How Do Experimental Physicists Measure Momentum?

[nonequilibrium]

How does an experimental physicist measure it, without going into too much detail?

EDIT: found an old topic on it ( https://www.physicsforums.com/showthread.php?t=227477 ) which wasn't very helpful, so I'll specify more:

Is the following procedure a "valid" momentum measurement?

Two position measurements at separate times; dividing the displacement vector between these two positions by the time waited in between gives the momentum.

 

[Vanadium 50]

This depends entirely on the object's mass and velocity. The techniques suitable for a planet may not be suitable for a proton.

 

[Meir Achuz]

Momentum is measured by the radius of curvature in a magnetic field, with p=qBR.

 

[nonequilibrium]

This depends entirely on the object's mass and velocity. The techniques suitable for a planet may not be suitable for a proton.

I'm thinking quantum-mechanically.

Anyway, what I was wondering about, if an experimental physicist were to measure momentum the way I described in the first post, would he be "allowed" to use the momentum operator to predict his results?

 

[nonequilibrium]

Momentum is measured by the radius of curvature in a magnetic field, with p=qBR.

Okay thank you. And I presume to measure the radius one has to do position measurements?

 

[Meir Achuz]

The uncertainty principle doesn't enter because you don't measure position in the direction of the momentum.

 

[nonequilibrium]

I don't know why you're starting about the uncertainty principle, but can you expound?

 

[kith]

There has been a lengthy discussion about momentum measurements in this thread.

To me, the relation between real momentum measurements (like in particle physics) and idealized measurements of the observable P is still unclear.

 

[nonequilibrium]

To me, the relation between real momentum measurements (like in particle physics) and idealized measurements of the observable P is still unclear.

Sad to hear that! That was indeed why I started this thread.

Are there people for whom the connection is clear? Or is it simply inherently vague?

 

[Fredrik]

I don't know why you're starting about the uncertainty principle, but can you expound?

Uncertainty is a concern because as you probably know, a state preparation procedure that gives the particle a sharply defined momentum gives it a poorly defined position and vice versa. (This is usually explained using Fourier transforms). Because of this, the suggestion that momentum can be measured by a series of position measurements can sound like crazy talk. If the first one determines the position accurately, it will put the particle in a superposition of a wide range of momenta, and everything will be messed up.

So what's really going on when a particle makes a track through a detector and a momentum is calculated from the curvature of the path? I think the only possible answer is that none of the interactions that leave records (e.g. bubbles in a bubble chamber) from which a particle track can be determined, gives the particle a really sharply defined position. These interactions must be position "measurements" that are inaccurate enough to not change the state of the particle by much.

Clem's comment about the position measurement not being in the direction of momentum was probably inspired by something like the following: (I can't know if this is really what he meant, so don't blame him for anything that I'm saying here).

Consider a particle with a well-defined energy (and therefore well-defined $\vec p^2$) moving along the x-axis in the direction of increasing x towards a tiny hole in a screen in the yz plane. Behind the screen, a distance L from the screen, there's a wall (parallel to the yz plane) covered with tiny particle detectors. Suppose that a detector in the xy plane signals detection. (Alternatively, suppose that an arbitrary detector signals detection, and now rotate the coordinate system in the yz plane so that the detector is in the yx plane). The angle of deflection θ satisfies tan θ=y/L. Define $p=|\vec p|$. At least one peer-reviewed article* has claimed that this y coordinate measurement can also be considered a p_y measurement with result p sin θ. The author used this to argue that it's possible to simultaneously measure y and p_y with margins of error δy and δp_y such that $\delta y\,\delta p_y\ll\hbar$. (This product can be made arbitrarily small by making L large).

*) Leslie Ballentine: The statistical interpretation of quantum mechanics. PDF link. (See in particular figure 3 on page 365).

