• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Uniqueness of eigenvector representation in a complete set of compatible observables

+ 3 like - 0 dislike

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 ]
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)
@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)

2 Answers

+ 6 like - 0 dislike

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 (3,575 points) [ no revision ]
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)
$\psi_i$ 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)
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)
+ 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)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights