# Showing that the Virial Theorem holds

by Ed Quanta
Tags: holds, showing, theorem, virial
 P: 297 So I have already calculated correctly that the expectation of the interaction potential for hydrogenic atoms is =-(uZ^2e^4)/((hbar^2)*n^2) Note that u= mass,and V(r)=-Ze^2/r I now have to calculate 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 is going to equal which the book hints at being true since and are said to satisfy the Virial Theorem. Help anyone?
PF Patron
Emeritus
P: 5,539
 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.
 HW Helper Sci Advisor P: 11,722 There's a piece of genius on my behalf: $$\hat{H}=\hat{T}+\hat{V}$$ (1) $$\langle nlm|\hat{H}|nlm\rangle = E_{n} =-\frac{Z^{2}\mu e^{4}}{2\hbar^{2}(4\pi\epsilon_{0})} \frac{1}{n^{2}}$$ (2) $$\langle nlm|\hat{V}|nlm\rangle =-\frac{Ze^{2}}{(4\pi\epsilon_{0})}\langle \frac{1}{r}\rangle _{|nlm\rangle}$$ (3) Compute (3) using the average of 1/r. Then: $$\langle nlm|\hat{T}|nlm\rangle =\langle nlm|\hat{H}-\hat{V}|nlm\rangle=E_{n}+\frac{Ze^{2}}{(4\pi\epsilon_{0})}\langle \frac{1}{r}\rangle _{|nlm\rangle}$$ (4) Tell where u get stuck. Daniel.
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.
P: 416
 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 $$( 4 \pi \epsilon_0 )^{\frac{1}2}$$ is "absorbed" into the unit of charge, so that Coulomb's law can be written as $$\vec{F} = \frac{e^2}{r^2}\hat{r}$$

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