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

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.)

• It's difficult to address this without knowing what you did. My problem is the opposite of yours. I can't imagine what you did if you did not calculate expectation values. Concerning "no hint", what do you make of the phrase "by continuously measuring the position of the electron"? Apr 19 '21 at 11:36
• This is the same with the thing that we do in quantum Zeno effect. The state of the electron will remain in the position eigenstate if we make continuous measurement of position. However, the point of my question is not about quantum Zeno effect, but the problem that I had to solve. Apr 19 '21 at 12:00
• Welcome to this community! May I suggest you take a look at the second part of this answer and the literature referred there? (apologies for linking my own answer). Apr 19 '21 at 13:34
• @pglpm I really appreciate your kindness to introduce helpful summary to understand the meaning of observable and measurement more clearly!! But still, I’m not sure about the answer to my original question... Apr 19 '21 at 13:58

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.
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 $$-i\nabla + eA$$.
• In the context of your problem, I think the observed momentum is intended (since you're doing measurements). A natural question you should ask is "how does uncertainty affect my measurements?" Indeed, when you are measuring position, you cannot assume every measurement is infinitely precise (this violates the uncertainty principle). The only resolution is to take expectation values, after which you can use the formula $\langle p \rangle = m\partial_t \langle x\rangle$ to obtain the expectation of momentum. Apr 19 '21 at 14:33