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Spherically symmetric potential and spherical harmonics

  1. Apr 8, 2010 #1
    When solving the time-independent Schrodinger equation for a spherically symmetric potential, using the separation of variables, we find that solutions of the form [tex]\psi =R(r)Y_l^m(\theta ,\phi)[/tex] where the [tex]Y_l^m[/tex] are the spherical harmonics. We apply this to the (idealized) electron in a Hydrogen atom and of course allow m to take on any integer value from [tex]-l[/tex] to [tex]+l[/tex]. However, I'm reading "Introduction to Quantum Mechanics, 2nd Ed" by Griffths and when he covers quantum scattering (by a spherically symmetric potential) he says "since we are assuming the potential is spherically symmetric, the wave function cannot depend on [tex]\phi[/tex]" (bottom of page 401). Afterwards he always assumes [tex]m=0[/tex]. I assume this is wrong?
     
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  3. Apr 8, 2010 #2
    WHAT is wrong?

    if the pot is spherical sym, then the wave fcn is indep of phi, hence m = 0.
     
  4. Apr 8, 2010 #3

    SpectraCat

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    No, it is not "wrong", why do you think it is? If a potential is spherically symmetric, then all of the m-states for a given value of l are degenerate. So therefore one is free to choose a mathematically convenient value of m .. m=0 is particularly nice, since the angular part of the wavefunctions then becomes purely real.
     
  5. Apr 8, 2010 #4
    ansgar: the [tex]Y_l^m[/tex] with [tex]m\neq 0[/tex] are solutions to the Schrodinger equation for a spherically symmetric potential, in fact that's exactly the way they get introduced in Griffths. The (idealized) electron in the H atom is in a spherically symmetric potential right, but we don't say that [tex]m=0[/tex] there right?

    SpectraCat: When you say degenerate here do you mean with repect to energy (same energy)? I realize the states with different [tex]m[/tex] have the same energy (and even the states with different [tex]l[/tex] also have the same energy). But I don't see how that implies that "the wave function cannot depend on [tex]\phi[/tex]." If the latter is true, how can H atom wave functions depend on [tex]\phi[/tex]?
     
  6. Apr 8, 2010 #5

    SpectraCat

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    Well, I guess I see your point. However, I am certain that the response I gave in my earlier post captures the correct aspects of the physics and mathematics of scattering off a spherical potential. I think the answer to your question above is, for a free H atom in the absence of any external potentials, the electronic wave functions really *don't* depend on phi. It is not until the spherical symmetry (EDIT: actually, even cylindrical symmetry is enough to enforce phi-independence) is broken that the phi-dependence becomes measurable, therefore it is physically consistent to say that it doesn't exist in the absence of such symmetry breaking. The spherical harmonics (and for that matter the Hamiltonian) are mathematical constructions that have been used to develop an abstraction so that we can better understand the physics of the H-atom, but they are not the *only* such construction that is possible.

    Note that the situation is different for the theta-dependence of the wavefunctions. Even in the absence of an external field, angular momentum must be conserved in a spectroscopic transition (the emitted/absorbed photon carries one unit of angular momentum that must be accounted for), therefore we have experimental evidence that there must be some angular dependence to the H-atom wavefunctions, even though the energy eigenvalues show no dependence on the l-quantum number.
     
    Last edited: Apr 9, 2010
  7. Apr 8, 2010 #6
    Thanks for the info and insights SpectraCat. I think I understand what you are saying, but I'll think some more. One problem I have that is somewhat related is that Griffths was initially talking about the particle coming in with an impact parameter b (i.e. not necessarily aimed at the "center" of the scattering potential, which seems like it would not give the symmetry you described, but then Griffths later said the "incoming wave function" was just e^ikz which doesn't have [tex]\phi[/tex] (or b) dependence and gives extra symmetry (not sure what happened to b though).
     
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