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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?


marked as duplicate by ACuriousMind quantum-mechanics Jun 30 '16 at 13:14

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  • $\begingroup$ 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 .. $\endgroup$ – Abhishek Pal Jun 30 '16 at 4:50
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    $\begingroup$ Related: physics.stackexchange.com/q/68686/2451 and physics.stackexchange.com/q/181078/2451 . Weyl quantization is e.g. discussed in this Phys.SE post. $\endgroup$ – Qmechanic Jun 30 '16 at 9:30

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.

  • $\begingroup$ 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". :) $\endgroup$ – 299792458 Jun 30 '16 at 5:59
  • $\begingroup$ 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? $\endgroup$ – David Zhang Jun 30 '16 at 6:01
  • $\begingroup$ @TheDarkSide - No, I wanted to write exactly this tautology. $\endgroup$ – Luboš Motl Jun 30 '16 at 6:04
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Jun 30 '16 at 6:07
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    $\begingroup$ @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. $\endgroup$ – Luboš Motl Jun 30 '16 at 6:40

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