- #1

- 38

- 1

## Homework Statement

The Hamiltonian for a particle in a harmonic potential is given by

[itex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{Kx^2}{2}[/itex], where K is the spring constant. Start with the trial wave function [itex]\psi(x)=exp(\frac{-x^2}{2a^2})[/itex]

and solve the energy eigenvalue equation [itex]\hat{H}\psi(x)=E\psi(x)[/itex]. You must find the value of the constant, a, which will make applying the Hamiltonian to the function return a constant time[sic, I assume he meant "times"] the function. Then find the energy eigenvalue.

## Homework Equations

[itex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{Kx^2}{2}[/itex]

[itex]\psi(x)=exp(\frac{-x^2}{2a^2})[/itex]

[itex]\hat{H}\psi(x)=E\psi(x)[/itex]

## The Attempt at a Solution

Application of the Hamiltonian gave me:

[itex]\hat{H}\psi(x)=[\frac{h^2}{2m}(\frac{1}{a^2}+\frac{x^2}{a^2})+\frac{1}{2}Kx^2]\psi(x)=E\psi(x)[/itex]

If I understand the problem correctly, since E and a must be constant, I must come up with an a, devoid of any x's or functions of x's so taking the Hamiltonian will yield something equally devoid of x's and functions of x's.

I mean, I get that I should have [itex][\frac{h^2}{2m}(\frac{1}{a^2}+\frac{x^2}{a^2})+\frac{1}{2}Kx^2]=E[/itex] for some constants E, a. The problem is, I have no clue how to go about solving that and getting a to not involve any x's, nor am I convinced that it's possible. Using the quadratic formula on it gives a big mess (involving x), so I'm pretty sure that's the wrong route.

Thanks for the help.