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Perturbation: First order correction to particle-in-box eigenstates

  1. Mar 29, 2010 #1
    First order correction to particle-in-box eigenstates for Dirac perturbation

    1. The problem statement, all variables and given/known data
    Calculate the first three nonzero terms in the expansion of the correction to the ground state [itex]\psi^{1}_{1}[/itex] for a Dirac delta perturbation of strength alpha at a/2 (box from 0 to a).

    2. Relevant equations
    [itex]\psi^{1}_{n} = \sum_{m\neqn} \frac{\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle}{E^{0}_{n}-E^{0}_{m}}\psi^{0}_{m}[/tex]


    3. The attempt at a solution
    I started with m=2 and tried to computer the integral [tex]\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle[/tex] by parts.

    Calculations:
    [tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{2 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]
    integration by parts
    [tex]\frac{2}{a}\alpha([sin(\frac{2 \pi x}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{2\pi}{a}\int^{a}_{0}cos(\frac{2 \pi x}{a})dx)[/tex]

    This clearly equals 0. Also, this equation indicates that the correction will be 0 for any m even.

    Trying m=3.
    [tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]
    integration by parts
    [tex]\frac{2}{a}\alpha([sin(\frac{\pi x 3}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{3\pi}{a}\int^{a}_{0}cos(\frac{3 \pi x}{a})dx)[/tex]

    However, this also is 0, and will be for any m odd.

    This clearly cannot be the correct result.
     
    Last edited: Mar 29, 2010
  2. jcsd
  3. Mar 29, 2010 #2

    kuruman

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    It is not. What is the integral

    [tex]\int^{\infty}_{-\infty}f(x) \delta(x-x_0)dx[/tex]

    ?
     
  4. Mar 29, 2010 #3
    [tex]f(x_{0})[/tex]

    In this case that gives:

    [tex]sin(\frac{\pi}{2}) = 1[/tex]

    when taking

    [tex]sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})[/tex]

    to be the part that gets integrated in the integration by parts.
     
  5. Mar 29, 2010 #4

    kuruman

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    Forget "by parts" and make your life simple. Identify

    [tex]
    f(x)=sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})
    [/tex]

    and see what you get.
     
  6. Mar 29, 2010 #5
    Oh. That's easy. Thanks.

    Just to confirm my answer, I've now got:

    for m = 3

    [tex]\frac{\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{3 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx}{E^{(0)}_{1}-E^{(0)}_{3}}[/tex]

    = [tex]\frac{\frac{2}{a}\alpha sin(\frac{3 \pi}{2})sin(\frac{\pi}{2})}{E^{(0)}_{1}-E^{(0)}_{3}}}[/tex]

    = [tex]\frac{-2 \alpha}{a(E^{(0)}_{1}-E^{(0)}_{3})}[/tex]

    = [tex]\frac{-2 \alpha}{a(\frac{\pi^{2} h^{2}}{2ma^{2}} - \frac{9 \pi^{2} h^{2}}{2ma^{2}})}[/tex]

    = [tex]\frac{ma \alpha}{2 \pi h^{2}}[/tex]

    Where h actually stands for h-bar and m stands for mass (rather than the index).
     
    Last edited: Mar 29, 2010
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