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In a first course on quantum mechanics, everybody learns some version of the following statement:

Postulate: To every classical observable $A$ of a physical system, there corresponds a Hermitian operator $\hat{A}$ such that a measurement of $A$ performed on a system in state $|\psi\rangle$ is expected (in the probabilistic sense) to return $\langle \psi | A\psi \rangle$. Possible outcomes of the measurement correspond to eigenvalues of $A$...

One then learns that $\hat{x} = x$ is the operator corresponding to position, $\hat{p} = -i\hbar\frac{\partial}{\partial x}$ is the operator corresponding to momentum, and from these, one can construct the operators for the kinetic and potential energies, angular momenta, etc.

But what happens if we want to find the operator corresponding to some more complicated classical observable, say, $$ A = \frac{e^{\sin(x^2 p)}}{x+xp} + \cos\left(\frac{p}{x}\right)? $$ (Assume here that we've scaled $x$ and $p$ by some characteristic length and momentum, so that they become dimensionless.) Granted, this bizarre quantity might not have a compelling physical interpretation, but it is still in principle a classical observable, that can be computed with knowledge of $x$ and $p$. It seems to me that if we are to take this postulate seriously, there ought to be a well-defined procedure that associates to any function of $(x,p)$ a Hermitian operator acting on the appropriate Hilbert space. What is this procedure?

More generally, if we have some classical system $S$ with configuration manifold $M$, it seems to me that any real-valued function $f: T^*\!M \to \mathbb{R}$ on the phase space of $S$ (i.e., cotangent bundle of $M$) defines (in principle) a classically observable quantity. (Or does $f$ need to be continuous/smooth?) Quantum mechanics ought to prescribe a mechanism that associates to any such $f$ some Hermitian operator $\hat{f}: L^2(M) \to L^2(M)$. What is this mechanism, and how does it work in general?

Qmechanic
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David Zhang
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  • Essentially we are taking the quantity in classical mechanics and then promoting it to an operator wherein its classical values become eigen values of the operator. Then we make the operator hermitian. A is an infinite dimensional matrix with continuous eigen values. I am interested in the mathematical answer though .. – Abhishek Pal Jun 30 '16 at 04:50
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    Related: http://physics.stackexchange.com/q/68686/2451 and http://physics.stackexchange.com/q/181078/2451 . Weyl quantization is e.g. discussed in this Phys.SE post. – Qmechanic Jun 30 '16 at 09:30

1 Answers1

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There doesn't exist any procedure to uniquely associate a Hermitian operator $L$ to a function of the phase space $f(x,p)$. Quantum mechanics is a theory that exists independently of classical physics. Quantum mechanics is not just a cherry on a classical pie that needs the classical theory to exist at every moment. If we want to define a quantum theory, we must define a quantum theory. The definition does not involve the finding of a classical theory first, and then finding a unique quantum theory associated with it.

More seriously, there is no natural isomorphism between the algebra of operators on the Hilbert space; and the algebra of functions $f(x,p)$. The simplest reason is that the latter is a commutative algebra while the former is not. For this simple reason, a naive identification of the elements on both sides simply has to be wrong.

The correct relationship between quantum mechanics and classical physics, whenever both of them may be relevant, is exactly the opposite one: classical physics is derived from quantum mechanics. It is derived as a limit, the $\hbar\to 0$ limit. But even this relationship isn't quite universal. There exist quantum theories without any classical limit.

We may ask what are the Hermitian operators $L$ such that their $\hbar\to 0$ limit produces a given function $f(x,p)$ on the phase space. But the answer isn't unique. The possible solutions may differ by terms that go to zero for $\hbar\to 0$.

For example, the classical observable $x^2 p^2$ might "generate" the quantum operator $\hat x^2 \hat p^2$. However, the latter operator isn't Hermitian. Its Hermitian conjugate is $\hat p^2 \hat x^2$ which isn't equal to the original one. If we want a Hermitian operator, we may talk about e.g. $$ \frac{ \hat x^2\hat p^2+ \hat p^2\hat x^2}{2} $$ but also e.g. about $$ \hat x \hat p^2 \hat x $$ Both are Hermitian and naively reduce to the classical $x^2 p^2$. However, these two Hermitian operators are different from each other. They differ by a numerical multiple of $\hbar^2$, in this case.

