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Eigenvaules and quantum numbers

  1. Nov 7, 2011 #1
    Eigenvalues and quantum numbers

    My professor was discussing the Schrodinger equation in 3-d today and mentioned that there are degeneracies within quantum numbers (for the hydrogen atom, the x, y, and z components are symmetric?). Does that imply that the eigenvalues of some operator are the quantum numbers of a system? If so, which operator is it?
     
    Last edited: Nov 7, 2011
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  3. Nov 8, 2011 #2

    tom.stoer

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    Studying the Hamiltonian H you see that it is rotationally invariant, i.e. it commutes with all components of orbital angular momentum Li acting as generators of rotations:

    [H,Li] = 0

    H contains an angular momentum term, namely ~L²/r² which is the familiar additional term in the effective potential (already present in the Kepler problem).

    Studying L² and Lz you get eigenfunctions, the so called spherical harmonics Ylm; they are labels by two quantum numbers l and m:

    L² Ylm = l(l+1) Ylm

    Lz Ylm = m Ylm

    The radial part of H is responsible for the quantum number n, so in total the states are labelled as |nlm> with n=1,2,..., l=0,1,...,n-1, m=-l, ...,l-1, l

    It's interesting (and I have to admit that I do not understand if there is some deeper reason!) that
    a) the radial equation results in a radial wave funtion Rnl(r) where l is due to the additional term in the effective potential ~L²/r² ~ l(l+1)/r², but
    b) the eigenvalues Enl do in the very end not depend on l, i.e. are labelled as En ~ 1/n²

    So in total you have

    H |nlm> = En |nlm>
    L² |nlm> = l(l+1) |nlm>
    Lz |nlm> = m |nlm>
     
  4. Nov 8, 2011 #3

    dextercioby

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    What are the so-called <quantum numbers> ? Eigenvalues of operators describing observables. So what are these observables which could have a countable number of spectral values ? Energy for once, then spin, then orbital angular momentum, then total angular momentum, then electric charge, then barionic charge, parity, etc.

    So for a conserved observable, (its operator, assumed time-independent in the Schrödinger picture) there's always a degeneracy of the discrete energy levels, since each closed subspace associated with an energy value is reduced by the symmetry generator, for example L2 for the spherically-symmetric potentials (free particle in 3D, isotropic oscillator in 3D, H-atom/hydrogenoid ions).
     
  5. Nov 8, 2011 #4

    kith

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    There's an additional conserved quantity in the Kepler problem -the Laplace–Runge–Lenz vector- which is due to a subtle SO(4) symmetry of the potential. This leads to the l-degeneracy of the hydrogen atom energies.
     
  6. Nov 8, 2011 #5

    tom.stoer

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    This was my conjecture, but I wasn't sure about it.

    That means that for V(r)~1/r with its SO(4) symmetry in dim=3 and for V(r)~r² with its SU(N) symmetry in dim=N we have l-degeneracy, but for no other V(r)~rz (for different z-values). For the harmonic oscillator in dim=N it's trivial to prove SU(N) symmetry and degeneracy, for the hydrogen atom it's more involved.
     
  7. Nov 9, 2011 #6
    This was much more in depth than I intended. Thanks for the info. When does this sort of material get introduced in a physics major?
     
  8. Nov 9, 2011 #7

    dextercioby

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    Depends on the curriculum the teacher has prepared for his QM class. You may also get a separate group theory course where applications of group theory to quantum mechanics are presented.
     
  9. Nov 9, 2011 #8

    kith

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    I didn't hear about this in lectures but stumbled upon it by studying symmetry operators for an oral exam. I think that most lecturers don't want to spend half the QM course on the hydrogen atom. ;-)
     
  10. Nov 9, 2011 #9

    Vanadium 50

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    Correct. This is also responsible for the fact that the Schrodinger equation is separable in two coordinate systems for those two potentials: spherical coordinates, and rectangulars for the harmonic oscillator and parabolic for the Coulomb potential.
     
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