Is quantum velocity generally equal to group velocity?

Question

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?

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).

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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.

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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.

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*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.