19.5: Eigenvectors of the Linear Chain

Let’s get back to our chain, with eigenfunction equation of motion the \(N\) dimensional equivalent of

\begin{equation}
-m \Omega^{2}\left(\begin{array}{c}
A_{1} \\
A_{2} \\
A_{3} \\
A_{4}
\end{array}\right)=\left(\begin{array}{cccc}
-2 \kappa & \kappa & 0 & \kappa \\
\kappa & -2 \kappa & \kappa & 0 \\
0 & \kappa & -2 \kappa & \kappa \\
\kappa & 0 & \kappa & -2 \kappa
\end{array}\right)\left(\begin{array}{c}
A_{1} \\
A_{2} \\
A_{3} \\
A_{4}
\end{array}\right)
\end{equation}

. We see the matrix is a circulant, so we know the eigenvectors are of the form

\begin{equation}
\left(1, \omega^{j},\left(\omega^{j}\right)^{2},\left(\omega^{j}\right)^{3}, \ldots,\left(\omega^{j}\right)^{N-1}\right)^{T}
\end{equation}

, which we’ll now write

\begin{equation}
\left(1, e^{2 \pi i j / N}, e^{2 \pi i 2 j / N}, e^{2 \pi i 3 j / N}, \ldots, e^{2 \pi i j(N-1) / N}\right)
\end{equation}

What does this mean for our chain system? Remember that the \(n^{\mathrm{th}}\) element of the eigenvector represents the displacement of the \(n^{\mathrm{th}}\) atom of the chain from its equilibrium position, that would be proportional to \(e^{2 \pi i j n / N}\)

The steady phase progression on going around the chain \(n=0,1,2,3, \ldots\) makes clear that this is essentially a (longitudinal) wave. (The actual \(n^{\mathrm{th}}\) particle displacement is the real part of the \(n^{\mathrm{th}}\) element, but there could be an overall complex factor fixing the phase.)