= Solution
For a lattice-periodic one-electron <Hamiltonian>, <Bloch's theorem> gives
$$
\boxed{\psi_{n\mathbf k}(\mathbf r)=e^{i\mathbf k\cdot\mathbf r}u_{n\mathbf k}(\mathbf r),\qquad
u_{n\mathbf k}(\mathbf r+\mathbf R)=u_{n\mathbf k}(\mathbf r).}
$$
Here $\mathbf R$ is any lattice translation and $n$ labels an <energy band>. Under that translation the <wavefunction> gains phase $e^{i\mathbf k\cdot\mathbf R}$. The <crystal momentum> $\hbar\mathbf k$ labels this translation eigenvalue and is defined modulo $\hbar\mathbf G$ for a <reciprocal lattice> vector $\mathbf G$.
The <mechanical momentum> is different. In zero <magnetic field> its operator is $-i\hbar\nabla$, which acts as
$$
-i\hbar\nabla\psi_{n\mathbf k}
=\hbar\mathbf k\psi_{n\mathbf k}
+e^{i\mathbf k\cdot\mathbf r}(-i\hbar\nabla u_{n\mathbf k}).
$$
A general <Bloch state> is therefore not a mechanical-momentum eigenstate. Expanding its periodic factor gives plane-wave components with momenta $\hbar(\mathbf k+\mathbf G)$, with probabilities determined by their <Fourier coefficients>. For a local <Hamiltonian> $p^2/(2m_e)+V(\mathbf r)$, its mean momentum satisfies
$$
\boxed{\langle\mathbf p\rangle=m_e\mathbf v_g=\frac{m_e}{\hbar}\nabla_{\mathbf k}E_n(\mathbf k),}
$$
not generally $\hbar\mathbf k$. The lattice exchanges actual momentum with the electron; crystal-momentum conservation in scattering is instead modulo reciprocal-lattice vectors, as in <Umklapp scattering>.
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