It is often said that the velocity operator in quantum mechanics identifies the group velocity. Here's the argument: for the free particle, we have $\omega = \hbar k^2/2m$ and the (quantum) velocity is defined by $$\langle v \rangle = \langle p \rangle/m = \frac{\hbar k}{m}.$$ Using the dispersion relation, we find $$v_p = \frac{\hbar k}{2m}, \quad v_g = \frac{\hbar k}{m}.$$ Since $\langle v \rangle = v_g$, we conclude that the quantum velocity operator measures group velocity as opposed to phase velocity.

I'm having a hard time showing this is true more generally. For example, consider a wavepacket $\psi(x, t)$ with mean wavenumber $k$ and group velocity $v_g(k) = \langle v \rangle$. Then consider the modified wavepacket $e^{iqx} \psi(x, t)$. This state has mean wavenumber $k+q$ and $$v_g = v_g(k+q), \quad \langle v \rangle = v_g(k) + \frac{\hbar q}{m}$$ and these are not necessarily the same! It looks like $\langle v \rangle = v_g$ only works for free particle wavepackets.

Is there something wrong with my argument? In general, how should $\langle v \rangle$ be interpreted in quantum mechanics?

  • $\begingroup$ You keep talking about the "velocity operator" without actually specifying exactly what you mean by the term. That makes it very hard to provide any sensible answer. $\endgroup$ Commented Sep 29, 2017 at 18:41
  • 1
    $\begingroup$ @EmilioPisanty That's part of the question. Evidently in the free particle case $\hat{v} = \hat{p}/m$ works, but if we preserve that definition then $\langle v \rangle$ corresponds to neither $v_p$ nor $v_g$ in general. So I'm basically also asking for a good definition of quantum velocity. $\endgroup$
    – knzhou
    Commented Sep 29, 2017 at 18:43
  • $\begingroup$ What makes you think that you're guaranteed the existence of such an operator in general? (As a counterpoint, neither $v_g$ nor $v_p$ exist in general, even in free space, and they only make sense for quasi-monochromatic wavepackets.) $\endgroup$ Commented Sep 29, 2017 at 18:47
  • $\begingroup$ And why, exactly, do you think that $\hat v=\hat p/m$ does not correspond to $v_g$ in the situations where the latter exists? When you say "we just added a constant to the dispersion relation", it seems that you indicate that the dispersion relation has changed, but you don't describe any such change ─ as currently written, you change the state but you keep the same dispersion relation. $\endgroup$ Commented Sep 29, 2017 at 18:47
  • $\begingroup$ @EmilioPisanty "What makes you think..."? Because people say that quantum velocity is group velocity all the time; such a statement is even in common undergraduate intro textbooks. $\endgroup$
    – knzhou
    Commented Sep 29, 2017 at 19:07

1 Answer 1


It seems to me that you're tying yourself up into knots here. To be clear:

  • defining $\hat v = \hat p/m$ is the only reasonable way* to go about velocity in quantum mechanics.

Group velocity, on the other hand, is a bad concept to pin all your understanding on, because it can only be defined in very specific situations, that is, for wavepackets that are very narrow around a single central frequency. For all other waves (starting with an even superposition of two narrow wavepackets about two different frequencies, let alone anything with an interesting spectrum) the concept makes little to no sense.

To be sure, introductory textbooks do use the concept quite a bit, in the form

for wavepackets for which $v_g$ exists, it coincides with $\langle \hat p/m\rangle$,

which is a good way to justify the introduction of the latter. However, the concept won't extend to a general definition of velocity in quantum mechanics, and you won't find the concept in more serious literature (with the exception of specific treatments of narrow wavepackets, that is).

Beyond that, you're doing some mistakes in what you're plugging into what slots (or rather, your formalism doesn't actually make sense). So, let's consider your example, where we have a free particle in one dimension, with dispersion relation $$ \omega(k)=\frac{\hbar k^2}{2m}, $$ and where we have two states, $|\psi_1\rangle$ and $|\psi_2\rangle=e^{i q\hat x}|\psi_1\rangle$, with wavefunctions $$ \langle x|\psi_1\rangle = \psi_1(x) \quad\text{and}\quad \langle x|\psi_2\rangle = e^{iqx}\psi_1(x)=\psi_2(x), $$ respectively, where $\psi_1(x)$ is a quasi-monochromatic wavepacket that's sharply peaked around central momentum $\hbar k_1$, and therefore $\psi_2(x)$ is a quasi-monochromatic wavepacket that's sharply peaked around central momentum $\hbar k_2=\hbar (k_1+q)$.

So, what do our different measures of velocity give us for these states? Let's find out.

  • For our first state, the operator definition returns $$\langle \psi_1 |\hat v|\psi_1\rangle = \langle \psi_1 |\hat p/m|\psi_1\rangle = \hbar k_1/m,$$ as a standard calculation of how quasi-monochromatic wavepackets work.

