Background
In introductory Quantum Mechanics textbooks, the algebraic method of determining the energy eigenstates of the harmonic oscillator involves factorising the Hamiltonian as $$H=\hbar\omega(a^\dagger a + \frac{1}{2})$$ and showing that the $a^\dagger$ and $a$ operators respectively raise and lower the energy by $\hbar\omega$. Thus, the spectrum of the harmonic oscillator is determined to be a ladder with a step-size of $\hbar\omega$.
However, there is rarely any discussion on why this is the end all and be all for the harmonic oscillator spectrum. How do we know that there aren't other operators or, equivalently, a different factorisation of $H$, that would yield a new ladder with a possibly different step-size. This question had been nagging me for a long time. Now, with the help of ChatGPT-5, I understand why.
Uniqueness of ladder operators
The harmonic oscillator Hamiltonian is $$H = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 x^2$$ with corresponding operator identities $$[x,p] = i\hbar, \quad [H,x] = -\frac{i \hbar}{m} p, \quad [H,p] = i m \omega^2 \hbar x.$$
Assume a general operator $$B = \alpha x + \beta p$$ in the linear span of $\{x,p\}$ and demand that it be an eigenoperator of the adjoint action $ad_H(\cdot) = [H,\cdot]$ of $H$. In other words $$[H,B] = \lambda B$$ which implies $B$ acts a ladder with fixed energy step $\lambda$.
Evaluating the commutator relationship on the left hand side gives $$[H,B] = i\hbar(\beta m \omega^2 x - \frac{\alpha}{m}p).$$
Equating coefficients on the right hand side of the above two equations we get $$i \hbar \beta m \omega^2 = \lambda \alpha, \quad -\frac{i\hbar}{m}\alpha = \lambda\beta.$$
Nontrivial solutions, with $\alpha$ and $\beta$ not both $0$, exist only if $\lambda^2 = (\hbar\omega)^2$, so $\lambda = \pm\hbar\omega$. Thus, at least within the linear span of $\{x,p\}$, there are exactly two nontrivial eigenoperators of the adjoint action $ad_H(.)$ with eigenvalues $\pm\hbar\omega$. The corresponding normalised operators, unique up to a phase, turn out to be the canonical ladder operators $$a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}x - i\frac{p}{\sqrt{2m\hbar\omega}}$$ and $$a = \sqrt{\frac{m\omega}{2\hbar}}x + i\frac{p}{\sqrt{2m\hbar\omega}}$$ and they shift energies by $\pm\hbar\omega$.
(Justification for the above claim: for $\lambda = \pm\hbar\omega$, we get $\beta = \mp\frac{i}{m\omega}\alpha$ and hence $B = \alpha(x \mp \frac{i}{m\omega}p)$. Equivalently, $B=\alpha\sqrt{\frac{2\hbar}{m\omega}}a^{\dagger}$ or $B=\alpha\sqrt{\frac{2\hbar}{m\omega}}a.$ To wit, $B$ is proportional to $a^\dagger$ and $a$.)
Multiplying a ladder $B$ with any function $f(H)$, which commutes with $H$, produces another operator with the same step size $\lambda$: $$ [H,f(H)B] = \lambda f(H)B. $$
