Variational Method - Gaussian Trial Wavefucntion

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Homework Statement

Use the variational method with a gaussian trial wavefunction ψ(x) = Ae^{\frac{-a^{2}x^{2}}{2}} to prove that in 1 dimension an attractive potential of the form shown, no matter how shallow, always has at least 1 bound state.

*Figure is of a potential V(x) that has a minimum on the negative y axis, and tends to 0 as x -> ∞ and -∞*

THERE ARE NO BOUNDARIES SHOWN ON THE DIAGRAM, THIS IS A GENERALISED POTENTIAL

HINT: prove that it is always possible to make the trial energy negative.

Homework Equations

E_{0} \geq \frac{\left\langle ψ | \hat{H} | ψ \right\rangle}{\left\langle ψ | ψ \right\rangle}

The Attempt at a Solution

We don't quite know that Hamilitonian, but we do know that it has a kinetic energy part. So I've written the Hamiltonian as:

\hat{H} = \frac{-h^{2}}{2m}\frac{d^{2}}{dx^{2}} + V(x) Since the potential is unknown at this point.

State should end up negative but larger than or equal to the bound state due to the above condition.

I tried to normal the scalar \left\langleψ|ψ\right\rangle to start things off so I could find the constant, but I have no clue on how to integrate this stupid gaussian function. I used wolfram alpha and I basically get A = \sqrt{\frac{2a}{\sqrt{\pi}}}

So I'm stuck at the first hurdle. I need to find the scalar product and then superimpose the hamiltonian and integrate to see if I meet the condition.

Otherwise, what do I need to do?

 

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Any ideas guys? It's the gaussian that's irritating me. How do I integrate it?

 

[dreamycam]

http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions