Showing that the Virial Theorem holds

  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?
     
  2. jcsd
  3. Tom Mattson

    Tom Mattson 5,539
    Staff Emeritus
    Science Advisor
    Gold Member

    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. dextercioby

    dextercioby 12,292
    Science Advisor
    Homework Helper

    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. 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. 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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