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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? :/

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- Thread starter mrausum
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- #1

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

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

- #2

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[tex]\int_{-\infty}^{\infty} x^2 e^{-b x^2} = \frac{\sqrt{\pi}}{2b^{3/2}}[/tex]

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- #4

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Did you normalize your wave test function?

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Did you normalize your wave test function?

Yeah that's the one, thanks.

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- #7

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- #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]

[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:

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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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