Showing that the Virial Theorem holds

[Ed Quanta]

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?

 

[quantumdude]

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]

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.

 

[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.

 

[kanato]

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 $$\vec{F} = \frac{e^2}{r^2}\hat{r}$$