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In terms of the energy defined by <br>
<br>
<img style="vertical-align: middle"
src="cid:part1.05000906.09020006@mercer.edu" alt="$\frac{e^4
\mu}{\pi^2 \epsilon_0^2 \hbar^2}$"><br>
<br>
I get the value of -1/128 for the trial function <img
style="vertical-align: middle"
src="cid:part2.00010304.07080303@mercer.edu" alt="$(2-cr)
e^{-cr/2}$">. That is more positive that the ground state
energy given in the problem.<br>
<br>
<br>
On 10/20/13 20:05, Connor Gregory Holt wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD735258278030266A720E5@MERCERMAIL.MercerU.local"
type="cite">
<pre wrap="">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</pre>
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<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
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