# Change in ground state energy due to perturbation

## 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>

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

DrClaude
Mentor
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>?

Yes.but I got zero as answer.
Have you tried calculating <Ψ| V'(x) |Ψ>?

DrClaude
Mentor
Yes.but I got zero as answer.
Were you careful to consider that it is |x| that appears in the potential?

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

DrClaude
Mentor