I have trying to show that the continuum limit of n quantum harmonic oscillators gives rise the the Klein-Gordon field. However, instead of a usual finite string, I want to do it on a ring.
Assume $n \gg 1$.
Hence, my Lagrangian is
$$L=\frac{m}{2}(\dot{q_1}^2+\dot{q_2}^2+.... \dot{q_n}^2)-\frac{m \omega^2}{2}[(q_1-q_2)^2+(q_2^2-q_3^2)+....(q_n-q_1)^2]$$
So that the matrix for V is
$V=\begin{pmatrix} 2 & -1 &0 &. &. &. &-1\\-1&2&-1\\0 &-1 &2 &-1 \\.\\.\\.\\-1 &&.&.&.&-1 & 2 \end{pmatrix}_{n\times n}$
So that $L=\frac{m}{2}[\dot{x}^2-\omega^2 x^{T}V{x}]$. All the quantities here are matrices.
How do I find the eigenvalues of this matrix?
I tried to find a recursion relation between the characteristic polynomial of $N$ and $N-1$ dimensional matrix, but I failed. Is this the correct method? What other method is there?
After finding the eigenvalues, the Lagrangian can be written separated into its normal modes, and the propagator or kernel can be found out easily using that of the free particle. The limit of this as $N\to \infty$ should be the Klein-Gordon field.
But I am stuck on this. Any help will be appreciated.
Note that your matrix is not just a Toeplitz matrix but also a circulant matrix. Those matrices have eigenvectors
$$v_j=(1,\omega_j,\omega_j^2,\ldots,\omega_j^{n-1})$$
with $\omega_j=\exp\left(2\pi i j/n\right)$ and corresponding eigenvalues
$$\lambda_j=c_0+c_{n-1}\omega_j+c_{n-2}\omega_j^2+\ldots+c_{1}\omega_j^{n-1} \; .$$
In your case, the eigenvalues thus become
$$\lambda_j=2-\omega_j-\omega_j^{n-1}= 2-2\cos\left(2\pi j/n\right) \; .$$
How to arrive at this result? Easiest way I can think of is to rewrite your eigenvalue equation as a difference equation with a periodicity condition $v_{j,k}=v_{j,n+k}$. Then you have
$$v_{j,k+1}+(\lambda_j-2)v_{j,k}+v_{j,k-1}=0$$
This is a homogeneous difference equation with constant coefficients that can be solved by guessing the solution to be of the form $v_{j,k}=\omega_j^k$. You then obtain a characteristic equation
$$\omega_j^2+(\lambda_j-2)\omega_j+1=0$$
with the additional condition that
$$\omega_j^{n}=1$$
to ensure periodicity. Since our equation is quadratic with real coefficients and we need periodicity, its solutions must be two complex conjugate numbers whose sum is $-(\lambda_j-2)$, therefore
$$\lambda_j-2=-\omega_j-\omega_j^* \; .$$
But from the periodicity condition we have that $\omega_j^*=\omega_j^{n-1}$ and thus
$$\lambda_j=2-\omega_j-\omega_j^{n-1} \; .$$
Finally, there are $n$ possible complex roots for the periodicity condition which implies that we have $n$ different eigenvalues and corresponding eigenvectors.