrivial products of x, p it's often a good first attempt to just symmetrize them, e.g., if a classical expression has a term xp, then try (xp + px)/2 and see how that goes. The classical Poisson brackets (which define the dynamical algebra of observables in the classical case) generally become something different when the corresponding commutators are evaluated in the quantum case. One must also ensure that the resulting quantum dynamical algebra does indeed close as a Lie algebra.

As to "why" we're allowed to do this -- well, it's just something which often works in terms of agreement with experimental results.

(Sorry I couldn't give an explanation without mentioning Poisson brackets and dynamical algebra -- they're kinda central to the dark art of quantizing classical theories in general.)

Last edited: Oct 16, 2012
3. Oct 16, 2012

### dextercioby

Historically,

*Weyl noticed the ambigity in the quantization recipe by Dirac. The need to symmetrize products of x and p. Finesse issues analyzed by von Neumann (existence of self-adjoint extensions of symmetric operators on separable Hilbert spaces).
*Wigner constructed his quantization recipe based on Wigner distribution.
*Weyl constructed his quantization recipe based on Weyl product.
*Jordan constructed his quantization recipe based on Jordan product (the so-called Jordan algebras).
*Groenewold and van Hove obstruction theorem: groundbreaking result on general quantization. Birth of geometric quantization.
*Dirac's theory of quantization of Hamiltonian systems with constraints by means of quantum Dirac bracket => Gribov's work on ambiguity of quantization of gauge systems.
*Kostant and others build on geometric quantization developing it into a full sound theory.

Last edited: Oct 16, 2012
4. Oct 16, 2012

### strangerep

Umm, did you mean "recipe" ?

5. Oct 16, 2012

### dextercioby

No, I meant "invoice" actually.

For the proficient ones in English, typos corrected.

6. Oct 16, 2012

### Staff: Mentor

If you read Chapter 3 of Ballentine - QM A Modern Development you will find a much better justification of this stuff than Griffiths - its based on symmetry and is very beautiful.

It does not solve the issue of quantising a system (in general no exact recipe exists even within geometric quantisation - but dont pick up a book on that because its HARD) but as Ballentine points out his approach does not seem to really suffer any problems because of it.

Thanks
Bill

7. Oct 16, 2012

### strangerep

Unfortunately, he stops at the case of a spinless particle interacting with an external field. He says (top of p89) that his eq(3.60) is "the most general case encountered in practice", which puzzles me since particles with spin can interact with external fields in more general ways, iiuc. Or am I missing something?

BTW, do you (or anyone else) know of any treatment that takes Ballentine's approach a bit further in this regard?

8. Oct 16, 2012

### HJ Farnsworth

Thanks for the replies, everyone.

I started looking at Ballentine's text, and although I haven't had much time to delve into it yet, I do like the symmetry approach much better than Griffiths' treatment, since it offers a justification for the construction of operators.

I'm not to the point where I can classify this stuff rigorously yet, but in hand-wavy terms, here's what I'm thinking now...

For different situations, different symmetrization requirements will be beneficial. Ballentine uses the Galilei symmetry rules, but in particle physics, for example, Lorentz symmetry would be more appropriate, and in statistical thermodynamics, permutation symmetry would be key.

Presumably, to find a quantum operator corresponding to a classical quantity, it will always be necessary to first choose appropriate symmetrization requirements, and derive operators based on the fact that they must obey those requirements.

It is common for introductory quantum texts to give the recipe that I was using before my first post, where you just replace Q(x,p) with $\hat{Q}$($\hat{x}$,$\hat{p}$). This is based on an isomorphism between commutators and Poisson brackets, and is ultimately a rule that comes from postulating a symmetrization requirement - something like, a quantity is invariant upon switching Q(xp) for Q(px) - ie., it comes from postulating that it doesn't matter in which order x and p are multiplied. This is, in general, false for quantum systems, since while classically the order in which they're multiplied doesn't matter, the order in which they operate on each other can matter.

No recipe exists, but still, for every classical quantity there must be a corresponding operator - otherwise QM would fall short of being a useful theory. These operators can be derived from symmetrization requirements, even if there's no standard algorithm to do it (just as there's no algorithm for proving a random theorem in math, even if we know that that particular theory must be valid or invalid. It can still be proved one way or the other, but there's not always an obvious way to do it).

Does this sound correct at all?

Also, does anyone know of a source that explains where the isomorphism between Poisson brackets and commutators comes from? If I could see the derivation (even a dumbed down version of it), I think I could make more sense of why it is sometimes acceptable to use the (x,p)->($\hat{x}$,$\hat{p}$) approach, and what assumptions it is based on that ultimately make it false when those assumptions are not met.

Thanks again.

-HJ Farnsworth

9. Oct 16, 2012

### Staff: Mentor

You are not missing something - his approach does not solve all the issues. However, as far as I can see, it does solve the spin case and all the others I can think of (he simply did not detail it) - but then again I am hardly an expert. For that though I suspect QFT would be better - but of course that is a whole new ball game.

The best approach is the Geometric approach mentioned previously. The book I have on that is Varadarajan - Geometry Of Quantum Theory. But be warned - it is mathematically non trivial (code word for HARD) even for a person whose background is math like me.

Thanks
Bill

10. Oct 16, 2012

### Staff: Mentor

You can do something similar and end up with stuff like the Dirac equation. See the book by Varadarajan for example (I don't suggest you have a look at that book right now unless your math is at graduate level - Ballentine is fine even though it is designed for a graduate level course - but Varadarajan is well beyond that).

That said relativistically I think QFT is a better approach - but again best to leave that alone for a while. But once you feel a bit more emboldoned I rather like Zee's book:
https://www.amazon.com/Quantum-Field-Theory-Nutshell-Zee/dp/0691010196

I guess I should point out why. Relativistically you should treat space and time on the same footing. But in standard QM time is a parameter but position an operator - they really need to be the same. In QFT they are both parameters.

Thanks
Bill

Last edited by a moderator: May 6, 2017
11. Nov 2, 2012

### A. Neumaier

Observables must be selfadjoint, but $$(pq)^*=q^*p^*=qp\ne pq.$$ This demonstrates the need for worrying about ordering. Any selfadjoint ordering (e.g., symmetrization = Weyl quantization) produces from a classical polynomial in p and q a physically meaningful Hamiltonian, but different such orderings may produce different Hamiltonians.

This just means that there is no canonical way to go from a classical Hamiltonian to a quantum Hamiltonian. (This is even the case with geometric quantization.) In reality things work anyway the other way around: Physical systems are always quantum, but in some cases one can take without harm a classical limit. Indeed, the classical limit is much better defined.