Momentum measurement and uncertainty

[hokhani]

TL;DR

How to measure momentum of a confined QM particle ?

Theoretical measurement of momentum of a QM particle is equivalent to action of momentum operator on the particle wave function. What is the experimental apparatus for this measurement? I would like to know if we confine a particle in a narrower region, how uncertainty principle results in less exact measurement?

 

[weirdoguy]

how uncertainty principle results in less exact measurement?

Uncertainty is not about how exact is a single measurement. Uncertainty is defined as: $\Delta p=\sqrt{\langle\hat{p}^2\rangle-\langle\hat{p}\rangle^2}$, so for a single measurement it gives 0. You have to have at least two measurements to make it non-zero. And how exact these measurements are is a technical question, not dircetly related to uncertainty principle. In principle you can make it very exact, even though uncertainty will be large - that's the answer to second question.

And I don't know the answer for your first question, since I'm a theoretical physicists, who hates everything experimental related.

 

[anuttarasammyak]

https://www.feynmanlectures.caltech.edu/III_02.html#Ch2-S2　might be of your help. Momentum is wave number multiplied by Dirac constant.

[edit]
In high energy physics experiments scientists do tracking the spatial trajectory and bending radius of subatomic particles inside a magnetic field to measure their momentum vectors accurately.

Scientists use Compton effect of X ray or gamma ray to explore Fermi surface in momentum space.

 

[hokhani]

Uncertainty is defined as: $\Delta p=\sqrt{\langle\hat{p}^2\rangle-\langle\hat{p}\rangle^2}$, so for a single measurement it gives 0.

I think this equation says that by a single measurement you would obtain the value of momentum in the interval $\Delta p$ around $\langle p \rangle$ while you belive it gives $\Delta p=0$!

 

[weirdoguy]

Because it gives zero for single measurement. Do the math, it's simple. For single measurement: $\langle\hat{p}\rangle^2=p^2=\langle\hat{p}^2\rangle$. What do you get when you insert that in the definition of uncertainty?

I think this equation says

It says no such thing. It's the definition of uncertainty.

 

[hokhani]

Because it gives zero for single measurement. Do the math, it's simple. For single measurement: $\langle\hat{p}\rangle^2=p^2=\langle\hat{p}^2\rangle$. What do you get when you insert that in the definition of uncertainty?

From theoretical point, in a single measurement, the wave function collapses on one of the momentum eigenstates, say $|p_o\rangle$. Do you mean that for this eigenstate we have $\Delta p=0$?

 

[Lord Jestocost]

TL;DR: How to measure momentum of a confined QM particle?

First of all, one needs a meaningful, physical definition of the term “momentum of a confined quantum mechanical particle.”

 

[anuttarasammyak]

Uncertainty is defined as: Δp=⟨p^2⟩−⟨p^⟩2, so for a single measurement it gives 0

I have thought that <> for uncertainty relation means average of large number of measurement data, ideally infinite.
[edit] more precisely <> for the Kennerd inequality(corrected)

 

[anuttarasammyak]

First of all, one needs a meaningful, physical definition of the term “momentum of a confined quantum mechanical particle.”

Is that more difficult to get than definition of the term “position of a confined quantum mechanical particle”? Fourier transform seems saying that they are dual.

 

[kuruman]

TL;DR: How to measure momentum of a confined QM particle ?

Theoretical measurement of momentum of a QM particle is equivalent to action of momentum operator on the particle wave function. What is the experimental apparatus for this measurement? I would like to know if we confine a particle in a narrower region, how uncertainty principle results in less exact measurement?

(Emphasis mine)
As far as I know, the only experimental apparatus that can make a theoretical measurement is found in a thought experiment.

You might find this reference helpful for visualizing (theoretically) the tradeoff between position and momentum measurement.

 

[weirdoguy]

Do you mean that for this eigenstate we have Δp=0?

Do you understand the definition of uncertainty? Do you know what variance is? What standard deviation is? I gave you the only definion we need, why won't you calculate yourself?

You are a victim of bad popularizations, which say a lot of nonsense about uncertainty principle. When to understand it one only needs to know what standard deviation is. In Poland it is taught in high-school! Standard deviation of single number is zero, per very definition. To use uncertainty principle you need A LOT of measurements performed on a particle that has always been prepared in the same state. THEN you can calculate uncertainty. The more measurements, the closer it should be to theoretically calculated value.

