Showing that the Virial Theorem holds

  1. Feb 9, 2005 #1
    So I have already calculated correctly that the expectation of the interaction potential for hydrogenic atoms is

    <nlm|V(r)|nlm>=-(uZ^2e^4)/((hbar^2)*n^2)

    Note that u= mass,and V(r)=-Ze^2/r

    I now have to calculate <nlm|T|nlm> where T is the kinetic energy operator, and

    T=p^2/(2u) + L^2/(2ur^2)

    Note that p is the radial momentum operator and L is the angular momentum operator

    I know hbar^2l(l+1)=L^2 and I know (pretty sure) that <1/r^2>=e^2/(2n^3hbar^2).

    However, I am unsure how to find the expectation value of the p^2/(2u) term, and don't see how <T> is going to equal <V> which the book hints at being true since <T> and <V> are said to satisfy the Virial Theorem.

    Help anyone?
     
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  3. Feb 9, 2005 #2

    Tom Mattson

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    You can do it by brute force (that's what I call working in function space). Insert the differential operator for p into the expression for T, sandwich it between &psi;n'l'm'* and &psi;nlm, and integrate.
     
  4. Feb 9, 2005 #3

    dextercioby

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    There's a piece of genius on my behalf:

    [tex] \hat{H}=\hat{T}+\hat{V} [/tex] (1)

    [tex] \langle nlm|\hat{H}|nlm\rangle = E_{n} =-\frac{Z^{2}\mu e^{4}}{2\hbar^{2}(4\pi\epsilon_{0})} \frac{1}{n^{2}} [/tex] (2)

    [tex] \langle nlm|\hat{V}|nlm\rangle =-\frac{Ze^{2}}{(4\pi\epsilon_{0})}\langle \frac{1}{r}\rangle _{|nlm\rangle} [/tex] (3)

    Compute (3) using the average of 1/r.

    Then:

    [tex] \langle nlm|\hat{T}|nlm\rangle =\langle nlm|\hat{H}-\hat{V}|nlm\rangle=E_{n}+\frac{Ze^{2}}{(4\pi\epsilon_{0})}\langle \frac{1}{r}\rangle _{|nlm\rangle} [/tex] (4)

    Tell where u get stuck.

    Daniel.
     
    Last edited: Feb 9, 2005
  5. Feb 9, 2005 #4
    My book seems to leave out the 4pi*epsilon term in the denominator for some reason. But yeah thanks, I am cool now. I was missing the expectation value for the Hamiltonian. Thanks for your genius my man.
     
  6. Feb 9, 2005 #5
    It must use the cgs unit system.. in cgs a factor of [tex] ( 4 \pi \epsilon_0 )^{\frac{1}2} [/tex] is "absorbed" into the unit of charge, so that Coulomb's law can be written as [tex] \vec{F} = \frac{e^2}{r^2}\hat{r} [/tex]
     
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