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I started a new thread from a side discussion in https://www.physicsforums.com/showthread.php?t=681625&page=2, since it seems very off topic, but I still had questions.
Is there a requirement for an operator that corresponds to an observable to be part of a complete set, ie. its eigenvectors span the vector space?
Would ##|u><u|## be part of a complete set?
atyy said:No, I mean a pure state. I suppose my question is whether every state is an eigenvector of some observable.
Fredrik said:This is true, but even the arbitrary vector is an eigenvector of some observable.
atyy said:Is there a short proof, or can you point me to a reference? (Anyway, I do think it's not very relevant to the question, since I know this is true in many cases.)
micromass said:A nonzero vector ##|u>## is an eigenvector of ##|u><u|##.
atyy said:Are all eigenvalues of ##|u><u|## real?
Fredrik said:That's a projection operator, so it only has eigenvalues 0 and 1.
Note that if ##P^2=P## and ##Px=\lambda x## where ##\lambda## is a complex number and ##\|x\|\neq 0##, we have ##P^2x=P(Px)=P(\lambda x)=\lambda Px=\lambda^2x##, and also ##P^2x=Px=\lambda x##. So ##\lambda(\lambda-1)x=0##, and this implies that ##\lambda## is 0 or 1.
atyy said:Ok, thanks. I have to say that it seems very unintuitive in the case of the wave function, but perhaps it's just very hard to measure such an observable, although it exists. (It seems quite intuitive to me for spin).
Is there a requirement for an operator that corresponds to an observable to be part of a complete set, ie. its eigenvectors span the vector space?
Would ##|u><u|## be part of a complete set?