# How to measure momentum in principle?

1. Aug 24, 2015

### pellman

If we devise a physical system and perform an observation of some physical quantity, how can we infer that this quantity is related to the eigenvalues of the momentum operator -ih d/dx ?

Another way to look at it. Suppose you were handed the theory of quantum mechanics and that you already had an understanding of how to measure the fundamental quantities mass, length and time. How would you design an apparatus to measure particle momentum, one that you were confident was associated with the eigenvalues of -ih d/dx ?

Furthermore, what were the early quantum theorists talking about when they said "particle's momentum?" When de Broglie proposed that the momentum of a particle is proportional to its wavelength, he wasn't defining momentum as p=hk. He had some prior idea of a (measurable) momentum in mind that he was relating to wavelength. What was it?

Previously posed this question back 2009 here https://www.physicsforums.com/threads/what-is-the-definition-of-a-momentum-measurement.328164/ but never got an answer.

Last edited: Aug 24, 2015
2. Aug 24, 2015

### bhobba

You can measure it exactly the same way as classically eg for a beam of particles you measure velocity and multiply by m.

Why is it the QM momentum operator? See Chapter 3 - Ballentine - QM - A Modern Development.

Its the deep and surprising reason for much of physics - symmetry.

If you haven't seen this before QM is probably not the best for a first exposure - Landau Mechanics is a much better choice:
https://www.amazon.com/Mechanics-Third-Edition-Theoretical-Physics/dp/0750628960
'If physicists could weep, they would weep over this book. The book is devastingly brief whilst deriving, in its few pages, all the great results of classical mechanics.' 'The reason for the brevity is that, as pointed out by previous reviewers, Landau derives mechanics from symmetry. Historically, it was long after the main bulk of mechanics was developed that Emmy Noether proved that symmetries underly every important quantity in physics.'

This truly is one of the great insights of modern physics. Unfortunately, its often not encountered until graduate school. One of the mentors here posted when he introduces it to students in class they sit there in stunned silence - it truly is deep and surprising.

Forget about why the early QM pioneers postulated it. It is known, for example, Schroedinger made an error in his original paper:
http://arxiv.org/pdf/1204.0653.pdf

The modern view from symmetry is much clearer.

Thanks
Bill

Last edited by a moderator: May 7, 2017
3. Aug 24, 2015

### Staff: Mentor

There are a bunch of questions here. Taking them a bit out of order here....
The early quantum theorists were of course starting from a background in classical mechanics, so they had a perfectly good prior idea of a measurable momentum - the classical momentum, established and well understood more than two centuries. As you say, de Broglie wasn;t defining momentum, he was associating a wavelength with the well-understood notion of momentum. It took many more years before the connection between momentum and the operator $i\hbar\frac{\partial{}}{\partial{x}}$ was understood.

I can collide it with something and infer its momentum from conservation laws and the behavior of the particles coming out of the collision. If it is charged I can observe how its path is curved by a magnetic field. Of course that just gives me a measurement of the particle's momentum that would be perfectly natural to a classical physicist; confidence that it has anything to do with $i\hbar\frac{\partial{}}{\partial{x}}$ comes from a theoretical understanding of QM.

That's where that theoretical understanding of QM comes in. The historical development of the theory was understandably somewhat ad hoc, involving some leaps of intuition and brilliant guesses whose correctness only became clear after the fact. However, once the theory was in place, it became possible (it's a lot easier to plot a direct route when you know the destination) to make the connection between the classically conserved quantity we knew as momentum and the operator $i\hbar\frac{\partial{}}{\partial{x}}$ on the basis of symmetry considerations.

Bhobba's two book recommendations are good. Neither is exactly light reading.

4. Aug 24, 2015

### andresB

And how are you would measure the velocity of the particles?

5. Aug 24, 2015

### blue_leaf77

Using magnetic field to deflect its orbit and measure the radius of curvature.

6. Aug 24, 2015

### atyy

That is wrong.

7. Aug 24, 2015

### atyy

One way to see that the quantum measurement is not "exactly" the same as the classical measurement is to understand what important point is omitted in these descriptions of quantum momentum measurement:

https://www.physicsforums.com/insights/misconception-of-the-heisenberg-uncertainty-principle/
http://arxiv.org/abs/quant-ph/0703126

Reading those descriptions, one may think that the quantum momentum measurement in the double slit experiment is "exactly" the same as in the classical case. But it is not. In order for the quantum and classical methods to agree, the screen must be placed at an infinite distance from the slit. At finite distance, the measurement using the classical method does not yield an accurate quantum momentum measurement.

8. Aug 24, 2015

### bhobba

As per another post using a magnetic field.

Its simply an example of how it can be done. Its not how you would do it in all instances.

Thanks
Bill

9. Aug 24, 2015

### Staff: Mentor

That is indeed how I did it long long ago with relativistic electrons, but atyy is correct that this does not produce a classical momentum measurement. It narrows the momentum spread down, with commensurate increase in the uncertainty of the position, but it does not place the particle in the momentum eigenstate that would be equivalent to a successful classical measurement - it cannot, unless I'm willing to place the screen an infinite distance from the source.

These concerns do not affect the answers to the original question in this thread - we can measure momentum, and we can be confident that the quantity we're measuring corresponds to conserved quantity we find from symmetry considerations.

10. Aug 28, 2015

### pellman

This is the point of my question. Is there a theoretical basis for this confidence (I mean a meta-theory about experiments, one that tells us how to relate experiment to the physical theory) or is it simply that the measured values match the theoretical values?