This was discussed in the thread that kith linked to. Unfortunately the thread was derailed by a new member who spent a couple of days doing nothing but posting insulting comments about me. So I wouldn't recommend that thread. It turned out that the main reason he thought I was wrong about everything was that he didn't know how to negate a "for all" statement.

However, I think something good eventually came out of that thread. I think Demystifier's argument for why the position measurement in Ballentine's thought experiment shouldn't be considered a momentum measurement was convincing. He actually posted it in another thread, here. (See my posts and his, in the 35-40 range. The main point is in post #40).

 

[kith]

Sad to hear that!

I think most people who do QM, carry around their personal set of open questions. I thought about opening a collection thread, but it would probably be one big mix-up without much insight. ;-)

 

[Fredrik]

There has been a lengthy discussion about momentum measurements in this thread.

To me, the relation between real momentum measurements (like in particle physics) and idealized measurements of the observable P is still unclear.

Sad to hear that! That was indeed why I started this thread.

Are there people for whom the connection is clear? Or is it simply inherently vague?

It's still not completely clear to me. I don't know if there are people to whom it's completely clear. I suppose that experimentalists who design momentum measurements must have a pretty good idea, but I doubt that there's a clear explanation that can be written down in a couple of pages. If there was such an explanation, I think it would have appeared in a standard textbook by now.

 

[Meir Achuz]

I don't know why you're starting about the uncertainty principle, but can you expound?

I just thought (apparently mistakenly) that was why you mentioned position measurement.

 

[Edgardo]

So what's really going on when a particle makes a track through a detector and a momentum is calculated from the curvature of the path? I think the only possible answer is that none of the interactions that leave records (e.g. bubbles in a bubble chamber) from which a particle track can be determined, gives the particle a really sharply defined position. These interactions must be position "measurements" that are inaccurate enough to not change the state of the particle by much.

I think this is the case. For example, consider an electron from a Beta-decay that leaves a track in the cloud chamber. Let's say the width of the track is 0.1 mm. Then we have made a position measurement.

                     /\             electron moving upwards in x-direction      
                     |
                     |
                     |
                     |
               |<--- e ---->|

                 Δy = 0.1mm

In other words we prepared the electron in a state whose uncertainty in y-direction is $\Delta y$ = 0.1mm
The Heisenberg uncertainty relation is $\Delta y \cdot \Delta p_{y} \geq \hbar/2$ and we get $\Delta p_y \geq \hbar/(2\Delta y$) = 1.05457148 × 10^-34 m² kg/s / (2*0.1*10^-3 m) = 5.2728574 × 10^-32 kg*m/s

With p = mv (non relativistic) we get:
v = p/m = 5.2728574 × 10^-32 kg*m/s / (9.10938×10^-31 kg) = 0.058 m/s = 5.8 cm/s.

This is small compared to the velocity in x-direction (I think we can say that although I didn't calculate the velocity v_x here).
This means the y-component or better the transversal component of the momentum is small and does not "deflect" the momentum vector very much.

--

I think the cloud chamber makes many measurements in sucession (or continuous?) and you could compare this to many single slit measurements.
I would compare this to a random walk or a drunk man whose step size is small due to small $\Delta p_{y}$

 

[nonequilibrium]

Thanks for the replies! I have some things to say (it isn't as much as it seems; a lot of space is filled by quote tags):

Consider a particle with a well-defined energy (and therefore well-defined $\vec p^2$) moving along the x-axis in the direction of increasing x towards a tiny hole in a screen in the yz plane. Behind the screen, a distance L from the screen, there's a wall (parallel to the yz plane) covered with tiny particle detectors. Suppose that a detector in the xy plane signals detection. (Alternatively, suppose that an arbitrary detector signals detection, and now rotate the coordinate system in the yz plane so that the detector is in the yx plane). The angle of deflection θ satisfies tan θ=y/L. Define $p=|\vec p|$. At least one peer-reviewed article* has claimed that this y coordinate measurement can also be considered a p_y measurement with result p sin θ. The author used this to argue that it's possible to simultaneously measure y and p_y with margins of error δy and δp_y such that $\delta y\,\delta p_y\ll\hbar$. (This product can be made arbitrarily small by making L large).

