Cosine perturbation to potential well

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SUMMARY

The discussion focuses on the calculation of perturbed energy in a quantum mechanical potential well using cosine perturbation. The user reports obtaining a result of zero for the perturbed energy, \Delta E_n, across all orders of perturbation, which they find unexpected. The integral evaluated is \frac{2}{L}U_0\int_0^L cos(\frac{2\pi}{L}x)sin^2(\frac{n\pi x}{L}), leading to a zero result due to the periodic nature of the cosine function over its defined interval. The user suggests that the integration of functions over their entire period can yield zero area, prompting further exploration of the integration for different values of n.

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Homework Statement



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Part (b): Find the perturbed energy.

Homework Equations


The Attempt at a Solution



I've solved everything, except part (b).
I got an answer of 0 for part (b) for all orders, which is kind of strange, as one would expect some perturbation.

[tex]\Delta E_n = \langle \psi_n |U|\psi_n\rangle[/tex]
[tex]= \frac{2}{L}U_0\int_0^L cos(\frac{2\pi}{L}x)sin^2(\frac{n\pi x}{L})[/tex]
[tex]= 0[/tex]

Integrating this leads to zero.

All the other orders give zero as well.
 
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I'm thinking since the period is L, integrating a function throughout its entire period gives 0 area?
 
not generally. for example, think of the function cos(x)+5, the curve lies above the x-axis for all values of x, so it definitely does not integrate to zero. Anyway, you should try to do the integration for the three values of ##n## and see what you get.
 

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