Easy variational principle question that I can't integrate

1. Nov 25, 2009

mrausum

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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

2. Nov 25, 2009

Feldoh

It's a Gaussian integral that comes out to be

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

3. Nov 25, 2009

mrausum

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. Nov 25, 2009

Feldoh

Did you normalize your wave test function?

5. Nov 25, 2009

mrausum

Yeah that's the one, thanks.

6. Nov 25, 2009

mrausum

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. Nov 25, 2009

Feldoh

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. Nov 25, 2009

mrausum

sure:

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

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

Last edited: Nov 25, 2009
9. Nov 26, 2009

Feldoh

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