# Analysis of Harmonic Oscillator [duplicate]

In griffiths,

Why was the lowest chosen like that while in the ladder analogy, the lower energies are like $a_-^n\psi$? That is, why didnt he chose the lowest rung to be $$a_-^n\psi$$

• Your question is very hard to follow because you do not explain your notation or anything. Of course we all know the harmonic oscillator in QM and can kind of guess what you mean, but please formulate your question a bit better and more clearly. – Sanya Sep 29 '16 at 9:20
• Related: physics.stackexchange.com/q/90051/2451 , physics.stackexchange.com/q/23028/2451 and links therein. – Qmechanic Sep 29 '16 at 10:49

The operator $\mathbf a_-$ is the annihilation operator and is normally just written as $\mathbf a$ (we distinguish the creation operator by writing it as $\mathbf a^\dagger$). The annihilation operator acts on a state $\vert \psi_n\rangle$ as follows:
$$\mathbf a\vert \psi_n \rangle = \sqrt{n}\vert \psi_{n-1} \rangle \tag{1}$$
so it converts the state $\psi_n$ to the state $\psi_{n-1}$. Obviously there is a state $\psi_0$ for which $n=0$, and acting on this state according to equation (1) produces zero not another state. Therefore the state $\psi_0$ is the lowest state that can exist.
It therefore makes sense to take $\psi_0$ as the lowest rung and the (infinite) series of states $\psi_n$ above it are obtained by repeatedly using the creation operator.