# Complete set of eigenvectors question

1. Jan 2, 2015

### Clear Mind

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?

2. Jan 2, 2015

### vanhees71

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.

3. Jan 2, 2015

### dextercioby

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).

4. Jan 2, 2015

### Clear Mind

Many thanks for the help guys :) !

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.

5. Jan 2, 2015

### ShayanJ

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?

6. Jan 2, 2015

### Clear Mind

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. Jan 4, 2015

8. Jan 4, 2015

### dextercioby

<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.