How can I tell if the spectrum of an operator in QM is degenerate? I know that the collection of all the eigenvalues of an operator $\hat{Q}$ is called its point spectrum, and sometimes two or more linearly independent eigenfunctions share the same eigenvalue, and in this case the spectrum is said to be degenerate.
My question is how I can find out if this spectrum is degenerate, for example for $\hat{Q}=i\frac{d^2}{d\phi^2}$?
 A: $\renewcommand{ket}[1]{|#1\rangle}$
As others have pointed out, you can go whole hog and solve the characteristic equation $\text{det}(\hat{Q} - \lambda \hat{I})=0$ and find repeated solutions for $\lambda$.
However, there is a simpler, more physically intuitive way to hunt for degeneracy: look for symmetry.
When an operator $\hat{Q}$ has a symmetry there is a "symmetry operator" $\hat{S}$ which commutes with $\hat{Q}$
$$ [\hat{Q}, \hat{S}] = 0\, . $$
Consider an eigenvector of $\hat{Q}$:
$$\hat{Q}\ket{\Psi} = q \ket{\Psi} \, .$$
$\hat{S}\ket{\Psi}$ is also an eigenvector with the same eigenvalue
$$
\hat{Q}(\hat{S}\ket{\Psi}) = \hat{S}(\hat{Q}\ket{\Psi}) = \hat{S}(q\ket{\Psi}) = q(\hat{S}\ket{\Psi})
$$
Since we've now found two vectors, $\ket{\Psi}$ and $\hat{S}\ket{\Psi}$ with the same eigenvalue, we have by definition found degeneracy.
Of course, this only really shows degeneracy if $\hat{S}\ket{\Psi} \neq \ket{\Psi}$.
So, for example, if $\hat{S} = \hat{I}$ we haven't actually done anything interesting (because the fact that $\hat{S}$ commutes with the identity is trivial).
Anyway, this construction gives you a lot of intuition.
Consider the case where $\hat{Q}$ is the momentum operator squared $\hat{p}^2$.
It's obvious that there's a symmetry there: you can flip left/right and $\hat{p}^2$ doesn't change.
The the eigenvectors of $\hat{p}^2$ are the plane waves $e^{ikx}$, so this symmetry idea tells you that this plane wave should be degenerate with the left/right flipped function $e^{-ikx}$.
You can easily check that this correct.
This idea works for any case where you have linear operators and care about degeneracy of eigenvalues, not just quantum mechanics problems.
(If someone would like to expand this answer, giving a more details review of the theory or more examples please do so)
A: Assuming that your operator has a spectrum consisting of isolated points you can look for all the independent solutions of the eigenvalue equation
$$(Q-\lambda I)\xi = 0$$
Let these solutions generate a vector space $V_\lambda$ and then compute the dimension of $V_\lambda$. If it is greater than 1 then the eigenvalue $\lambda$ is degenerate.
A: As Phoenix87 said, you need to solve the asociated eigenvalue problem. $\lambda$ is an eigenvalue if and only if $\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator $\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side:
$$(\hat{Q}-\lambda\hat{I})|\Psi>=0\iff\det(\hat{Q}-\lambda\hat{I})=0$$
Using the above equation you can calculate the $\lambda$ values, simplifying the determinant and making the result equal 0. The spectrum will be degenerate if you find something like $(\lambda-1)^2(\lambda-2)=0$ because solution $\lambda=1$ appears twice. (And the associated subspace will be 2-dimensional).
If the operator has a differential expresion, then I recommend to write $|\Psi>$ as a function and solve the eigenvalue problem as a differential equation.  In the example you've used,
$$\hat{Q}|\Psi>=\lambda|\Psi>\Rightarrow i \frac{d^2\Psi(\phi)}{d\phi^2}=\lambda\Psi(\phi)$$
Remember that the function must be bounded:
$$\int_{-\infty}^{+\infty}{|\Psi(\phi)|^2d\phi<+\infty}$$
Also, you may want to apply boundary conditions. This usually leads to a quantization of $\lambda$, which are the eigenvalues. Once you have the eigenvalues, you have to take a look to the eigenfunctions and check the dimension of the subespaces associated with every $\lambda$.
