# How do you measure momentum?

1. Jul 7, 2012

### Fiziqs

I realize that this would probably be patently obvious if I understood the nature of particles, but how is it possible to measure the momentum of a particle with just one measurement? Doesn't determining velocity require multiple measurements?

2. Jul 7, 2012

### Fiziqs

I'm sorry, but I have been giving this problem some further thought, and as oftentimes happens, my thought processes have become confusing, but please have the patience to bear with me.

It seems to me that it takes two, and only two measurements to determine the velocity of a particle. The measured momentum only applies between the last two measurements. The momentum going forward from any measurement becomes unknown, and can only be determined with another measurement. This is due to the fact that the measurement affects the momentum. So we can only know what a particle's momentum was at the time that we measured it. After that its momentum becomes unknown again.

But it occurs to me that the same can be said for the particle's position. We only know what the particle's position was at the time that we measured it. Like the momentum, a particle's position is affected by the measurement, so that we only know the particle's position at the moment that we measure it. After that, just as with momentum, its position becomes unknown again.

But it seems to me that this means that HUP is wrong. We can know the position and momentum of a particle at the same time. If we make a measurement of the particle's position at point "A" and then we measure its position again at point "B" then we know both its momentum and its position at point "B". Any change in either position or momentum will occur at exactly the same time. So if one argues that we no longer know its momentum because the measurement has changed it, then the measurement has changed its position as well, so in fact we don't know either one. The only thing that we do know is what its position and momentum were at the point that we made the measurement.

So isn't it true that when we make the measurement at position "B" we know both the particle's position and its momentum? And that after that we don't know either one?

If it seems like I'm trying to say that I have discovered something that no other physicist has noticed, that's not what I'm saying at all. What I'm saying is, what in the heck have I missed? I know that there's something simple that I'm just too dense to see, but I don't know what it is.

What in the bloody H*%# am I missing?

3. Jul 7, 2012

4. Jul 7, 2012

### Fiziqs

Thank you dlgoff, that was exactly what I was looking for. HUP makes perfect sense to me now.

Of course I have my own personal Uncertainty Principle, I may understand it now, but that doesn't mean that it will still make sense to me ten minutes from now. But I'll try to enjoy the moment.

5. Jul 7, 2012

### Fredrik

Staff Emeritus
There's been a number of threads about what exactly a "momentum measurement" is, but I don't think any of them really answers your question. In the thread that dlgoff linked to, we discussed (among other things) a thought experiment proposed by Ballentine in 1970, and specifically if the process that he described should be considered a momentum measurement. In the end, I was convinced (by an argument posted by Demystifier) that it shouldn't be considered a momentum measurement.

My view is that a theory isn't fully defined without a specification of how to test its predictions. So if you take a quantum theory and make a small change to its definition of "momentum measurement", you are now dealing with a new theory. If one of those theories makes better predictions (either more accurate predictions, or just the same predictions plus a few new predictions that have roughly the same degree of accuracy), then we should say that this is a better theory, or a better version of the same theory.

This is kind of weird, because it means that we have to use experiments to determine some of the details in the definition of the theory. First we guess what sort of thing should be considered a "momentum measurement". (This could be a pretty intelligent guess that relies on what we have learned from other theories). Then we test the accuracy of the theory's predictions. Then we can change the guess somewhat, to see if this makes the theory better or worse. Eventually we should reach some sort of consensus about what a momentum measurement is. I'm pretty sure that this has already happened, even though the experimentalists who did it probably weren't thinking about it in these terms.

So the best people to answer the question are (not surprisingly) experimentalists who actually do momentum measurements. Unfortunately we don't seem to have many of those around. I think ZapperZ has been involved in something like this, but I don't really know. In one of these threads he posted a link to an article that describes an experimental technique called ARPES. I'm going to have to take a closer look at it some time.

One specific thing that's been discussed in several of these threads is the kind of momentum measurement that involves measuring the curvature of a particle track in a detector. The particle track consists of the results of a series of approximate position measurements, and the information you get from the whole series can be considered an approximate momentum measurement. This will only work if the particle has interacted with something that puts it into a state of approximately (but not exactly, because that would be impossible) well-defined position and momentum, and if the approximate position measurements don't change this state too much.

6. Jul 7, 2012

### Fiziqs

Thanks Fredrik, I'll go back and read the rest of that thread. I had read the first few pages but the discussion appeared to be getting a bit contentious, and I had what I thought was a clearer understanding of the problem, so I didn't take the time to read all of the thread. But I have time now, so I'll go back and read the rest, and make a point of reading the argument posted by Demystifier.

