5
$\begingroup$

I'm studying quantum mechanics without ever properly having studied analytical mechanics. Thus, I lack some basics to build upon, which is something I'm trying to fix. In particular, I realized the canonical commutation relations are something coming from classical mechanics rather than quantum mechanics. The thing is that I don't have a feeling for what it actually means/represents and why one needs that. What's the reasoning behind promoting position and momentum to hermitian operators obeying those Heisenberg commutation relations?

$\left[\hat{x}_{i},\hat{x}_{j}\right]=0,\quad\left[\hat{p}_{i},\hat{p}_{j}\right]=0,\quad\left[\hat{x}_{i},\hat{p}_{j}\right]=\boldsymbol{i}\delta_{ij}$

And what exactly is an operator in the classical sense? What does it mean to quantize something? What's the need for the commutator? I'm looking for a dedicated answer that can give me a lot of intuition. Thanks for considering it.

$\endgroup$
1
  • 1
    $\begingroup$ There must be multiple questions on why $[x,p]=i\hbar$... You should start there. $\endgroup$ Commented Jun 1, 2017 at 18:36

1 Answer 1

14
$\begingroup$

The answer(s) to your question(s) could easily fill a book (and in fact have). In short, quantization is a heuristic procedure that, given a classical system (Usually taken to be a Lagrangian or Hamiltonian theory of particles and or fields), produces a quantum theory (A Hilbert space and a Hamiltonian operator, more generally time evolution or even a unitary representation of the Symmetry group of spacetime (Gallilei-group in the non-relativistic case and Poincaré group for a relativistic theory)). In my opinion it is impossible to understand what quantization is whithout understanding the very basics of analytical mechanics, since one always starts from a classical system.

There are many "quantization methods" and even having chosen one such method one usually can't apply it to completely arbitrary classical systems without running into problems or ambiguities. Few of these methods are formulated in a mathematically rigorous manner as being a well defined "map" from "classical systems" to "quantum systems". The idea is to try and guess a quantum theory that has a certain classical limit and obeys the same (or closely related) symmetries as that classical limit (i.e. the system one wants to quantize). It is a problem of modeling: Given some real world phenomenon find its quantum description. Our intuition works fairly well in suggesting a classical description so we usually start from there rather than guessing the quantum system right away. A quantization procedure can be called successfull if it produces a mathematically well-defined quantum theory with predictions that match experiments.

It is worth noting that in general it is not obvious what "the classical limit" of a given quantum system is supposed to be. Naively, the limit is $\hbar\to 0$ but that is certainly not the full story; a somewhat more meaningfull attempt at defining such a limit would be large energies and the presence of decoherence, but that is a different topic. Suffices to say, the classical system should be the classical limit of your quantized version of it, whatever your notion of classical limit. Clearly quantum mechanics is more general than classical mechanics,so many different quantum systems might have the same classical limit.

The relation between commutators and Poisson brackets is best understood via "deformations of algebras" but that topic is somewhat technical and I can't give a good description of it due to lack of knowledge. Most physicists are not familiar with those details. It is the next step to what is called deformation quantization. The most commonly used heuristic procedure of Canonical Quantization involves identifying canonical coordinates and the conjugate momenta of a system (to understand what these are you can't get around learning some analytical mechanics), promoting them to operators on some Hilbert space to be guessed later and replacing the poisson bracket relations between them $$\{q_i,p_j\}=\delta_{ij}$$ by $$\frac{1}{i\hbar}[\hat{q}_i,\hat{p}_j]=\delta_{ij}.$$ Many introductory textbooks only cover this method. It "works" (can be attempted) for theories of particles as well as classical field theories.

Guessing the Hilbert space systematically amounts to finding all Hilbert spaces that have operators satisfying those commutation relations. This is called looking for representations of the Heisenberg algebra. The Stone-von-Neumann theorem says that for systems of particles (which have finitely many degrees of freedom as opposed to fields, which have infinitely many) all reasonable choices of such a Hilbert space are equivalent. This is why usually no great care is taken in this choice. Choosing the Hilbert space of square integrable functions over your space (parametrized by the possible values of the $q_i $'s) is usually good enough and gives you wave function pictures to play with. For field theories, inequivalent representations exist but one still doesn't care about them in most physics courses.

Effectively, this means that we replace Poisson brackets $\{,\}$ by commutators $\frac{1}{i\hbar}[,]$ and put hats on the variables. Then we guess the Hamiltonian operator by putting hats on the $p$'s and $q$'s in the expression for the Hamiltonian function. Here there is an ambiguity in terms of ordering to deal with, since in the Hamiltonian function the $p$'s and $q$'s are just numbers and commute while the operators do not. Usually in introductory courses one does not delve too deeply into either the meaning or general implications of this procedure and just applies it to examples where it works. The mathematical approach trying to systematize this is called geometric quantization and of course goes much further in terms of concepts. Although the formalisms of classical mechanics and quantum mechanics looks somewhat similar in these respects, the conceptual meaning of the equations is completely different, as are the mathematical tools employed. A way to represent classical mechanics in a mathematical formalism closer to quantum mechanics is called "Koopman–von Neumann classical mechanics".

