[CHM 331] var prob

[Andrew J. Pounds]

In terms of the energy defined by

$\frac{e^4 \mu}{\pi^2 \epsilon_0^2 \hbar^2}$

I get the value of  -1/128 for the trial function $(2-cr) e^{-cr/2}$.  
That is more positive that the ground state energy given in the problem.


On 10/20/13 20:05, Connor Gregory Holt wrote:
> Dr. Pounds,
>
> When looking at Table 10.1, I noticed  the trial function three that you gave us resembles the radial portion of the 2s orbital, so this gives us the ground state energy for the 2s orbital when I apply the variational method, not the 1s orbital. Was that intentional? Because that energy is lower than the 1s orbital, which would explain why I can't get the energy to be higher than the ground state for the 1s orbital. For every other trial function of a form not similar to the third, I am getting an energy that is either exact or greater.
>
> Thanks!!
>
> Connor


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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