Uncertainty Principle in the context of tracking particle paths

[WWCY]

TL;DR Summary

How does momentum-position uncertainty come into play in a detector (eg http://cms.web.cern.ch/news/muon-detectors) that allows you to trace out a particle's path?

I have a bit of confusion regarding the application of the uncertainty principle in the context of experiments.

If a detector allows you to measure a particle's path through said detector, does that mean that you know a particle's position at all points in time, and are able to work out its momentum at any point by analysing its trajectory as a function of time?

 

[PeterDonis]

Have you looked at the position uncertainty of the detectors? And tried to figure out what limits the uncertainty principle sets on the precision to which position and velocity can be known, given the mass of the muon?

 

[Nugatory]

... does that mean that you know a particle's position at all points in time, and are able to work out its momentum at any point by analysing its trajectory as a function of time?

Only to the limits of accuracy implied by the uncertainty principle. The detectors do not give us an exact position.

 

[PeterDonis]

Only to the limits of accuracy implied by the uncertainty principle.

Actually, I think real detectors are usually orders of magnitude less accurate than the limit imposed by the uncertainty principle. I think people often fail to realize just how small Planck's constant actually is.

 

[WWCY]

Thank you for your response

Have you looked at the position uncertainty of the detectors? And tried to figure out what limits the uncertainty principle sets on the precision to which position and velocity can be known, given the mass of the muon?

No I have not, I have only just started to read up a little on particle physics, and this was the first bit of confusion that jumped out at me.

Also, how would device uncertainty affect the position-momentum uncertainty inherent in a particle?

Finally, are the concepts you mentioned covered in a "standard" particle physics text? Thank you.

 

[WWCY]

Only to the limits of accuracy implied by the uncertainty principle. The detectors do not give us an exact position.

Does this mean that (in layman terms) the non-exact position measurements allow us to measure momentum up to some precision and thus gives us a particle-trajectory with some uncertainty?

How does this apply to muons in cloud chambers though?

Thank you.

 

[Nugatory]

How does this apply to muons in cloud chambers though?

The cloud chamber won’t nail down the position with any more precision than the size of a single droplet, and that is an enormous uncertainty at the scale that we’re working with here.

 

[PeterDonis]

how would device uncertainty affect the position-momentum uncertainty inherent in a particle?

What do you mean by "the position-momentum uncertainty inherent in a particle"?

 

[WWCY]

What do you mean by "the position-momentum uncertainty inherent in a particle"?

$\Delta x$ and $\Delta p$ in the Heisenberg relation are uncertainties related to the particle itself, rather than any measurement device, is this right?

If so, why would a device's spatial measurement resolution affect the particle's momentum measurement uncertainty? Which seems to be what your initial reply seemed to imply, if I haven't misinterpreted it.

Thank you for your patience.

 

[PeterDonis]

$\Delta x$ and $\Delta p$ in the Heisenberg relation are uncertainties related to the particle itself, rather than any measurement device, is this right?

They are both. They are properties of the particle's wave function, but the particle's wave function tells you about the probabilities for different possible results of measurements, so they are also properties of different possible measurements you could make. However, those "measurements" are highly idealized ones in which the only limitation is that imposed by the uncertainty principle. Real measurements almost never come anywhere close to that level of accuracy. See below.

why would a device's spatial measurement resolution affect the particle's momentum measurement uncertainty?

Because if your device can only measure position to some particular accuracy, then that also limits the accuracy to which your measurements give you the time derivative of position, i.e., velocity, which is how you are measuring the momentum.

Which seems to be what your initial reply seemed to imply, if I haven't misinterpreted it.

You have misinterpreted it (although there is also an answer, given above, to what you thought I was implying). What I was saying is that the uncertainty in measurements like the ones on muons you described, due to purely classical properties of the measuring devices (things like the sizes of droplets in the cloud chamber that @Nugatory referred to, for example), are typically orders of magnitude larger than the uncertainty due to the uncertainty principle, so in practice the uncertainty principle is irrelevant to the actual uncertainties in the actual measurements. So the muon detectors, in practice, are not "measuring the particle's path" to anywhere near the kind of precision they would have to for the uncertainty principle limitations to come into play when you take the derivative of the particle's measured position with respect to time to get a measured velocity.

 

[PeroK]

Summary: How does momentum-position uncertainty come into play in a detector (eg http://cms.web.cern.ch/news/muon-detectors) that allows you to trace out a particle's path?

I have a bit of confusion regarding the application of the uncertainty principle in the context of experiments.

If a detector allows you to measure a particle's path through said detector, does that mean that you know a particle's position at all points in time, and are able to work out its momentum at any point by analysing its trajectory as a function of time?

There's another common misconception about the HUP that the uncertainty is in terms of individual measurements of position and momentum. The uncertainty in the HUP is the standard deviation of a hypothetical set of measurements of an ensemble of identically prepared particles.

Let's take as an example the single slit. (If you have time you might like to watch this lecture on the HUP. The single slit experiment starts from about 32 mins. )



If the particle passes through the slit, the slit acts as a position measurement in the transverse direction. This implies a position measurement for every particle with a small standard deviation ("uncertainty") in that direction, which implies a large standard deviation in a subsequent measurement of momentum in that direction. The narrower the slit, therefore, the more the particles tend to diffract. The point at which they impact the screen is effectively a measurement of their momentum/velocity in the transverse direction.