On the other hand, expressions such as your complicated functions – but with hats – are well-defined and calculable (possibly except for the singularity at $x=0$ and $p=-1$ in the case of your particular function). For example, the exponential of an operator may be calculated via Taylor series $$\exp(\hat L) = \sum_{n=0}^\infty \frac{\hat L^n}{n!} $$ Even more complicated functions of operators are calculable. The function $g(\hat L)$ of an operator $\hat L$ may be calculated e.g. by diagonalizing $\hat L$ i.e. writing $$\hat L = U \hat D U^\dagger $$ where $D$ is diagonal. Then $$g(\hat L) = U g(\hat D) U^\dagger $$ However, $g(\hat D)$ is simple to calculate: we just apply the function $g$ to each diagonal element of $\hat D$.

For this reason, even your function defines an operator, except for the singularity problems near $x=0$ and $p=-1$. Well, we must also refine what you mean by $p/x$ – there is no simple division of operators. If you define it as $px^{-1}$, it's something else than $x^{-1}p$ etc. because the operators don't commute.

However, it's clear that aside from all these small issues, your operator won't be Hermitian because $\hat x+\hat x\hat p$ isn't Hermitian and $\hat x^2 \hat p $ isn't Hermitian and its sine isn't Hermitian, either. You would have to take the Hermitian part(s) at some moment to correct the Hermiticity but there wouldn't be a unique way to do so, as explained above.

There is no natural way to find an operator for a function $f(x,p)$ that is given by its values, i.e. without an explicit formula. This is particularly manifest if we imagine that each $f(x,p)$ is a continuous superposition of functions such as $\delta(x-x_0)\delta (p-p_0)$ supported by one point in the phase space.

This $\delta(x-x_0)\delta (p-p_0)$ has no good quantum counterpart because it wants to be localized both in position and the momentum. But the uncertainty principle prohibits such a localization. One might associate a minimum-uncertainty Gaussian with this product of the delta-functions but it is not really a "canonical choice".

If we sacrifice most of the algebraic properties, there exists a one-to-one map between functions and the matrices, the mathematics used in the Wigner quasiprobability distribution. But this map has some other properties one may find undesirable. The product gets mapped to the "star product". Also, a positive definite operator is generically mapped to a function that goes negative for some values of $(x,p)$, and so on.

