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I am struggling with the concept of completeness in Quantum Mechanics. I have studied the theory of Hilbert spaces and I know that an orthonormal set is defined to be complete if it is maximal.

However, my QM professor stated that an orthonormal set of eigenstates is complete if and only if by definition its spectrum is non-degenerate. This is where I am lost, because it is the same as requiring that all eigenspaces be 1-dimensional, whereas there are plenty of examples of degenerate spectra. Can anyone help me? What is the definition of completeness?

EDIT: I think I got it. An orthonormal subset of a Hilbert space $\mathcal{H}$ is complete if it is maximal. There is another notion of completeness though. A set of observables, represented by the commuting selfadjoint operators $A_1, \dots, A_n$ on $\mathcal{H}$, is complete if its spectrum is non-degenerate. In this case, there exists a unique (up to normalization) orthonormal basis of eigenstates.

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  • $\begingroup$ Commuting self-adjoint operators have the same eigenstates. So you only need one of them. The non-degeneracy implies that these eigenstates are orthogonal. The non-degeneracy does not impose the completeness. You can also have a complete basis when there are degeneracies, but then the basis is not unique. $\endgroup$ Apr 24 '21 at 4:23
  • $\begingroup$ This is what I thought initially, but then I can't work out why my QM professor defines a set of operators to be complete if there is no degeneracy. He may just be wrong at this point. $\endgroup$
    – fresh
    Apr 24 '21 at 8:33
  • $\begingroup$ Get yourself a good book on functional analysis. Then you can find out what actual definitions of these concepts are. $\endgroup$ Apr 25 '21 at 4:44
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The degeneracy only affects the orthogonality in the sense that the basis elements of an orthonormal set would not be unique. When the spectrum is non-degenerate then the orthonormal set would be unique. One can still have an orthonormal set for the case where there are degeneracies, but then any two elements with degenerate eigenvalues can be replaced by suitable linear combinations of them.

Does it make sense? If not I can add some math.

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  • $\begingroup$ It makes perfect sense, but I don't understand why we should limit ourselves to the non-degenerate case. I see nothing wrong with having multiple bases. We just choose one and move on, right? Or is that physically unacceptable? $\endgroup$
    – fresh
    Apr 23 '21 at 9:31
  • $\begingroup$ The issue comes with the fact that you refer to eigenstates. That begs the question of some Hermitian operator, perhaps the Hamiltonian, which would then introduce the possibility for degeneracies. When we talk about a Hilbert space, there is no Hermitian operator per se. Hence, we can select any orthonormal basis. But it may not be convenient to use when studying a specific Hermitian operator and its spectrum. $\endgroup$ Apr 23 '21 at 13:02
  • $\begingroup$ I still don't get it. It seems to me that there are two definitions of completeness. Dirac says that a complete set of eigenstates is a Hilbert basis, without ever mentioning degeneracy. I'm a bit confused... $\endgroup$
    – fresh
    Apr 23 '21 at 14:32
  • $\begingroup$ Ok, I think I got it. I have edited the question to include my answer. $\endgroup$
    – fresh
    Apr 23 '21 at 14:42
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A set of $n$ orthonormal eigenstates if there are $n$ eigenvalues obviously is only complete if there is no degeneracy. However such a statement is not very useful in my opinion. A set of $n$ orthonormal eigenstates matches the description of an orthonormal set regardless of $n$, but if $n$ is less than the number of eigenstates, including degeneracies, it is also not complete. This is another rather useless statement. A true, useful statement is that not all true statements are useful.

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  • $\begingroup$ I don't agree with the first statement you made. I know that if the $n$ eigenvalues are distinct, then the corrisponding eigenvectors are linearly independent. The converse is not true though. $\endgroup$
    – fresh
    Apr 23 '21 at 14:23
  • $\begingroup$ I don't see your point. The statement is that the $n$ states are orthonormal, so they are independent as well. $\endgroup$
    – my2cts
    Apr 23 '21 at 15:13
  • $\begingroup$ Yeah you're right, sorry. I still think what you wrote is wrong though. Anyway, I found a solution and I have edited my question. $\endgroup$
    – fresh
    Apr 23 '21 at 15:22
  • $\begingroup$ Why? Anyway good luck with your mathematicsl quest. $\endgroup$
    – my2cts
    Apr 23 '21 at 18:59
  • $\begingroup$ I think a set of $n$ orthonormal eigenstates could form a basis even if it is degenerate. But I might just be tripping. $\endgroup$
    – fresh
    Apr 23 '21 at 21:32

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