What is Velocity in Quantum Mechanics?

[AJ Bentley]

Nobody urges you to talk about where the photon really "is"

I agree wholeheartedly.
The whole concept of location seems to me to be a hangover from a very mechanistic 17th Century view of the world.
The only way we know where anything 'is' is by the effect it's presence has on our detectors. At best that location is inferred from our mathematical projection of 'fields of force', themselves inferred from other experiments where we think we are confident about where similar objects 'are'.
I see no need for statistical interpretations and have no difficulty with the idea that the apparent centre of action of a particle's mass and or charge (or any other aspect) might be indeterminate.

 

[tom.stoer]

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, $$\hat{\vec{p}}/m \neq \hat{\vec{v}}$$!

As I said (several times): I am perfectly aware of the fact that kinematical momentum and canonical momentum may differ and that I simplified my arguments for the case that H = p²/2m + V(x).

The original question was whether velocitiy in quantum mechanics can be defined w/o measuring twice the position. I explained (for a very simple system) that this is possible. In more complicated situations (electromagnetic field, Dirac equation, ...) the very simpy equation v=p/m is no longer valid and one has to use the commutator [x,H]. But still this definition avoids to define velocity by dx/dt (measuring twice the position) and allows one to introduce v w/o measuring position.

 

[sweet springs]

Hi. tom.stoer

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.

But still this definition avoids to define velocity by dx/dt (measuring twice the position) and allows one to introduce v w/o measuring position.

I would like to comment on your idea here.

#1
Relativistic QM says (dx/dt)_i=cα_i where α_i i=1,2,3 are 4x4 matrices. Eigenvalues are +c,-c. This result corresponds to measuring twice the position, because the first precise measurement makes momentum uncertain with no limit of magnitude, so it is almost probable that the speed of particle becomes +c,-c. Well made, isn't it?

#2

x = Zitterbewegung trembling motion + pc^2/H t + const.

This equation comes from integrating dx/dt formula in Heisenberg picture. You see

{(x - Zitterbewegung trembling motion) - x0 }/ t = c^2 p H^-1

Zitterbewegung trembling motion is of order h'/mc, Compton wave length. If you would like to call the left hand side of the formula VELOCITY, though I myself reserve to call it so, it is given by measuring right hand side quantity as you say. See Dirac IX-69 for all here.
Regards.

 

[tom.stoer]

I don't exactly understand your idea. Again: try to construct a velocity operator w/o using the position operator; perhaps the problem with the Zitterbewegung goes away.

 

[sweet springs]

Hi,tom.stoer.

Again: try to construct a velocity operator w/o using the position operator; perhaps the problem with the Zitterbewegung goes away.

To construct a velocity operator in consonant to the position operator is achieve by [H,x]/ih'=cα_i.
This is logical and correct but not interesting because the measurement when achieved destroys the original nature of the system that we would like to explore.

About constructing a velocity operator w/o using the position operator, I am showing

{(x - Zitterbewegung trembling motion) - x0 }/ t = c^2 p H^-1

The rhs operator corresponds to the velocity in classical mechanics, I agree. So you call it velocity in QM but I am reserving because it is different from original mathematical definition of velocity.

In classical limit of h'→0, Zitterbewegung motion→0, HUP does not apply and infinitesimal time interval can be taken. dx/dt, the mathematical definition of velocity, and p/m mechanical correspondent coincide. However in QM, mathematical definition of velocity i.e. successive two position measurements in infinitesimal time, cα_i and classic mechanical correspondent c^2 p H^-1 do not coincide. I read from lhs rough displacement measurement with errors much more than compton wave length divided by finite time interval much more than h'/mc^2. Which one, or the third one, is velocity in QM? It depends on what we expect velocity to be.
Regards.

 

[tom.stoer]

1) Let's summarize non-rel. qm.: In general one can define v by using the canonical formalisms with [H,x]. This works e.g. for the case of the magnetic field; one finds v = (p - eA)/m.

2) I tend to agree that the interpretation of the velocity operator in the Dirac theory is not so clear :-(

 

[alxm]

If you don't mind, I'd add another possible (non-rel) definition, which is to take the square root of the kinetic-energy operator.

If you're dealing with a stationary state, <p> = 0, so you can't use that to get the magnitude of the average momentum (say, if you want to use it to get an idea of the magnitude of relativistic corrections). Or simply to answer the question 'What's the average speed of an electron in an atom?'.

 

[tom.stoer]

If you don't mind, I'd add another possible (non-rel) definition, which is to take the square root of the kinetic-energy operator.

(... times some factor cancelling the mass)

That's fine in case T=p²/2m. But as pointed out correctly it does no longer work (unambiguously) if you couple a charged particle to a magnetic field. That's why the definition via [H,x] is more appropriate.

I don't think that there is any problem with defining velocity in non-rel. qm. The problem is
a) how to measure it directly (= w/o measuring x) and
b) how to generalize it for rel. qm.

 

[vanhees71]

Of course, the question how to measure something is something else than to define the operators representing the corresponding observable. One should always distinguish between the abstract objects of the theory (state vectors or Statististical operators; operators associated with observables) and the "real world". An observable is not an operator in Hilbert space but defined by some real-world measurement and/or preparation process. That the description of the so defined observable within the quantum-theoretical formalism is indeed described by the operators in Hilbert space can only be verified by comparison of the predictions of the theory with experiment. That's the case for all scientific theories and models. The difference between "classical physics" and "quantum physics" is that in the former case the relation between observables (defined operationally by a measurement prescription) and the theoretical objects (components of position or momentum for particles wrt. a frame of reference, field components in classical electrodynamics etc.) are much more direct than in the latter. The correspondence between the quantum-theoretical objects (state vectors, operators) is given by Born's probability interpretation.

One way to measure velocities (at least in principle), even in an ideal von Neumann measurement sense, where with the measurement particles are even prepared such as to have a certain velocity, is to use a velocity filter, i.e., a crossed electric and magnetic field and a little hole, where only those particles come through with a certain velocity (within some accuracy only of course which can be made arbitrarily small but never absolutely 0 due to the Heisenberg Uncertainty Principle).

That such a setup really prepares particles of a definite velocity (component in one direction) can be verified theoretically (at least in principle ;-)) by solving the Schrödinger equation for a particle running through such an electromagnetic field.

 

[tom.stoer]

This is what we already discussed above; it means that one can define and measure x-velocity by measuring yz-position. Fine.