Change in ground state energy due to perturbation

  • #1
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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>
 

Answers and Replies

  • #2
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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.
 
  • #3
DrClaude
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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.
Have you tried calculating <Ψ| V'(x) |Ψ>?
 
  • #5
DrClaude
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Yes.but I got zero as answer.
Were you careful to consider that it is |x| that appears in the potential?
 
  • #6
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Yes ofcourse...
Were you careful to consider that it is |x| that appears in the potential?
 
  • #7
DrClaude
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Then please show your work, because that integral is not zero.
 
  • #8
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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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