# Fundamentals of operators

I have some questions about operators in quantum mechanics. Hopefully these are similar enough that they be asked together.

1. We "derive" the operator for momentum in position space by proving that it works for plane waves. How does that show that we've chosen the correct operator for any given wavefunction?
2. Is the canonical commutation relation more fundamental than the forms of the momentum and position operators themselves? That is, should one (or could one at least) start from the canonical relation and then prove that $\hat r = r\hat{Id}$ and $\hat p = -i\hbar \nabla$?
3. Assuming the we've got the right forms of the position and momentum operators, why should it be that the same relationships that apply to the values of the observables should apply to the operators of the observables? For instance, how do we really know that the velocity operator $\hat v$ should be $\frac{d}{dt}\hat r$ or $\frac{\hat p}{m}$?
4. This is less conceptual and more computational, how do we show that $$\frac{d}{dt}\hat r = \frac{dr}{dt}\hat{Id} = \frac{\hat p}{m} = \frac{-i\hbar}{m}\nabla\ ?$$
• 1. Plane waves are a "basis" (although they do not live in the Hilbert space $L^2(\mathbb R)$, hence the quotation marks). 2. Yes they are. What you propose is the Schroedinger representation, that can be characterised as the unique irreducible representation of the CCR. 3. There is no "right" form, and those relation follows, e.g., by applying them to the functions of point 1, or by considering the expectation values. 4. See point 3. – Phoenix87 Sep 15 '16 at 21:15

(2) The answer to this question is a bit philosophical; see my answer to Uncertainty Relations, Conjugate Quantities, and Fourier Transforms. But if you want to sweep the subtleties under the rug, then the short version of the answer is yes, the canonical commutation relations are more fundamental. For example, $\hat{X} \to x$ and $\hat{P} \to -i\hbar\, \nabla$ is not the only valid representation of the commutation relations in position space; for example, $\hat{P} \to f(x) - i\hbar\, \nabla$ for any function $f(x)$ would work as well. This is called a "gauge transformation;" it would change the wave function, but it turns out that if you're careful, it doesn't actually change any observable physical quantities.
(1) There are many ways to "derive" the momentum operator, depending on your teacher's taste. Theoretically, you can choose any operator that satisfies the canonical commutation relations - the choice $-i \hbar \nabla$ is usually the simplest, but is not necessary.
(4) Your equation is not quite correct: in the Schrodinger picture $\hat{r}$ is a time-independent operator whose time derivative is zero. You need to add expectation values with respect to a specific quantum state: $d\langle \hat{r} \rangle/dt = \langle \hat{p} \rangle / m$. The result can then be proven by Ehrenfest's theorem, as mentioned above.