Easy variational principle question that I can't integrate

  • Thread starter mrausum
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  • #1
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Homework Statement



Use trial wavefunction exp(-bx^2) to get an upper limit for the groundstate energy of the 1-d harmonic oscillator

The Attempt at a Solution



This is always going to give an integral of x^2*exp(-x^2). How do you do it? :/
 

Answers and Replies

  • #2
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It's a Gaussian integral that comes out to be

[tex]\int_{-\infty}^{\infty} x^2 e^{-b x^2} = \frac{\sqrt{\pi}}{2b^{3/2}}[/tex]
 
  • #3
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yeah, I know that from wolfram alpha too. But that can't be right, because then my upper limit for the ground energy would depend on 1/b^3/2, which would mean if b was very large it would violate the uncertainty principle.
 
  • #4
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Did you normalize your wave test function?
 
  • #5
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Did you normalize your wave test function?

Yeah that's the one, thanks.
 
  • #6
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Wait..that still doesn't work. The norm.const = b^.5/(pi/2)^.5, which doesn't quite cancel with the b^3/2 in the denominator from the integral?
 
  • #7
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Can you show me some of your work? When I do it I get the appropriate contributions from the kinetic and potential expectation values.
 
  • #8
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sure:

[tex]<H> = <(b/\pi).exp(-bx^2)|H|(b/\pi).exp(-bx^2)>[/tex]


[tex]<H> = b/\pi\int^{inf}_{-inf}exp(-2bx^2).\left(\frac{2h^2x}{m}+\frac{1}{2}mw^2x^2\right)dx[/tex]
 
Last edited:
  • #9
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The normalization constant is [tex]A = (\frac{2b}{\pi})^{1/4}[/tex]

From there the Hamiltonian of a 1D harmonic oscillator:

[tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m \omega x^2[/tex]

I'm not sure if you all ready dealt with the second partial with respect to x do your expectation value of the Hamiltonian looks wrong so far.

Remember it's [tex]|A|^2\int_{-\infty}^{\infty} (e^{-b x^2})^* (-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m \omega x^2) e^{-b x^2} dx[/tex]
 

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