Change in ground state energy due to perturbation

[Shelly1]

Homework Statement

Consider a quantum particle of mass m in one dimension in an infinite potential well , i.e V(x) = 0 for -a/2 < x < a/2 , and V(x) =∞ for |x| ≥ a/2 . A small perturbation V'(x) =2ε|x|/a , is added. The change in the ground state energy to O(ε) is:

Homework Equations

The ground state wave function for a particle an symmetric potential well is Ψ=√(2/a) cos(πx/a)
change in energy= <Ψ| V'(x) |Ψ>

The Attempt at a Solution

change in ground state energy can be calculated using this equation <Ψ| V'(x) |Ψ>

 

[Shelly1]

The correct answer to this question is ε(π^2 -4)/ 2π^2. I don't know how they got this.

 

[DrClaude]

The correct answer to this question is ε(π^2 -4)/ 2π^2. I don't know how they got this.

Have you tried calculating <Ψ| V'(x) |Ψ>?

 

[Shelly1]

Yes.but I got zero as answer.

Have you tried calculating <Ψ| V'(x) |Ψ>?

 

[DrClaude]

Yes.but I got zero as answer.

Were you careful to consider that it is |x| that appears in the potential?

 

[Shelly1]

Yes ofcourse...

Were you careful to consider that it is |x| that appears in the potential?

 

[DrClaude]

Then please show your work, because that integral is not zero.

 

[Shelly1]

You were right.unfortunately i have made a mistake with integral and it is never be zero.fIinally i got the same answer.Thank you Dclaude