# Easy variational principle question that I can't integrate

## 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? :/

It's a Gaussian integral that comes out to be

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

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.

Did you normalize your wave test function?

Did you normalize your wave test function?

Yeah that's the one, thanks.

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?

Can you show me some of your work? When I do it I get the appropriate contributions from the kinetic and potential expectation values.

sure:

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

$$<H> = b/\pi\int^{inf}_{-inf}exp(-2bx^2).\left(\frac{2h^2x}{m}+\frac{1}{2}mw^2x^2\right)dx$$

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The normalization constant is $$A = (\frac{2b}{\pi})^{1/4}$$

From there the Hamiltonian of a 1D harmonic oscillator:

$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m \omega x^2$$

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 $$|A|^2\int_{-\infty}^{\infty} (e^{-b x^2})^* (-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m \omega x^2) e^{-b x^2} dx$$