# Perturbation theory infinite well

• Dammes
In summary, the conversation discussed the calculation of total energy in an infinite well with small potential using the time independent Schrodinger equation and the integration of the equation of the slope. It was determined that the addition of a small potential does not affect the total energy of the system. However, there was some confusion with the integration, which was eventually resolved through the use of integration by parts.
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.

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• 1-D_Box.jpg
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Dammes said:
i
when integrating ∫ ψkψ dx between 0 and L
I got Zero, ∫ ψkψ dx=0

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

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

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.

the pictures in the attachment shows some of my working

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• Photo on 10-7-12 at 6.37 PM.jpg
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• Photo on 10-7-12 at 6.38 PM #2.jpg
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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!

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 I am correct this time
sorry and thank you

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

## 1. What is perturbation theory in the context of an infinite well?

Perturbation theory is a mathematical method used to approximate solutions to complex systems by breaking them down into simpler, solvable parts. In the case of an infinite well, perturbation theory is used to calculate the energy levels of a particle trapped inside the well when a small perturbation is applied.

## 2. Why is perturbation theory useful in studying an infinite well?

Perturbation theory allows us to calculate the energy levels of a particle in an infinite well with much greater accuracy than other methods. It is especially useful when the perturbation is small, as it allows us to calculate the effects of the perturbation without having to solve the entire system again.

## 3. Can perturbation theory be applied to other systems besides an infinite well?

Yes, perturbation theory can be applied to a wide range of physical systems, including quantum and classical mechanical systems. It is particularly useful when the system contains a small parameter that can be treated as a perturbation.

## 4. What are the assumptions made in perturbation theory for an infinite well?

The main assumption in perturbation theory for an infinite well is that the perturbation is small, meaning its effects on the system are negligible. Additionally, perturbation theory assumes that the system is linear, meaning the perturbation can be added to the original system without causing significant changes to the dynamics.

## 5. Are there any limitations to perturbation theory in the context of an infinite well?

While perturbation theory is a powerful tool in solving complex systems, it does have its limitations. For example, it may not accurately predict the behavior of the system when the perturbation is large or when the system is highly nonlinear. Additionally, perturbation theory relies on the assumption that the perturbation is small, which may not always be the case in real-world systems.

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