Analysis of Harmonic Oscillator

Question

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 $$a_-^n\psi$$

Answer 1

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.