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Change in ground state energy due to perturbation

  1. Apr 1, 2016 #1
    1. The problem statement, all variables and given/known data
    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:

    2. Relevant 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>
    3. The attempt at a solution
    change in ground state energy can be calculated using this equation <Ψ| V'(x) |Ψ>
     
  2. jcsd
  3. Apr 1, 2016 #2
    The correct answer to this question is ε(π^2 -4)/ 2π^2. I don't know how they got this.
     
  4. Apr 1, 2016 #3

    DrClaude

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    Staff: Mentor

    Have you tried calculating <Ψ| V'(x) |Ψ>?
     
  5. Apr 1, 2016 #4
    Yes.but I got zero as answer.
     
  6. Apr 1, 2016 #5

    DrClaude

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    Staff: Mentor

    Were you careful to consider that it is |x| that appears in the potential?
     
  7. Apr 1, 2016 #6
    Yes ofcourse...
     
  8. Apr 1, 2016 #7

    DrClaude

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    Staff: Mentor

    Then please show your work, because that integral is not zero.
     
  9. Apr 1, 2016 #8
    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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