Perturbation of Potential (Particle in a Box)

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Homework Statement



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.

Homework 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.]

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..
 

Answers and Replies

  • #2
kuruman
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[tex]E_{n}^{0} = <\psi_{n}^{0}|V_{1}|\psi_{n}^{0}> = V_{1}<\psi_{n}^{0}|\psi_{n}^{0}> = V_{1}[/tex]

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.
 
  • #3
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[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?
 
  • #4
kuruman
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It is. Now do the integral and see what you get.
 
  • #5
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[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
 
  • #6
kuruman
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What are your limits of integration?
 
  • #7
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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]

?
 
  • #8
kuruman
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Does the box extend from minus infinity to plus infinity?
 
  • #9
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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]
 
  • #10
kuruman
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You forgot the normalization constant in front of the wavefunction. Also, can you simplify your expression?
 
  • #11
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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.
 
  • #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.
 
  • #13
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[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]
 
  • #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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