What is Velocity in Quantum Mechanics?

[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?

 

[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

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

 

[orienst]

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

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

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.

 

[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.

 

[tom.stoer]

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.

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.

 

[tom.stoer]

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.

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."

 

[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!

 

[tom.stoer]

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!

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 :-)

 

[nnnm4]

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

 

[tom.stoer]

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

 

[reilly]

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 :-)

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

 

[akhmeteli]

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.

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

 

[orienst]

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!

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

 

[sweet springs]

Hi, 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.

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.

 

[sweet springs]

Hi, tom-stoer

Of course one can define a velocity operator

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

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.

 

[tom.stoer]

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.

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.

 

[tom.stoer]

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

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.

 

[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.

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 :-(

 

[Naty1]

I want to know what does velocity really mean in quantum mechanics

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.

 

[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.

 

[sweet springs]

Hi, tom.stoer,

It is enough to measure the x-position of the particle in order to calculate the y-momentum.

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.

 

[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.

 

[sweet springs]

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

 

[yossell]

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.

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 traveled 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 traveled 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 traveling 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.

 

[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 allows 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.

 

[yossell]

If one defines velocity by v = p/m there is no reason to worry about position.

Sure, one can redefine concepts.

 

[tom.stoer]

It's not a redefinition.

p=mv
v=p/m

What's wrong with it?

 

[Antiphon]

If someone show me the case that completely no position information is necessary for momentum measurement, I appreciate it very much.
Regards.

You can measure the momentum of a photon by bouncing it off of a diffraction grating. The direction it recoils into is a measurement of the momentum. There is no position information needed and the larger the grating (the less you know about the position) the more accurate will be the momentum measurement.

 

[Helios]

Now for waves, there's phase velocity and group velocity. I think group velocity corresponds to classical velocity.

 

[sweet springs]

Thanks, Antiphon.
Any geometrical arrangement has to be prepared for any experiment. Here two positions i.e. of grating and of screen or detector must be arranged. It may not be physical but philosophical concern I have.
As I said in my second last post, it is nothing to do with HUP. Here diffraction grating disturbs original orientation of momentum. Momentum used to be prior to entrance to grating and position on the screen/detector are not HUP partners.
Regards.

 

[sweet springs]

Hi.

Q. In QM how much is the speed of particle?
A1. pc^2/H
A2. speed of light
Both answers have their grounds.

In QM we may decompose velocity to instantaneous velocity and average velocity. Magnitude of instantaneous velocity is almost speed of light. Integrating instantaneous velocity by time, we get Zitterbewegung motion after Schroedinger. Average velocity is pc^2/H that is p/m in non relativistic limit. Two velocities commute. Integrating sum of two kinds of velocities by time, x = Zitterbewegung trembling motion + pc^2/H t + const. See Dirac IX-69.

Regards.

 

[yossell]

It's not a redefinition.

p=mv
v=p/m

What's wrong with it?

Nothing is wrong with it - nothing at all. Where did I say something was wrong with it?

It is a different definition from `distance traveled divided by time taken', the concept Naty1 was using.

 

[tom.stoer]

OK.

In classical mechanics (Hamiltonian mechanics) one eliminates v = dx/dt and uses p instead. Of course there is a one-to-one correspondence.

In qm it is all about a definition of an apropriate operator. That's why my conclusion is that v=p/m is natural. Of course deriving it from [H,x] is more general and I agree that in some cases it must be used.

All these definitions do not require to measure x twice and to calculate dx/dt = (x'-x)/(t'-t). This can be done in classicsal mechanics but I see no way to use it as a qm operator.

 

[vanhees71]

OK.

In classical mechanics (Hamiltonian mechanics) one eliminates v = dx/dt and uses p instead. Of course there is a one-to-one correspondence.

In qm it is all about a definition of an apropriate operator. That's why my conclusion is that v=p/m is natural. Of course deriving it from [H,x] is more general and I agree that in some cases it must be used.

All these definitions do not require to measure x twice and to calculate dx/dt = (x'-x)/(t'-t). This can be done in classicsal mechanics but I see no way to use it as a qm operator.

The funny (and dangerous!) thing is that it is often but not always right. You should be aware that quantum theory is based on the Hamilton formulation of mechanics (or field theory for that matter), and the momenta appearing there are the canonical momenta.

The operator, representing the observable velocity is, as already said several times in this thread unambiguesly given by

$$\hat{\vec{v}}=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}},\hat{H}]$$.

This is $$\hat{\vec{p}}/m$$ if and only if the Hamiltonian is of the form

$$\hat{H}=\frac{\op{\vec{p}}^2}{2m} + V(\hat{\vec{x}})$$.

Already for a particle in an external magnetic field, $$\hat{\vec{p}}/m \neq \hat{\vec{v}}$$!

 

[orienst]

Anyone who majored experimental physics can tell me how to measure the momentum in the laboratory ?