 Solution

The single-particle Hamiltonian $H_0=-{\hslash }^2{\partial }_x^2/(2m)+V(x)$ 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 $H_0{T}_d=T_d{H}_0$.

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 provesThen $u_{jk}(x)=e^{-ikx}{\psi }_{jk}(x)$ 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 $\hslash k$. In a degenerate energy eigenspace an arbitrary superposition need not itself have one translation phase; the result refers to the simultaneous Bloch states.