# 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.