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We define the translation operator, $T_{op}$ corresponding to the primitive lattice translation vector, $T$, such that $T_{op}f(r)=f(r+T)$. Similarly, $T_{op}\psi_n(r)=\psi_n(r+T)$, where $\psi_n(r)$ is an eigenfunction of the Hamiltonian. We then have that, $$HT_{op}\psi_n(r)=H\psi_n(r+T),$$ while, $$T_{op}H\psi_n(r)=E_{n}T_{op}\psi_n(r)=E_{n}T_{op}\psi_n(r)=E_{n}\psi_n(r+T).$$ Naturally, in an isolated system, the potential is periodic. One can further intuit that the kinetic energy term of the Hamiltonian must be periodic by law of conservation of energy, (i.e. because potential is periodic, then kinetic energy must be periodic in that there is no energy flux in an isolated system). While this is palatable to me, I would like to know if there is a nice mathematical proof that, $$E_n\psi_n(r+T)=H\psi_n(r+T),$$ or if my purely physical argument is sufficient.

One could use Bloch's theorem. That is, $\psi_n(r+T)=e^{ik\cdot T}\psi_n(r)$, however, my book uses the invariance of the Hamiltonian under translations to derive Bloch's theorem, and it seemed a bit circular to use it to prove the converse.

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The first thing to note is that the statement you are trying to prove is that $H$ commutes with $T_\mathrm{op}$ $$ HT_\mathrm{op}=T_\mathrm{op}H $$

An operator function of $x$ will commute with $T_\mathrm{op}$ if and only if $$ V(x) = T_\mathrm{op}^{-1}V(x)T_\mathrm{op} = V(x+T) $$ where in the second equality we have used the defintion of $T_\mathrm{op}$ as a translation. In other words if $V$ commutes with $T_\mathrm{op}$ if it is periodic with period $T$.

The momentum operator is generator of translations; that is a translation operator for a translation by $a$ can be written as $$ T_a = \exp\left(-\imath \frac{a p}{\hbar}\right) $$ which clearly commutes with $p$. In particular this inculdes our particular translation operator $T_\mathrm{op}$. Since $p$ commutes with $T_\mathrm{op}$, so does the kinetic energy operator $\frac{p^2}{2m}$.

Since the kinetic and potential terms individually commute with $T_\mathrm{op}$, so does $H$.

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