Question About Momentum and Position Operators and the Postulates of Quantum Mechanics

Question

I've surmised that there are four big facts about the relationships between the position and momentum operators, their Hilbert spaces, and their eigenstates in QM. I think I am just about at a point where -- given one of them -- I'm able to derive the other three. I suppose my question is which of them is fundamental? None of the four are explicitly postulates of QM, so I'm wondering which of the four are axiomatic, take-it-on-faith statements. The four facts I'm referring to are:

1. In position space, momentum eigenstates are plane waves $\psi_p(x)=e^{ikx}$.

2. Position and momentum wave functions are related by Fourier transforms $\psi(p)=\int_{-\infty}^{\infty}\psi(x)e^{ikx}dx\Rightarrow\psi(x)=\int_{-\infty}^{\infty}\psi(p)e^{-ikx}dp$.

3. The momentum operator in position space is given as $\hat{p}=-i\hbar\frac{\partial}{\partial{x}}$.

4. The momentum operator is the generator of translations in position space $e^{i\hat{p}\Delta{x}/\hbar}|x\rangle=|x+\Delta{x}\rangle$.

Obviously (1) is a direct consequence of (2), and by solving a differential equation (3) can easily be derived from (1). I haven't derived for myself yet how (4) follows from any of the others which is what I'm trying to do now and what actually led me to this question. I don't know that (2) can be rigorously derived from (1), but assuming it can, it seems to me that the most likely candidate to be an axiom of QM is (1). Am I correct? Some things I've read have suggested to me that it's actually (4) that is fundamental. Does it even matter in the end?

Answer 1

Out of the 4 statements you've given, (3) and (4) are the most "fundamental"... they are equivalent to just the commutation relation $[x,p]$ which is the most common to take as a postulate of the theory. This means that you can replace $[x,p]=i\hbar$ with your choice of (3) or (4) in the postulates, and still get the same theory.

From there, see that (3) implies (1) by solving the eigenvalue problem:

$$\hat{p} \psi_p(x) = p \psi_p(x) \implies -i \hbar \frac{d}{dx} \psi_p(x) = p \psi_p(x) \implies \psi_p(x) = e^{ipx/\hbar}$$

and then (1) can be used to show (2):

$$\psi(p) = \langle p|\psi \rangle = \int_{-\infty}^\infty \langle p|x\rangle\langle x|\psi\rangle dx = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty e^{-ipx/\hbar}\psi(x) dx$$

Finally, as already noted, (3) and (4) are equivalent so regardless of your choice, the other one follows. This proof is a bit longer and can be found here.

As for your last paragraph there are some things I wouldn't agree with, so I answered each part separately:

Obviously (1) is a direct consequence of (2)

Starting from the postulates, you would actually get (1) before (2), and it would be an ingredient in proving (2). That is because you need the expression $\langle x|p\rangle = \frac{1}{\sqrt{2 \pi \hbar}}e^{ipx/\hbar}$ in order to switch between position or momentum space representation.

by solving a differential equation (3) can easily be derived from (1).

I think you meant that (1) can be derived from (3)?

I haven't derived for myself yet how (4) follows from any of the others which is what I'm trying to do now and what actually led me to this question.

The proof in the reference I gave above demonstrates this nicely.

I don't know that (2) can be rigorously derived from (1)

I showed this above for $\psi(p)$. Doing this for $\psi(x)$ works the same way, just starting from $\psi(x) = \langle x | \psi \rangle$ instead, and inserting the identity in momentum eigenstates rather than position eigenstates.

Does it even matter in the end?

In this case yes, not all of these are equivalent. But between (3) and (4) you can choose your favorite, or else replace either one with $[x,p] = i\hbar$

Answer 2

1. This is a result. No one tried to make momentum eigenstates planewaves.

2. This is a property. The derivation is the following:

$$\psi(p)=\langle p| \Psi \rangle = \int\langle p | x \rangle \langle x | \Psi \rangle dx = \int \frac{1}{\sqrt{2\pi}}e^{-ipx}\psi(x) dx $$

Where in the second equality, I inserted the identity operator (and using natural units).

3. This is my go-to choice, and what was historically made. It comes from Planck's equation for a photon: $\vec{p}=\hbar \vec{k}$. The thing is, a defined frequency or $k$-vector is only for pure sine waves (or plane waves). One way to extract the frequency of a plane wave is to take the spacial derivative:

$$-i\frac{\partial}{\partial x} e^{ikx}=k e^{ikx}$$

This means that the $k$-vector is an eigenvalue of the derivative operator. Using Planck's Equation, we can show that the momentum operator is:

$$-i\hbar\frac{\partial}{\partial x} e^{ikx}=\hbar k e^{ikx} = pe^{ixk}$$

And so the momentum is an eigenvalue of this operator. By using this definition, you can extend the concept of frequency (and so momentum) to other functions.

4. This is an argument. Some people would like to use it, but personally, it looks more like a nice way to gut-check we used the correct momentum operator. I don't understand it completely (How being a translation generator is like being conserved as a result of symmetry), but from what I've seen, the people who use it say it works. I think the other comment points that out.

5. I would like to add another definition: $[x,p]=i\hbar$. Even though this is derived, because our understanding of quantum mechanics has expanded, we know that there is uncertainty between momentum and position. This is how QFT starts. Nevertheless, this is a result in quantum mechanics and not a definition.

Answer 3

Instead of explicitly postulating properties of "position space" or "momentum space" or talking about Fourier transforms, it is much more common in axiomatic treatments to - in the spirit of canonical quantization where we turn classical Poisson brackets into quantum commutators - just axiomatize that we have two operators $x$ and $p$ on a Hilbert space with the canonical commutation relation $[x,p] = \mathrm{i}\hbar$.

Due to the Stone-von Neumann theorem, this then directly implies that we have a position space and a momentum space of wavefunctions in which these operators act as we're used to, the Fourier relationship and all the rest.