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Hydrogen atom 1/r^2 expectation value

  • Thread starter Unkraut
  • Start date
  • #1
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Homework Statement


Using the Feynman-Hellman theorem, determine the expectation values of 1/r and 1/r^2 for the hydrogen atom.


Homework Equations


Hamiltonian: [tex]H=-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}[/tex]
energy eigenvalues: [tex]E_n=-\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2(N+l+1)^2}[/tex]
N is the largest power of the Laguerre polynomial associated with the solution of [itex]u_{nl}[/itex] (and I have actually no idea what that is and I hope I don't need it), l is the azimuthal quantum number and n the principal quantum number.
n=N+l+1

Feynman-Gellman theorem: [tex]\frac{\partial E_n}{\partial \lambda}=\langle \psi_n | \frac{\partial H}{\partial \lambda} | \psi_n \rangle[/tex] for a Hamiltonian which depends on a parameter [tex]\lambda[/tex]

The Attempt at a Solution


For <1/r^2>:
Take the derivative of the Hamiltonian with respect to l:
[tex]\frac{\partial H}{\partial l}=\frac{\hbar^2}{2m}\frac{2l+1}{r^2}[/tex]
[tex]\Rightarrow \frac{1}{r^2}=\frac{2m}{\hbar^2(2l+1)}\frac{\partial H}{\partial l}[/tex]
[tex]\Rightarrow \langle\frac{1}{r^2}\rangle=\frac{2m}{\hbar^2(2l+1)}\langle\frac{\partial H}{\partial l}\rangle[/tex]
[tex]=\frac{2m}{\hbar^2(2l+1)}\frac{\partial E_n}{\partial l}[/tex] (Feynman-Gellman theorem used here)
[tex]=...[/tex] ... straightforward direct calculation which is missing now because texing is so tedious and I think maybe I have made an error in my notes which I must check later.

So, now I have the expectation value of 1/r^2.

But I can't do the same for 1/r. I cannot express 1/r in terms of the derivative of the Hamiltonian with respect to some parameter because in the term [tex]\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}[/tex] there is no parameter.
Seems I need the expectation value of [tex]\frac{d^2}{dr^2}[/tex], then I can calculate the expectation value of 1/r from the Hamiltonian. My uneducated intuition tells me that maybe that is zero, but that is not a good argument and often my intuition is wrong. I have no idea. Any hint appreciated.
 

Answers and Replies

  • #2
525
7
You're seeing too many boundaries ;)

Why wouldn't the electric charge serve as a parameter..?
 
  • #3
30
0
Oh, now that you say it, that seems reasonable. Thank you!
In quantum mechanics I always have problems determining what kind of reasoning is actually reasonable :)
 

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