Velocity in quantum mechanics

1. Jul 16, 2010

orienst

I want to know what does velocity really mean in quantum mechanics. Since the particle doesn’t have exact position, how can we talk about the velocity and momentum?

2. Jul 16, 2010

tom.stoer

One can have a "particle" in a momentum eigenstate

$$\hat{p}|\psi\rangle = p|\psi\rangle$$

Of course one can define a velocity operator

$$\hat{v} = \frac{\hat{p}}{m}$$

And the above mentioned eigenstate will be a velocity eigenstate as well:

$$\hat{v}|\psi\rangle = v|\psi\rangle =$$

with

$$v = \frac{p}{m}$$

If the state is not an eigenstate of the momentum operator the particle will not have an exact velocity; instead one has to use the expectation value

$$\langle v \rangle_\psi = \langle\psi|\hat{v}|\psi\rangle$$

And of course one can write down an uncertainty relation for the velocity

$$\Delta x\; \Delta p \ge \frac{\hbar}{2}\;\; \to \;\; \Delta x\; \Delta v \ge \frac{\hbar}{2m}$$

Last edited: Jul 16, 2010
3. Jul 16, 2010

orienst

If a "particle" in a momentum eigenstate, then according to the uncertain principal, the particle’s position has the biggest uncertainty. It may appear everywhere. Then you say that the particle has an exact velocity… I can’t understand that.

4. Jul 16, 2010

DrDu

In general the velocity operator is $$\frac{i}{\hbar}[H,x]$$, where H is the hamiltonian.
This is the analog of the classical Poisson bracket formula for x dot.

Even if the position x of a particle is indetermined, the correlation function of finding the partile at position x+d at time t had it been at position x at 0 may have a sharp value. Then d/t yields the velocity of the particle.

5. Jul 16, 2010

tom.stoer

In most cases your definition coincides with mine as the commutator will just produce a p from the p² in H; yours is more general than mine.

6. Jul 16, 2010

tom.stoer

You cannot understand it. Nobody does. You can only apply the formalism.

Niels Bohr said, "Anyone who is not shocked by quantum theory has not understood it."

7. Jul 16, 2010

DrDu

If I tell you that my home is 8 km from my work, and that it takes me 30 min to go there you can calculate my speed although you have no clue where I am located.
Have a nice weekend guys!

8. Jul 16, 2010

tom.stoer

But according to QM you can be both at home and in your office at the same time and with the same probability, even so you are moving forward with constant velocity :-)

9. Jul 16, 2010

nnnm4

No QM says that when an ensemble of similarly prepared systems are measured, different values of the measurement will occur with different probabilities.

10. Jul 16, 2010

tom.stoer

This is the ensemble interpretation; there are others as well.

11. Jul 16, 2010

reilly

Not so. There is absolutely no evidence that I, or you, or most anything can be in two or more distinct places at the same time. The joint probability that I can be in Seattle and Chicago at the same time is zero; any theory that describes Nature must honor this fundamental property of things. In fact,the very structure of QM, and classical dynamics as well, requires that a dynamical variable can have one and only one value per measurement.

So, assertions that QM says we can be in two places at the same time are at best confusing and confused.
Regards,
Reilly Atkinson

12. Jul 16, 2010

akhmeteli

The Dirac equation presents an important example where these two definitions do not coincide

13. Jul 16, 2010

orienst

You are calculating the velocity using the classical mechanics. I don't think you can use macroscopic objects to explain the microscopic phenomenon.

14. Jul 17, 2010

sweet springs

Hi, orienst.
I share inquiry with you because the definition of speed or velocity needs position measurements, i.e. definition of speed is v(t)= {x(t+dt) - x(t)}/dt as we have learned. How we can get value of v(t) without measuring x(t) and x(t+dt)? Does QM apply another definition of v(t)?
Regards.

15. Jul 17, 2010

sweet springs

Hi, tom-stoer
Following my previous post, can you measure momentum without position measurement? For example when we use magnetic field to measure momentum by bending the trajectory of charged particle, we need to know where the particle beam pass through. If someone show me the case that completely no position information is necessary for momentum measurement, I appreciate it very much.
Regards.

16. Jul 17, 2010

tom.stoer

This is fundamentally wrong.

If you look at the double-slit experiment and try to explain what's going one when one single photon goes through the slits the formalism (e.g. the path integral) tells you that the only interpretation consistent with reality (with the measurement of interference pattern) is that the photon goes through both slits and that these two (classically mutually inconsistent) paths lead to interference.

Of course you can insist on the fact that talking about photons and humans is different. I agree that you are not a quantum object and that the same reasoning does not apply w/o modifications. But the same is true the other way round. The reasoning that because you have never been in two places at the same time does not mean that this must be true for photons.