11. Aug 28, 2015

### Staff: Mentor

There is such a theoretical basis and it is, as Bhobba said above, covered by Ballentine.

12. Aug 28, 2015

### bhobba

Its subtler than that.

By definition in classical mechanics momentum is defined as mv. From a symmetry assumption (specifically Galilean relativity) it can be shown that in QM the momentum operator is mv where v is the velocity operator defined, pretty obviously, as the time derivative of the position operator. If you take the expectation of the QM operator since it is mv you end up with the classical definition of momentum so experimentally it must be compatible with the classical definition.

In fact, it must be like that due to a very important and beautiful theorem, Noethers Theorem, due to the great female mathematician Emmy Noether:
https://www.amazon.com/Noethers-Wonderful-Theorem-Dwight-Neuenschwander/dp/0801896940
https://en.wikipedia.org/wiki/Emmy_Noether
http://www.vox.com/2015/3/23/8274777/emmy-noether

Thanks
Bill

Last edited by a moderator: May 7, 2017
13. Aug 28, 2015

### pellman

I've been scanning through Ballentine and I don't see it. If you could provide a page number, that would be very appreciated. However, I suspect that the answer is not really in there. See my next reply in a few minutes.

14. Aug 28, 2015

### blue_leaf77

I remember it's somewhere in the beginning of chapter 4.

15. Aug 28, 2015

### bhobba

See page 81 where its proven for a free particle the quantum momentum operator is mV where V is the velocity operator.

There are even deeper reasons - see the book - Noether's Wonderful Theorem for further details.

Thanks
Bill

16. Aug 28, 2015

### bhobba

Its in Chapter 3 page 81.

Thanks
Bill

17. Aug 28, 2015

### Staff: Mentor

A fair amount of chapter three is devoted to this question. Classical momentum, the $p=mv$ thing that we measure with the experimental techniques discussed above, is the conserved quantity associated (via Noether's Theorem) with translation symmetry. Ballentine shows in chapter three that that the quantum mechanical observable corresponding to the Hermitian operator $i\hbar\frac{\partial}{\partial{x}}$ is also the conserved quantity associated with this symmetry.

18. Aug 28, 2015

### pellman

I think I have not clearly stated my question. I will post it soon. Thanks, guys.

19. Sep 5, 2015

### pellman

In classical theories (except perhaps thermodynamics) the theory itself incorporates a model of the system. We can model billiard balls as point particles or as hard spheres constrained to a 2-D plane. The physical position of a billiard ball measured with values X,Y corresponds to a values x,y in our model, velocity to those quantities' time derivative. Momentum is straightforward. The theory tells us what to measure. It is built in. If I measure the mass of the jth billiard ball and its velocity, I know that corresponds to the quantity $m_j$ in my theory and the derivative of $x_j, y_j$.

In quantum theory we have a wave-function $\Psi(x_1 , x_2, ... ,x_n )$for n particles. It doesn't come with a picture of the physical system. The wave function is merely a function from R^3n to the complex plane. It is a formula for calculating probabilities, not a model of the system. We have a prescription that says $| \Psi(x_1 , x_2, ... ,x_n )|^2$ is the pdf for observing a particle at x_1, a particle at x_2, etc. But this by itself gives us no clue at all what "observing particle" physically involves. Or if we decompose psi into contributions by energy eigenfunctions, the coefficients-squared give us a probability distribution for energy measurements... "energy" here being the theoretical quantity associated with symmetry under translations of the equations' parameter (which we denote as "time"). This tells us nothing about how to measure physical energies... A physical system may involve measuring the wavelengths of photons emitted during transitions... while the wavefunction we are using may not even contain information about photons.

And similarly, the coefficients-squared of the contributions to psi from the eigenfunctions of the operator -ih d/dx give us probabilities for observing the various values of the eigenvalues of -ih d/dx . We call it "momentum" because of its relation to translations of the position variables.. which in the classical theory is mv. But the function $\Psi(x_1 , x_2, ... ,x_n )$ gives us no information about what kind of measurement of our system would correspond to those eigenfunctions of -ih d/dx.

People DO measure momenta of particles.. and the measurements correspond to theoretical values. But it seems to me that they do so by designing their apparatus using an additional mental model as a guide, a mental model that is not provided by quantum mechanics itself. It seems that there is on the one-hand quantum mechanics (which is math whose quantities are given physically suggestive labels such as position, energy, etc.) and on the other a set of guidelines for how to relate those quantities to physical lab equipment.

But is there any other reason to believe in the mental model that relates $\Psi(x_1 , x_2, ... ,x_n )$ to measurements other than the correspondence of measured values with theoretical values? In Newtonian mechanics the relation is obvious. Is there a general set of rules that says, "If you use such-and-such apparatus arranged in such-and-such way, then your measurements correspond to the eigenvalues of -ih d/dx," a set of rules which seem reasonable prior to the confirmation that the observed values do in fact match the eigenvalues?

20. Sep 5, 2015

### bhobba

As has been explained in this thread the reason momentum is as it is in QM has nothing to do with a mental model, or even experimental considerations, but with a deeper understanding of what classically momentum is.

In modern times momentum is not defined as mv but via Noether's theorem. It is the conserved quantity from spatial homogeneity that Noethers theroem says must exist. If you take the same definition and apply it to QM you get the momentum operator. And, as you would expect, when you take the expectation values you end up with the classical definition so it is theoretically and experimentally compatible.

Thanks
Bill