# Momentum in QM

1. Jan 21, 2009

### xboy

In CM, we have a prescription for measuring momentum. Velocity is defined, and we can measure it, and we can find out the momentum. Now, does quantum mechanics have the same prescription for measuring momentum (for a single free particle at least) ? I mean, for a single free particle, can we define a velocity operator first and define momentum operator as mass multiplied by that?

We have this definition of momentum in quantum mechanics that it is the generator of translations. Roughly, all arguments regarding momentum seem to be logically deducible from that. Also, we know that in the classical limit, according to Ehrenfest's theorem, mean of the momentum evolves according to Newton's Law.

But from these considerations, what can we say about the observable momentum's relation with the simple idea of momentum being something to do with how fast a particle moves.

And how do we measure momentum in practice?

2. Jan 21, 2009

### muppet

I put this in another thread yesterday:
In common or garden QM, it's almost meaningless to talk about the particle moving continuously anywhere like in a classical picture; the whole point is that that's not what the system is doing when you're not looking at it. If you are looking at it (taking a large number of measurements to see how the position of the particle changes with time) then the ordinary conception of velocity will do; you're just working with average values.
There is, however, a path integral approach, even to ordinary QM, which describes the time evolution of a state in terms of all possible trajectories linking two moments in time, weighted by their actions. I'm afraid I'm not conversant enough with the formalism to tell you if the average velocity one might associate with such a path differs from the "common sense" definition of a time-averaged velocity (displacement/time) in cases when the classical trajectory is not the only non-neglible one -perhaps someone else can help?

Of course, the Bohmian approach to QM does afford a particle a definite trajectory at all times, so I'd guess that it gives them a definite velocity (albeit one determined by wholly non-classical dynamics), but the best person to ask about that would be Demystifier.

3. Jan 21, 2009

### clem

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

4. Jan 22, 2009

### xboy

'The way I think about ordinary momentum [In QM] is that the gradient of a function is a vector that points in the direction in which the function increases fastest; so the momentum operator picks out the direction in space in which you're most likely to find the particle.'

Yes, but do we come to any conclusions just from knowing the grad of the wave function? We have to solve the eigenvalue equation, and then get the eigenvalues which actually carry some physical meaning ( we can measure them).

Another thing. In CM, we can take two position measurements and find out the momentum/ velocity. In QM, we can't do that, or at least we can't associate the 'velocity' found with a momentum measurement because after a position measurement a particle can't be in a momentum eigenstate.

So how does one measure momentum? Clem says through measuring the curvature in a magnetic field. But can that be done w/o making position measurements? Shouldn't QM come with a prescription 'measure momentum zis way' or something?

5. Jan 22, 2009

### malawi_glenn

what is happening is that particles interact with atoms, creating ionizing tracks, then you measure the radius by means of a ruler..

6. Jan 22, 2009

### DeepSeeded

The way I understand it from the book I am reading currently (Quantum Theory by David Bohm) we measures the momentum of the wave packet associated with the particle.

The particle is thought to be a wave of "different frequencies that interfere constructively over only a small region of space, outside of which they produce an amplitude that reduces to zero rapidly." In other words, an electron is only an resonance phenomenon.

Because of this, the relation momentum = h/wavelength holds for particles as well as electro-magnetic waves in quantum physics.

Correct me if I am wrong.

7. Jan 22, 2009

### muppet

Do we come to any "conclusions" from knowing the classical momentum? It's a vector- a direction in space with an associated magnitude. It still is in QM. But in QM the concept of momentum is different, because at its heart QM is fundamentally and intrinsically different to CM. The best we can hope for is to make an educated guess as to where it will be next, given roughly where it is now.
No, we can't do that. You're right.
The point I was trying to make above is that in convential QM the whole concept of velocity has absolutely no meaning whatever. A particle simply isn't thought to be moving about from definite place to definite place. Anyone who describes a particle as being in anything other than a momentum eigenstate who talks about it moving talks about it moving along several paths at once*, and anyone who talks about a momentum eigenstate has no idea AT ALL of where along that direction in space the particle is to be found. (As I said above, it might be possible to associate a definite "classical" velocity with each of these paths- I'm afraid I don't know.)
Sometimes a "velocity" is defined in the way you suggest, by multiplying the momentum operator by the reciprocal of the mass. In solid state physics, for example, the effects of interference etc. are often limited, and electrons can be described as having a velocity in the classical sense, behaving in a vaguely sensible way when they experience some applied potential. But in QM's theoretical guts there's no real idea of motion per se at all.

The electron is *not* only a resonance phenomenon. This kind of idea was how Schroedinger interpreted his equation. But it was shown (by Sommerfeld, I think) that such a wave packet would inevitably demonstate dispersion- it would spread out as the constituent waves travelled at different speeds.
However, a particle localised to some finite region of space with some small spread in momenta would be described by such a superposition of plane waves, which are the eigenstates of the momentum operator. The group velocity of this wave packet can be identified with the classical velocity, and for a narrow spread of momenta the wave packet will approximately retain its shape for an appreciable period of time.

