# Ground state of wavefunction

A particle has mass 'm' and potential $V(x) = x^2$. Find the ground state of this wavefunction.

Should I use the Hamiltonian operator here?

## Answers and Replies

Yes. You should get something very similiar to the harmonic oscillator.

Thanks, inha! I compared it to the PE function of a harmonic oscillator.

$$V(x) = {1\over 2} m \omega^2 x^2$$

Comparing it with [itex]V(x) = x^2[/tex],

$${1\over 2} m \omega^2 = 1$$

$$\omega = \sqrt{2\over m}$$

And I substitute this value in the formula:
$$E_n = \hbar \omega \left(n + {1\over 2}\right)$$

Am I going right? Reshma said:
Thanks, inha! I compared it to the PE function of a harmonic oscillator.

$$V(x) = {1\over 2} m \omega^2 x^2$$

Comparing it with [itex]V(x) = x^2[/tex],

$${1\over 2} m \omega^2 = 1$$

$$\omega = \sqrt{2\over m}$$

And I substitute this value in the formula:
$$E_n = \hbar \omega \left(n + {1\over 2}\right)$$

Am I going right? It looks good (Of course, n=0 for the ground state, don't forget that!), but make sure you know how to do the problem from scratch. I'm guessing that's what your professor wanted you to do.

-Dan