A 1D quantum harmonic oscillator with an additional degree of freedom I was given a Hamiltonian and asked to find the energy levels. I've managed to find the alpha parameter by using the fact that H is hermitian. Afterwards I tried to express an eigenstate as a superposition of regular eigenstates (ones from a Hamiltonian that doesn't have this color operator in it) and the color states and it gave a result but it feels wrong to me and also I can't think of a way to extract information about degeneracies from that result.
I would be really grateful if someone could point out what I'm doing wrong or missing. Thanks a lot! :)

 A: Note that $[H,A]=0$ and so, these two operators can be simultaneously diagonalized. Now consider the eigenvalues and eigenstates of $A$. You have to compute
$$
  [|g\rangle\langle g|+i\alpha|g\rangle\langle b|-i\frac{\sqrt{5}}{2}|b\rangle\langle g|+3|b\rangle\langle b|]|\chi\rangle=\lambda|\chi\rangle.
$$
This is a two-state operator and you can look for solutions in the form
$$
   |\chi\rangle =a|b\rangle+b|g\rangle
$$
with $a$ and $b$ two coefficients to be computed. Put this into the equation above and obtain
$$
  a(i\alpha|g\rangle+3|b\rangle)+b(|g\rangle-i\frac{\sqrt{5}}{2}|b\rangle)=
\lambda(a|b\rangle+b|g\rangle) 
$$
giving the equations for the coefficients
$$
   i\alpha a+b(1-\lambda)=0
$$
and
$$
  (3-\lambda)a-i\frac{\sqrt{5}}{2}b=0.
$$
With $\alpha=\frac{\sqrt{5}}{2}$ this yields $\lambda_1=1/2$ and $\lambda_2=7/2$.
Now, the generic eigenstate of your Hamiltonian has the form $|n;k\rangle=|n;\lambda_k\rangle|\chi_k\rangle$ with $n$ spanning the states of harmonic oscillator and $k=1,2$. This will yields
$$
   E_{n,k}=H|n;k\rangle=\left[\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2A\right]|n;\lambda_k\rangle|\chi_k\rangle=
\left[\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2\lambda_k\right]|n;\lambda_k\rangle|\chi_k\rangle
$$
and finally
$$
   E_{n,k}=\left(n+\frac{1}{2}\right)\hbar\omega\sqrt{\lambda_k}.
$$
