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First order perturbation question

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose we put a delta function bump in the center of the infinite square well:
    [tex]
    H' = \alpha \delta(x -a/2),
    [/tex]
    where [itex] \alpha [/itex] is constant.

    a) Find the first order correction to the allowed energies.

    b) Find the first three non-zero terms in the expansion of the correction to the ground state, [itex] \psi_1^1 [/itex].

    2. Relevant equations

    First order correction to the energy:
    [tex] E_n^1 = \langle \psi_n^0 | H' | \psi_n^0 \rangle. [/tex]

    First order correction to the wave function:
    [tex]\psi_n^1 = \sum_{m \neq n} \frac{\langle \psi_m^0 | H' | \psi_n^0 \rangle}{(E_n^0 - E_m^0)}\psi_m^0. [/tex]

    Wavefunction for the infinite square well:
    [tex] \psi_n(x) = \sqrt{2/a} \sin{\frac{n \pi}{a}x} [/tex]

    Energy for the infinite square well:
    [tex] E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2} [/tex]


    3. The attempt at a solution

    I think I got part a right. It worked out to [itex]\frac{2 \alpha}{a}[/itex] for odd n and 0 for even n.

    Part b has me a little confused. In the equation for the first order correction to the wavefunction, what am I supposed to sum over? If it is m, then the first three terms are m=3, m=5, m=7 since m is not n and m must be odd. But then I have [itex] \langle \psi_3^0 |H'|\psi_1^0\rangle, \langle \psi_5^0 |H'| \psi_1^0 \rangle, \text{ and } \langle \psi_7^0 |H'| \psi_1^0 \rangle [/itex] in the numerators of the first three terms. But wouldn't orthogonality then make all of these terms zero so there would be no non-zero terms in the expansion? I'm not understanding how to evaluate this sum. Any help would be appreciated.
     
  2. jcsd
  3. Mar 19, 2013 #2

    TSny

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    Gold Member

    What does [itex] \langle \psi_3^0 |H'|\psi_1^0\rangle[/itex] look like when written out as an integral?
     
  4. Mar 19, 2013 #3
    [tex] \langle \psi_3^0|H'|\psi_1^0 \rangle = \frac{2 \alpha}{a}\int \sin{\frac{3 \pi x}{a}} \sin{\frac{\pi x}{a}} \delta(x - a/2) = \frac{2 \alpha}{a} \sin{\frac{3 \pi}{2}} \sin{\frac{\pi }{2}} = - \frac{2 \alpha}{a}[/tex]

    I worked out something like this for all three terms (m=3, m=5, m=7). Is that the correct approach? I thought orthogonality might make this go to zero.
     
  5. Mar 19, 2013 #4

    TSny

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    Homework Helper
    Gold Member

    Looks right to me.
     
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