 

[weirdoguy]

I have thought that <> for uncertainty relation means average of large number measurement data, ideally infinite.

Yes, it means average of measurement data. And for only one measurement, this average is the measured value. Hence uncertainty is zero.

 

[PeterDonis]

Uncertainty is not about how exact is a single measurement.

"Uncertainty" as in the uncertainty principle is not even about the variance of a large number of measurements of a single observable, as your post #2 is claiming. Uncertainty is about the product of the variances of a large number of measurements of two non-commuting observables (for example, momentum and position) on an ensemble of identically prepared systems.

You are a victim of bad popularizations, which say a lot of nonsense about uncertainty principle.

Unfortunately, your post #2 also gives incorrect information about the uncertainty principle. See above.

 

[PeterDonis]

if we confine a particle in a narrower region, how uncertainty principle results in less exact measurement?

Mathematically, the uncertainty principle sets a lower limit on the product of the variance in position and the variance in momentum. Confining a particle in a narrower region means reducing the variance in position, and that must result in an increase in the variance in momentum.

Note that, as I said in post #13, this is not about making a single measurement less exact, at least not as far as the uncertainty principle is concerned. It is about the variance in a large number of measurements on an ensemble of identically prepared systems. For example, if we prepare a large number of particles, all confined in a narrow region of the same size, and then make momentum measurements on all of them, the uncertainty principle sets a lower limit on the variance of those momentum measurements, based on the fact that the variance in position cannot be larger than the size of the narrow region each of the particles is confined in. If we call that size $\Delta x$, and the variance in the momentum measurements (after we do a large number of them and do the statistics) $\Delta p$, then the uncertainty principle says that $\Delta x \Delta p \ge \hbar$.

Note also that this is not a claim about how the uncertainty principle gets "enforced"--what is going on "behind the scenes" to make $\Delta p$ obey the above inequality. It's only a claim about what you will find when you do the statistics.

 

[PeterDonis]

From theoretical point, in a single measurement, the wave function collapses on one of the momentum eigenstates

This has nothing to do with the uncertainty principle, because we are not looking at the probabilities of results of future measurements on the same particle. We are looking at the variances of position and momentum for single measurements on a large number of particles.

 

[weirdoguy]

Unfortunately, your post #2 also gives incorrect information about the uncertainty principle. See above.

Unfortunately, I'm not talking about uncertainty principle besides this sentence:

And how exact these measurements are is a technical question, not dircetly related to uncertainty principle.

"Uncertainty" as in the uncertainty principle is not even about the variance of a large number of measurements of a single observable

I gave explicitly the definition of what I (and most textbooks) call uncertainty, and to use it in experimental context (as OP wants) one calculates the variance of a large number of measurements of single observable. And then move on to measure another observable, and calculate variance.

What about it is incorrect? I think you are overcorrecting. Especially when your next post agrees perfectly with what I wrote.

 

[PeterDonis]

I'm not talking about uncertainty principle

That was not at all clear, and the OP question is about the uncertainty principle.

 

[PeterDonis]

I gave explicitly the definition of what I (and most textbooks) call uncertainty

Statistical uncertainty, which is not the same as the "uncertainty" part of the uncertainty principle. I'm not sure the OP grasps that distinction, which is all the more reason to make an extra effort to be very clear about it.

 

[FactChecker]

I have thought that <> for uncertainty relation means average of large number measurement data, ideally infinite.
[edit] more precisely <> for Robertson inequality

That sounds wrong or unclear. The average of many may have a very small uncertainty, while the uncertainty (variance/spread) of the individuals that make up the average is large.
I am only a casual amateur on this subject, but the latter is how I interpret the uncertainty principle.

 

[PeterDonis]

The average of many may have a very small uncertainty, while the uncertainty (variance/spread) of the individuals that make up the average is large.

I'm not sure what you mean by this.

the latter is how I interpret the uncertainty principle.

As has already been pointed out, the uncertainty principle does not constrain the variance of measurements of a single observable. It only constrains the product of the variances of two non-commuting observables.

