About the velocity of Bloch electrons In a crystal, an electron's wavefunction satisfies the equation
\begin{eqnarray}
H\psi_{n,k}(x)=E_n(k)\psi_{n,k}(x),
\end{eqnarray}
where $n$ is the band index and $k$ is the reduced wave vector. Hence, the wavefunction satisfying the Schrodinger's equation should be
\begin{eqnarray}
\psi(t)=\psi_{n,k}(x)e^{-iE_n(k)t/\hbar}.
\end{eqnarray}
The mean velocity of the electron is defined as
\begin{eqnarray}
v=\frac{d}{dt}\left<{\psi(t)}\right|x\left|\psi(t)\right>.
\end{eqnarray}
In the textbook, the velocity is related to the gradient of the $E_n(k)$ with respect to the wave vector, i.e., 
\begin{eqnarray}
v=\frac{1}{\hbar}\partial_kE_n(k).
\end{eqnarray}
Therefore, the electrons moves forever if the partial is not zero.
However, if we directly calculate the velocity
\begin{eqnarray}
v=\frac{d}{dt}\int dx \psi_{n,k}^*(x)e^{iE_n(k)t/\hbar}x\psi_{n,k}(x)e^{-iE_{n}(k)t/\hbar}=0.
\end{eqnarray}
Where am I wrong?
 A: I think that the simple answer here is that $m\frac{d}{dt} \langle x\rangle\ne\langle p\rangle$. The first is proportional to the rate of change of the expectation value of the position, and the second is the expectation value of the quantum-mechanical momentum. They are different in subtle but important ways.
Look at the simplest example of what could be a bloch wavefunction: the momentum eigenstate $|\psi\rangle~e^{i(kx-\omega t)}$.  It's a momentum eigenstate with momentum eigenvalue $\hbar k$ as expected:
$$
p|\psi\rangle=pe^{i(kx-\omega t)} = -i\hbar \frac{d}{dx}e^{i(kx-\omega t)}=\hbar k|\psi\rangle
$$
However,
$$
\frac{d}{dt}\langle x\rangle=\frac{d}{dt}\int_{\text{all}}e^{i(kx-\omega t)}xe^{-i(kx-\omega t)}dx=\frac{d}{dt}\int_\text{all}xdx=0
$$
The integral is clearly not well behaved, which might have you woried, but it's also clearly not time-dependent, so I think it's fair to just call it zero. More mathematically versed physicists might have something more intelligent to say about the divergence of the integral. Regardless, it's clearly not $\propto k$. This is also not a surprise for several reasons.


*

*It is generically true that for any energy eigenstate $|\psi\rangle$, $\frac{d}{dt}\langle x\rangle=0$, this follows from the definition of an energy eigenstate. For more well-behaved cases this usually implies that the momentum distribution of the eigenstate must be even (that can probably be traced to only $p^2$ appearing in the kinetic energy), but in the case the wavefunction extends to infinity the relation above holds even for non-even momentum distributions.

*Bloch wavefunctions oscillate off to infinity, so the expected position (i.e. your integral before taking the derivative) is not well-defined. It therefore seems ill-suited to try to define a velocity based on the change in the expected position. This definition is a very classical one: It describes how your measurements evolve with time. The particle's position don't evolve with time because at t=0, the particle is everywhere, and at $t=t_0$, the particle is still just everywhere. 

