- #1
latentcorpse
- 1,444
- 0
Obtain a variational estimate of the ground state energy of the hydrogen atom by taking as a trial function [itex]\psi_T(r) = \text{exp } \left( - \alpha r^2 \right)[/itex]
How does your result compare with teh exact result?
You may assume that
[itex]\int_0^\infty \text{exp } \left( - b r^2 \right) dr = \frac{1}{2} \sqrt{ \frac{\pi}{b}}[/itex]
and that [itex] 1 \text{Ry} = \left( \frac{e^2}{4 \pi \epsilon_0} \right)^2 \frac{m}{2 \hbar^2}[/itex]
so i obviously need to minimise
[itex]E[r] = \frac{ \langle \psi \vline \hat{V} \vline \psi \rangle}{ \langle \psi \vline \psi \rangle}[/itex]
i think I'm getting the wrong answer because my V is wrong.
so i want a coulomb potential between a proton and a electron surely?
[itex]V=\frac{-e^2}{4 \pi \epsilon_0 r^2}[/itex]
clearly I'm going wrong since i have this [itex]r^{-2}[/itex] term that i don't know how to integrate (the only r dependence in hte given integral is in the exponent) and also i haven't used the rydberg thing (although i guess that might not be needed until the end of the question).
thanks for any help.
How does your result compare with teh exact result?
You may assume that
[itex]\int_0^\infty \text{exp } \left( - b r^2 \right) dr = \frac{1}{2} \sqrt{ \frac{\pi}{b}}[/itex]
and that [itex] 1 \text{Ry} = \left( \frac{e^2}{4 \pi \epsilon_0} \right)^2 \frac{m}{2 \hbar^2}[/itex]
so i obviously need to minimise
[itex]E[r] = \frac{ \langle \psi \vline \hat{V} \vline \psi \rangle}{ \langle \psi \vline \psi \rangle}[/itex]
i think I'm getting the wrong answer because my V is wrong.
so i want a coulomb potential between a proton and a electron surely?
[itex]V=\frac{-e^2}{4 \pi \epsilon_0 r^2}[/itex]
clearly I'm going wrong since i have this [itex]r^{-2}[/itex] term that i don't know how to integrate (the only r dependence in hte given integral is in the exponent) and also i haven't used the rydberg thing (although i guess that might not be needed until the end of the question).
thanks for any help.