Perturbation theory infinite well

[Dammes]

in the infinite well with small potential shown in the attachment.
I calculated the total energy by using the time independent Schrodinger equation and adding the correction energy to the equation of the slope k=(Vo/L)x.

E=h^2/8mL^2 +∫ ψkψ dx

ψ=√(2/L) sin⁡(∏/L x)

when integrating ∫ ψkψ dx between 0 and L
I got Zero, ∫ ψkψ dx=0
∴ total energy=h^2/8mL^2 +0

so what i don't understand is when adding a small potential it doesn't affect the total energy of the system? that is what it shows when i integrated it.

 

[The_Duck]

I suspect you've done the integral wrong. That said, it's perfectly possible for a perturbation to have no effect on the energy of any given state, especially if you only compute the change in energy to first order in perturbation theory.

 

[mfb]

With a symmetric density and an antisymmetric additional potential, the first order correction will be 0. The second order includes that the wave function changes (based on the modified potential), and that will give a non-zero correction.