# How to arrive at the Bloch equation $H(k)u(k) = E(k)u(k)$?

Bloch's theorem states that in the presence of a periodic potential solutions to the Schrödinger equation take the following form:

$$Ψ(k) = exp(ik•r)*u(k)$$

I am trying to show that using this ansatz results in the Bloch equation $$H(k)u(k) = E(k)u(k)$$ but something is lost upon me...

I tried using the Bloch Hamiltonian in the following way:

$$H(k) = exp(ik•r)*H*exp(-ik•r))$$

Time independent Schrödinger equation is:

$$HΨ = EΨ$$

giving

$$[exp(ik•r)*H*exp(-ik•r)][exp(ik•r)*u(k)] = E[exp(ik•r)*u(k)]$$

simplifying to

$$exp(ik•r)*H*u(k) = E[exp(ik•r)*u(k)]$$

From here can I "pull out" the exp(ik•r) term to cancel them? It doesn't feel right. Please point me in the right direction... resources would be immensely helpful!

• Please use MathJax. Commented Oct 4, 2022 at 5:16

A couple of remarks are in order. The periodic potential is periodic in space typically, since $$V \equiv V(r)$$. In other words, we start with knowing that $$V(r) = V(r + R)$$ for some $$R \in \mathbb{R}$$. Then Bloch's Theorem states that, $$$$\psi_k(r) = e^{ikr}u_k(r)$$$$ are the solutions to the Schrödinger equation with $$u_k(r) = u_k(r+R)$$. These solutions are called Bloch functions. Such a situation may be observed in a perfect crystal, where the periodicity would correspond to its atomic structure. As to why I restate the Theorem in position space instead of momentum is because the Bloch functions $$\psi_k(r)$$ aren't generally periodic in the reciprocal space. More details on that in the answer here.
Now, applying the Time-Independent Schrödinger Equation $$H\psi_k(r) = E_k \psi_k(r)$$ to the Bloch function, we get $$$$H_k u_k(r) = \left[ \frac{p^{2}_{eff}}{2m} + V(r) \right] u_k(r) = E_k u_k(r)$$$$ where, $$p_{eff} := -i\hbar\partial_r + \hbar k$$. In the context of a crystal, one may identify this effective momentum as the crystal momentum $$\hbar k$$ (a quasimomentum really) added to the usual momentum operator $$-i\hbar\partial_r$$. I leave the proof to you as an exercise.
Side-note: There's nothing wrong with pushing out the translation operator $$T_r = e^{ikr}$$ to one side in your equations, so long as it commutes (must take care of -ve signs if it anticommutes) with the operators that you're pushing them through. In fact, for a perfect crystal with period $$R$$, we must have $$[H, T_R] = 0$$. As far as cancellations go, since $$T_r$$ is unitary, all you have to do is (pre/post-)multiply by the adjoint $$T^{\dagger}_{r}$$ on both sides after pushing out $$T_r$$ to the same end of either side.