You can also avoid to talk about any interpretation at all. Nobody urges you to talk about where the photon really "is" in the double slit experiment. It is enough to calculate the amplitudes w/o thinking about any interpretation. That's Ok. But as soon as you start to think about an interpretation you get into these weird reasonings. No way out.

17. Jul 17, 2010

tom.stoer

I agree. I only had the feeling that orienst wanted to have a rather elementary explanation. Of course your reasoning is perfectly valid but in most cases the commutator [H, x] does nothing more but giving you a p from the p² in H. orienst's problem does exist already for the simplest case H= p²/2m.

18. Jul 17, 2010

tom.stoer

The problem in qm is that you can neither measure nor assign values to two non-commuting observables at the same time. But you can for communiting observables. If you set up a measurement with a magnetic field you do not need the bending of the trajectory. It is enough to measure the x-position of the particle in order to calculate the y-momentum.

In addition if you have a free particle w/o external forces, the non.-rel. Hamiltonian is just H=p²/2m. So measuring energy is the same as measuring momentum.

If you want to have a detailed experimental setup I can't tell you more. I am focussed o theoretical physics :-(

19. Jul 17, 2010

Naty1

I want to know what it "really" means in classical mechanics....the best we can say is that, for example, d = vt; but nobody knows what space (distance) "really" is nor, especially, what time "really" is. So how can anyone "really" understand velocity?? Everybody thought the answer was 'obvious' until Einstein.

So the best we can do so far, after several thousand years of scientific progress, is explain what we observe with mathematics; that's gotten us a good way but we likely have an even longer way to go.

20. Jul 17, 2010

DrDu

I just wanted to point out that for the determination of a distance it is not necessary to measure two positions. Consider e.g. an atom which travels in a constant potential gradient. Now it gets ionized at t1 and takes up an electron at t2. I could measure somehow the potential difference of the two electrons which is proportional to the distance of the two events but contains no information about the absolute position of the atom.

21. Jul 17, 2010

sweet springs

Hi, tom.stoer,
Please take a look at mass spectrometer experiment configuration below.

y direction

0→x direction

○○○○●: detector for the measurement of position (x,y).
○○○/
○○/ Long distances so that the magnetic field area can be regarded as a point in calculation of bending angle
○/
□area of z-magnetic field. The beam must pass through this area.

beam of particle with y momentum

Not only y but also x should be measured to know bending angle i.e y momentum that the particle had befre it entered the magnetic field area, shouldn't it? I would like to know the experiment configuration of your case.

We can easily see that precise measurement of position (x,y) at detector brings precise measurement of y-momentum. It does not break HUP because position "now" and momentum "use to be prior to interaction with magnetic field" are not HUP partners.

Regards.

Last edited: Jul 17, 2010
22. Jul 17, 2010

tom.stoer

Think about a spherical screen with radius R around (x,y) = (0,0) instead of a planar one. The electrons start from a source at (0,0), of course with some uncertainty. Again we have a magnetic field in z-direction.

At each point of the screen you have a local coordinate system with its origin in (x,y) with x²+y² = R² which is rotated by a certain angle which you can calculate geometrically. At (x,y) = (R,0) the rotation angle of this local coordinate system is zero.

Now once an electron hits the screen at (x,y) one exactly knows the rotation angle. This corresponds to measuring the "x' coordinate" which according to the local coordinate system defines the tangent space of the spherical screen. But the electron has moved along the "y' direction" which is radial = perpendicular to the screen.

So by measuring the point where the electron hits the screen you measure its radial velocity.

23. Jul 17, 2010

sweet springs

Hi, tom.stoer.
Thanks. I see you mean ρ-direction and φ-direction in cylindrical coordinate.
Regards.

24. Jul 17, 2010

yossell

I agree one can always ask the `really' question ad nauseam; but do you *really* think velocity in classical mechanics is as puzzling as in QM? Yes - velocity is distance travelled divided by time taken; yes, it turned out that time taken turned out to be a relative rather than absolute notion; many had long suspected that velocity was only a relative notion too, and that distance travelled was relative to a frame, always part of the concept.

Applying the concept to an object which lacks determinate position and does not travel along a definite trajectory seems problematic as it undermines the idea that the object is travelling a certain distance during its flight. Now, I think there are ways to combat this worry, and many have been mentioned in this thread, but it seems a decent worry in this context.

25. Jul 17, 2010

tom.stoer

If one defines velocity by v = p/m there is no reason to worry about position. It is of course no realistic interpretation and v = dx/dt seems to make more sense, but in qm things need not make sense any more; all that is required is that predictions are confirmed by experiment.

What one will find out in certain cases is that v = p/m allowes one to derive something like <dx/dt> and that this is rather close to a classical velocity. But that depends on the specific quantum state we are talking about; it need not work for all states.

Look at a qm oscillator: the Heisenberg operator x(t) "oscillates" with the classical frequency, but <x(t)> doesn't oscillate at all.