in the infinite well with small potential shown in the attachment.(adsbygoogle = window.adsbygoogle || []).push({});

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 dont understand is when adding a small potential it doesnt affect the total energy of the system? that is what it shows when i integrated it.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Perturbation theory infinite well

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