Momentum measurement paradox in a box

Question

We have seen that a momentum eigenstate can be expressed as sum of position eigenstates as: $${| {p} \rangle}= \int{Ae^{i(p/\hbar)\hat{x}}| {x} \rangle}dx$$ Where $A$ is a constant . Suppose we measure the momentum of a particle in a box and the wavefunction collapses to: $${| {p'} \rangle}.$$ We know that the particle cannot be found outside the box (infinite square well) . Therefore $${Ae^{i(p'/\hbar)\hat{x}}}=0$$ for any $x$ outside the box or at the boundary of the box. And the wavefunction also have to be continuous . The only way it can be guaranteed is by making $A=0$ but then it will be 0 everywhere inside the box also which is not physically possible . So can position be measured for a particle ina box?

EDIT:-

I recently noticed an issue with the phenomenon of wave function collapse . After collapse the wavefunction collapses to a single eigenstate of its respective observable like position or momentum or something else. This means that the wavefunction takes the form of a delta function in that observable space . But we all know that Delta functions are not continuous thus failing the law of QM that wavefunction needs to be continuous

Answer 1

The real answer is that the QM framework was designed for ideal situations of measurements at fixed, chosen times $t$. Real measurement situations don't work like that, and the state-after-measurement postulate is known to be in contradiction with experiment. The state after a momentum measurement is not actually a $|p\rangle$ eigenstate, although that might in some way approximate the state after measurement. There are a million issues that can be found with a perfect momentum eigenstate, one of those issues being that it predicts that the particle can be found on the opposite side of the universe faster than light could travel there. $|p\rangle$ gives equal probability everywhere in the universe in position space.

That being said, we don't just "measure momentum" by applying a "momentum measurement" at a fixed time to a particle in whatever state we want. There are a variety of momentum measurement techniques, all of which would require the particle to interact with something more complicated than just an infinite square well box. For example, if the particle is charged, you could let the particle fly out of the box and into a magnetic field, and measure the curvature of the path it flies along. Yes, this is fundamentally a position measurement, but that's how momentum is measured in real life. Actually this one is directly from the ATLAS inner detector.

So, in order to ask questions about states after measurement, you have to have an awareness of what your method of measurement is. Then you can respond just based on the physical system.

I think it's time to re-evaluate those von Neumann postulates & set up more truthful postulates that define the theory.

Answer 2

States in an infinite square well satisfy boundary conditions $\psi(0)=\psi(L)=0$. On the other hand, the state on which you are trying to project does not satisfy these boundary conditions. This is a contradiction in the setup, not in the result of a measurement (or the QM) - it means that your box is not really infinite (not when you measure at least).

In other words: the state on which we project/measure must be a possible state of the system, i.e., satisfying the same boundary conditions (but not necessarily an eigenstate of the Hamiltonian.) If it is not, it means that measurement modifies the system (as we indeed expect in QM), so one shouldn't make the claims based on what system was before the measurement.

Answer 3

I don't see why you think the ket for a state with definite momentum $|p'\rangle$ is a sum over some position kets $|x\rangle$. Positions can have any value from some real interval, so this ket is sometimes written as integral over all positions on the real line: $$ |p'\rangle = \int_{-\infty}^{\infty}|x\rangle~ e^{ip'x}\frac{dx}{\sqrt{2\pi}}. $$

Then, you're applying this ket as if it was a valid quantum state that the system can be in. This is not correct; the ket $|p\rangle$ is a mathematical object that we can work with in calculations, e.g. one can give a sense to equations such as

$$ \hat{p} |p'\rangle = p' |p'\rangle, $$ but it does not represent a realizable quantum state. The reason is this ket is not normalizable to 1, as the expression $\langle p' | p'\rangle $ is not defined.

Realizable quantum states describing a single system have to be normalizable to 1. (There are other normalizations where the psi function need not be normalizable to 1, but those are used e.g. as statistical description of a stream of infinite number of particles with finite arrival rate).

Not only is $|p'\rangle $ not a normalizable momentum eigenket, the momentum operator does not have such an eigenket (because of the continuous character of index $p'$). So we cannot apply the usual formulation of the projection postulate to measurement of momentum(and similarly for any continuous variable). The usual formulation works for discrete variables only, e.g. spin projection.

Instead, the best we can do, is to take the result of the measurement, which is some probability distribution for $p'$, or interval $p'-\Delta p/2,p'+\Delta p/2$, and collapse to some normalized state that obeys all constraints: it is zero outside the box, and the probability distribution of momentum it implies is consistent with this result of measurement. This means the state after momentum measurement will be a superposition of many different normalized states, implying a range of momenta.

If the potential barrier is infinite, the psi function outside has to remain zero. I'm not sure it is necessary to require it is continuous on the walls; but if it is, it has to vanish on the wall also inside, so we have the restriction $\psi(x) = 0$ for $x$ on a wall.