Closure relation for degenerate eigenkets Consider an observable in quantum mechanics, with a degenerate eigenvalue in a continuous spectrum.


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*Is it possible for such an eigenvalue to have a finite degeneracy? 

*If the degeneracy is infinite, can it have countably infinite eigenvectors? (that is, can its eigenvectors be listed?)
Now suppose we have a degenerate eigenvalue in a discrete spectrum. 


*Is it possible for such an eigenvalue to be infinitely degenerate? If so, are the corresponding eigenvectors countable or uncountable? 

*I am also interested in how you would write the unit operator (the completeness relation) in each of these cases. 
 A: (1) Yes, take ${\cal H} = L^2(\mathbb R, dx)\oplus L^2(\mathbb R, dx)$ and thereon $\left(X (\psi, \phi)\right)(x,y) := (x\psi(x),y\phi(y))$. We have  $\sigma(X)=\sigma_c(X)$ and the degeneracy is just $2$.
(2) Yes, use the example (1) with a countably infinite copies of $L^2(\mathbb R, dx)$ and use the Hilbertian direct sum of Hilbert spaces. (There are infinitely many linearly independent eigenvectors.)
(3) Yes, referring to the Hilbertian direct sum, take ${\cal H} = \oplus_{k=1}^{+\infty} {\cal H}_k$ with ${\cal H}_k = L^2(\mathbb R, dx)$ and consider the self-adjoint operator (with natural domain) $H = \oplus_{k=1}^{+\infty} H_k$, where $$H_k:= \frac{1}{2m}P_k^2+ \frac{k}{2}X^2_k$$ with $P_k$ and $X_k$ the momentum and position operator in ${\cal H}_k$ and define $\omega = 2\pi\sqrt{k/m}$. It turns out that $\sigma(H)= \sigma_p(H)= \omega(n + \frac{1}{2})$, $n=0,1,2,\ldots$ and the degeneracy is countably infinite for every $n$ .
In principle it is possible to construct  examples with $\sigma(H)=\sigma_p(H)$ and  the degeneracy is uncountable, but in QM the Hilbert space is assumed to be  separable, therefore these examples have no much physical meaning.
