It is often stated that Bloch's theorem and Floquet's theorem are equivalent, even the Bloch's theorem is often referred as Floquet-Bloch theorem. However, it seems quite confusing to me since the former involves a second order differential equation (Schroedinger equation with a periodic potential) while the latter is defined for a first order one.

Can someone clarify this to me? Also it would be nice if I can get references that connect the two.


2 Answers 2


No worries about the order of the differential equation. You can always transform a second or higher order equation to a system of first order differential equations.

The Bloch theorem is dealing particularly with the Schrödinger equation, while Floquet's theorem holds for any homogeneous, linear system of first order differential equations with a periodic coefficient matrix.

You can for instance start with the stationary, one-dimensional Schrödinger equation $$-\frac{\hbar^2}{2m}\Psi''(x)+V(x)\Psi(x)=0$$ with an $L$-periodic potential $V(x+L)=V(x)$.

Now substitute $\varphi_{1}(x):=\Psi(x)$ and $\varphi_{2}(x):=\Psi'(x)$ to obtain the first order system $$\begin{pmatrix} \varphi_{1}'(x)\\ \varphi_{2}'(x) \end{pmatrix}= \begin{pmatrix} 0 & 1\\ \frac{2m}{\hbar^2}V(x) & 0 \end{pmatrix} \begin{pmatrix} \varphi_{1}(x)\\ \varphi_{2}(x) \end{pmatrix}=A(x)\vec{\varphi}(x)\text{ .} $$ Since $A(x)=A(x+L)$, the conditions of Floquet's theorem are met. Floquet theory states that the fundamental matrix solution $\Phi(x)$ of this system reads $$\Phi(x)=P(x)\mathrm{e}^{xB}\text{ ,}$$ with $P(x)=P(x+L)$. If you compute the matrices $P$ and $B$, it should become obvious that this is a Bloch function and $\Phi(x+L)=\Phi(x)$. Unfortunately I don't know any reference where this somewhat elaborate calculation is carried out in detail.

  • $\begingroup$ In fact, Bloch's theorem is actually a bit more general: it is about crystals in any number of spatial dimensions, while Floquet's theorem is about ODEs. $\endgroup$
    – Ruslan
    Dec 29, 2014 at 10:04

Here's a small addition of how I've always understood it:

In the band theory of solids basically every problem is dealt with by prescribing a non-zero periodic potential energy $\mathcal{V}(r)$ in Schroedinger equation for an electron in a crystal, of course along with a set of assumptions:

  1. The wave-functions are calculated for perfect lattices and scatterings are introduced later as perturbations.

  2. In principle one just takes a single electron system and treats everything else in the crystals as an effective potential energy $\mathcal{V}(r).$

  3. All this allows to adopt the one electron Schroedinger equation: $$[(-\hbar^2/2m)\nabla^2+\mathcal{V}(r)]\psi =\epsilon \psi$$ (with $\epsilon$'s chosen in accordance with the Fermi-Dirac distribution.)

The first step of the works of Felix Bloch entailed considering the total potential in two parts, the electrostatic potential of atomic cores with translational periodicity and the potential due to all other electrons, assuming a constant charge density. He then proposes to substitute the total $\mathcal{V}(r)=-eV(r)$ into the Schroedinger equation above, which has also the periodicity of the lattice. All of which led to his conclusion of Bloch wavefunctions which satisfy the above Schroedinger equation: $$\psi_k (\mathbf{r})=U_k (r)e^{i\mathbf{k}\mathbf{r}}$$

Finally, note that the Schroedinger equation including the mentioned assumptions above belong to the family of differential equation known in mathematics as Hill's equation: where the periodicity of $\mathcal{V}(r)$ requires the existence of solutions of Bloch-type, and this is usually known (more to mathematicians) as Floquet's theorem.

So you see there's not much difference, and just think of Floquet's theorem as the general case governing such differential equations, and whenever one discusses electrons in lattices, the theorem leading to the equation of $\psi_k$ is described as the Bloch theorem.

For a more detailed discussion:

J.S Blakemore discusses this distinction in his book of Solid State Physics, second edition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.