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Homework Help: Perturbation of Potential (Particle in a Box)

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Assume that the particle in the box is perturbed by a potential [tex]V_{1}(x) = x [/tex].

    Calculate the energy shift of the ground state and the first excited state in first-order
    perturbation theory.

    2. Relevant equations

    Unperturbed wave functions for the particle given by:

    [tex]\psi_{n}^{0}(x) = \sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})[/tex]

    [Hint: This energy shift is given by the expectation value of the perturbation.]

    3. The attempt at a solution

    Perturbation: [tex]H' = V_{1}(x) = \gamma x[/tex]

    Correction to energy of 'n'th state is:

    [tex]E_{n}^{0} = <\psi_{n}^{0}|V_{1}|\psi_{n}^{0}> = V_{1}<\psi_{n}^{0}|\psi_{n}^{0}> = V_{1}[/tex]

    Therefore corrected energy levels defined as:

    [tex]E_{n} \approx E_{n}^{0}+V_{1}(x)[/tex]

    Don't know where to go from here..
     
  2. jcsd
  3. Feb 21, 2010 #2

    kuruman

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    What you have here is incorrect and meaningless. To begin with

    [tex]E_{n}^{0} = <\psi_{n}_{0}|H^{0}|\psi_{n}^{0}>[/tex]

    is the unperturbed energy.

    Secondly, V1 is a function equal to γx and cannot be set equal to a constant, i.e. it cannot be pulled of the integral. Don't forget that

    [tex]<\psi_{n}^{0}|V_{1}|\psi_{n}^{0}> = \int (\psi_{n}^{0})^*\:\gamma x \;\psi_{n}^{0}\;dx[/tex]

    You need to do the integral correctly.
     
  4. Feb 21, 2010 #3
    [tex]\psi_{n}^{0}(x) = \sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})[/tex]

    [tex](\psi_{n}^{0}(x))^{2} = \left(\sqrt{\frac{2}{L}}sin(\frac{n \pi x}{L})\right)^{2} = \frac{2sin^{2}(\frac{n \pi x}{L})}{L}[/tex]

    [tex]<\psi_{n}^{0}|V_{1}|\psi_{n}^{0}> = \int \left( (\gamma x)(\frac{2sin^{2}(\frac{n \pi x}{L})}{L}) \right) dx[/tex]

    [tex]<\psi_{n}^{0}|V_{1}|\psi_{n}^{0}> = \frac{2\gamma}{L} \int \left( xsin^{2}(\frac{n \pi x}{L}) \right) dx[/tex]

    Is this now going along the right way?
     
  5. Feb 21, 2010 #4

    kuruman

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    It is. Now do the integral and see what you get.
     
  6. Feb 21, 2010 #5
    [tex]\int \left( xsin^{2}(\frac{n \pi x}{L}) \right) dx = \frac{\left(\frac{x}{2}\right) -sin(\frac{2n \pi x}{L})} {(\frac{4n\pi}{L})}

    [/tex]

    [tex]

    \int \frac{\left(\frac{x}{2}\right) -sin\left(\frac{2n \pi x}{L}\right)} {(\frac{4n\pi}{L})} = \left(\frac{-L}{4n\pi}\right) \int \left(\frac{x}{2}\right) -sin \left(\frac{2n \pi x}{L}\right) dx

    [/tex]

    [tex]

    \int \left(\frac{x}{2}\right) -sin\left(\frac{2n \pi x}{L}\right) dx = \left( \frac{x^{2}}{2} \right) + cos\left(\frac{2n \pi x}{L}\right)

    [/tex]

    Therefore:

    [tex]\int \left( xsin^{2}(\frac{n \pi x}{L}) \right) dx = \left(\frac{-L}{4n\pi}\right)[\left( \frac{x^{2}}{2} \right) + cos\left(\frac{2n \pi x}{L}\right)
    ]


    [/tex]

    .. where do I go from here? :S
     
  7. Feb 21, 2010 #6

    kuruman

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    What are your limits of integration?
     
  8. Feb 21, 2010 #7
    This:

    [tex]\int_{-\infty}^{\infty} \left( xsin^{2}(\frac{n \pi x}{L}) \right) dx = \left(\frac{-L}{4n\pi}\right)\left[\left( \frac{x^{2}}{2} \right) + cos\left(\frac{2n \pi x}{L}\right)\right]_{-\infty}^{\infty}[/tex]

    ?
     
  9. Feb 21, 2010 #8

    kuruman

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    Does the box extend from minus infinity to plus infinity?
     
  10. Feb 21, 2010 #9
    Ah! No.. it goes from x=0 to x=L. Better put the limits in then! ..

    [tex]\int_{0}^{L} \left( xsin^{2}(\frac{n \pi x}{L}) \right) dx = \left(\frac{-L}{4n\pi}\right)\left[\left( \frac{x^{2}}{2} \right) + cos\left(\frac{2n \pi x}{L}\right)\right]_{0}^{L}[/tex]

    So:

    [tex]
    \left[\left( \frac{x^{2}}{2} \right) + cos\left(\frac{2n \pi x}{L}\right)\right]_{0}^{L}
    =

    \left[\left( \frac{L^{2}}{2} \right) + cos\left(2n \pi \right)\right]
    -

    \left[1\right]


    [/tex]
     
  11. Feb 21, 2010 #10

    kuruman

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    You forgot the normalization constant in front of the wavefunction. Also, can you simplify your expression?
     
  12. Feb 21, 2010 #11
    Just multiplied everything together and simplified the expression, and got this:

    [tex]

    \left< \psi_{n}^{0}|V_{1}|\psi_{n}^{0}\right> = \left(\frac{2\gamma}{L}\right)\left( \frac{\left(2L-L^{3}\right)-2Lcos\left(2\pi x\right)}{8n\pi}\right)

    [/tex]

    And then just put in n=1 and n=2 to get the 2 expressions for energy shifts.
     
  13. Feb 21, 2010 #12

    kuruman

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    Try again. You are evaluating an integral from 0 to L. There should be no "x" in your expression. Also, the 2L - L3 term is dimensionally incorrect.
     
  14. Feb 21, 2010 #13
    [tex]\left< \psi_{n}^{0}|V_{1}|\psi_{n}^{0}\right> = \left(\frac{2\gamma}{L}\right)





    \left( \left(\frac{\left-L^{3}+L}{2}\right) -cos\left(2\pi n\right)}\right)\right)[/tex]
     
  15. Feb 22, 2010 #14

    kuruman

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    Your answer is still dimensionally incorrect. When you say (- L3 + L) you adding something that has dimensions of length cubed to something that has dimensions of length. You cannot do that and have an answer that makes physical sense. Check (or perhaps post) your integration steps; you must be dropping something along the way. Also, can you simplify cos(2πn)?
     
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