# Perturbation theory infinite well

1. Oct 6, 2012

### 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 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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2. Oct 6, 2012

### TSny

I don't believe that integral is zero. Check your work to see if you might have made a mistake.

3. Oct 7, 2012

### Dammes

i have looked over and did it again, i still get zero.
could you please try out the integral
∫ ψkψ dx

where ψ=√(2/L) sin⁡((∏/L) x)
and
k=(Vo/L)x

4. Oct 7, 2012

### TSny

Maybe you can show some of the steps of how you integrated $\int_0^L sin^2(\pi x/L)\; x \;dx$. Note that the integrand is positive throughout the range of integration. So, it's got to yield a positive result.

5. Oct 7, 2012

### Dammes

the pictures in the attachment shows some of my working

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6. Oct 7, 2012

### Dammes

looking over it again i see where i went wrong, i used integration in the product rule. sorry for wasting your time and thank you very much!

7. Oct 7, 2012

### Dammes

sorry, please don't read my previous comment. i see where i acutely went wrong, i have to use integration by parts. i have a problem of mixing up integration and differentiation.
hope im correct this time
sorry and thank you

8. Oct 7, 2012

### TSny

No need to be sorry. Glad to be of help.