Finding the coefficients of a spinor

From the Schrödinger equation of a system I'm investigating, where the wave function is a 4-component spinor of coefficients $C_1, C_2, C_3, C_4$, I am able to obtain the expression

$\begin{pmatrix} m \sin(\alpha) - E & m \cos(\alpha) & -ik & 0 \\ m \cos(\alpha) & - m\sin(\alpha) - E & 0 & -ik \\ ik & 0 & -m \sin(\alpha) - E & -m \cos(\alpha) \\ 0 & ik & -m\cos(\alpha) & m \sin(\alpha) - E \\ \end{pmatrix} \begin{pmatrix} C_1 \\ C_2 \\ C_3 \\ C_4 \end{pmatrix} = 0$

($\alpha$ is constant).

I can obtain the dispersion relation by calculating the determinant of the matrix and setting it to zero: $\det(A) = E^2 - m^2 - k^2 = 0\to E = \pm \sqrt{m^2 + k^2}$.

My question is: how do I obtain the coefficients $C_1, C_2, C_3,C_4$? When I try to solve it as a system of linear homogeneous equations I just end up with $0 = 0$.

• You have $\left(M-E_i\right)\Psi=0$ which is just finding the eigenvalues and eigenvectors of a matrix. You've found the eigenvalues $E_i$. Now how do you find the corresponding eigenvectors? If you're working a problem like this you've probably done this before, at least for a 2x2 matrix. :-) – Michael Brown Apr 24 '13 at 0:55
• @MichaelBrown when I write out the corresponding system of linear equations and attempt to find the coefficients, I end up with $0 = 0$. I understand that since this is a linear homogeneous system, there are an infinite amount of solutions, so I set $C_4 = 1$ (for example), but I still don't get anywhere. – Calavera Apr 24 '13 at 1:28
• Hmm, seems that the only way I can obtain eigenvectors is if one of the coefficients is equal to zero – Calavera Apr 24 '13 at 2:22
• It's not clear what you are doing... you put in one of the eigenvalues for $E$ and there are two linearly independent solutions for each eigenvalue (yes, infinitely many linear combinations of solutions the two are solutions as well). For each eigenvalue you can start finding the two solutions by taking $C_3=1,C_4=0$ and $C_3=0,C_4=1$. Mathematica pulls out the eigensystem in .05 seconds so there is nothing screwy with your matrix. :) – Michael Brown Apr 24 '13 at 8:06
• The matrix $M$ is 4x4 Hermitian so you know it has a complete set of four eigenvectors with 4 real eigenvalues, and you've worked out that the eigenvalues are $\pm\sqrt{p^2+m^2}$. You also know that $det(M)$, which is the product of the eigenvalues, is positive. So there must be an even number of negative eigenvalues. You should also know that $tr(M)$ is the sum of the eigenvalues and is zero. So there must be two degenerate positive eigenvalues and two degenerate negative eigenvalues. The corresponding eigenvectors can still be orthonormalised and used as a basis as usual. – Michael Brown Apr 24 '13 at 12:26