What is the meaning of the word 'momentum' in quantum physics? (Especially in the problem that I will describe below.)

Question

In undergraduate quantum physics class that I am taking in this semester, I had to solve a problem about quantum Zeno effect. The problem describes the thought experiment which is trying to determine the position and the momentum of the electron simultaneously by continuously measuring the position of the electron as $x(t)$ and differentiating the function $x(t)$ to get the momentum as $m x'(t)$. (This is exactly what we did in quantum Zeno effect. Of course, $x(t)$ will be constant.) And then, the problem asks to identify whether the argument that we can determine the momentum by this way in the thought experiment is correct or incorrect. (Of course, the answer is obviously 'incorrect'.)

Is the above argument correct or incorrect? Provide your own reasoning to support it or dispute it. If you think the above is not correct, what is the correct way of deducing the momentum from the measurement of the position?

I understood the word 'momentum' in the last sentence as the quantum momentum described by the operator $-i\hbar\nabla$ and wrote the answer based on this. (My answer was, in summary, "there is no way to deduce momentum from the position measurement".) However, the professor and TA of the class asserted that the word 'momentum' means the expectation value of the momentum in the last sentence of this problem and came up with the answer which is to get the expectation value of the momentum as $\langle p\rangle = m \frac{\text d\langle x\rangle}{dt}$ from the measurement of the position. If the problem asked to find the way of deducing any information related to the momentum, then I might describe this way to deduce expectation value based on Ehrenfest theorem. But the problem clearly mentioned the word ‘momentum’ only as you can see above, hence I could not agree with the instructor’s answer.
My question is,

Does it make sense to understand the word 'momentum' in the last sentence of the problem as the expectation value of the momentum? (In personal, it is hard to accept this idea for me.)

Answer 1

In quantum mechanics, the momentum operator is a linear function $\hat{p}: \mathcal{H} \to \mathcal{H}$ on a Hilbert space. If the state $\psi \in \mathcal{H}$ is an eigenvector, then the eigenvalue is interpreted as the momentum of $\psi$. If $\psi$ is not an eigenvalue of the momentum operator, then the momentum of $\psi$ is the expectation value of the resulting linear combination, weighted by the amplitudes of each contribution.

I think it comes down to being precise with your definitions. If you want the observed momentum, then you are probably referring to the classical notion, in which case you need to take expectation values (this has a nice statistical interpretation). If you're referring to the quantum mechanical momentum, then you are probably referring to the operator.

To give you more context, what you wrote as $-i\nabla$ is the definition of the canonical or conjugate momentum of the particle, which is a mathematical object constructed by quantization of the classical phase space. In the case of a free particle, the two notions coincide (as shown in the proof of Ehrenfest's theorem). However, this is in general not true. A simple example is the coupling of an external electromagnetic field $A$ to the particle $\psi$, which is now charged with electric charge $e$. The momentum of the particle is the same, but its canonical momentum will now be $-i\nabla + eA$.