
#1
Feb905, 02:22 PM

P: 297

So I have already calculated correctly that the expectation of the interaction potential for hydrogenic atoms is
<nlmV(r)nlm>=(uZ^2e^4)/((hbar^2)*n^2) Note that u= mass,and V(r)=Ze^2/r I now have to calculate <nlmTnlm> 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
Feb905, 02:39 PM

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PF Gold
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#3
Feb905, 02:41 PM

Sci Advisor
HW Helper
P: 11,863

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. 



#4
Feb905, 02:49 PM

P: 297

Showing that the Virial Theorem holds
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.




#5
Feb905, 11:52 PM

P: 416




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