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?
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 ψ_{n'l'm'}* and ψ_{nlm}, and integrate.
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.
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]