Let's say I have a single particle hamiltonian in a periodic potential, for example a 1D lattice such that: $$H = -\frac{\partial_x^2}{2m} + V(x) $$ with $ V(x+a) = V(x)$ where $a$ is the lattice spacing between the atoms or sites. It is known by Bloch's theorem that a solution to such a system will have the form $$\psi_{k}(x)=e^{ikx}u_k(x)$$ where $u_k(x+a)=u_k(x)$.

My questions is about the boundary conditions. If we take $$\psi(x+Na) = \psi(x)$$ we get, if $N$ is large enough, a lot of different values for $k$ in the first Brillouin zone: $$k=\frac{2\pi n}{N} \text{ with }-\frac{\pi}{a}<k<\frac{\pi}{a},$$ so we get a band of possible states.

In this case we can define Wannier functions which using Fourier over the wave-functions: $$\phi(x-R) = \sum_k e^{-ik R} \psi_k(x)$$ where the summation is over all the $k$'s in the first Brillouin zone.

But if I take the B.C $$\psi(x+a) = \psi(x)$$ I get a single value for the momentum in each Brillouin zone $$k = 0, \pm 2\pi , \pm 4\pi,...$$ Is it still possible to define a Wannier function for such a state? I mean what will the Fourier be like if we have a single possible value of $k$??

  • 2
    $\begingroup$ The relation $\psi(x + a) = \psi(a)$ is more than a boundary condition. A boundary condition tells you the values of the wave function at the edges of the system: for $N$ atoms at $x = a$, $2a$, $3a$, etc., the boundary consists of just two points, $x = 0$ and $x = Na$. But $\psi(x + a) = \psi(a)$ specifies the values of the wavefunction at more than just those two points: it tells you its value at $N$ points! The solutions you get are a subset of the $-\pi/a < k < \pi/a$ band of states which also satisfies the extra constraints. $\endgroup$
    – Ted Pudlik
    Apr 30, 2014 at 17:47

1 Answer 1


From the wikipedia article linked in your question, we see that Wannier functions can also be defined as $$ \underbrace{\phi_{\textbf{R}}(\textbf{x})}_{\textbf{R} \text{ Bravais lattice vector}} \equiv \frac{1}{\sqrt{N}} \underbrace{\sum_{\textbf{k}}}_{\substack{\text{summed over first}\\\text{Brillouin zone}}} e^{-i\textbf{k}\cdot\textbf{R}} \psi_{\textbf{k}}(\textbf{x}) $$ which is the same as "your" $\phi$ because thanks to the periodicity properties of Block wave functions it is easily seen that $$ \underbrace{\phi_{\textbf{R}}(\textbf{x})}_{\text{"mine" }\phi} = \underbrace{\phi(\textbf{x}-\textbf{R})}_{\text{"your" }\phi} $$ Once we note this, it is easier to realize that going from $\psi_\textbf{k}$ to $\phi_\textbf{R}$ it's just a way to change the basis set of functions used to describe the system: instead of characterizing the states with the crystal momentum, we characterize them throught the Bravais lattice vectors. We can do this because the number of crystal momentum vectors $\textbf{k}$ it's the same as the number of Bravais lattice points $\textbf{R}$ (which is $N$).

Back to your question: I don't think it makes much sense to define a "Wannier function" of a particular $\psi_\textbf{k}$ because Wannier functions are defined as a "redistribution" of all Bloch wave functions, as you can also see rewriting them in functional form as $$ \underbrace{\phi}_{\text{"your" }\phi} = \frac{1}{\sqrt{N}}\sum_\textbf{k} \psi_\textbf{k} $$ But even if you could do such thing, hence for some reason allowing your system to have a single crystal momentum vector $\textbf{k}$, you would have from the definition that $$ \phi_\textbf{R}(\textbf{x}) = e^{-i\textbf{k}\cdot\textbf{R}} \psi_\textbf{k}(\textbf{x}), \qquad \text{in "my" notation} $$ $$ \phi(\textbf{x}) = \psi_\textbf{k}(\textbf{x}), \qquad \text{in "your" notation} $$ which as you can probably see is really not that useful.


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