How did the operators come about?

Question

This relates a little bit to my previous question (Experimentally, what categorizes a measurement as corresponding to a certain observable?), but it's different in a way and more historical.

One of the postulates of quantum mechanics is:

To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics (Sherrill).

However, absent in the list of postulates is what these operators are. So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms? Specifically, where did these expressions come from?

Maybe the question can better be answered historically. If I had to guess, I'd say it began with the wavefunction. Some scientists thought perhaps the squared modulus of the wavefunction corresponded to charge distribution or something, but Born was the one who got it right. He essentially made a guess that the squared modulus corresponded to the probability of finding a particle at a certain location. What's funny is that he initially wrote down the wrong expression in his paper (Born), and his paper even got rejected by the first journal he submitted it to, but the right expression is in a corrective footnote. But again, Born simply guessed what the squared modulus meant. There was no procedure; he just thought of an idea and then experiment verified it to be correct.

Thus:

$$\langle\psi|\hat{X}|\psi\rangle = \langle\psi|x\rangle\langle x|\hat{X}|x\rangle\langle x|\psi\rangle = \int d\vec{x}\ \psi^*(\vec{x})\ \psi(\vec{x})$$

So basically in the position basis, the position operator is simply $\hat{X} = x$, which seems pretty intuitive.

I believe the momentum operator came about next; and it was "derived" by fiddling around with the de Broglie relation and just substituting classical expressions for momentum into the Schrodinger equation (can't remember where I read that). The fact that plugging these classical ideas into a quantum equation just happens to match experiment for momentum measurements seems like a pure coincident to me (it's along the same lines as trying to make the Schrodinger equation relativistic; one way of shoehorning relativity into it gives you the Dirac equation and a different way gives you the Klein-Gordon equation).

I imagine the spin operator came about by simple observation. An electron goes up or electron goes down in a magnetic field. Not hard to guess what the expression for that operator might be.

So anyway, am I on the right track here? Have all of the operators come about simply by guessing? And if I'm wrong, how did they actually come about? How would I come up with a new operator for a new type of measurement?

Answer 1

Quantum mechanical postulates

So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms?

The mathematical expression for each individual operator is sort of a postulate, but it should not be listed. The postulates define a (more or less) complete theory in that I can derive mathematical facts from it without really knowing anything about the physics.

One of the main problems is that the site you link to (following McQuarrie) is in my opinion a very bad site for understanding the postulates of quantum mechanics as they nowadays understodd and which let you penetrate deeper into the theory. Your postulates are in between the first postulates of quantum mechanics, when people were still figuring out the basics and current versions of the postulates that allow for better separation of mathematics and physics.

So let me first state them in a more modern form and explain the differences (I'm copying some of this from Wikipedia):

• Postulate 1: Each physical system is associated with a (topologically) separable complex Hilbert space $H$ with inner product . Rays (one-dimensional subspaces) in $H$ are associated with states of the system.

The difference in this with your postulate 1 is that we do not refer to "space". The wave function is an object in an abstract Hilbert space, not necessarily some "object" $\psi(x,t)$ with position and time. Your first postulate already incorporates the notion of "space" and thereby needs to have a few operators "fixed" (see further down). Let's go on:

• Postulate 2: Physical observables are represented by self-adjoint linear operators on $H$. The expectation value (in the sense of probability theory) of the observable $A$ for the system in state represented by the unit vector $|\psi\rangle\in H$ is $\langle\psi| A |\psi\rangle$.

There we go. This is your postulate 2 and 4 combined. Your postulate 2 is the first sentence, the Born rule is essentially the second system.

• Postulate 3: The Hilbert space of a composite system is the Hilbert space tensor product of the state spaces associated with the component systems.

Now, your system doesn't have such a postulate, if I see correctly. If you already operate in space - and since each particle lives in the same space - you do not need this postulate. However, it's easier to consider each particle with a separate wave-function in a separate Hilbert space and combine them by introducing the tensor products (for interactions, see below). Note that this can no longer be uphold in quantum field theory, as particle number is not conserved, but this is unimportant here.

This way, you can easily accomodate spin into the theory, which is not possible with your wave-function $\psi(x,t)$. It turns out that to properly describe spin, you have to introduce another parameter that your wave-function depends on. That's not very nice and means that your postulates will not work proberly. The abstract picture has no such problem.

• Postulate 4: The time evolution of the state is given by a (weakly-)differentiable function from the real numbers, representing instants of time, to the Hilbert space of system states. This map is characterized by the Schrödinger equation.

This is your postulate 5. Postulate 6 is not present - it cannot be derived from this set of postulates, but if we enhance quantum mechanics to quantum field theory (note that axiomatic QFT is problematic), then it can be, so I don't really want to focus on this and leave it out.