*) Leslie Ballentine: The statistical interpretation of quantum mechanics. PDF link. (See in particular figure 3 on page 365).

I've heard of this argument before, and it's interesting to say the least. Besides the objections by Demystifier (in the thread you linked to) --and which I'll comment on shortly-- I want to express my concern about the energy: the argument relies on the fact that energy is conserved, but is this certain? The initial state is an eigenfunction of energy in vacuum (i.e. also an eigenfunction of the momentum operator) but this eventually interacts with a strong potential, where the energy eigenfunctions are of a totally different shape, hence--according to me!--I do not directly see conservation of energy and hence I don't know whether one can conclude that the final state (before measurement) has the same momentum squared as the initial state. But this is a minor remark, perhaps.

I move on to Demystifier's arguments, which I give titles for clarity:

A) "MEASURING" VS. CALCULATING
As you quote Demystifier saying

I haven't seen the debate on this forum, but I can present a simple reason why the Ballentine's experiment is NOT a measurement of momentum p_y. The point is that in this experiment p_y is NOT MEASURED but CALCULATED.

upon which you replied with

I reject that argument. A full definition of QM would include a definition of "momentum measurement". When we write down such a definition, we can choose to include Ballentine's technique among the things we call "a momentum measurement", and we can choose not to. These choices define two slightly different theories. The only thing that matters to me is which one of these theories has the best agreement with measurements.

So let's call QM without this measuring technique QM1, and QM with it QM2.

I understand your reply, but I have an even more fundamental objection: isn't every measurement to some extent deduced from something else? Even a position measurement: if I read off the value indicated by the pointer on my position measuring device, I'm relying on the information that the pointer deduces from the spring which it is attached to, a chain of command that tracks all the way back to the system under inspection. The justification for each step is justified by an element of theory and is thus a form of deduction.
But I realize this objection is a bit abstract, so let me state it alternatively: let's presume one can define QM1 and QM2 as you did, what kind of measurement does belong to QM1? Let's take position measurements for a moment (despite my earlier reasoning) as belonging to QM1: can you think of any momentum measurement that belongs to QM1? Isn't every momentum measurement deduced from (a) position measurement(s)?

B) CORRECT MOMENTUM PROBABILITIES?
I quote:

what result of an experiment would make you conclude that QM2 is wrong while QM1 is still potentially right?

If we perform the experiment that Ballentine describes, and find that the distribution of results isn't given by $p_y\mapsto|\langle p_y|\psi\rangle|^2$, then I would say that QM2 has been falsified.

Fine!

I haven't worked out the calculation in detail, but it is rather straightforward to see that in QM2 the probability of p_y is NOT $|\langle p_y|\psi\rangle|^2$. Namely, in QM2 p_y is determined by the position y, so the probability of p_y is related to the probability of y, which is $|\langle y|\psi\rangle|^2$. In other words, the probability amplitude of p_y in QM2 is related to $\langle y|\psi\rangle$ and not to its Fourier transform $\langle p_y|\psi\rangle$. Thus, QM2 is already falsified, as it is already experimentally tested that the probability of y is given by $|\langle y|\psi\rangle|^2$.

Despite its broadstroked nature, Demystifier's argument makes sense. It is however not conclusive: the thing which makes me think otherwise is a proof showing that if one does two position measurements (separated by a time t) on a free particle and calls $\textbf p := \frac{m \textbf r}{t}$ (where $\textbf r$ is of course the displacement between the two measurements) then the probability distribution for this $\textbf p$ (calculated using usual quantum formalism) is the same as the probability for momentum, i.e. the squared modulus of the Fourier transform of psi! The idea of this experiment is very comparable to Ballentine's measurement (I can't see a fundamental/conceptual difference) yet Demystifier's conclusion (and thus also his argument) seems to be incorrect in this case? For those interested, the proof can be found on page 34 (section 2.6).

