Showing that the Virial Theorem holds


by Ed Quanta
Tags: holds, showing, theorem, virial
Ed Quanta
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#1
Feb9-05, 02:22 PM
P: 297
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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Tom Mattson
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#2
Feb9-05, 02:39 PM
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Quote Quote by Ed Quanta
However, I am unsure how to find the expectation value of the p^2/(2u) term,
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.
dextercioby
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Feb9-05, 02:41 PM
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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.

Ed Quanta
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#4
Feb9-05, 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.
kanato
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#5
Feb9-05, 11:52 PM
P: 416
Quote Quote by Ed Quanta
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.
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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