Operators in quantum mechanics

Note that in this description, we do not even define "space". The above theory describes a very abstract theory. To what does the position operator correspond? Well, we don't know, because we don't even know what "space" means. However, this describes a full theory and I can derive mathematically meaningful statements from it. For example, I can say that by virtue of postulate 2 (in principle) the Hamilton operator in the system must itself be a self-adjoint operator, because time evolution must preserve $\langle \psi |\psi \rangle$, as this constitutes a probability measure. However, what about physics? The idea of these postulates is that for any experiment, you now have to specify:

• the Hilbert space
• the Hamilton operator
• the possible observables

only if you do that, you can start calculating anything at all. This is somewhat reminiscent of classical theory, where you would normally first define the phase space (maybe, your system is constraint to a one-dimensional motion such as a pendulum. In this case, your phase space would have only one $x$ and one $p$-coordinate). Then, you go on and define the Lagrangian or Hamiltonian according to your problem and then you can calculate everything. You have to do the same in quantum mechanics.

This means also that a priori, there is no need whatsoever to have the square of the momentum operator be represented by the negative Laplacian (for example). And - guess what - it's also not true in any system (see e.g. discrete systems, where a "momenumt operator" can still make sense). Let's make an example. Suppose you want to consider a particle in a three dimensional box. It is somehow intuitive to take the Hilbert space $L^2([0,1]^3)$ of functions in the box $[0,1]^3$. The position of the particle should then be just the "position" $x$ in the box, i.e. the position operator as you know it. However, there is a mathematical theorem that tells us that all infinite dimensional (separable) Hilbert spaces are the same, so I could also go on describe the particle in a the box $[0,1]^3$ by a function in the Hilbert spae $L^2(\mathbb{R})$. This would of course be highly counterintuitive. What would the position operator look like? Probably different than the "intuitive" one.

This is why the operators are not posutlated: They are part of what constitutes the actual physical system you want to consider! The specific mathematical expression of an operator is therefore not independent of how you choose to describe the theory and therefore, they do not form postulates.

However, given your postulates here again, this is no longer true. The first postulate identifies a specific Hilbert space, which basically fixes (for most systems) how the position and momentum operators should look like. This might explain your confusion and hopefully helps you in your quest of understanding quantum mechanics better!

How to find and attribute operators?

However, it doesn't really answer the real underlying question: How do you actually find these observables? And what about the history?

As I said, for any system, you have to find the triple Hilbert space, Hamiltonian, Observables. All of this must somehow come from physics. And here is, where your historical picture is somewhat accurate (and this relates also to Mark Mitchison's comment):

Finding Hilbert space, Hamiltonian and observables is usually/often done by analogy. Your historical picture gives the right idea in that it tells a story of how people coming from classical mechanics arrive at the observables by analogy.

An example: Given a classical particle in a box, you can write down a Hamilton function in classical mechanics. For a quantum particle, knowing that it has something like "momentum" and "position", you want to take a space that looks rather similar to the classical configuration space, hence $L^2([0,1]^3)$ seems a very good choice. Choosing the position operator as you know it is then a "natural" choice, knowing what "position" means in the classical picture. You then choose the momentum operator such that the basic commutation relations are fulfilled. These commutation relations are "postulated" from the Poisson brackets in Hamiltonian mechanics. In short, it turned out that in classical mechanics, we have $\{p,q\}=1$ with the Poisson brackte, and in quantum mechanics, we have $[P,Q]=i\hbar$ with the commutator. This leads to postulate the rule "take any Poisson bracket to a commutator modulo $i\hbar$", which is not well-defined, but works pretty well in defining operators for many systems (this is called canonical quantization). This is "identifying operators by analogy", but of course, it can only work if also the Hilbert space is identified by analogy.

What to do with observables like spin that do not have a classical counterpart? Well, in this case the abstract postulates are very powerful: They allow you to just make something up. People knew that the spin works somewhat like angular momentum so it makes sense to postulate that there is a "spin-part" in the Hilbert space that is just $\mathbb{C}^2$ and the spin-measurements are just $\sigma_i$, the Pauli-matrices. Since the experiments confirm your choice, all is well.

Answer 2

I believe that the important thing to remember is the basic difference between quantum and classical description of a system. In classical mechanics you describe it by defining some sort of equation of motion which gives you a position of all the constituents of the system described by function of time but for each of these particles you have one function. In the quantum case there is no such thing as a well defined position and the main object is a wave function or state function which describes what is a probability for a certain configuration to take place. When you act upon this system with any kind of measurement you change it so you get a new state. If you consider a state to be a vector you can surely say that you are acting on a vector and then producing a new one. It is the same thing an operator does. Main operator in a Schro equation is, I guess, Hamiltonian operator or energy operator. It is called that way because its eigenvalues are obviously energy values.After that, you can proceed with momentum or something else. I have to point out that measurements define what an operator does to the basis, and by this description, operator is defined id est, its action upon any kind of vector is defined. I hope that it is clear what I wanted to say. If anyone has more to say, I would be grateful too.. :-)