# Quantum Particles on a circle and Circulant matrices [closed]

I have $$n$$ particles on a circle with the Hamiltonian $$\begin{equation} H = \sum_{n=1}^N \frac{p_n^2}{2m} + \frac{1}{2}m\omega^2 \sum_{n=1}^N (x_{n+1}-x_n)^2 \end{equation}$$

I need to find the energy eigenvalues for this system. The potential can be written in terms of a Circulant matrix $$\begin{equation} V = \frac{1}{2} m\omega^2 X^T\underbrace{\begin{pmatrix} 2 & -1 & 0 &\cdots &0 & -1\\ -1 & 2 & -1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots& \ddots &\vdots & \vdots \\ -1 & 0 & 0 &\cdots & -1 & 2 \end{pmatrix}}_{N\times N \text{ matrix } C}X \end{equation}$$

Where $$X =\begin{pmatrix}x_1 & x_2 & \cdots & x_n \end{pmatrix}$$. I now have to diagonalize the matrix $$C$$. I know that the eigenvectors of a circulant matrix are given as $$\begin{equation} X^{[q]}= \begin{pmatrix} 1 & \exp \frac{2\pi q}{N} & \exp \frac{4 \pi q}{N} &\cdots & \exp\frac{2(n-1)q\pi}{N} \end{pmatrix} \end{equation}$$ And the corresponding eigenvalues of $$m\omega^2 C$$ are $$\begin{equation} \lambda_q = 4 m \omega^2 \sin \frac{\pi q}{N} \end{equation}$$

But these eigenvectors are not real and I am therefore having trouble to get the eigenenergies of the system. Can someone help me with this problem?

• What do you mean with "the eigenvectors are not real"? Have you checked if $HX=\lambda X$? Nov 30, 2022 at 12:08
• What you need is to rewrite the Hamiltonian in terms of eigenmodes - for simplicity you can start with classical problems and the Newton's equations (you can look up in the books on classical mechanics, how it is done). Then you will have just uncoupled oscillators. Nov 30, 2022 at 12:14
• Elaborating on what @TobiasFünke said: there's no particular problem with having complex eigenvectors, so long as the corresponding eigenvalues are real. Nov 30, 2022 at 13:03
• This is equivalent to the Hamiltonian of $1D$ phonon, see Quantum Treatment and Interpretation sections in en.wikipedia.org/wiki/Phonon Nov 30, 2022 at 13:29