Uniqueness of Ladder Operators in the Harmonic Oscillator

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, 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 $\lambda$: $[H,f(H)B] = \lambda f(H)B$.

Nonexistence of fractional ladders

For monomials $(a^\dagger)^r a^s$ where $r,s\in\mathbb{N}$ $$[H,(a^\dagger)^r a^s] = (r-s)\hbar\omega(a^\dagger)^r a^s \quad (r,s \in \mathbb{N}).$$ Thus, monomial operators change the energy by $(r-s)\hbar\omega$ which is always an integer multiple of $\hbar\omega$. 

Any operator $B$ that is a suitably convergent polynomial in $a^\dagger$ and $a$ can be written as $$B = \sum_{r,s>0} c_{r,s}(a^\dagger)^r a^s,$$ with complex $c_{r,s}$. The action of $[H,\cdot]$ then is $$[H,B] = \sum_{r,s>0} c_{r,s} (r-s)\hbar\omega(a^\dagger)^r a^s.$$ Imposing the eigenoperator condition $[H,B] = \lambda B$ $$\sum_{r,s>0} c_{r,s} (r-s)\hbar\omega(a^\dagger)^r a^s = \lambda \sum_{r,s>0} c_{r,s}(a^\dagger)^r a^s$$ or $$\sum_{r,s>0} c_{r,s} ((r-s)\hbar\omega - \lambda)(a^\dagger)^r a^s = 0.$$

The monomials $(a^\dagger)^r a^s$ are linearly independent and hence each coefficient must vanish: $$c_{r,s} ((r-s)\hbar\omega - \lambda) = 0, \quad \text{for all }r,s\in\mathbb{N}.$$

Thus, for any pair $(r,s)$ with $c_{r,s} \ne 0$ we get $$\lambda = (r-s)\hbar\omega.$$

That is, every nonzero monomial in the expansion must share the same integer difference $r-s$. Hence $\lambda$ equals an integer multiple of $\hbar\omega$ $$\lambda = n\hbar\omega, \quad n\in\mathbb{Z}.$$

Thus, there is no operator that changes the energy by a non-integer multiple of $\hbar\omega$. If $\lambda/(\hbar\omega)$ were not an integer, then necessarily all $c_{r,s} = 0$, and the only solution is $B=0$.

References

Per ChatGPT the above argument can be reconstructed using the following sources:

1. J. J. Sakurai & J. Napolitano, Modern Quantum Mechanics, 2nd ed. (2011).
2. E. Merzbacher, Quantum Mechanics, 3rd ed. (1998).

From Sakurai we get the linear uniqueness result: solving $[H,B]=\lambda B$ with $B=\alpha x + \beta p$ gives $\lambda = \pm\hbar\omega$ and operators proportional to $a^\dagger$ and $a$.

From Merzbacher we get the general monomial commutator identity $[H,(a^\dagger)^r a^s] = (r-s)\hbar\omega (a^\dagger)^r a^s$ which nails down why higher ladders only step in integer multiples.