Thanks

7. Jul 7, 2012

### nonequilibrium

I suppose (all) the orthodox interpretation implies is that a momentum measuring device is a device that interacts in such a way with the system (that is being measured) such that only the eigenfunctions of the momentum operator are left unperturbed; I'm not sure if this condition is sufficient, I would suspect so but it would need to be proven.

Note that a time of flight measurement does not satisfy this necessary condition, so this at least suggests that a TOF measurement is not a momentum measurement.

8. Jul 8, 2012

### Fredrik

Staff Emeritus
I'm not sure how useful it will be to read the entire thread. As you have noticed, a new guy showed up and spent a couple of days doing nothing but yelling insults at me. It might be helpful to read the part of Ballentine's article that argues that you can measure position and momentum simultaneously. See my post #287 for more about that (and for the link to Demystifier's argument, which was posted in another thread).

9. Jul 8, 2012

### Fiziqs

Fredrik, I really had to give this a lot of thought, and I hope that I can explain myself in a manner that makes sense. Anyway, I'll do my best. I can definitely see Demystifier's point when he says that the momentum is calculated and not measured. Now I may be mistaken, but I believe that momentum will always be calculated, and not measured. I can't think of any way to determine momentum using only one measurement. Thus momentum will always necessitate a calculation.

But after further thought I decided that the whole debate about whether momentum is calculated or not, is a red herring. It really is beside the point. HUP is describing a completely different situation. HUP states that the more precisely we know a particle's position, the less precisely we will know its momentum. Yet when we measure a particle's position, we can also calculate its momentum based upon a previous measurement. Granted this calculation may not be exact, but it should be very close. You could indeed argue that we really don't know what path the particle took from point "A" to point "B", and thus we can't define its momentum precisely. But this is true regardless of how well we define its position. More accurately defining a particle's position does nothing to change the paths the particle could have taken, and so doesn't affect the accuracy with which we can determine its momentum. If anything, the more precisely we measure a particle's position, the fewer the number of paths the particle could have taken. Which would imply that the more precisely we know its position, the more precisely we should know its momentum also. Any inherent inability to define momentum has nothing to do with HUP. In any case we may not know its momentum exactly, but we should be able to come pretty darn close.

But according to HUP we shouldn't even be able to come close. Since we know the particle's position precisely, the wave function describing its momentum should be massive, but it's not. Even using a debatably imprecise calculation, we can still come pretty close, and HUP states that we shouldn't even be able to come close. After a measurement, position and momentum definitely don't act like conjugate variables.

So what exactly is HUP saying? I think HUP is saying that if you try to define the wave functions of two conjugate variables at the same time, (if you try to confine them to some arbitrarily small value) then the more precisely you define the wave function of one, the larger the wave function of the other will be. But when you actually make a measurement you collapse the wave functions for them both. The wave function defining the particle's position will collapse, and the wave function defining the particle's momentum between its last interaction and this one will also collapse. I think that HUP is about defining the wave functions of two conjugate variables, but does not apply once a measurement has been made, and the wave function has collapsed.

I hope that this explanation has made sense, sometimes my thoughts aren't as clear on paper (or monitor) as they are in my head.

Whether the momentum is calculated or measured, is irrelevant.

At least that's my ninth grade, Wikipedia educated opinion.

But I'm definitely interested in yours or anyone else's ideas.

10. Jul 8, 2012

### Fredrik

Staff Emeritus
I wasn't referring to his original argument ("it's calculated, not measured"). I rejected that in #35, and presented my own view, which is that the two possible answers (yes or no) to the question of whether Ballentine's procedure should be considered a momentum measurement, give us two slightly different theories. Demystifier then explained (in #40) why the theory that does consider it a momentum measurement is worse than the one that doesn't. That's the argument I had in mind.

Yes, if the wavefunction is sharply peaked around some point in its domain, then its Fourier transform isn't, and vice versa.

To collapse them both would be to make both the wavefunction and its Fourier transform sharply peaked around a single point. This is impossible.

A position measurement that doesn't absorb the particle is also a state preparation procedure that leaves the particle in a state represented by a wavefunction with a peak that's located at the result of the measurement and has a width that's roughly the size of the margin of error. So if the accuracy is very high (=small margin of error), the wave function is sharply peaked, and this means that its Fourier transform is not. So the Fourier transform (i.e. the "momentum space wavefunction") is very far from being "collapsed". It's even more spread out than it was before (if the position measurement was very accurate).

This means that if e.g. the interaction that produces a bubble in a bubble chamber could be thought of as a very accurate position measurement, the direction to the location of the next bubble would be very unpredictable. So the fact that the actual result is a particle track must mean that the interactions don't squeeze the wavefunction enough to make it sharply peaked at the location of the bubble.

