Momentum measurement and uncertainty

[anuttarasammyak]

Do you mean the sample variance of the sample distribution of a large number of measurement data, or the variance of the sample average? Wouldn't the Central Limit Theorem apply to the average and give it a small variance as the sample size grows?

For infinite set of measurement x data collected from infinite independent labs where scientists do single time measurement for equally prepared systems, x^n averages are $$<x^n>=\int \psi^+ x^n \psi dx$$
with
$$\int \psi^+\psi dx=1$$
I expect there exists such $\psi$.
Does it answer your question?

 

[FactChecker]

Suppose I prepare a large number of particles in the same state (i.e., same wave function), and then I measure the momentum of each one of them and do statistics. Which of the two things you describe here am I calculating?

"do statistics" is vague.
If you determine the variance of the single random momentum value of a single sample (possibly using a large sample to estimate the distribution), that is one thing. But if you "average" several sample momentum results and talk about the variance of that average, that is different.

 

[FactChecker]

For infinite set of measurement x data collected from infinite independent labs where scientists do single time measurement for equally prepared systems, x^n averages are $$<x^n>=\int \psi^+ x^n \psi dx$$
with
$$\int \psi^+\psi dx=1$$
I expect there exists such $\psi$.
Does it answer your question?

Thanks. The terminology is tricky. That is the expected value for a single sample from that distribution. That is not the same as the average of several results, which would be expected to have a much smaller variance. I will assume that in all the posts, "average" meant what you say, and leave this discussion to others.

UPDATE: I have decided that I do not know how to interpret this and will leave this discussion to others.

 

[bhobba]

I don't have Ballentine handy at the moment, but he gives an excellent treatment of the uncertainty relations. It is well worth reading. Even Heisenberg got it wrong, and Bohr had to correct him, so some confusion is quite possible.

Thanks
Bill

 

[anuttarasammyak]

I observe that the Kennard inequality for variances of conjugate observables is well established. As for errors-disturbances there seems some progress since Heisenberg and still in progress or controversial.
An example https://journals.aps.org/pra/abstract/10.1103/PhysRevA.88.022110

 

[PeterDonis]

If you determine the variance of the single random momentum value of a single sample (possibly using a large sample to estimate the distribution), that is one thing

What do you mean by "sample" here? What I described was preparing a large number of particles in the same state, and measuring the momentum of each of them. Is that entire set of particles a "sample"?

 

[PeterDonis]

"do statistics" is vague.

What I meant was, we have a large number of results of momentum measurements on the particles that were prepared as I described, and we compute the mean and variance of that set of numbers.

 

[Matterwave]

That product is what appears in the quantum uncertainty principle. But each $\Delta$ by itself is not constrained by that principle; only their product is.

Yes, this much is clear.

From a purely terminological perspective, I would call each $\Delta$ an "uncertainty" while the uncertainty principle constrains a product of uncertainties $\Delta x \Delta p$ of non commuting operators (giving $x$ and $p$ as examples). Would you agree with this terminology?

 

[hokhani]

That's why the uncertainty principle is an inequality--it gives the minimum possible product of the two $\Delta$s, assuming that the only thing that contributes to those $\Delta$s is the unavoidable quantum uncertainty from the wave function. If there are other contributions as well, that can make the product larger.

To approve further, the variance of wave functions changes by time and so the product of the two $\Delta$s may change by time.

 

[hokhani]

You can get momentum wave function from coordinate wave function of the confined state. That is a straightforward way to know all about momentum of the system. After that you may be able to investigate uncertainty of your favor, variances, measurement error, etc. Uncertainty relation might be philosophical principle of QM, but we don’t have to use it in mathematical construction of QM.

As a peculiar thing, suppose that we have constricted a particle in a 1D box with the length $L$. By measurement of the momentum, the particle state collapses to one of the momentum eigenstates, say $|p_0\rangle$ which is extended in all the infinite region while the particle itself is confined in the box!

 

[Matterwave]

...which is extended in all the infinite region while the particle itself is confined in the box!

This is a self contradictory statement. A particle can't be everywhere and also confined to a box. :)

Food for thought: if you have an infinite square well of length L, you will find that a momentum eigenstate is not a valid state of the system as it does not obey the relevant boundary conditions.

 

[hokhani]

This is a self contradictory statement. A particle can't be everywhere and also confined to a box. :)

Food for thought: if you have an infinite square well of length L, you will find that a momentum eigenstate is not a valid state of the system as it does not obey the relevant boundary conditions.

To resolve this problem, Born-Von-Karman boundary conditions is used.

 

[FactChecker]

What do you mean by "sample" here? What I described was preparing a large number of particles in the same state, and measuring the momentum of each of them. Is that entire set of particles a "sample"?

Yes.

 

[anuttarasammyak]

As a peculiar thing, suppose that we have constricted a particle in a 1D box with the length $L$. By measurement of the momentum, the particle state collapses to one of the momentum eigenstates, say $|p_0\rangle$ which is extended in all the infinite region while the particle itself is confined in the box!

Minor comment: ket $|p_0>$ is not normalized state so it is not physical state. We have to be satisfied with the states whose momentum wave function lies only in a narrow region such as $[p_0-\delta,p_0+\delta]$ where $\delta$ can be small as we like but not zero.

Main comment: Coordinate wave function of such prepared state distributes outside the box, yes. The particle got energy by interacriton with measurement apparatus so it can go beyond the box wall energy barrier. By measurement some component of the original state is picked up and this picked up state has different energy from the original in general. Inputting enegy by measuring apparatus to the observed system is necessary for the pick up.

 

[vela]

As a peculiar thing, suppose that we have constricted a particle in a 1D box with the length $L$. By measurement of the momentum, the particle state collapses to one of the momentum eigenstates, say $|p_0\rangle$ which is extended in all the infinite region while the particle itself is confined in the box!

You're assuming the usual momentum operator is valid for a particle confined to a finite region.

 

[Matterwave]

To resolve this problem, Born-Von-Karman boundary conditions is used.

Unfortunately you can not resolve a logical contradiction via boundary conditions.

 

[PeterDonis]

From a purely terminological perspective, I would call each $\Delta$ an "uncertainty" while the uncertainty principle constrains a product of uncertainties $\Delta x \Delta p$ of non commuting operators (giving $x$ and $p$ as examples). Would you agree with this terminology?

I think that's reasonably standard terminology, yes.

 

[PeterDonis]

As a peculiar thing, suppose that we have constricted a particle in a 1D box with the length $L$. By measurement of the momentum, the particle state collapses to one of the momentum eigenstates, say $|p_0\rangle$ which is extended in all the infinite region while the particle itself is confined in the box!

No, if the particle is confined in a box, the momentum eigenstates are different. You have to actually solve the Schrodinger Equation to see what they are. The eigenstates $e^{ipx}$ are only momentum eigenstates for a free particle, not confined anywhere.

 

[PeterDonis]

To resolve this problem, Born-Von-Karman boundary conditions is used.

No, those boundary conditions are for a particle on a lattice, not a particle confined to a finite region.

 

[PeterDonis]

Thread closed for moderation.