= Solution
In position space,
$$
\begin{aligned}
\Psi_k(x)
&=\sum_{n=0}^2e^{ikna}\psi(x-na)\\
&=e^{ikx}\sum_{n=0}^2\psi(x-na)e^{-ik(x-na)}.
\end{aligned}
$$
Hence
$$
\Psi_k(x)=e^{ikx}u_k(x),
\qquad
u_k(x)=\sum_{n=0}^2\psi(x-na)e^{-ik(x-na)},
$$
up to the common normalization factor $1/\sqrt3$.
Under $x\mapsto x+a$, reindexing $m=n-1$ gives the same three terms modulo three sites. The term shifted by three sites is unchanged because $\psi_{m+3}=\psi_m$ and $e^{3ika}=1$. Therefore
$$
\boxed{u_k(x+a)=u_k(x)}.
$$
The one-dimensional <Bloch theorem> states that an eigenfunction in a potential of period $a$ can be chosen in the form
$$
\Psi_k(x)=e^{ikx}u_k(x),
\qquad u_k(x+a)=u_k(x).
$$
It applies because the three-atom ring is invariant under translation by one lattice spacing and obeys periodic boundary conditions.
Back to article page
