# Harmonic Oscillator

1. Jul 17, 2008

Griffiths solves the Schrodinger equation using ladder operators, and he then says that there must be a "lowest rung," or $$\psi_{0}$$, such that a_$$\psi_{0}$$ = 0. I'm guessing this also means that E = 0 for a_$$\psi_{0}$$, which is why it wouldn't be allowable.

Because a_$$\psi_{0}$$ is (mathematically, at least) a solution to the Schrodinger equation, it corresponds to the energy ($$E_{0}$$ - $$\hbar\omega$$), which equals zero. Doesn't this mean that $$E_{0}$$ = $$\hbar\omega$$?

But then this conflicts with Griffith's statement that $$E_{0}$$ = $$\frac{1}{2}\hbar\omega$$. HELP!

2. Jul 17, 2008

### CompuChip

Watch out! The statement $a_- \psi_0 = 0$ just says that when you apply the ladder operator to this state, you get nothing, i.e. the vacuum. Physicists have a tendency to use "natural" notation, and you will often see things like $a_- |0\rangle = 0$. Note that here the $|0\rangle$ on the left hand side is a state (wavevector) and the 0 on the right hand side is the zero vector, which is not a state. Griffiths is just kind enough to call it $\psi_0$ instead of $|0\rangle$ so you don't get completely confused

You say that $a_- \psi_0$ is a solution to the Schrodinger equation: that's right. But it isn't a physical solution. It's just the solution $\psi = 0$. The energy of $\psi_0$ is $\frac12 \hbar \omega$ (formula 2.61) which can be seen by applying the Hamiltonian in the form (2.56) to this state and using that $a_-$ annihilates it.

Hope that makes it more clear.

3. Jul 18, 2008