A relation between the classical and the quantum pictures can be to some extent seen from the concepts of "symmetry" or "conserved quantity" as well as "generator of symmetry". In classical (Hamiltonian) mechanics, observables are functions of the canonical coordinates and conjugate momenta (which are all together coordinates for phase space, the space of all possible states the system may take). Consider observables $A(q_1,\dots,q_n,p_1\dots p_n), B(q_1,\dots,q_n,p_1\dots p_n)$. The Poisson bracket is defined as $$\{A,B\}:=\sum_{i=1}^n\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial B}{\partial q_i}\frac{\partial A}{\partial p_i}$$ and is itself again a function on phase space; an observable just like $A,B$. A symmetry transformation is some reasonable function mapping the phase space into itself.

Let's take the example of a single particle flying around in space. Its state at any given time is described by positions and momenta $(x,y,z, p_x,p_y,p_z)$. These variables (coordinates of phase space) can be thought of at the same time as observables, e.g. $$y[(x_0,y_0,z_0, p,q,r)]=y_0.$$ Now to say that "the canonical momentum $p_x$ generates translations in $x$-direction" is heuristically understood via $p_x$'s action on an observable $A(x,y,z,p,q,r)$ via the Poisson bracket. Suppose $\epsilon$ is a very small number. Then $$A(x+\epsilon,y,z,p,q,r)\approx A(x,y,z,p,q,r)+\epsilon\frac{\partial A}{\partial x}(x,y,z,p,q,r)=A+\epsilon \{A,p_x\}.$$

Momentum generates translations, position generates "boosts", angular momentum generates rotations in this way (there is much more to be said about this in terms of vector fields and flows). The Hamiltonian function (an observable) generates time translations, as expressed in the equation $$\frac{d A}{dt}=\{A,H\}+\frac{\partial A}{\partial t}$$ (where so far I only considered observables, which do not explicitly depend on time, so the last partial derivative term would be zero). Continuous symmetries (continuous here means that it makes sense to "perform the transformation only a little bit". For example you can translate or rotate something a little bit. You cannot mirror something a little bit.) give rise to conserved quantities (this is called Noether's theorem). By above equation a conserved quantity $A$ does not depend on time if it does not explicitly depend on time and also $\{A,H\}=0$. If your hamiltonian is "translation invariant" or "rotation invariant" etc. you get momenta, angular momenta, etc. as conserved quantities.

In the quantum picture, the observables are operators on a Hilbert space and the commutator is $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$ with composition of linear maps (matrix multiplication in finitely many dimensions). Nevertheless, here still the position operator generates boosts, the momentum operator generates translations, the angular momentum operator generates rotations and the Hamiltonian generates time evolution. The analogon of the above evolution equation is called the Heisenberg equation in Quantum mechanics and is easily obtained by the "quantization replacement" $\{\}\mapsto\frac{1}{i\hbar}[ \ ]$: $$ \frac{d\hat{A}}{dt}=\frac{1}{i\hbar}[\hat{A},\hat{H}]+\frac{\partial \hat{A}}{\partial t}$$ where again most observables don't explicitly depend on time in practice, so the last partial derivative term is zero.

Suppose you have an observable $\hat{A}$ and a state $\psi$. The expectation value of $\hat{A}$ in the state $\psi$ at a time $t$ is given as $$\langle\psi|\hat{A}|\psi \rangle.$$ Now suppose you move everything in $x$-direction by $\epsilon$ again. Then $$\langle\psi_{\text{moved}}|\hat{A}|\psi_{\text{moved}} \rangle\approx \langle\psi|\hat{A}|\psi\rangle+\epsilon \langle\psi|\frac{1}{i\hbar}[\hat{A},\hat{p}_x]|\psi\rangle.$$ in direct analogy to the above consideration for classical translations. In quantum mechanics, a conserved quantity $\hat{A}$ is an operator with no explicit time dependence satisfying $[\hat{A},\hat{H}]=0$. An important point to make is that $[\hat{A},\hat{B}]\neq 0$ implies that it is not always possible to measure both observables at the same time in a given state (there is an uncertainty relation associated to those observables!). In classical mechanics you can measure anything you want and $\{A,B\}\neq 0$ has no obvious intuitive meaning I know off.

After these general remarks, let me answer some of your questions directly: Classical mechanics has no operators, except in the Koopman–von Neumann formulation mentioned above, which is probably the only meaningfull answer to "What is an operator clasically". The reason for promoting position and momentum to hermitian operators with those commutation relations is, to some extent, that it works and gives you a theory that matches experiment and has many analogies to the formulas of classical mechanics, even if these are merely visual analogies (the equations looks similar and mean fairly different things). Deeper conceptual insigts can be found associated to some of the key words I gave, none of which are usually covered in detail in introductory texts.

There are many other quantization methods and I want to mention at least two. The most important one after canonical quantization is path integral quantization, which takes a somewhat different approach not relying on commutation relations and involving some very beautifull concepts coming from the physical intuition that "quantum particles take any possible path and the result is the sum of each possible path result". Another related method is called stochastic quantization where one in some sense tries to "quantize a differential equation by artificially introducing noise into it" via a stochastic process. (pardon the uninformed oversimplification)

$\endgroup$
1
  • 1
    $\begingroup$ Nice answer. Just to add: the procedure replacing poisson brackets with commutators and adding hats does not work for general canonical variables, probably only for cartesian coordinates. Already a point transformation to cylindrical coordinates seems problematic. $\endgroup$
    – lalala
    Commented Jun 1, 2017 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.