Change in ground state energy due to perturbation

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Homework Help Overview

The discussion revolves around a quantum mechanics problem involving a particle in an infinite potential well and the effect of a small perturbation on the ground state energy. The original poster presents the problem and relevant equations, focusing on calculating the change in ground state energy due to the perturbation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the change in energy using the integral <Ψ| V'(x) |Ψ>. Some express confusion about arriving at the correct answer, while others question the handling of the absolute value in the potential.

Discussion Status

The conversation includes attempts to clarify the calculation process, with some participants noting mistakes in their integrals. There is a recognition of the importance of correctly interpreting the potential, and guidance is offered to ensure proper evaluation of the integral.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. The discussion reflects varying levels of understanding and interpretation of the problem setup.

Shelly1
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Homework Statement


Consider a quantum particle of mass m in one dimension in an infinite potential well , i.e V(x) = 0 for -a/2 < x < a/2 , and V(x) =∞ for |x| ≥ a/2 . A small perturbation V'(x) =2ε|x|/a , is added. The change in the ground state energy to O(ε) is:

Homework Equations


The ground state wave function for a particle an symmetric potential well is Ψ=√(2/a) cos(πx/a)
change in energy= <Ψ| V'(x) |Ψ>

The Attempt at a Solution


change in ground state energy can be calculated using this equation <Ψ| V'(x) |Ψ>
 
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The correct answer to this question is ε(π^2 -4)/ 2π^2. I don't know how they got this.
 
Shelly1 said:
The correct answer to this question is ε(π^2 -4)/ 2π^2. I don't know how they got this.
Have you tried calculating <Ψ| V'(x) |Ψ>?
 
Yes.but I got zero as answer.
DrClaude said:
Have you tried calculating <Ψ| V'(x) |Ψ>?
 
Shelly1 said:
Yes.but I got zero as answer.
Were you careful to consider that it is |x| that appears in the potential?
 
Yes ofcourse...
DrClaude said:
Were you careful to consider that it is |x| that appears in the potential?
 
Then please show your work, because that integral is not zero.
 
You were right.unfortunately i have made a mistake with integral and it is never be zero.fIinally i got the same answer.Thank you Dclaude
 

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