# Proof that the QHO annihilation operator has nullity of 1

In a traditional analysis of a quantum harmonic oscillator (QHO), operators $$a$$ and $$a^\dagger$$ are introduced and it is shown that

$$H a |{n}\rangle = (E_n - \hbar \omega_0)a|{n}\rangle,$$ $$H a^\dagger|n\rangle = (E_n + \hbar\omega_0)a^\dagger|n\rangle.$$

Next, after deducing that $$a|0\rangle = 0$$, the ground state energy, $$E_0$$ is found to be $$E_0 = \frac{\hbar\omega_0}{2}.$$

Finally, it is stated that because $$H a^\dagger|n\rangle = (E_n + \hbar\omega_0)a^\dagger|n\rangle$$, All adjacent eigenstates are separated by energy $$\hbar\omega_0$$, therefore $$E_n=(n+\frac{1}{2})\hbar\omega_0.$$

The previous statement relies on the fact that $$a^\dagger|n\rangle \propto|n+1\rangle$$. In other words, it assumes that the set $$F = \left\{\left(\left(a^\dagger\right)^n|0\rangle\right)\propto|n\rangle\right\}_{n\in\mathbb{W}}$$ includes all eigenstates of $$H$$.

A simple proof (attempt) is as follows:

We know that $$E_0 < E_1 \le E_0 + \hbar\omega_0$$ because $$a^\dagger|0\rangle$$ is an eigenstate with energy $$E_0+\hbar\omega_0$$. In other words, we can say that

$$a^\dagger|0\rangle=|i\rangle, E_i = E_0+\hbar\omega$$

The question becomes: does there exist an eigenstate $$|k\rangle$$ such that

$$E_0 < E_k < E_i?$$

If not, then $$E_i = E_1 = E_0 + \hbar\omega$$, and the remaining eigenstates are in the set $$F$$.

The approach is to assume such a state $$|k\rangle$$ exists and see if a contradiction arises. Knowing that $$E_0 = \frac{\hbar\omega_0}{2}$$ and $$E_i = \frac{3\hbar\omega_0}{2}$$, this assumption means that

$$E_k = \frac{1}{2}\hbar\omega_0 + \epsilon, 0 < \epsilon < \hbar\omega_0.$$

Since $$|k\rangle$$ is an eigenstate of the QHO, $$a|k\rangle$$ is also an eigenstate with energy $$E_k-\hbar\omega_0=\epsilon-\frac{1}{2}\hbar\omega_0.$$

For any $$\epsilon < \hbar\omega_0$$, $$\epsilon - \frac{1}{2}\hbar\omega_0 < \frac{1}{2}\hbar\omega_0$$. Since $$\frac{1}{2}\hbar\omega_0$$ is by definition the lowest allowable energy, we have a contradiction.

However, this conclusion relies on the implicit assumption that $$a|k\rangle\ne 0$$. If, instead, we impose $$a|k\rangle= 0$$ as was done for $$|0\rangle$$ when defining it to be the ground state, then we no longer arrive at a contradiction, and have not yet disproved the existence of the state $$|k\rangle$$.

Note that it is not sufficient to show non-degeneracy of the QHO Hamiltonian because $$a|0\rangle = a|k\rangle = 0$$ does not imply degeneracy of the Hamiltonian.

An alternative that does complete the above proof is to verify that the null space of $$a$$ is 1 dimensional ($$\textbf{Nullity}(a) = 1$$). Then, because $$|0\rangle\in\textbf{ker}(a)$$, $$a|k\rangle=0$$ is not possible and $$|k\rangle$$ does not exist.

My question, then, is this: how can we verify that $$\textbf{Nullity}(a) = 1$$?

Abstractly speaking, from the point of view of creation and annihilation operators, one cannot prove that the ground state is non-degenerate with OP's current list of assumptions, cf. OP's title question. The Hilbert space $${\cal H}$$ could be in principle be a direct sum of several Fock spaces. OP's sought-for non-degeneracy is an additional assumption, cf. e.g. my Phys.SE answer here.