- #36
PeterDonis
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redtree said:In this context, the derivation we have been discussing is as follows
No, it isn't. You pulled ##\omega## out of nowhere, and you switched from operators back to vectors halfway through. Here is the derivation I gave earlier, restated for clarity (I've left the vector arrows out, but it is assumed that ##\hat{p}##, ##\hat{k}##, and ##\hat{v_G}## are vector operators, while ##\hat{H}## is a scalar operator):
$$
\hat{p} = \hbar \hat{k}
$$
$$
\hat{H} = \frac{\left( \hat{p} \right)^2}{2 m}
$$
$$
\hat{v_G} = \frac{d \hat{H}}{d \hat{p}} = \frac{\hat{p}}{m}
$$
Therefore,
$$
\hat{p} = m \hat{v_G} = \hbar \hat{k}
$$
Now, the above is just operators, and all of those equations will hold in any representation and for any state; the only assumption we have made is that we are dealing with a free particle, since there is no potential term in the Hamiltonian. Previously in this thread, we were specifically talking about momentum eigenstates, but as the above shows, we don't actually need to do that to derive ##\hat{p} = m \hat{v_G}##.
However, since the above should apply to any state, we should be able to see how it applies to a momentum eigenstate. In the momentum representation, this is of course trivial: for a momentum eigenstate ##\phi(p)##, we have ##\hat{p} \phi(p) = \hbar \hat{k} \phi(p) = \hbar \vec{k} \phi(p)##, where now I've put the vector arrow back on in the last step to make it clear that ##k## there is an eigenvalue, not an operator. Therefore, since ##\hat{v_G} = \hat{p} / m##, we have ##\hat{v_G} \phi(p) = \left( \hbar \vec{k} / m \right) \phi(p)##.
In the position representation, the same operator and eigenvalue equations should hold, and we know that they do hold for a state ##\psi(x) = exp \left( i \vec{k} \cdot \vec{x} \right)##, because we have ##\hat{p} = - i \hbar \partial / \partial \vec{x}##, and therefore ##\hat{H} = - \left( \hbar^2 / 2 m \right) \partial^2 / \partial \vec{x}^2##.
redtree said:In order to complete the derivation, one must show the following
Before one can even talk about ##\omega## and ##\vec{x}##, one needs to first show what operators correspond to them. For ##\vec{x}##, this is easy: in the position representation, we have ##\hat{x} \psi(\vec{x}) = \vec{x} \psi(\vec{x})##. But what operator corresponds to ##\omega##?
Your suggestion, as far as I can see, is that we should define ##\hat{H} = \hbar \hat{\omega}##, by analogy with ##\hat{p} = \hbar \hat{k}##. If we do that, then we have ##d \hat{\omega} / d\hat{k} = d \hat{H} / d \hat{p}##, and the expression for ##\hat{v_G}## stays the same as in the above derivation, which is still perfectly valid.
You then, however, say that we should have ##d \hat{\omega} / d\hat{k} = d \hat{x} / dt##. But for a free particle, none of the operators we have given are time-dependent, so ##d\hat{x} / dt## is zero trivially; yet we still have a well-defined group velocity operator, as has been demonstrated above. So I'm not sure where this claim of yours is coming from.
To put this another way: you seem to be saying that we should have a "velocity operator" ##\hat{v} = d \hat{x} / dt##. But what we really want is to justify the idea that, in the classical limit, whatever "velocity" we are getting from quantum mechanics should become the ordinary classical velocity we are used to for a free particle, which is ##d\vec{x} / dt##. But that's not a statement about operators; it's a statement about expectation values, since those are what should correspond to classical quantities in the classical limit. So what we really want to show is that ##\langle \hat{v_G} \psi \rangle = d \langle \hat{x} \psi \rangle / dt##.
We can't actually do this for a momentum eigenstate, because it doesn't have a well-defined expectation value for position. But we can do it for a wave packet, and I think (based on what I remember of reading in QM textbooks quite a while back, but I have not had time to check) that for a wave packet, the expectation value equation comes out as I have just said.