Evaluate the ground state energy using the variational method

[Urvabara]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

1 = |A|^{2}\int_{-a}^{a}\frac{1}{(x^{2}+c^{2})^{2}}\,\text{d}x. Is this a correct way to start? How can I calculate this? I used Mathematica, but it only gives some weird-looking answers.

 

[Gokul43201]

Yes, that's how you normalize the trial function. And what do you mean by "weird"?

Also, why do you say a=infty?

 

[Urvabara]

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?

Also, why do you say a=infty?

In the paper, it says: "Assume that a=\infty and use the trial \psi(x)=...."

 

[Gokul43201]

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 a=\infty and use the trial \psi(x)=...."

Then please write this down as part of the question.Always write down the complete question. Do not summarize or reword in any way.