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 $\textbf{Nullity}(a) = 1$?


1 Answer 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.


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