Now, let's compare this to the group velocity, which is defined as $$ v_g(k) = \frac{\mathrm d\omega(k)}{\mathrm dk} = \frac{\hbar k}{m}, $$ defined only for quasi-monochromatic wavepackets and evaluated at the central wavevector. With that then,

  • For our first state, the group velocity is $$v_g(k_1) = \frac{\hbar k_1}{m},$$ and it coincides with the operator definition.

OK, so what about the second state?

  • The nontrivial part where you're getting bollixed up is with the group velocity. Here's the important part: the dispersion relation doesn't change, and neither does $v_g(k)$ as a function of wavevector, but you do change the inputs to both of them. Thus, the group velocity of the second wavepacket is $$v_g(k_2) = \frac{\hbar k_2}{m} = \frac{\hbar( k_1+q )}{m},$$ higher by $\hbar q/m$ than the previous value.
  • The operator definition also increases by the same amount; to introduce some amount of nontriviality into this answer, let's calculate this via the operator algebra, within which one can show that $$e^{-iq\hat x} \hat p e^{iq\hat x} = \hat p + \hbar q,$$ so that $$\langle \psi_2 |\hat p/m|\psi_2\rangle = \langle \psi_1 |e^{-iq\hat x} \hat p e^{iq\hat x}/m|\psi_1\rangle = \hbar (k_1+q)/m,$$ which coincides with the group-velocity value.

OK, having said that, let me add a general comment:

how should $⟨v⟩$ be interpreted in quantum mechanics?

You shouldn't try much. Velocity is about as useful an observable in quantum mechanics as window wipers on a submarine: there are some specific situations in which it's useful, but for everyday usage you really do need to shift all the conceptual weight over onto momentum instead.

This is analogous the the shift from $\dot x$ to $p$ in the passage from lagrangian to hamiltonian (classical) mechanics, where you make $p$ a full-blown dynamical variable at the expense of $\dot x$ as a variable of interest. If you're doing QM and you're not doing something very specific (namely, path-integral formalisms, or QFT with quantization all the way from a lagrangian density) then the classical analog is hamiltonian mechanics, not the lagrangian or newtonian formalisms.

This is why people just don't talk much about velocity in QM ─ we use momentum instead.

*Important caveat: That's only true in the absence of a vector potential, i.e., that statement is gauge dependent. If you do have a vector potential, then the velocity needs to be defined as $\hat v = (\hat p-qA(\hat x,t))/m$ or its equivalent. This is exactly the same issue as in classical mechanics and it requires extreme care if you want to not mess up.

  • $\begingroup$ Adding to your *Important caveat: and because p is gauge dependent and v not, you can actually only measure v and not p (no measurement apparatus can give you the value of a gauge dependent quantity). Ballentine's book seems the only one discussing this. $\endgroup$
    – lalala
    Commented Sep 29, 2017 at 20:00
  • $\begingroup$ Thanks for the thorough answer! I still am not sure we're considering the exact same setup -- I'm thinking of a more general dispersion relation $\omega = \omega(k)$ which can be realized by $H \sim \omega(p)$. Then assuming the first state has $v_g = \langle v \rangle$, for the second state I find group velocity $v_g(k_2)$ and $\langle v \rangle = v_g(k_1) + \hbar q/m$, which are not the same. $\endgroup$
    – knzhou
    Commented Sep 29, 2017 at 20:02
  • $\begingroup$ That is, I find a shift in the argument, and a shift in the value, respectively. They only coincide in the free case since $v_g(k)$ is linear in $k$. Did I make a mistake here? $\endgroup$
    – knzhou
    Commented Sep 29, 2017 at 20:03
  • $\begingroup$ @knzhou The result is correct, but I'll have to think some more to see whether it's relevant, because you really don't see those hamiltonians much. Ultimately, you could just define $$\hat v = \frac{\mathrm d\omega}{\mathrm dk}(\hat p/\hbar),$$ but you'd only do that if you're fulfilling some specific need ─ and there really isn't a need for a velocity operator very often. $\endgroup$ Commented Sep 29, 2017 at 20:24
  • 1
    $\begingroup$ No, that's a newtonian view of momentum and you need to be using hamiltonian mechanics. In a classical system with hamiltonian $H=\hbar \omega(p/\hbar)$, the velocity is given by Hamilton's equations as $$\dot x = \frac{\partial H}{\partial p} = \frac{\mathrm d\omega}{\mathrm dk}(p/\hbar),$$ exactly as above, and this is the physical velocity you should use. It feels like you're drastically underestimating the differences between newtonian and hamiltonian mechanics and between kinematic and canonical momentum, even in classical mechanics. $\endgroup$ Commented Sep 29, 2017 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.