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:

$$HΨ = EΨ$$


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

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!


1 Answer 1


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, $$ \begin{equation} \psi_k(r) = e^{ikr}u_k(r) \end{equation} $$ 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 $$ \begin{equation} H_k u_k(r) = \left[ \frac{p^{2}_{eff}}{2m} + V(r) \right] u_k(r) = E_k u_k(r) \end{equation} $$ 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.

PS: My descriptions are for a 1D case. The generalisation to higher dimensions is straightforward if necessary.


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