Luboš Motl
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  • Sorry LuMo, but you perhaps intended to write something else, in the sentence "If we want to define a quantum theory, we must define a quantum theory". :) – 299792458 Jun 30 '16 at 05:59
  • Thanks for your answer! I suppose, based on the remarks in your third paragraph, that the "postulates" I was taught were really more like heuristic principles than foundational axioms, since they explicitly emphasized the passage from a classical system to a corresponding quantum system. If this is not the right way to think about things, then what is the correct way to state the relationship between operators and observables? – David Zhang Jun 30 '16 at 06:01
  • @TheDarkSide - No, I wanted to write exactly this tautology. – Luboš Motl Jun 30 '16 at 06:04
  • Dear @DavidZhang - it is definitely incorrect to use the word "postulate of quantum mechanics" for anything that uses mathematical objects from the classical theory (e.g. phase space or functions on it). In quantum mechanics, (Hermitian) operators and observables are exactly the same thing. That's the relationship! The set of Hermitian operators is isomorphic to the set of Hermitian matrices with respect to a given orthonormal basis. A QM theory is defined once one has something like the matrix elements of the Hamiltonian (or S-matrix) with respect to any basis. – Luboš Motl Jun 30 '16 at 06:07
  • Typically, such a Hamiltonian is defined as a function of some other operators, like $x,p$ in the simplest models of quantum mechanics, and other functions of these operators may be defined as well, as I discussed in the answer. But in all these cases, the operators are noncommuting so all the functions are functions of noncommuting operators where the ordering in the product matters and so on. To imagine some classical commuting functions behind everything is just fundamentally wrong. – Luboš Motl Jun 30 '16 at 06:10
  • @LubošMotl: You said "Quantum mechanics is a theory that exists independently of classical physics". I do not agree with this. In fact QM is based on classical physics. Can you develop your own QM which do not take any help from classical physics? Schrodinger & Heisenberg both started their developments from classical physics only. – atom Jun 30 '16 at 06:16
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    @atom - well, you're just fundamentally wrong. What can I say? The people sociologically were building on their knowledge of classical physics because it was the approximate theory used in the previous 3 centuries. But the new theory doesn't logically depend on the previous one, just like special or general relativity isn't just a refinement of old Newton's mechanics, either. And yes, there are theories with no classical limit at all, e.g. (2,0) superconformal CFT in 6 dimensions on the conformal point. – Luboš Motl Jun 30 '16 at 06:40
  • One may heuristically guess the right form of a particular form of a quantum mechanical theory while using its classical limit. But the set of possible quantum theories is completely different than the set of possible classical theories, there's in no way a 1-to-1 map here. Also, in QFT, some classical theories don't allow any quantum cousin e.g. because of gauge anomalies. On the other hand, those that do admit a QFT are non-unique and all the higher-dimension operators reflect the ambiguities. If someone thinks classically about the fundaments, he's just not doing QM yet. – Luboš Motl Jun 30 '16 at 06:41
  • Since no one said the names yet: Groenewold and Van Hove – AHusain Jun 30 '16 at 06:49
  • @LubošMotl I see! So if observables are simply synonymous with Hermitian operators in quantum mechanics, could you elaborate upon the way in which classical observables arise in the $\hbar \to 0$ limit? For example, if we start with the premise that the momentum operator is defined by $\hat{p} = -i\hbar\frac{\partial}{\partial x}$, how is it that we obtain the classical notion of momentum $p = m\frac{\mathrm{d} x}{\mathrm{d} t}$ when we let $\hbar \to 0$? – David Zhang Jun 30 '16 at 07:28
  • Dear @DavidZhang - the formula $\hat p = m\cdot d \hat x / dt$ is exactly true (in the simple mechanical models) even in quantum mechanics - it's the Heisenberg equation of motion for $\hat x$ - so it obviously remains valid in any limit such as $\hbar\to 0$, too. If you formulated a different question, a different answer could be needed, but for your question, I gave you the obvious answer. – Luboš Motl Jun 30 '16 at 07:32
  • @LubošMotl I see. Is there a similar relationship that reveals how the Poisson bracket ${f,g}$ arises as an $\hbar \to 0$ limit of the commutator bracket $[\hat{A}, \hat{B}]$? Here, there is a critical piece of geometry built into the Poisson bracket (the symplectic structure on $T^*!M$) which I suppose must be hidden somewhere in the underlying quantum theory. – David Zhang Jun 30 '16 at 07:43
  • Yes, that's exactly the most meaningful "mathematical demonstration" of the limit. If one identifies functions $f,g$ on the phase space as the $\hbar\to 0$ limits of operators $F,G$ in a quantum theory, then the $\hbar\to 0$ limit of $[F,G]/i\hbar$ is equal to the Poisson bracket ${f,g}$. The symplectic structure on the phase space is nothing else than the information about the commutators of the corresponding operators. – Luboš Motl Jun 30 '16 at 07:46
  • This fact may be shown at various levels of assumptions. But if you doubt it, you should calculate the Poisson bracket and commutators of $x^a p^b$ and $x^c p^d$. You will always get $(ad-bc)$ times $x^{a+c-1} p^{b+d-1}$ plus subleading corrections, perhaps times $i\hbar$ for the commutator. The Poisson bracket gives you the same structure. So those agree. All "nice" functions of $x,p$ may be expanded in similar power series, so the conclusion is general. (The powers aren't the only way how to prove it, but a particularly transparent one.) – Luboš Motl Jun 30 '16 at 07:48
  • An even simpler demonstration of the relationship: for the flat-space phase space with coordinates $x_i$ which include both $q,p$. The commutator $[x_i,x_j]$ is exactly ($i\hbar$ times) the $ij$-antisymmetric tensor that defines the symplectic form. – Luboš Motl Jun 30 '16 at 07:50
  • @LubošMotl Wait, how is it that we perform the identification between functions $f,g$ on the phase space and quantum operators $\hat{F}, \hat{G}$? Given some Hermitian operator $\hat{F}$, is it always the case that $\hat{F}$ can be expanded in a power series in $\hat{x}$ and $\hat{p}$? (If the answer is "yes," then I follow everything afterwards.) – David Zhang Jun 30 '16 at 07:56
  • For the 1D QM Hilbert space, there are infinitely many ways to write a given operator in terms of functions of $\hat x,\hat p$. A trivial way to see the degeneracy: you may always add the factor of $\exp(2\pi [x,p]/\hbar)$ to any form of your operator because it's one. ;-) But it's in no way natural to try to write a given operator as a function of $\hat x,\hat p$ - this very temptation is just another proof that you're still thinking classically, not quantum mechanically. Operators are intrinsically matrices relatively to a basis on the Hilbert space. – Luboš Motl Jun 30 '16 at 08:01
  • Let us continue this discussion in chat. – David Zhang Jun 30 '16 at 08:03