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Degeneracy of Hydrogen atomic orbitals with different l-values but same n-value

  1. Mar 17, 2012 #1
    I am terribly confused. I have always been hearing that in the hydrogen atom, 2s and 2p orbitals have the same energy. Similarly, the 3s, 3p and 3d orbitals have the same energies. This is also suggested by the hydrogen spectrum, my professor also believes the same, and I am unable to find anything against this on the internet.

    But what is the basis for this degeneracy?
    Upon solving the radial equations for 2s and 2p orbitals, we get the same eigenvalue for Energy, that depends only on the principal quantum number n. However, the wave functions also have an angular part and upon solving the angular equations for 2s and 2p we get a zero value for the 2s (angular momentum=0) and a finite value for 2p (angular momentum=root(2)*hbar). This angular momentum will contribute an extra value of root(2)*hbar/(2*I) to the energy. This will immediately give 2s and 2p different energy values, so they cannot be degenerate.

    Have I gone wrong somewhere?
     
  2. jcsd
  3. Mar 17, 2012 #2

    tom.stoer

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    The l-degeneracy, i.e. the fact that E(n,l) = E(n) is l-independent is due to a hidden dynamical symmetry of the 1/r potential which results in an additional conserved quantity, the so-called Laplace-Lenz-Runge vector. The 1/r potential has not only the obvious SO(3) symmetry for spatial rotations but a larger SO(4) symmetry. The existence of the Laplace-Lenz-Runge vector and the l-degeneracy allowes one to solve the energy eigenvalue problem algebraically w/o solving the Schrödinger equation (W. Pauli)

    Just google for hydrogen atom SO(4) and you will find numerous articles, scripts and presentations. I am pretty sure that we had this disucussion here a couple of times.

    Your reasoning regarding the additional l-term in Veff(r) giving the Ylm(Ω) functions a different energy is not correct b/c different l-values also affect the Rnl(r) functions. I think you can't understand the l-degeneracy by just solving the Schrödinger equation (you can derive it, but you don't see the deeper reason)
     
  4. Mar 19, 2012 #3
    Thanks! :)
    I get it, it has got something to do with the symmetry, I'll go and look that up.

    However, I still find my reasoning contradictory to this, and I am unable to see any flaw in it.

    I agree that different l-values affect the R(r) functions, however they don't affect the eigenvalues obtained when the separable hamiltonian acts on R, which depend only on 'n'.
     
  5. Mar 19, 2012 #4

    DrDu

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    To be more specific about the difference of radial functions for orbitals with different l, the 2s orbital has 1 radial node, 2p zero nodes. In general ns has n-1 nodes, np n-2, nd n-3 etc. These additional nodes make up for the lower centrifugal potential of states with lower l as compared to states with higher l.
     
  6. Mar 21, 2012 #5
    Oh....
    thanx! :)
    I suppose the hamiltonian need not be separable, after all
     
  7. Mar 21, 2012 #6

    Vanadium 50

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    Tom is correct. The symmetry is more explicit in parabolic coordinates. However, working in parabolic coordinates is not simple.
     
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