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What is the definition of velocity in quantum mechanics? is it an operator?

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    $\begingroup$ Just a brief note (FWIW) that, in the context of Bohmian mechanics, there are actual particles with actual velocities yet, the velocity of a particle is not an observable. $\endgroup$ Sep 22, 2018 at 2:14

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  1. The $j$th velocity operator in the Heisenberg picture can be calculated using Heisenberg's EOM $$ \dot{\hat{q}^j}~=~\frac{1}{i\hbar}[\hat{q}^j, \hat{H}],$$ where $\hat{q}^j$ is the $j$th position operator.

  2. In the Schrödinger picture, the $j$th velocity operator is defined as the commutator $\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$.

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The problem with thinking about velocity in QM is that particles do not have a defined position until being measured. So when you have the classical definition of velocity $$\mathbf{v}=\frac{d\mathbf{x}}{dt}$$

The time derivative of position cannot be determined.

You can, however, look at the time derivative of the expectation value of the position, and you can get to a familiar looking relation with $H$ as the Hamiltonian $$\frac{d\langle x\rangle}{dt}=\frac{i}{\hbar}\langle [H,X]\rangle=\frac{\langle p \rangle}{m}$$

And you could define this to be the velocity in QM. But it isn't an operator.

You usually just have the momentum operator rather than a velocity operator. Although I guess you could define a "velocity operator" as the momentum operator divided by the mass of the particle. You would just have issues with makes massless particles. The momentum operator is much more useful.

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Since $$\hat p = \frac{\hbar}{i} \frac{\partial}{\partial x},$$ you can define a velocity operator $$\hat v \equiv \frac{\hat p}{m} = \frac{\hbar}{im} \frac{\partial}{\partial x}.$$ This is of course only valid for particles that have mass ($m \ne 0$).

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  • $\begingroup$ And technically you wouldn't even need to define the operator in the position basis. $\endgroup$ Sep 21, 2018 at 23:53
  • $\begingroup$ You mean, like in a momentum basis? $\hat v = p/m$? Or just in general? $\hat v$ being defined such that for definite momentum states we have $\hat v \ket p = p/m \ket v$? $\endgroup$ Sep 22, 2018 at 1:44
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    $\begingroup$ I just meant to could define the operator without reference to any basis. $\endgroup$ Sep 22, 2018 at 1:46
  • $\begingroup$ That is very true! $\endgroup$ Sep 22, 2018 at 1:48
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The answer by Qmechanic is quite standard and here I will try to offer a further explanation by relating it to group velocity of wavefunctions. We already have(1D case as an example)

$$ \frac{d\langle x\rangle}{dt}=\frac{i}{\hbar}\langle [H,\hat x]\rangle.$$ Not necessarily $\frac{d\langle x\rangle}{dt}=\frac{\langle p \rangle}{m}$, for example for an electron in magnetic field $H=\dfrac{1} {2m}\left(p+eA\right)^{2}$ we get $\frac{d\langle x\rangle}{dt}=\frac{\langle p+eA \rangle}{m}$, it's acutally mechanical momentum over mass.

Using momentum eigenbasis(suppose $[\hat{k},\hat{H}]=0$) we get
$$ \frac{i}{\hbar}\langle [H,\hat x]\rangle=\frac{i}{\hbar}\langle [H,i\hbar\ \frac{\partial}{\partial p}]\rangle=\langle \frac{\partial}{\partial p} H\rangle=\frac{1}{\hbar}\frac{\partial E(k)}{\partial k}.$$ I think the above result can be interpreted as velocity only when we are trying to describe a wave packet satisfying minimum uncertainty relation. Such a wave packet corresponds to a real particle in the classical world. The speed of this wave packet, i.e group velocity, represents a particle's velocity in the real world. For other types of wavefunctions, for example, the plane wave,$\frac{\partial E(k)}{\partial k}=\frac{\hbar k}{m}$ can't be directly recognized as velocity since now the position is completely uncertain.

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In ordinary quantum mechanics, of course we have the wavefunction of a particle traveling in one dimension,

$$ \langle x \lvert \Psi \rangle = \Psi(x) $$

which contains all the possible information we could ever want to know. To extract information that is testable, we take expectations values of observables which exist as operators.

What is the definition of velocity in quantum mechanics? is it an operator ?

So, one can call,

$$v = \frac{dx}{dt}$$

a velocity operator for a dynamical particle, and then take its expectation value. But this is awkward, however, since in quantum mechanics we use the canonical conjugate variables of position, $x$, and momentum, $p$, which are related via an uncertainty relation (they do not commute).

Alternatively, we take the velocity to be the time derivative of the expectation value of position,

$$v = \frac{d\langle x \rangle}{dt} = \langle \frac{p}{m} \rangle$$

which is not an operator.

More generally, the notion of velocity is encapsulated in the current density operator, also called the probability current, which is the rate of flow for a probability density that is treated as a heterogeneous fluid.

It does appear latently, for instance, in the quantum mechanical Ehrenfest's Theorem.

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