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- Thread starter Clear Mind
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- #1

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- #2

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E.g., for a single non-relativistic particle with spin, a complete basis of generalized eigenvectors ##|\vec{p},\sigma_z \rangle## with ##\vec{p} \in \mathbb{R}^3## the momentum of the particle and ##\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}##, where ##s \in \{0,1/2,1,\ldots \}## is the spin of the particle (##s=1/2## for an electron). Thus the three components of the momentum of the momentum and the spin and magnetic quantum number are a complete set of observables for a single non-relativistic particle.

- #3

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(Are you reading Dirac's book? Only Dirac denotes observables/operators by lower case Greek letters).

- #4

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No, i'm reading an italian book "Lessons of Quantum Mechanics" written by Luigi E. Picasso, it's usually used by the University of Pisa. It's a pity that doesn't exist an english version, the book it's really understandable.(Are you reading Dirac's book? Only Dirac denotes observables/operators by lower case Greek letters).

- #5

ShayanJ

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- #6

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I'm not sure if it's true ... I know that if you have three observables ##\xi, \eta,\chi## so that:

##[\xi,\eta]=0##

##[\xi,\chi]=0##

##[\chi,\eta]\neq 0##

than you have that ##\xi## is a degenerate observable

- #7

ShayanJ

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No answer to #5 @dextercioby and @vanhees71 ? I searched myself But I found nothing!

- #8

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So if the spectrum of A is degenerate, there has to be an operator B strongly commuting with it, with respect to whose spectral values one can compute the denegeracy (dimension of a vector space) of each spectral value of A.

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