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Complete set of eigenvectors question

  1. Jan 2, 2015 #1
    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?
     
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  3. Jan 2, 2015 #2

    vanhees71

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    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.
     
  4. Jan 2, 2015 #3

    dextercioby

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    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).
     
  5. Jan 2, 2015 #4
    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.
     
  6. Jan 2, 2015 #5

    ShayanJ

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    Its simple to prove that if [itex] [\xi,\eta]=0 [/itex], 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?
     
  7. Jan 2, 2015 #6
    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
     
  8. Jan 4, 2015 #7

    ShayanJ

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  9. Jan 4, 2015 #8

    dextercioby

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    <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.
     
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