# Variational Method - Gaussian Trial Wavefucntion

Unto

## Homework Statement

Use the variational method with a gaussian trial wavefunction ψ(x) = A$e^{\frac{-a^{2}x^{2}}{2}}$ 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?