Is there a physical explanation why benzene's pi molecular orbitals in a Frost circle are degenerate but pyridine's are not?
My first thoughts are along the lines of your second-last paragraph. Here is a quick sketch of how to formalise it. From simple Hückel theory, you can obtain the coefficients of the AOs in the MOs:

$$|\psi_i\rangle = \sum_a c_{n,i} |n\rangle$$

where $c_{n,i}$ denotes the coefficient of AO $|n\rangle$ in the MO $|\psi_i\rangle$.

In Hückel theory the main difference between benzene and pyridine is that one atomic orbital is shifted in energy, so one of the $\alpha$ terms turns into, say, $\gamma$. Without loss of generality we will label the nitrogen with the index $n = 1$. In theory, the resonance integrals $\beta$ involving the nitrogen should change as well, but to a first approximation we assume they are the same.

In this case therefore the Hamiltonian matrix, when cast in the AO basis, changes from $\mathbf{H}_0$ to $\mathbf{H}$:

$$\mathbf{H}_0 = \begin{pmatrix}
\alpha &amp; \beta &amp; 0 &amp; 0 &amp; 0 &amp; \beta \\
\beta &amp; \alpha &amp; \beta &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; \beta &amp; \alpha &amp; \beta &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; \beta &amp; \alpha &amp; \beta &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; \beta &amp; \alpha &amp; \beta \\
\beta &amp; 0 &amp; 0 &amp; 0 &amp; \beta &amp; \alpha 
\end{pmatrix} \qquad \longrightarrow \qquad \mathbf{H} = \begin{pmatrix}
\gamma &amp; \beta &amp; 0 &amp; 0 &amp; 0 &amp; \beta \\
\beta &amp; \alpha &amp; \beta &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; \beta &amp; \alpha &amp; \beta &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; \beta &amp; \alpha &amp; \beta &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; \beta &amp; \alpha &amp; \beta \\
\beta &amp; 0 &amp; 0 &amp; 0 &amp; \beta &amp; \alpha 
\end{pmatrix}$$

Most of the Hamiltonian is unchanged, so we could consider this within the framework of perturbation theory, where we have a perturbed Hamiltonian $\hat{H} = \hat{H}_0 + \hat{V}$. The relevant perturbation is therefore

$$\mathbf{V} = \begin{pmatrix}
\gamma - \alpha &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 
\end{pmatrix}$$

or to condense this back into operator form, $\hat{V} = (\gamma - \alpha)|1\rangle\langle 1|$.

The proper mathematical analysis is slightly annoying because the (unperturbed) eigenstates that we're interested in looking at are degenerate, which then necessitates that you cast the matrix $\hat{V}$ in the basis of the degenerate states of interest. Here, the two states we're interested in are $|\psi_2\rangle$ and $|\psi_3\rangle$. The application of Hückel theory to benzene gives the formulae for these unperturbed states as

$$\begin{align}
|\psi_2\rangle &amp;= \frac{\sqrt 3}{3}|1\rangle + \frac{\sqrt 3}{6}|2\rangle - \frac{\sqrt 3}{6}|3\rangle - \frac{\sqrt 3}{3}|4\rangle - \frac{\sqrt 3}{6}|5\rangle + \frac{\sqrt 3}{6}|6\rangle \\
|\psi_3\rangle &amp;= \frac{1}{2}|2\rangle + \frac{1}{2}|3\rangle - \frac{1}{2}|5\rangle - \frac{1}{2}|6\rangle
\end{align}$$



Note that this choice of coefficients is not unique, as the two states are degenerate: any linear combination of these two MOs is also a permissible MO with the same energy. In any case, the relevant matrix elements of $\hat{V}$ in the new basis $\{|\psi_2\rangle,|\psi_3\rangle\}$ are given by

$$V_{i-1,j-1} = \langle \psi_i | \hat{V} | \psi_j \rangle$$

and if you assume that $\langle n | n' \rangle = \delta_{nn'}$ (i.e. the AOs have no overlap with each other, which is a key assumption of simple Hückel theory anyway), then it becomes very simple:

$$\mathbf{V} = \begin{pmatrix} \frac{\gamma-\alpha}{3} &amp; 0 \\ 0 &amp; 0\end{pmatrix}$$

Perturbation theory tells us that the eigenvectors of $\hat{V}$ are the "states which are stable towards the perturbation". What this means is that, since the degeneracy is lifted by the perturbation, you also remove the free choice of having any linear combinations be a permissible MO: the coefficients of AOs in the new MOs must be uniquely determined (ignoring the phase factor). Since $\hat{V}$ is diagonal in the basis $\{|\psi_2\rangle,|\psi_3\rangle\}$, it conveniently means that both $|\psi_2\rangle$ and $|\psi_3\rangle$ are the eigenvectors of $\hat{V}$.

More interesting here is the first-order corrections to the eigenvalues of $\hat{H}_0$, which are simply the eigenvalues of $\hat{V}$. In this case, since $\hat{V}$ is diagonal, the eigenvalues are just $(\gamma-\alpha)/3$ and $0$: that is to say, the MO $\psi_2$ is shifted in energy by $(\gamma-\alpha)/3$ and the MO $\psi_3$ is unshifted in energy.