= Solution
The single-particle <Hamiltonian> $H_0=-\hbar^2\partial_x^2/(2m)+V(x)$ has a <periodic potential> $V(x+d)=V(x)$. Define the unitary <spatial translation operator> $T_d f(x)=f(x+d)$. Its derivatives commute with translation, and the potential's periodicity gives $H_0T_d=T_dH_0$.
Each energy eigenspace is therefore preserved by $T_d$, so within it one may choose simultaneous energy and translation eigenstates. On a large periodic interval of $M$ cells, $T_d^M=1$ and its eigenvalues are phases $\lambda=e^{ikd}$, with $k=2\pi j/(Md)$; the infinite-volume limit gives continuous quasimomentum modulo $2\pi/d$. For a simultaneous state in band $j$, this proves
$$
\boxed{\psi_{jk}(x+d)=e^{ikd}\psi_{jk}(x).}
$$
Then $u_{jk}(x)=e^{-ikx}\psi_{jk}(x)$ obeys $u_{jk}(x+d)=u_{jk}(x)$, which constructs the usual <Bloch state> rather than assuming its form. The label $j$ distinguishes <energy bands> at the same <crystal momentum> $\hbar k$. In a degenerate energy eigenspace an arbitrary superposition need not itself have one translation phase; the result refers to the simultaneous <Bloch states>.
Back to article page
