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(I asked a very similar question already, but the core idea is very different in both)

Here are the two equations Im concerned with

$$\psi = \sqrt{2\over a}\sin\Big(n\pi {x\over a}\Big)$$

$$E = {n^2\hbar^2\pi^2\over 2ma^2}.$$

If we have a particle with some energy in the infinite potential well, then we get a probability density function which doesnt change with time.

So, if we put an electron in such potential well, then wouldn't it move? I mean if it has energy then it must be moving, and hence have some velocity. And the probability of finding it as certain places should change with time. But the equations dont seem to pertain to any kind of motion of electron. The equations just say that electron can be here or there.

Is the electron even moving?

I mean, is it like a ball bouncing between two walls?

OR

A ball between two walls that teleports around in space (based on its probability equation)?

Is velocity even a thing then? Is there any velocity associated with our electron in the 1D potential well?

Qmechanic
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Rohit Shekhawat
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    In what state did you notionally insert your particle in the box? An eigenstate of energy, or not? – Cosmas Zachos Apr 19 '23 at 14:45
  • In which part of the world is this considered a 'homework' question? Well, whatever that part it, they are actually giving good homework questions. – Rohit Shekhawat Apr 19 '23 at 17:14

2 Answers2

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The best way to think about this problem is only through the wavefunction viewpoint. In quantum mechanics, the electron "is" its wavefunction. The first equation you list is the energy eigenfunctions in position basis, and the second are the corresponding energy eigenvalues. For the infinite square well potential, there are discrete energy eigenstates labeled by the integer $n$. If your electron is in an energy eigenstate of the square well hamiltonian, then its wavefunction will not evolve with time.

Let's consider the ground state, $n=0$ wavefunction: $\psi = \sqrt{\frac{2}{a}}\sin(\pi \frac{x}{a})$. This function fits 1/2 of a period of a sin wave into the box and has no nodes. The probability amplitude for the particle is: $|\psi|^2$. This tells you the probability of finding the particle at any point in the box. Thus, until you make a position measurement, the particle is "everywhere" in the box with a nonzero probability. If you measure the position, the wavefunction collapses to a delta function in position space--a spike located at one value of $x$ and the momentum becomes completely unspecified.

Essentially, you should not be thinking about this problem with an electron particle bouncing around in the box. This is misleading. Thinking in terms of wave functions makes quantum mechanics more intuitive.

If you are interested in momentum, you can measure the particle's momentum rather than its position via the momentum operator (in position space), or you could compute the expected value of the momentum for a given eigenstate with the momentum operator.

The particle won't have a defined position or momentum until you make a measurement. But, you can write its wavefunction in either position or momentum basis and then compute the probability density function from there--from which you can compute the probability of the particle having a certain position or momentum.

klippo
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A free particle plane wave solution is moving, but its probability density function does not change with time.

``if it has energy then it must be moving" is false. It can have just rest mass energy and no more.

What you really can ask is if the electron has a momentum. If you want to ask that, then you need to express the wavefunction in terms of momentum eigenstates. I am not sure if there is complication coming from the fact that this system is bounded in space and thus cannot have an infinite plane wave. But if we proceed naïvely, then yes, the expansion is possible.

You should think hard about Cosmas Zachos's comment.

You can also compute the probability current.

naturallyInconsistent
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