 

[kuruman]

As explained to me in QM class decades ago, the uncertainty principle is inherent in the wave-matter mathematical superposition. Expectation values, variances and "many measurements" are peripheral to this basic idea. How so?

The figure on the right is lifted from the link in post #10. At the very top, the packet is quite well localized in $x$. If someone asked you, "look at the top left frame and tell me where the particle approximately is", you would easily be able to find a region $\Delta x$ where the particle is likely to found. However, as seen in the top right frame, that requires the superposition of many wavenumbers $k.$

As one move down the figure and use fewer momenta in the superposition, the packet becomes increasingly delocalized. The region $\Delta x$, within which the packet approximately is, increases.

At the very bottom, the packet looks like an almost-perfect sinusoidal in position space which means that it is well localized in momentum space. However, if someone asked you, "look at the bottom left frame and tell me where the particle approximately is", you would have difficulty figuring it out.

The rest is details and I see no proverbial Devil in them.

 

[PeterDonis]

the uncertainty principle is inherent in the wave-matter mathematical superposition.

This intuitive picture works nicely for position and momentum. But the uncertainty principle is not limited to position and momentum. It applies to any pair of non-commuting observables--for example, spin measurements in different directions. For other observables, the nice wave packet visualization unfortunately doesn't work. You have to fall back on the math, which is simply that the minimum uncertainty is given by the non-vanishing commutator of the observables.

 

[Matterwave]

Statistical uncertainty, which is not the same as the "uncertainty" part of the uncertainty principle.

Could you elaborate on what exactly you mean with the "uncertainty" in quotes? Do you mean the product itself? For example $\Delta x\Delta p$?

 

[kuruman]

This intuitive picture works nicely for position and momentum. But the uncertainty principle is not limited to position and momentum. It applies to any pair of non-commuting observables--for example, spin measurements in different directions. For other observables, the nice wave packet visualization unfortunately doesn't work. You have to fall back on the math, which is simply that the minimum uncertainty is given by the non-vanishing commutator of the observables.

Agreed. However, OP's question was about position and momentum.

 

[PeterDonis]

Could you elaborate on what exactly you mean with the "uncertainty" in quotes? Do you mean the product itself? For example $\Delta x\Delta p$?

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.

Each $\Delta$ by itself can be measured (with caveats, see below) by making measurements of that observable on a large number of identically prepared systems and computing the variance. This could be called the "statistical uncertainty" for that observable by itself (but see further comments below). The wave function that is prepared will make a prediction for what that $\Delta$ should be. And if you do it for both observables, the wave function will make a prediction for both $\Delta$s, and therefore for their product. The quantum uncertainty principle says that it is impossible for any wave function to predict a product of the two $\Delta$s that is less than $\hbar$ (for the case of position and momentum).

Note that in any real experiment, the actual statistical variance might be larger--in some cases much larger--than what would be predicted from the wave function alone. That's because real experiments have many other limitations that can be much coarser than the limitations imposed by QM alone. 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.

 

[anuttarasammyak]

TL;DR: How to measure momentum of a confined QM particle ?

I would like to know if we confine a particle in a narrower region, how uncertainty principle results in less exact measurement?

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.

 

[anuttarasammyak]

From theoretical point, in a single measurement, the wave function collapses on one of the momentum eigenstates, say |po⟩.

Momentum, coordinate eigenstate kets are not normalized so they are not realizable physical state. Quantum field theory tells us that attempt of some extremely precise measurements, e.g., position measurement of an electron in narrower measure than its Compton wavelength, make a mess on the system.

 

[PeterDonis]

Momentum, coordinate eigenstate kets are not normalized so they are not physical state.

Yes, but we can still use them mathematically to help understand how actual physically realizable states like wave packets work.

Quantum field theory tells us that attempt of extremely precise measurement make a mess on the system.

This is much too vague. If you have some specific math and a reference, please give it. Otherwise, please do not clutter the thread with vague statements like this.

 

[FactChecker]

I have thought that <> for uncertainty relation means average of large number of measurement data, ideally infinite.
[edit] more precisely <> for the Kennerd inequality(corrected)

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?

 

[PeterDonis]

Do you mean the sample variance of the sample distribution of a large number of measurement data, or the variance of the sample average?

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?