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Hydrogen atom ground state with zero orbital angular moment question.

  1. Jan 27, 2009 #1
    Hi all:
    As we know, if we solve the schrodinger equation, the ground state wavefunction is independent of theta and psi. We find the expectation value of ground state orbital angular momentum is zero. But if we don't do any mathematical calculation, can we conlude that?
    For example, Due to symmetry of coulomb potential, Hydrogen atom (one proton plus one electron) wavefunction must be symmetric or antisymmetric (for ground state, there is no degeneracy). Then I am stuck. How can I conclude that only by virtue of physical concept or symmetry? Does symmetry means spherical symmetry?
    thanks

    xf
     
  2. jcsd
  3. Jan 28, 2009 #2

    malawi_glenn

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    well symmetric with respect to partiy : spatial inversion. C.f with particle in box solutions.
     
  4. Jan 28, 2009 #3
    Thank you. What is C.f? parity property only determines psi(-x)=+/-psi(x). where x is vector. It doesn't mean it is spherical symmetry. Could you explain more? thanks
     
  5. Jan 28, 2009 #4

    malawi_glenn

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    c.f means "compare with"

    No, the wavefunctions just need to have positive or negative parity eigenvalue. I don't think there is a theorem which states that the symmetry of the ground state wave function must have the same symmetry as the potential.
     
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