---

And when all else is said, I just want to note: do we want QM2 to be wrong? After all, as e.g. Edgardo's post demonstrates, experimentally we do measure momentum by deduction/calculation of position measurements, so if QM2 is wrong, we're not only in trouble, we should be very confused as to how come we get sensible results!

I eagerly await replies :)

 

[Demystifier]

For those interested, the proof can be found on page 34 (section 2.6).

The proof is done with the assumption that the wave function is a Gaussian wave packet. Indeed, it is well known that the Fourier transform of a Gaussian wave packet is also a Gaussian wave packet, so it is not a surprise that the proof works for that case. However, the proof does not work for more general wave functions.

In fact, I have seen many similar "proofs" that quantum mechanics can be explained by classical mechanics, based on the assumption of Gaussian probability distributions. But all such "proofs" fail when more general probability distributions are considered.

 

[Demystifier]

And when all else is said, I just want to note: do we want QM2 to be wrong? After all, as e.g. Edgardo's post demonstrates, experimentally we do measure momentum by deduction/calculation of position measurements, so if QM2 is wrong, we're not only in trouble, we should be very confused as to how come we get sensible results!

I don't think we are in trouble without QM2. For example, in his book "Quantum Mechanics: A Modern Development", Ballentine himself describes how the momentum is correctly measured in accordance with QM1 (not QM2). See Sec. 5.2 "Momentum Distribution in an Atom" of this book.

In fact, ALL measurements in QM (momentum, energy, spin, ...) eventually reduce to measurements of positions of something. And yet, this is not a problem for QM1 (standard QM). To see why, you need to understand the GENERAL theory of quantum measurements. It is presented in many books and papers. For example, see Sec. 2.1 of my
http://xxx.lanl.gov/pdf/1112.2034v1

 

[nonequilibrium]

The proof is done with the assumption that the wave function is a Gaussian wave packet.

Where do they assume that?

I don't think we are in trouble without QM2. [...] In fact, ALL measurements in QM (momentum, energy, spin, ...) eventually reduce to measurements of positions of something. And yet, this is not a problem for QM1 (standard QM). To see why, you need to understand the GENERAL theory of quantum measurements.

I'm glad to hear you're so convinced of your view, it means I might be close to understanding it myself. Thanks for replying. I will check out those sections in Ballentine and your paper!

 

[Demystifier]

Where do they assume that?

Sorry, they don't. I have not been reading their proof carefully. :shy:

 

[nonequilibrium]

Sorry, they don't. I have not been reading their proof carefully. :shy:

No problem! But then, do you still stand behind your earlier argumentation (e.g. my last quote of you in this post) as the argument seems to carry over to this case, while its conclusion seems invalid?

Also, I've read the section you referred to in Ballentine's book (next up: your paper), about which you said

For example, in his book "Quantum Mechanics: A Modern Development", Ballentine himself describes how the momentum is correctly measured in accordance with QM1 (not QM2). See Sec. 5.2 "Momentum Distribution in an Atom" of this book.

But I don't quite see it. First of all I don't quite understand how he can claim equation (5.11): this is the classical variant of conservation of momentum; does it make sense in QM? Why? But maybe this is a minor point. The general idea of the section seems to be: how to infer the momentum from the electron in the atom from other measurements (i.e. the momenta of the scattered and ejected electrons), hence we're "calculating and not measuring", as you call it, so isn't this QM2?
Another thing that bothers me is that they don't say how they measure the momenta of the scattered and ejected electrons. I can only presume that they do a series of position measurements, much like the case we were just discussing?

