Evaluate the ground state energy using the variational method

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Urvabara
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Homework Statement



[tex]V(x) = k|x|, x \in [-a,a], V(x) = \infty, x \notin [-a,a][/tex]. Evaluate the ground state energy using the variational method.

Homework Equations



[tex]a = \infty[/tex] and [tex]\psi = \frac{A}{x^{2}+c^{2}}[/tex].

The Attempt at a Solution



[tex]1 = |A|^{2}\int_{-a}^{a}\frac{1}{(x^{2}+c^{2})^{2}}\,\text{d}x.[/tex] Is this a correct way to start? How can I calculate this? I used Mathematica, but it only gives some weird-looking answers.
 
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Gokul43201 said:
Yes, that's how you normalize the trial function. And what do you mean by "weird"?

This is what I put:
In[2]:=
\!\(∫\_\(-a\)\%a\( 1\/\((x\^2 + c\^2)\)\^2\) \[DifferentialD]x\)

This is what I got:
Out[2]=
\!\(2\ a\ If[Im[c\/a] ≥ 1 || Im[c\/a] ≤ \(-1\) || Re[c\/a]
≠ 0, \(c\/\(a\^2 + c\^2\) + ArcTan[
a\/c]\/a\)\/\(2\ c\^3\),
Integrate[1\/\((c\^2 + \((a - 2\ a\ x)\)\^2)\)\^2, {x, 0, 1}, \
Assumptions \[Rule] Re[c\/a] \[Equal] 0 && \(-1\) < Im[c\/a] < 1]]\)

I do not understand. What are those imaginary things?

Gokul43201 said:
Also, why do you say a=infty?

In the paper, it says: "Assume that [tex]a=\infty[/tex] and use the trial [tex]\psi(x)=...[/tex]."
 
Urvabara said:
This is what I put:
In[2]:=
\!\(∫\_\(-a\)\%a\( 1\/\((x\^2 + c\^2)\)\^2\) \[DifferentialD]x\)

This is what I got:
Out[2]=
\!\(2\ a\ If[Im[c\/a] ≥ 1 || Im[c\/a] ≤ \(-1\) || Re[c\/a]
≠ 0, \(c\/\(a\^2 + c\^2\) + ArcTan[
a\/c]\/a\)\/\(2\ c\^3\),
Integrate[1\/\((c\^2 + \((a - 2\ a\ x)\)\^2)\)\^2, {x, 0, 1}, \
Assumptions \[Rule] Re[c\/a] \[Equal] 0 && \(-1\) < Im[c\/a] < 1]]\)

I do not understand. What are those imaginary things?
I can't read that easily, but I believe it allows for values of c that are not real. For the problem, you could chose to limit yourself to real c.

In the paper, it says: "Assume that [tex]a=\infty[/tex] and use the trial [tex]\psi(x)=...[/tex]."
Then please write this down as part of the question.Always write down the complete question. Do not summarize or reword in any way.