# Harmonic oscillator

1. Dec 4, 2005

### asdf1

why is the lowest allowed energy not E=0 but some definite minimum E=E0?

2. Dec 4, 2005

### siddharth

If you solve the Time Independent Schrodinger equation for the Harmonic Oscillator, that is
$$-\frac{\hbar^2}{2m} \frac{d^2\Psi}{dx^2} + \frac{1}{2}kx^2 \Psi = E \Psi$$

The quantization of energy comes from the boundary conditions (ie, $\Psi = 0$ when $x= \infty$ or $x = -\infty$).

The permitted energy levels will be

$$E_n = (n+\frac{1}{2}) \hbar \omega$$

So the lowest Energy is not E=0.

Last edited: Dec 4, 2005
3. Dec 4, 2005

### Galileo

I could give a hand-wave argument. We have E=1/2mv^2+1/2kx^2.
If E=0 both x and v are zero, which contradicts Heisenberg.

4. Dec 4, 2005

### asdf1

thank you very much!!! :)