 

[Demystifier]

Sorry, they don't. I have not been reading their proof carefully. :shy:

In the meantime, I have found another mistake in their derivation. The mistake is present in the first line of Eq. (2.43). Namely, in the Fourier transform they treat energy as an explicit function of momentum, so that they integrate over this momentum-dependent function when they calculate the Fourier transform in the 3-space. But this is NOT how <x|psi> and <p|psi> in QM1 are defined. Instead, energy should be treated as an independent quantity, i.e., E=p^2/2m should be exploited only AFTER integration over p has been performed.

 

[nonequilibrium]

That's an interesting note, but I don't understand why that it's true. Can you justify it? Or rather, can you tell me what's wrong about how I see it:

Since phi is the Fourier transform of psi, we have that $\psi(x,t) = \frac{1}{\sqrt{2 \pi }} \int \phi(p,t) e^{ipx} \mathrm dp$ (taking $\hbar = 1$).
Filling this in in the Schrödinger equation for a free particle, we get that $i \frac{\partial \phi(p,t)}{\partial t} = \frac{p^2}{2m} \phi(p,t)$ and hence $\phi(p,t) = \varphi(p) e^{-i \frac{p^2}{2m}t}$.
This gives us that $\boxed{ \psi(x,t) = \frac{1}{\sqrt{2 \pi }} \int \varphi(p) e^{i \left( px - \frac{p^2}{2m}t \right)} \mathrm dp }$.

 

[Demystifier]

Despite its broadstroked nature, Demystifier's argument makes sense. It is however not conclusive: the thing which makes me think otherwise is a proof showing that if one does two position measurements (separated by a time t) on a free particle and calls $\textbf p := \frac{m \textbf r}{t}$ (where $\textbf r$ is of course the displacement between the two measurements) then the probability distribution for this $\textbf p$ (calculated using usual quantum formalism) is the same as the probability for momentum, i.e. the squared modulus of the Fourier transform of psi! The idea of this experiment is very comparable to Ballentine's measurement (I can't see a fundamental/conceptual difference) yet Demystifier's conclusion (and thus also his argument) seems to be incorrect in this case? For those interested, the proof can be found on page 34 (section 2.6).

During all this time, I still thought that something must be wrong with this result, and yet I was not sure what exactly it was. Today I finally decided to look at it once again carefully, to finally resolve the puzzle. Now I think I can definitely say what exactly is wrong.

First, let me confess that what I previously thought was wrong, now I understand is OK. The wrong thing is something else.

The wrong thing is Eq. (2.42). Why? Observe that in this equation, $\psi$ is evaluated at t=0. The time dependence of the wave function is carried by the exponential factor in front of $\psi$. But this can be traced back to the assumption that
$$\psi(x,t)=e^{iEt/\hbar} \psi(x,0) ... (Eq.1)$$
where $E=p^2/2m$. At first sight this looks quite innocent, but is actually wrong. Namely, (Eq.1) is valid ONLY for energy-eigenstates, not for general states. (Actually, there is also a wrong sign involved, but that's not important). In the case of free particles, typical energy-eigenstates are trivial plane waves, for which the Fourier transform (and hence momentum probability density) is trivial. But for general free wave packets this is not so. For general free wave packets, (Eq.1) is invalid implying that Eq. (2.42) in the book is invalid too.

 

[Demystifier]

I have also asked the author of the book, Prof. Dalibard, to explain it to me. Now I finally understand it completely, so let me share it with you.

Eq. (2.42) is a DEFINITION, so it cannot be wrong. However, this quantity tilde{phi} is NOT the Fourier transform phi. (They are equal only under certain approximations in the equation above (2.44).) Therefore, Eq. (2.41) does NOT say that probability in the position space is given by the Fourier transform of Psi.

 

[Demystifier]

the probability distribution for this $\textbf p$ (calculated using usual quantum formalism) is the same as the probability for momentum, i.e. the squared modulus of the Fourier transform of psi!

From my post above, now it should be clear that this is not so, except under certain approximations (large t).