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 space given to a 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 $\lambda$: $$ [H,f(H)B] = \lambda f(H)B. $$
