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Consider an observable in quantum mechanics, with a degenerate eigenvalue in a continuous spectrum.

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

  2. 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.

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

  2. I am also interested in how you would write the unit operator (the completeness relation) in each of these cases.

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  • $\begingroup$ Strictly speaking there is no cardinality in physics (other than the trivial case of everything measurable being finite). If we are talking about physically unrealizable systems that are oversimplified, then I believe can imagine an example for 1), so mathematically I would answer affirmatively. I don't see why 2) should be ruled out but I would love to hear why/if you think that it can. I have no idea for 3). Doesn't the most general form of the unit operator follow from a fairly general statement about linear operator spectra? $\endgroup$ – CuriousOne Nov 1 '15 at 8:27
  • $\begingroup$ Assume that the Hilbert space is separable. 1) Yes, think of the Coulomb (hydrogen atom) excited states where the eigenvalue for different m quantum numbers is the same. 2) Yes. The identity operator has eigenvalue 1 And each orthonormal base consists of a countable number of mutually orthogonal eigenvectors. 3) Yes, see 2). $\endgroup$ – Urgje Nov 1 '15 at 10:24
  • $\begingroup$ @CuriousOne Two is a problem because once you have two distinct eigenvectors of the same eigenvalue then any linear combination of them is also an eigenvector with the same eigenvalue and there are uncountably many linear combinations of two distinct vectors. Now having an uncountable orthonormal basis of eigenvectors is different, but you generally assume your space is separable. $\endgroup$ – Timaeus Nov 1 '15 at 17:57
  • $\begingroup$ @Timaeus: Thanks! Now I am getting a better idea of what the OP might have meant. $\endgroup$ – CuriousOne Nov 1 '15 at 18:00
  • $\begingroup$ Thank you for your comments. Of course I was referring to the number of linearly independent eigenvectors (not the total number of eigenvectors, which is clearly uncountable). $\endgroup$ – HeisenbergsDog Nov 2 '15 at 7:39
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(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.

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  • $\begingroup$ For number two, degeneracy is different than the number of eigenvectors, degeneracy refers to the dimension of the subspace of eigenvectors. Clearly once there is degeneracy there are uncountably many eigenvectors even if you only select only one from every group of eigenvectors that are proportional to each other. Because there are uncountably many directions in a subspace of more than one dimension. $\endgroup$ – Timaeus Nov 1 '15 at 18:03
  • $\begingroup$ Yes, indeed I was referring to the number of linearly independent eigenvectors....However in the case (2) a further problem concerns the fact that the spectrum is continuous. There is in fact a generalized notion of degeneracy related to a version of spectral representation theorem (see e.g. Dumford Schwartz, second volume I guess) which is valid both for point and continuous spectrum. In my example (2) this notion it is 2. Indeed in the case (2) there are no eigenvectors at all just because the spectrum is continuous. $\endgroup$ – Valter Moretti Nov 1 '15 at 18:22
  • $\begingroup$ Thank you for your answer. However I do not quite have the mathematical tools yet to understand everything you wrote. So basically the answer is yes to all my questions. $\endgroup$ – HeisenbergsDog Nov 2 '15 at 7:43
  • $\begingroup$ As for the identity operator, when expanding it in the eigenbasis: In (2) for example, would you include a discrete sum inside the integral, summing over the degeneracies? $\endgroup$ – HeisenbergsDog Nov 2 '15 at 7:46
  • $\begingroup$ Yes, I would do. $\endgroup$ – Valter Moretti Nov 2 '15 at 7:47

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