19.7: Finding the Eigenvalues

The eigenvalues are found by operating on the eigenvector we just found with the matrix, meaning the \(N\) dimensional generalization of

\begin{equation}
-m \Omega^{2}\left(\begin{array}{c}
1 \\
e^{i k_{n} a} \\
e^{i k_{n} 2 a} \\
e^{i k_{n} 3 a}
\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}
1 \\
e^{i k_{n} a} \\
e^{i k_{n} 2 a} \\
e^{i k_{n} 3 a}
\end{array}\right)
\end{equation}

Applying the matrix to the column vector

\begin{equation}
\left(1, e^{i k_{n} a}, e^{2 i k_{n} a}, e^{3 i k_{n} a}, \ldots, e^{i(N-1) k_{n} a}\right)^{T}
\end{equation}

, and cancelling out the common \(e^{i k_{n} n a}\) factor, we have

\begin{equation}
-m \Omega_{n}^{2}=\kappa\left(e^{i k_{n} a}+e^{-i k_{n} a}-2\right)
\end{equation}

(Of course, this same result comes from every row.)

The complete set of eigenvalues is given by inserting in the above expression

\begin{equation}
k_{n}=2 \pi n / N a, \quad n=0,1,2, \ldots, N-1 \text { so } e^{i k_{n} a}=e^{2 \pi i n / N}
\end{equation}

so \(n=0\) is displacement of the system as a whole, as is \(n=N\).

Wavenumber values \(k_{n} \text { beyond } n=N\) repeats the eigenstates we already have, since

\begin{equation}
e^{i k_{N+n} a}=e^{i \frac{2 \pi(N+n) a}{N a}}=e^{2 \pi i} e^{2 \pi i n / N}=e^{2 \pi i n / N}=e^{i k_{n} a}
\end{equation}

k are restricted to

\begin{equation}
\Omega_{n}=2 \sqrt{\frac{\kappa}{m}} \sin \left(\frac{k_{n} a}{2}\right)=2 \sqrt{\frac{\kappa}{m}} \sin \left(\frac{n \pi}{N}\right)
\end{equation}

\begin{equation}
0 \leq k<2 \pi / a
\end{equation}

or equivalently

\begin{equation}
-\pi / a<k \leq \pi / a
\end{equation}

The eigenvalue equation is

\begin{equation}
\Omega_{n}^{2}=2(\kappa / m)\left(1-\cos k_{n} a\right)
\end{equation}

or

\begin{equation}
\Omega_{n}=2 \sqrt{\frac{\kappa}{m}} \sin \left(\frac{k_{n} a}{2}\right)=2 \sqrt{\frac{\kappa}{m}} \sin \left(\frac{n \pi}{N}\right)
\end{equation}

To see the dynamics of this eigenstate

\begin{equation}
\left(1, e^{i k_{n} a}, e^{2 i k_{n} a}, e^{3 i k_{n} a}, \ldots, e^{i k_{n}(N-1) a}\right)
\end{equation}

, we need to multiply by the time dependence \(e^{i \Omega_{n} t}\), then finally take the real part of the solution:

\begin{equation}
\left(\cos \Omega_{n} t, \quad \cos \left(k_{n} a+\Omega_{n} t\right), \quad \cos \left(2 k_{n} a+\Omega_{n} t\right), \quad \cos \left(3 k_{n} a+\Omega_{n} t\right), \ldots, \cos \left((N-1) k_{n} a+\Omega_{n} t\right)\right)
\end{equation}

Notice that in the continuum limit, meaning large N and small a, the atom displacement as a function of position has the form \(\cos (k x+\Omega t)\) in other words we’re looking at a sinusoidal wave disturbance with wavenumber \(k_{n}\) here.

Now, \(-k_{n}\) is also a solution, but that is the same as \(n^{\prime}=N-n\) so one must be careful not to overcount. The two frequencies \(\pm \Omega_{n}\) correspond to waves going in opposite directions.