# Momentum of a stationary particle/wave?

• I
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.

entropy1

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.

vanhees71
PeterDonis
Mentor
2020 Award
lets say I have a machine that measures the position very very precisely. Then the uncertainty of this non-moving particles momentum reach infinity

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.

vanhees71
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.
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.

Nugatory
Mentor
Which is strange because it's fairly straightforward to calculate the energy-levels for each quantum number n.
Yes, but the state isn't any one of those energy eigenstates. It's a superposition of many of them.

PeterDonis
Mentor
2020 Award
Then I could take the limit until the particle no longer can move in space.

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.

A. Neumaier
Then I could take the limit until the particle no longer can move in space.

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

Nugatory
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).

bhobba and Philip Land
vanhees71
Gold Member
For any size of the cavity, the particle would oscillate very fast inside the tiny cavity, so that its momentum is unpredictable.
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

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

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.

vanhees71
Gold Member
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.

weirdoguy

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• Bonneau, G. - Self-adjoint extensions and the teaching of quantum mechanics (arxiv preprint).pdf
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A. Neumaier
There's no way to define a momentum observable for this case anyway.
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.

mrinal arandhara
A. Neumaier
it's a completely unphysical situation. I've never had a clue what these examples are supposed to teach me.
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.

vanhees71
Gold Member
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.
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...

hilbert2
Gold Member
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.

vanhees71
Gold Member
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).

A. Neumaier
To make it become a stationary state you should measure the total energy observable instead.
But there is no experiment that would allow you to do that....

hilbert2
Gold Member
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).

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".

vanhees71
Gold Member
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.

mrinal arandhara, dlgoff and hilbert2
But there is no experiment that would allow you to do that....
That's how I also understand "stationary"

I have one essential question. The uncertainty in momentum and space, is that the "size" of the interval we can find it within OR number of potential states in the given interval? Because I assume the former in my original question below:

If we restrict the space that a particle can be in, this case by a theoretical infinite square well. The uncertainty in position is no longer big, since the particle has to be in the well. And the smaller we make the well, the more accurately we know its position without actually measuring it.

This mean that uncertainty in momentum gets bigger as we shrink the well.

Hence the momentum (##\alpha## Energy) could be 'extremely' large. How do we explain a particle that is in rest (since it must me in the very small well) having almost infinite momentum? Where does all that momentum comes from? Vacuum fluctuation we have 'forced' the presence of?

I do know that high uncertainty in momentum does not equal high momentum, but potentially high momentum. So the case above is possible. And sorry for my abuse of language, just trying to make this as intuitive as possible.

Nugatory
Mentor
How do we explain a particle that is in rest (since it must me in the very small well) having almost infinite momentum?
As several posters have already pointed out, the particle is NOT at rest. You can imagine it bouncing back and forth between the walls of the well as a way forming an intuitive picture of how this might be.

Be aware, however, that this classical image is misleading in several ways. First, it suggests that the particle has a position within the box and a momentum even when neither of these quantities have been measured. Second, it suggests that reducing the width of the well to zero would give you an immobile particle with a fixed position, and that doesn't work - set the width to zero and there is no solution.
Where does all that momentum comes from?
The total momentum of the system comprised of the particle and the measuring device is conserved.

bhobba and Philip Land
vanhees71
Gold Member
I have one essential question. The uncertainty in momentum and space, is that the "size" of the interval we can find it within OR number of potential states in the given interval? Because I assume the former in my original question below:

If we restrict the space that a particle can be in, this case by a theoretical infinite square well. The uncertainty in position is no longer big, since the particle has to be in the well. And the smaller we make the well, the more accurately we know its position without actually measuring it.

This mean that uncertainty in momentum gets bigger as we shrink the well.

Hence the momentum (##\alpha## Energy) could be 'extremely' large. How do we explain a particle that is in rest (since it must me in the very small well) having almost infinite momentum? Where does all that momentum comes from? Vacuum fluctuation we have 'forced' the presence of?

I do know that high uncertainty in momentum does not equal high momentum, but potentially high momentum. So the case above is possible. And sorry for my abuse of language, just trying to make this as intuitive as possible.
The standard deviation of a quantity ##A## is defined by
$$\Delta A^2=\langle A^2 \rangle-\langle A \rangle^2.$$
This is a socalled cumulant of the distribution function. If you know all cumulants or moments, you can reconstruct the distribution function. If you have only the mean value and the standard deviation, you have incomplete knowledge about the distribution function. Then information theory tells you that the best you can do with the given information is to use a Gaussian distribution with this mean value and standard deviation. This then tells you that the probability to find a value for ##A## in an interval of twice the width of the standard deviation is around 68%. Have a look at the first figure in

https://en.wikipedia.org/wiki/Standard_deviation

As I already stressed, the infinite-square well is a very bad example to argue about localizability and the position-momentum uncertainty since you cannot define a proper momentum observable in this case.

The most simple case, you should carefully study, are Gaussian wave packets, where you can easily tune the position uncertainty (uncertainty=standard deviation; it's just the jargon of the Copenhagen gang trying to confuse the issue) and then calculate the standard deviation ("uncertainty") of momentum. Then you can also analytically solve the time evolution via the time-dependent Schrödinger equation. I'm sure that with this exercise, you'll get a pretty good understanding what's going on concerning the position-momentum uncertainty relation.

Philip Land
A. Neumaier
How do we explain a particle that is in rest (since it must me in the very small well) having almost infinite momentum?
Did you ever solve a problem with a square well? In the most general state the expectation of the position and the velocity oscillates with a superposition of very high frequencies, corresponding to the energy levels obtained by diagonalizing the Hamiltonian. Thus the particle is never at rest though it is confined to a very small region, and the expected momentum (mass times velocity) is almost always huge and oscillating in sign.

bob012345
Gold Member
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
Can you explain the physical implications of the difference between Hermitian and self-adjoint operators. The terms seem to be used interchangeably. There may be different mathematical nuances in linear algebra but don't they both refer to operators with real observables? Thanks.

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