Easy variational principle question that I can't integrate

[mrausum]

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? :/

 

[Feldoh]

It's a Gaussian integral that comes out to be

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

 

[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.

 

[Feldoh]

Did you normalize your wave test function?

 

[mrausum]

Did you normalize your wave test function?

Yeah that's the one, thanks.

 

[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?

 

[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.

 

[mrausum]

sure:

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


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

 

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