8. Jan 22, 2009

### DeepSeeded

Does this mean that the equation p = h/wavelength is not correct? Somewhat correct? Is there another equation?

9. Jan 22, 2009

### RedX

In QM, you can solve for the wavefunction as a function of time, square it to get probability, and take the derivative with respect to time to get a flow-field of probability? In fluid mechanics that's the Eularian description of a fluid, but not a Lagrangian description of a fluid.

All the paths you sum over are perfectly normal classical paths (except the velocity can exceed the speed of light, corresponding to paths like when a particle travels around the world many times over, in a short time - so both time-like and space-like paths in the terminology of special relativity). So if you want you can calculate the average velocity for each path. But like you said, I think the final result of such a summation is only meaningful when the classical trajectory is the only non-neglible trajectory (when $$\hbar[\tex/] is much smaller than the action so that only the stationary-path contributes, which is the classical path). In that case the probablity is 0 or 1, depending on if the endpoint of the path is on the path determined classically. 10. Jan 22, 2009 ### muppet It is correct, but our understanding of what it means has changed with time. Originally, De Broglie made his radical proposal of a wave associated with matter, which could be thought of as a "pilot wave", guiding the particle along trajectories that demonstrate interference patterns. This suggestion predated (and in large part inspired) modern quantum mechanics; so the Schroedinger equation, uncertainty relation and probabalistic interpretation of QM were unknown to De Broglie, so he was working with the concept of a particle having a definite momentum and hence a definite wavelength. However, most physicists no longer believe this picture to be true. Instead, plane waves described by complex exponentials are the eigenfunctions of the momentum operator. So if a particle is described by a pure wave with a definite single wavelength, then it has a single well-defined momentum; you just don't know where it is at all. In order to localise it, you have to construct a wave packet by a superposition of several waves, and you no longer have a single-well defined momentum. I'm not sure that the classical trajectories would be the only meaningful ones. In a diffraction experiment, for example, you could have several trajectories with a non-negligible probability- in fact you'd have to in order to produce the interference pattern! What I'm wondering is if each of the path integrals connecting two given points in space and time (e.g. describing a particle's flight to a particular maxmima) are such that the result is a single well-defined trajectory (which I guess would have to look like a Bohmian pilot wave -inspired trajectory!), with which one could associate a velocity, or whether there'd be many non-negligible paths between two possible places the particle could find itself at time t. 11. Jan 24, 2009 ### xboy And what does an ionized track correspond to - a momentum eigenvalue? But then, each interaction with an atom is a position measurement, no? @ Muppet With 'conclusion', what I was trying to say was : this is not the vector we measure. What we measure are eigenvalues, which we theoretically evaluate by solving the eigenvalue equation. Grad psi doesn't by itself tell us anything, it's one side of an equation we need to solve. Why is the direction defined by it important is not clear to me. About 'velocity', I agree with you that it doesn't mean anything in QM. So how then does one measure momentum? We know that we can measure momentum, and that process destroys all information about position. That's intriguing, so I am wondering what are the actual mechanisms of extracting information about momentum. 12. Jan 24, 2009 ### muppet The gradient tells you in which direction in space the function defined on that space increases fastest. For example, think about the function that defines the surface of a sphere. [tex]f(x,y,z)=x^2+y^2+z^2$$
This function gives the square of the radius of the sphere centred at the origin on which the point at which you evaluate it resides. Its gradient
$$\nabla\\f=2(x,y,z)$$
is a vector that points towards the point on the surface of the sphere- it points in the direction in which the function is increasing fastest, which is radially outwards. Have a look at the images on here and see if they help.
So the momentum vector in QM points in the direction the wavefunction increases fastest.

Regarding the measurement- the tracks correspond to a large number of position measurements, yes. The curvature of the tracks tells us the momentum because we can calculate the velocity using electrodynamics given that we know the applied field strength. This is really working things out in a classical sort of way, because it involves averaging out over a large number of measurements of the same particle repeatedly.

13. Jan 25, 2009

### xboy

I understand what the gradient tells us :) the part I have a problem with is your equating the (vector) function momentum operator maps the wave function to (i.e grad psi) as the momentum of the particle. What we measure is not this direction (which would be a function of space), but momentum eigenvalues. Grad psi certainly plays a role in theoretically calculating momentum eigenvalues, what I am contending is that it is not particle momentum.
Regarding measurement, averaging over a lot of momentum measurements gives us the classical momentum. But here we are making a number of position measurements. The whole procedure seems too classical.

What I would want to know is how to make a single momentum measurement. How to do this should be defined within the quantum theories.

14. Jan 25, 2009

### RedX

Does the gradient pointing in the direction of greatest change, apply to complex wavefunctions?

15. Jan 27, 2009

### xboy

^ No.

And I searched this forum and found out how momentum can be measured - crystal diffraction and similar experiments. So I am answered for the nonce. If I have any further questions about those experiments, I'll start a new thread. Thanks everybody.