# Eigenvalue for harmonic oscillator

1. Oct 1, 2013

### Habeebe

1. The problem statement, all variables and given/known data
The Hamiltonian for a particle in a harmonic potential is given by
$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{Kx^2}{2}$, where K is the spring constant. Start with the trial wave function $\psi(x)=exp(\frac{-x^2}{2a^2})$
and solve the energy eigenvalue equation $\hat{H}\psi(x)=E\psi(x)$. 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.

2. Relevant equations
$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{Kx^2}{2}$
$\psi(x)=exp(\frac{-x^2}{2a^2})$
$\hat{H}\psi(x)=E\psi(x)$

3. The attempt at a solution
Application of the Hamiltonian gave me:
$\hat{H}\psi(x)=[\frac{h^2}{2m}(\frac{1}{a^2}+\frac{x^2}{a^2})+\frac{1}{2}Kx^2]\psi(x)=E\psi(x)$

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 $[\frac{h^2}{2m}(\frac{1}{a^2}+\frac{x^2}{a^2})+\frac{1}{2}Kx^2]=E$ 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.

2. Oct 1, 2013

### ehild

You have a sign error: How is the operator $\hat{p}$ defined?

To remove x from the equation, collect the terms with x2. You have to choose the parameter a properly, so as the two terms cancel each other.

ehild

3. Oct 1, 2013

### TSny

Also check to see if you have enough factors of $a$ in the denominator of your $x^2/a^2$ term.

4. Oct 1, 2013

### Habeebe

Thanks, I found the sign error and got it worked out. As for the factors of a, you're right, I actually typed it in wrong.