# Complete set of eigenvectors question

Suppose you have two observables ##\xi## and ##\eta## so that ##[\xi,\eta]=0##, i know that there exists a simultaneous complete set of eigenvectors wich make my two observables diagonal. Now the question is, if ##\xi## is a degenerate observable the complete set of eigenvectors still exist?

## Answers and Replies

vanhees71
Science Advisor
Gold Member
2019 Award
Yes! The point is if ##\xi## is denerate there should be another observable ##\eta## that's compatible with it, i.e., the self-adjoint operators commute. It should be also independent of ##\xi##, i.e., it shouldn't simply be a self-adjoint operator function of ##\xi##. Then there's a common set of eigenvectors. If the simultaneous eigenspaces are all one-dimensional, then you have a complete compatible set of observables, otherwise there must be more compatible observables.

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.

dextercioby
Science Advisor
Homework Helper
If your xi is degenerate, it means there is another observable zeta which commutes with it, other than ita. You can then expand the CSCO to 3 members: ita, xi, zeta.

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

Many thanks for the help guys :) !

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

ShayanJ
Gold Member
Its simple to prove that if $[\xi,\eta]=0$, then both operators are degenerate. But I doubt about the converse.I remember having experiences that made me believe the converse is not true, but can't remember them exactly. Also I searched a lot but I never found a proof. Can you guys point me to a proof?

Its simple to prove that if $[\xi,\eta]=0$, then both operators are degenerate. But I doubt about the converse.I remember having experiences that made me believe the converse is not true, but can't remember them exactly. Also I searched a lot but I never found a proof. Can you guys point me to a proof?
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

dextercioby
Science Advisor
Homework Helper
<Operators are degenerate> is an inaccurate statement. The spectral values of that operator are actually degenerate which means (choose pure point spectrum for simplicity) the set of 'psi's satisifying the spectral eqn.: A psi = a psi for any particular value 'a' is at least a 2 dimensional vector space. Now what usually determines that sigma(A) is degenerate? The fact that A is included into some CSCO, that is there exists at least one operator B so that AB phi - BA phi =0 on the domain of the commutator. Example: what makes the discrete part of the energy spectrum of the H-atom's Hamiltonian (*the virtual particle assigned to the CoM*) degenerate? The fact that there's L2 with the property that HL2 phi - L2H phi =0 forall phi for which the operator products make sense. In a similar fashion: what makes the spectrum of L2 degenerate? The fact that there is an Lz etc.

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.