The following is taken from Kittel's solid state physics.

An electron moving in a one-dimensional lattice of lattice constant $a$, suffers a periodic potential $U(x)$ where $U(x+a)=U(x)$. The periodicity of $U(x)$ implies $$U(x)=\sum_G U_Ge^{iGx}.$$ where $G=G_n=2\pi m/a$, the reciprocal lattice vectors. Due to periodic boundary condition, the wavefunction satisfies the periodicity condition $\psi(x+L)=\psi(x)$ so that $$\psi(x)=\sum_k C_ke^{ikx}$$ where $k=k_n=2\pi n/L$. Substituting these expansions into time-independent Schrodinger equation for the electron in the lattice, we get $$\Big(\frac{\hbar^2 k^2}{2m}-E_k\Big)C_k+\sum_GU_GC_{k-G}=0.$$

How does the equation above mean that $\psi(x)$ given above can be rewritten as $$\psi(x)=\sum_GC_{k-G}e^{i(k-G)x}?$$


1 Answer 1


Let $\sim$ be an equivalence relation on the set of possible $k$'s, where $k\sim k' \iff k-k' = G$ for some reciprocal lattice vector $G$. That is, $\{k_0,k_{\pm N},k_{\pm 2N},\ldots\}$ is one equivalence class, $\{k_1,k_{1\pm N},k_{1\pm 2N},\ldots\}$ is another, and $\{k_{-1},k_{-1\pm N},k_{-1\pm 2N},\ldots\}$ is a third. We can label each equivalence class by the smallest $k$ (in absolute value) contained within it; then $k_0=0$ is the representative of the first class, $k_1 = \frac{2\pi}{L}$ is the representative of the second, and $k_{-1} = -\frac{2\pi}{L}$ is the representative of the third.

By convention, we choose the set of $N$ distinct representatives to be $$FBZ := \left\{-\frac{\pi N}{L}, -\frac{\pi (N-2)}{L},\ldots,\frac{\pi (N-2)}{L}\right\}$$ Noting that $L=Na$, this can also be written $$FBZ:= \left\{-\frac{\pi}{a},-\frac{\pi}{a}+\frac{2\pi}{L}, \ldots ,-\frac{\pi}{a} + (N-1)\frac{2\pi}{L}\right\}$$ which spans the range $\big[-\frac{\pi}{a},\frac{\pi}{a}\big)$. $FBZ$ is called the first Brillouin zone. In particular, observe that every possible value of $k$ is related to exactly one $k\in FBZ$ via translation by some reciprocal lattice vector.

This being the case, a generic wavefunction $\Psi$ can be decomposed as

$$\Psi(x) = \sum_{\text{all }k}C_k e^{ikx} = \sum_{k\in FBZ}\left( \sum_{G} C_{k-G} e^{i(k-G)x}\right) = \sum_{k\in FBZ} \psi_k(x)$$ where $$\psi_k(x) \equiv \sum_{G} C_{k-G} e^{i(k-G)x} = e^{ikx}\sum_G C_{k-G}e^{-iGx} \equiv e^{ikx} u_k(x)$$ with $u_k(x)\equiv \sum_G C_{k-G}e^{-iGx}$ a manifestly periodic function with period $a$. Furthermore, if we look carefully at the equation

$$\left(\frac{\hbar^2 k^2}{2m}-E\right)C_k = -\sum_G U_G C_{k-G}$$

we see that it only couples $C_k$'s from within the same equivalence class. That is, the Hamltonian operator is "block diagonal" and does not couple these $\psi_k$'s together:

$$H \Psi(x) = \sum_{k\in FBZ} H_k \psi_k$$

As a result, we can solve the Schrödinger equation for one $k\in FBZ$ at a time. The functions $\psi_k$ are called Bloch waves.

  • $\begingroup$ Once you define $\psi_k$, doesn't the proof complete at "... the manifestly periodic function of $a$." ? Doesn't the argument end there? Why do we need the Schrodinger equation? $\endgroup$ Sep 4, 2020 at 9:17
  • $\begingroup$ @mithusengupta123 Without the bit about the Hamiltonian, we've just shown that a generic periodic function can be written as the sum of these $\psi_k$'s. That's not particular surprising and also not particularly interesting. What is interesting is that the Hamiltonian doesn't mix these $\psi_k$'s together. With a bit of thought, it follows that you can choose an eigenbasis of Bloch waves for the Hamiltonian, which is Bloch's theorem. $\endgroup$
    – J. Murray
    Sep 4, 2020 at 16:28
  • $\begingroup$ @mithusengupta123 If you have a big matrix, then if you can find a way to make it block diagonal you can proceed to diagonalize the whole thing by diagonalizing each block individually. This is the same idea in infinite dimensions - if you want to find eigenstates of $H$, you can simply find the eigenstates of $H_k$ for each $k$. The resulting eigenstates will be a Bloch wave eigenbasis for $H$. $\endgroup$
    – J. Murray
    Sep 4, 2020 at 16:32
  • $\begingroup$ This is the best explanation I have chanced upon, thank you! $\endgroup$ Jan 8, 2021 at 19:56
  • $\begingroup$ I have a question regarding the convention of using the first Brillouin zone: in the (almost) free electron model we usually need more than the FBZ to describe electron motion (unlike in phonons where the FBZ is enough), then why is it any better to solve $N$ TISE's corresponding to each $k\in \text{FBZ}$ than say, the first two Brillouin zones? $\endgroup$
    – Jono94
    Dec 27, 2023 at 12:48

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