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Why isn't p^4 Hermitian for hydrogen-like l=0 wavefunctions?

  1. Dec 28, 2011 #1
    Sorry if this question has been asked a million times.

    Either way, I'm working my way through Griffiths. It's a fantastic book--he doesn't try to slip anything past the reader. He is completely honest, and he doesn't abuse mathematics the way most authors do (screwing around with the Dirac delta function to force a normalization constant on the free particle wavefunctions when it's really just there for convenience, etc)

    On that note, I'm working on the perturbation theory chapter. He attempts to correct the hydrogen-like wavefunction energies by applying a relativistic perturbation term. He writes down the relativistic energies, expands it in a power series in the momentum operator, and then applies the usual canonical substitution. The footnote of the page says that [itex]{\hat{p}}^4[/itex] is actually NOT a Hermitian operator when l = 0 (and the hermicity of the perturbation term was assumed in the derivation of the theory), so he says that pertubration theory isn't actually valid in this case. My question is.... why isn't [itex]{\hat{p}}^4[/itex] Hermitian??

    If I understand correctly, the momentum operator is Hermitian, and therefore,
    [itex]<ψ|\hat{p} ψ>=<\hat{p}ψ| ψ>[/itex]

    If this is true, then why isn't this?
    [itex]<ψ|{\hat{p}}^4 ψ> =< ψ|{\hat{p}}^2 {\hat{p}}^2 ψ>= < {\hat{p}}^2 ψ|{\hat{p}}^2 ψ>=< {\hat{p}}^4 ψ| ψ>[/itex]

    After all, of course [itex]{\hat{p}}^2[/itex] is Hermitian. At the end of the chapter, there is an exercise that attempts to answer my question. From skimming it, it appears that this is so because a boundary term doesn't vanish. Still, the proof that I wrote above should work for any wavefunction, and I'm not seeing where it fails. I'm really missing something here!
     
  2. jcsd
  3. Dec 28, 2011 #2
    No, it doesn't, your proof involves dropping the boundary terms. Just think about how to prove p is Hermitian in the first place, and why we need wavefucntion to vanish at infinity for p to be Hermitian.
     
  4. Dec 28, 2011 #3
    I see. Thanks!
     
  5. Dec 29, 2011 #4

    DrDu

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    I am not sure what p is. I suppose you mean the radial momentum operator, which is not self-adjoint on the half line 0 to infinity.
     
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