# I Momentum of a stationary particle/wave?

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1. Oct 9, 2018

### Philip Land We are all familiar with Heisenbergs uncertainty principle. When we determine the position of a particle or wave, the uncertainty of momentum reach infinity.

So lets say I have a machine that measures the position very very precisely. Then the uncertainty of this non-moving particles momentum reach infinity, the momentum could have big magnitude. But how can a particle that don't move have momentum? Or does the particle oscillate with a de broglie wavelength within the discrete finite volume we have measured it to be within (because we can not measure exactly)?

Or does the uncertainty principle don't apply once measurement is done? It just tells uncertainty if I "would" make a measurement?

Feel free to go complex on me.

2. 3. Oct 10, 2018

### A. Neumaier The uncertainty relation only applies to a fixed moment in time. The position of your particle will be very uncertain the very next moment.

4. Oct 10, 2018

### PeterDonis ### Staff: Mentor

Who said the particle wasn't moving? You measured its position, not its momentum. You can't say it isn't moving just because you measured its position very precisely one time.

5. Oct 10, 2018

### Philip Land Let me rephrase my question.

Let's say we trap it in a well with infinite potential such that the particle cant escape, the let the size of this well be small. Then I could take the limit until the particle no longer can move in space.

I can easily solve Schrödinger for this case, $\psi(x) = Constant * sin(k_n x)$ for even n, where $k=n \pi / 2a$. where a is half the length of a simple square well.

Decreasing $a$ makes the frequency very high.

As writing this I realize that the only thing that will happen if you trap a particle is that the energy / unit length (in this 1 dim case) will increase and we concentrate the energy which cannot escape.

But dosen't this also mean that now the position is always given at any time, hence we can never know the momentum of this trapped particle? Which is strange because it's fairly straightforward to calculate the energy-levels for each quantum number n.

6. Oct 10, 2018

### Nugatory ### Staff: Mentor

Yes, but the state isn't any one of those energy eigenstates. It's a superposition of many of them.

7. Oct 10, 2018

### PeterDonis ### Staff: Mentor

No, you can't, because an "infinite square well" with $a = 0$ is no longer a square well; it's an infinite potential everywhere, which has no solution.

8. Oct 10, 2018

### A. Neumaier For any size of the cavity, the particle would oscillate very fast inside the tiny cavity, so that its momentum is unpredictable.

9. Oct 10, 2018

### Nugatory ### Staff: Mentor

Original poster might also want to consider the negative delta function potential, which is as close as we can get to a particle that "can only be one place". This is not, of course, the zero-width limit of the infinitely deep square well, for which no solution exists (although solutions do exist for arbitrarily small non-zero widths).

10. Oct 11, 2018

### vanhees71 There's no way to define a momentum observable for this case anyway. You can define the Hamiltonian, but not the momentum observable. We have discussed this at length once in this forum. The infinite-square-well potential is not an easy problem despite it's treated by many textbooks as if it were. They never carefully explain the important difference between Hermitean and self-adjoint operators!

For a handwaving treatment, see

https://www.physicsforums.com/threa...uous-in-an-infinite-well.916171/#post-5773830

11. Oct 11, 2018 Staff Emeritus
Or maybe it's a delta function (which has one bound state).

This is an example of where one needs to be very careful about explaining not just that one is taking a limit, but exactly how they are taking the limit.

Classically, if I take ball bouncing between two walls and bring the walls closer together, the ball speeds up. In the limit that the walls get very close, the position of the ball is well determined, but its momentum is not.

12. Oct 11, 2018

### vanhees71 On top, it's a completely unphysical situation. I've never had a clue what these examples are supposed to teach me. It's a nice mathematical exercise, but no more. Even finite potential steps or wells are not very illuminating for me, although you can study easily some basic notions of scattering theory in 1D motion with them. From the particle-point of view, they don't help much. Note that such potentials imply $\delta$-distribution valued forces for the analogous classical problem.

13. Oct 11, 2018

### Philip Land This was indeed very tricky. For everyone interested in reading about this particular problem in detail, here's a pdf that goes in depth on self adjoint operators.

#### Attached Files:

• ###### Bonneau, G. - Self-adjoint extensions and the teaching of quantum mechanics (arxiv preprint).pdf
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14. Oct 12, 2018

### A. Neumaier Only if the cavity has hard walls. Momentum is perfectly well defined in a cavity with steep wall where position can take any value but absolutely large values have an extremely low probability. The prime example is a harmonic oscillator with very large stiffness and hence very large frequency.

15. Oct 12, 2018

### A. Neumaier It is an excellent approximation for a particle in a tiny cavity viewed from far away. The details of the cavity potental no longer matter, and the (renormalized) delta-potential says all there is to say. It is also the simplest (and hence useful) toy model where to study renormalization.

16. Oct 12, 2018

### vanhees71 Sure. It's much more valuable to study the harmonic oscillator in great detail than to deal with mathematically complicated artificial problems. Of course, you could as well argue that the harmonic-oscillator potential is also always an approximation...

17. Oct 12, 2018

### hilbert2 To me, the original question seems to be based on a belief that the property $\left<p\right>=0$ for a stationary state of 1D quantum system would imply that $\left<p^2 \right>$ and the variance of momentum should also be zero, which isn't true.

And I don't think the state that a quantum system collapses into upon a position measurement is very likely to be a stationary state. To make it become a stationary state you should measure the total energy observable instead.

18. Oct 12, 2018

### vanhees71 Despite the fact that I don't believe in the collapse religion, I don't understand your statement. Usually stationary states (eigenstates of the Hamiltonian) do not imply well localized particles (see the stationary states of the hydrogen atom as an example).

19. Oct 12, 2018

### A. Neumaier But there is no experiment that would allow you to do that....

20. Oct 12, 2018

### hilbert2 Ok, I think I misinterpreted the "stationary" in the threat title to mean "solution of time-independent SE", while it probably means "something with very small variance of position".

21. Oct 12, 2018

### vanhees71 That's how I also understand "stationary". That's why I was puzzled by your claim that then the particle's position should be pretty well determined. Of course, the OP #1 is a common misunderstanding of the meaning of the uncertainty relation. This is due to Heisenberg, who didn't understand his own uncertainty relation right in the beginning. He submitted a paper during Bohr's absence from Copenhagen, and the damage was done. Bohr immediately corrected the misunderstanding by Heisenberg, but it was too late, and the wrong interpretation stuck. Errors are particularly persistent by diffusing rapidly through the textbook literature. That's a kind of Murphy's law of science communication.

The correct interpretation according to Born's Rule and the minimal statistical interpretation is that the uncertainty relation is a general property of all quantum states of a particle, i.e., due to its preparation. If you prepare the particle well localized, i.e., with small $\Delta x$, then necessarily the momentum of the particle is pretty undetermined, i.e., $\Delta p$ is large (and vice versa). This has nothing to do with how accurately you can measure position or momentum, which you can do in principle with arbitrary precision (of course the technical difficulty rises with the demand of higher precision, but that's not a principle obstacle for measuring either position or momentum precisely).

Now take a free particle. Its energy $E=\vec{p}^2/(2m)$. So for a stationary state the momentum should be completely determined (which is impossible since the corresponding plane-wave solution of the Schrödinger equation is not a proper state, because it's not square integrable) and thus the position is completely undetermined. Physicswise this means to get an approximately stationary state for a free particle the standard deviation of momentum should be very small, implying through Heisenberg's uncertainty relation that the standard deviation of position must be very wide.

This is also true, if there are bound states, i.e., true stationary states like the electron in a hydrogen atom in its ground state. It's not very well localized too.