- #1
Ed Quanta
- 297
- 0
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?
<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?