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Uniqueness of eigenvector representation in a complete set of compatible observables

+ 3 like - 0 dislike
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Sakurai states that if we have a complete, maximal set of compatible observables, say A,B,C... Then, an eigenvector represented by |a,b,c....> , where a,b,c... are respective eigenvalues, is unique. Why is it so? Why can't there be two eigenvectors with same eigenvalues for each observable? Does maximality of the set has some role to play in it?

I asked this question on Physics SE and was not satisfied with answers. Hope that I get help here.

This post has been migrated from (A51.SE)
asked Dec 6, 2011 in Theoretical Physics by user15291 (75 points) [ no revision ]
@Moshe: I didn't bother looking at the Physics.SE link before answering, but now you've pointed it out I agree that genetth's answer was perfect.

This post has been migrated from (A51.SE)
I am not sure this question is a right fit for this site, I think physics.se is a much better fit. Besides, seems to me that genneth already gave an excellent answer over there.

This post has been migrated from (A51.SE)

2 Answers

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Yes, since it is the maximal set of compatible observables, it includes all observables for which $|a\rangle$, $|b\rangle$, $|c\rangle$, etc. are the eigenvectors (I'll use the notation $|\psi_1\rangle$, $|\psi_2\rangle$, $|\psi_3\rangle$ etc instead). Hence this includes the observable $D = \sum_k k |\psi_k\rangle \langle \psi_k|$ . However $D$ has a unique set of eigenvectors, and hence so does the any compatible set of observables which contains $D$.

This post has been migrated from (A51.SE)
answered Dec 6, 2011 by Joe Fitzsimons (3,555 points) [ no revision ]
Secondly, in quantum mechanics observable and Hermitian operator are synonymous. You can construct a physical measurement (in principle at least) for any Hermitian operator, and any physical observable is Hermitian.

This post has been migrated from (A51.SE)
$\psi_i$ above are a basis for the Hilbert space in which all measurements are diagonal. If the set of measurements is maximal then it necessarily contains $D$ for some specific choice of basis. Since you specify the set of observables by their eigenvectors, you can explicitly construct $D$.

This post has been migrated from (A51.SE)
Two questions. Are your psi_1, psi_2 etc. all different numbers? If yes, then what if the spectrum is degenerate? If no, then I think your proof won't work. Second, why is D an observable? Isn't the definition of observable that it is a physical quantity that can be measured? All I can see is that D is a Hermitian Operator. Why should it be an observable?

This post has been migrated from (A51.SE)
+ 0 like - 1 dislike

You can and you do sometimes have degenerate eigenvalues.

This might just be a question of definitions: "a complete set of commuting observables (CSCO) is a set of commuting operators whose eigenvalues completely specify the state of a system (Gasiorowicz 1974, p. 119)." [1]

That said, if you give me some cooked up Hamiltonian with at least one degenerate eigenvalue, perhaps one might be able to prove no observable commutes with it.

This post has been migrated from (A51.SE)
answered Dec 9, 2011 by user129 (5 points) [ no revision ]
That is incorrect. The Hamiltonian itself is an observable. Further, if you assign an arbitrary set of unique eigenvalues to the same eigenvectors (picking a basis for each degenerate subspace), thus lifting the degeneracy, this produces an observable which is simultaneously diagonalizable with the Hamiltonian, and hence commutes with it, but which has no degenerate eigenspaces.

This post has been migrated from (A51.SE)

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