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I have a problem on finding wave functions solutions for the 6-carbon ring system - benzene. To get the energy levels it is necessary to solve secular determinant equal zero equation. Avoiding some steps, the latter is

\begin{array}{|cccccc|} x & 1 & 0 & 0 & 0 & 1 \\ 1 & x & 1 & 0 & 0 & 0 \\ 0 & 1 & x & 1 & 0 & 0 \\ 0 & 0 & 1 & x & 1 & 0 \\ 0 & 0 & 0 & 1 & x & 0 \\ 1 & 0 & 0 & 0 & 1 & x \end{array} = 0

where x = a-E/b a are all the diagonal elements, b are diagonal adjacent elements and $H_{1,6}$ and $H_{6,1}$elements of Hamiltonian, E is the energy, The equation has solutions x = -2 and 2 and -1 (doubly degenerate) and 1 (doubly degenerate) As for x = -2 and 2 they correspond to maximum and minimum energy states.

My question rises for x = -1 and +1 states. So, they are doubly degenerate, it means that if one wants to find wave function, one should use the same secular matrix to do so and the solution will have two parameters. If one supposes that the wave function is normalized, then there is still one parameter left.

So the question is: where is the catch? Why there is no exact solution for wave function? Why is there a parameter left?

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When there are degeneracies (as in the energy here), a linear combination of the vectors with the same eigenvalues is also an eigenvector. There is no reason why any one linear should occur. In general, this is resolved by using a second operator that commutes with the Hamiltonian, and choosing the eigenvectors to be eigenvectors of both operators.

In the hydrogen atom, for instance, there are 4 states with energy $-13.6/4$eVs. To refine the labelling of the states and pinpoint the $4$ states completely, one uses the total angular momentum $L^2$ and the $z$-projection $L_z$ to supply additional quantum numbers $\ell$ and $m$ so the states can be fully labelled. Both $L^2$ and $L_z$ commute with the Hamiltonian.

In your example, you need another operator to "split" the degeneracy of energy. This operator must commute with the Hamiltonian. One possible operator to investigate is as follows.

Since the ring benzene system is unchanged by "rotating" the ring by one atom to the right, the transformation $X_i\to X_{i+1}$, where $X_i$ is the position of atom $i$ in the ring, may supply the commuting operator you need.

The matrix representation $P$ of this operation is a band matrix with $1$ just below the diagonal and a $1$ in the upper right position, where the cyclicity condition is enforced. By definition it will commute with the Hamiltonian matrix, and so will have with it common eigenvectors. Since all eigenvalues of $P$ are not degenerate, the eigenvectors of $P$ are uniquely determined and will also be eigenvectors of $H$.

Physically, this amounts to enforcing the condition that your solutions should be invariant (up to an overall phase) when all atoms are "rotated" to their immediate neighbors. The eigenvalues of $P$ come in conjugate pairs: you could redo this using the symmetry $X_i\to X_{i-1}$ and the appearance of conjugate pairs of eigenvalues of $P$ can be intuitively understood as a consequence of the two possible $X_{i}\to X_{i\pm 1}$ symmetry transformations.

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