I agree with this. My view is that all a measuring device can really do is to tell us that an interaction has taken place. The rest is a matter of how we interpret that information.

There is however a difference between state preparation and measurement, so it's not at all obvious that a theorem about state preparation like the uncertainty relation for position and momentum can be applied to situations where the measurement absorbs the particle. (If the particle is absorbed, no new state is prepared). This is why Ballentine thought it might be possible to measure position and momentum at the same time. But as far as I can tell, his attempt to prove that it is fails because a theory that considers his procedure a momentum measurement is less accurate than one that doesn't.

11. Jul 8, 2012

### Fiziqs

Thank you very much for taking the time to answer my questions. I can see that I have a lot of work to do to understand this stuff. But I have taken up enough of your time and I don't want to wear out my welcome. You have helped me immeasurably, and I appreciate your efforts. I know that I'll never achieve your level of understanding, but hopefully I won't lose my curiosity about physics. But for now I'm going to spend some time educating myself on the basics, so that perhaps in the future I'll be able to discuss this stuff without feeling so far out of my element.

Thanks Fredrik for your patience, and hopefully at some point I'll be back with somewhat less irritating questions.

12. Jul 8, 2012

### jmcelve

Hi Fredrik,

I've been following your posts quite a bit, particularly in the other thread. You've mentioned several times the situation where you measure both the position and momentum to arbitrary accuracy (up to instrumental error).

I've been wondering: how do we reconcile this with the inability to simultaneously diagonalize the position and momentum operators? I suspect this is why you keep mentioning the absorption of particles in certain measurements, because I think in the other thread you mentioned that as a consequence of measuring momentum and position simultaneously for an individual particle. Wouldn't the simultaneous measurements be an attempt at collapsing the state into both a position and momentum eigenstate? I think the answer lies somewhere in something you've already said, but I'm just having trouble putting the pieces together myself. I apologize if I end up making you repeat yourself.

Could you also maybe expand on *why* the particle absorption happens? This just sounds really strange, and actual measurements aren't something we've talked about yet in our QM courses.

As background, I'm new to QM. I just took my first course in quantum, so I'm still trying to work out all the conceptual details for myself. I appreciate any and all help, and please be sure to inform me if I'm being imprecise with the language.

Last edited: Jul 8, 2012
13. Jul 9, 2012

### Fredrik

Staff Emeritus
You're welcome. You haven't been irritating at all.

Hi jmcelve. At the start of that thread, I thought you could do that, because Ballentine (the author of one of the best textbooks on QM) said that you can, in a peer-reviewed article that's still being referenced 40 years after it was written. But as you can see in #287 I eventually changed my mind. The main reason is that the discussion with Demystifier convinced me that what Ballentine described shouldn't be considered a momentum measurement. So Ballentine's "proof" doesn't seem to work.

I do however stand by my comment in my previous post:
...it's not at all obvious that a theorem about state preparation like the uncertainty relation for position and momentum can be applied to situations where the measurement absorbs the particle.​
I think it's probably true that you can't simultaneously measure both, but since the uncertainty relations are really about state preparation, not measurement, and since there are measurements that don't prepare new states, this result is not an immediate corollary of the uncertainty relation for position and momentum.

An "eigenfunction" of position is a delta function. An "eigenfunction" of momentum is a plane wave, exp(ipx). This is a function with constant magnitude, so it's as far from a delta function as you can get. So a simultaneous eigenstate can't exist, and it wouldn't make sense to try to put the particle in such a state. The best you can do is to prepare a state represented by a wavefunction like exp(-a(x-y)^2) where a>0 is a real number that determines the width of the peak at y.

The details depends on the type of particle and the type of detector. Since I don't know solid state physics, I won't be able to explain them very well (e.g. the details of how an electron is absorbed by a conductor). But consider e.g. an experiment that uses a photographic plate to detect a particle. The partice interacts with a small part of the plate, that changes color as a result of the interaction. The physicist interprets the location of the discoloration as the result of a position measurement. If the particle doesn't pass through the plate and emerge on the other side, then we can't view it as an isolated quantum system anymore. It's now part of the plate.

14. Jul 9, 2012

### ZapperZ

Staff Emeritus
I wonder if people realize that my avatar is a photoemission spectrum that measures energy distribution curves (EDC) in the horizontal axis AND momentum distribution curves (MDC) in the other axis.

See Valla et al., http://arxiv.org/abs/cond-mat/9904449 for a similar spectrum.

Zz.

15. Jul 9, 2012

### jmcelve

It seems there are a few nuances I'll have to ruminate over in the coming days. I hope you don't mind if I come back with a few more questions. I really appreciate your response though! Thanks for the thoughtful discussion.