Now, nothing is stopping you measuring each individual point on the screen as accurately as you like. The HUP does not stop you measuring the momentum of each particle. Theoretically, you could give the momentum of each particle to any desired accuracy.

What the HUP is saying is that you get a wide range of momentum measurements. Not that each of your momentum measurements themselves must have some sort of measurement error in them (*).

You might like to think about what would happen if you make your cloud chamber more and more refined in terms of tracking the position of a particle. Like the single slit, beyond a certain point of accuracy you would start to see diffraction rather than a "crude" classical trajectory.

(*) Note, finally, that in orthodox QM a particle does not have a defined position until it is measured. If you have an ensemble of identically prepared particles (all in the same initial state), compare the following two statements:

1) They all have the same position (really) but you get an error in position measurements that leads to different recorded positions. This is wrong.

2) They each have the same probability of being found in a given position at a given time. The position measurements result in a probabilistic range of values, the standard deviation of which represents the uncertainty in position implied by the initial state. This is what orthodox QM says.

 

[vanhees71]

Also, how would device uncertainty affect the position-momentum uncertainty inherent in a particle?

Finally, are the concepts you mentioned covered in a "standard" particle physics text? Thank you.

I think this common misunderstanding to the uncertainty principle reaches back to Heisenberg's unfortunate misunderstanding of his own discovery. He wrote the first (partially flawed) paper in Copenhagen, but at the time Bohr wasn't around, and that's why there is this confusion. Bohr had to correct Heisenberg, and unfortunately some textbookwriters obviously didn't realize it.

The misunderstanding is that the uncertainty relation refers to the accuracy of measurements and to the perturbation of the measured system by the measurement. This is utterly wrong! The uncertainty relation refers to properties of the state of the system and thus to the preparation of a system in a given state. It's very general since all it needs to be derived is that observables are represented by self-adjoing operators and that the scalar product of a Hilbert space is positive definite.

If $A$ and $B$ are observables, represented by the self-adjoint operator $\hat{A}$ and $\hat{B}$ then for any (pure or mixed) state
$$\Delta A \Delta B \geq \frac{1}{2}|\langle [\hat{A},\hat{B}] \rangle|,$$
where the expectation values are calculated with the state the particle is prepared in, which is represented by a statistical operator $\hat{\rho}$, cf.
$$\langle f(\hat{A}) \rangle=\mathrm{Tr} (\hat{\rho} \hat{A}).$$
The standard deviations are defined by
$$\Delta A^2=\langle \hat{A}^2 \rangle - \langle \hat{A} \rangle^2.$$
Let's discuss the meaning using the position component $x$ of a particle and the momentum component in the same direction $p$. Then we have $[\hat{x},\hat{p}]=\hbar \hat{1}$, and the uncertainty relation reads
$$\Delta x \Delta p \geq \frac{\hbar}{2}.$$
This tells us that if the particle is prepared in a state with a pretty accurate localization, i.e., such that $\Delta x$ is pretty small, then necessarily $\Delta p$ is pretty large, i.e., not very well determined. So the Heisenberg uncertainty relation tells you about certain limits of accurate determination of two non-compatible observables by state preparation.

This doesn't tell you anything about the accuracy with which you might be able to measure the one or the other observable. This is not determined by the state the particle is prepared in but by the construction of the measurement apparatus. You can always, at least in principle measure the particles position or its momentum as accurately as you like. So you can choose to measure very accurately the position to figure out whether its standard deviation is really $\Delta x$ for a given state you prepared the particle in. For that purpose you just have to use a large enough ensemble of equally such prepared particles and make a position measurement that is much more accurate than the uncertainty given by the standard deviation $\Delta x$ due to the particle's preparation in the given state.

Of course there is indeed always some disturbance of the measured system by the measurement since the system must interact with the measurement device to perform this measurement, but this disturbance is much less simple to analyze and it's not such a general property of the QT formalism as the Heisenberg uncertainty relation since it must take into account the individual properties of the measured system and the used measurement device.

Of course, also measurement devices are after all quantum (many-body) systems and thus also are subject to the general laws described by QT, and particularly also to the uncertainty relations. Thus the position measurement of a particle is of course at least limited by the possibility to localize the measurement device itself, and as the uncertainty principle for position and momentum tells you, you can localize the measurement device in principle as accurate as you like, but $\Delta x=0$ is impossible, i.e., there's always some uncertainty in measuring the position of the particle (the same of course also holds for momentum).

Though very difficult to hit this fundamental quantum limit in accuracy is very hard since usually measurement devices, as macrscopic many-body systems, are very hard to control, and there's always thermal motion (fluctuations) which contribute much more to the uncertainty of a position or momentum measurement than the quantum limit, today there are some devices that reach it. One example are the gravitational-wave detectors. Here some limit is the quantum fluctuations of light, and today they even use socalled "squeezed states" (a generalization of the coherent states):

https://arxiv.org/abs/1109.2295
https://www.repo.uni-hannover.de/handle/123456789/3807

 

[WWCY]

Hi all, thanks for your responses! I've come to believe that my understanding of the Heisenberg principle is flawed, I shall go over each of your posts in detail.