I'm currently reading the book Solid State Physics by Neil W. Ashcroft and N. David Mermin. In Chapter 12 they introduce the "Semiclassical Model of Electron Dynamics". In short: After having solved the equations for a periodic crystal potential (solutions are Bloch-states $|\Psi_\vec{k} \rangle $), they tinker a wave packet $|\Psi_{\vec{k}_0} \rangle $ made out of Bloch-states, which then has a mean value for $\vec{k}_0$ and a mean value for the position $ \vec{r}_0$ (I added the index 0 here, the book doesn't, but its just renaming things).

The equations of motion for this wave package are then given as: $$ \dot{\vec{r}}_0 =\frac{1}{\hbar} \frac{\partial \epsilon(\vec{k}_0)}{\vec{k}} $$ Which is explained in Appendix E: There it says that for a given Bloch-State $ | \Psi_\vec{k}\rangle $ (one!) , the partial derivative of the eigen-energy is the mean velocity.

Question 1: Since I'm having a wave packet here, wouldn't I have to take the average $ \int d\vec{k} \frac{1}{\hbar} \frac{\partial \epsilon(\vec{k}}{\vec{k}}) g^*(\vec{k}) g(\vec{k})$ here? (g denotes the percentage by that $ | \Psi_\vec{k}\rangle $ contributes to $|\Psi_{\vec{k}_0} \rangle $).

The second equation of motion is given as: $$ \hbar \dot{\vec{k}_0} = -e[E(\vec{r}_0, t) + \frac{1}{c} \dot{\vec{r}}_0 \times H(\vec{r}_0, t)] $$

For justification the reader is referred to some Papers, which calculate these eqations of motion (more or less approximately), and as it is said in Appendix H, this Equations of motion look just like the usual classic hamiltonian equations of motion: If you interpret $\epsilon(\vec{k})$ as some kind of kinetic energy of a classical particle, and $\vec{k}$ as a momentum, then the equations that are written above are just the equations described by hamilton dynamics.

Question 2 (the Important one): Why can I interpret $\epsilon(\vec{k})$ as a kinetic energy term of a classical hamiltonian? More important: Why can I interpret $\vec{k}$ as momentum? $\vec{k}$ describes the crystal-momentum, which is not the same as the momentum ($\vec{k}$ is not an eigenvalue of the momentum operator). Of course, I can just say "well, assuming those things gives the right equations that were allready derived", but is there also an explanation that explains why this is approach is valid, starting from the quantum mechanical equations for $\dot{\langle \vec{r} \rangle}$ and $\dot{\langle \vec{p} \rangle}$ and ending at $H(\vec{r}, \vec{p}) = \epsilon(\frac{1}{\hbar}[\vec{p} + \frac{e}{c} \vec{A}(\vec{r}, t)]) - e\Psi(\vec{r},t)$ with $\vec{p} = \hbar \vec{k} - \frac{e}{c} \vec{A}(\vec{r}, t)$?


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