# Motivation for introducing ladder operators for the simple harmonic oscillator in quantum mechanics

I am teaching a quantum mechanics course and I have to explain the simple harmonic oscillator. I am familiar with the introduction of ladder operators and the consecutive proofs that show that we can build a ladder of eigenstates of the Hamiltonian with them. However, I find that I lack motivation for the first step. We start with the Hamiltonian

$$$$\hat{H}=\frac{1}{2}\big(\hat{X}^2+\hat{P}^2\big)$$$$ where I set $$m=\omega=\hbar=1$$ for simplicity. At this point, we ask the question: Can we factorize this square? After using the commutation relations for $$x$$ and $$p$$ we find that we almost can... we get

\begin{aligned} \hat{H}=&\frac{1}{2}\big(\hat{X}-i\hat{P}\big)\big(\hat{X}+i\hat{P}\big)+\frac{1}{2}\\ =&a^\dagger a + \frac{1}{2} \end{aligned}

My question is: Is there some motivation for wanting to factorize the Hamiltonian? I would like some intuitive a priori motivation. Of course after doing all the math, the usefulness of the ladder operators becomes clear after we prove that they increase/decrease the energy of any eigenstate. But I'm wondering if, in general, factorizing the Hamiltonian serves any purpose or if there's any intuitive motivation behind taking that first step.

• You can also use the ladder operators to compute eigenfunctions in a more fast way. When I thought the similar question, I have convinced myself that factorization of the Hamiltonian looks like we analy the differential operator with help of group theoretical approach for ODE Nov 16, 2023 at 9:28
• I think this note may be helpful : unioviedo.es/hepth/people/Patrick/fysica/zooi/factorisation.pdf Nov 16, 2023 at 13:30
• I always "justify" this post-hoc by making my students calculate the matrix elements of the position operator in the oscillator eigenstates, once in the position-space representation (using a recurrence relations for the Hermite polynomials) and once using the oscillator algebra. Every year, I hear exclamations about how much easier/nicer the latter calculation is, and at that point they're basically sold on the idea of the algebraic viewpoint in QM. (Even so, they still struggle mightily with angular momentum, but at least they're primed to accept that it's a good idea to do the algebra.) Nov 16, 2023 at 17:37

The motivation for this becomes quite clear in the Heisenberg picture using which, Dirac first introduced the ladder operators (p. 136, The Principles of Quantum Mechanics) and derived their properties in the context of the Quantum Harmonic Oscillator (QHO). Of course, using this as a motivation in a classroom would require your students to be familiar with the Heisenberg picture. I will stick with your notation and conventions $$(m = \hbar = \omega = 1)$$ in this answer.
Given the Hamiltonian in the Schrödinger picture for the QHO, $$\hat{H} = \frac{1}{2} \left( \hat{X}^2 + \hat{P}^2 \right),$$ The Heisenberg equations of motion are the coupled differential equations, $$\partial_t \hat{X}_t = [\hat{X}_t, \hat{H}_t] = \hat{P}_t \quad \text{and} \quad \partial_t \hat{P}_t = [\hat{P}_t, \hat{H}_t] = -\hat{X}_t,$$ where, the subscript indicates that the operators are time dependent in the Heisenberg picture. This is relatively straightforward to decouple. However, the key is to decouple it in a way that is reminiscent of the classical oscillator. To this end, we introduce an operator $$\hat{a}^\dagger = (\hat{X} + i \hat{P})/\sqrt{2}$$. This leads to the equation, $$\partial_t \hat{a}^\dagger_t = i\hat{a}^\dagger_t \implies \hat{a}^\dagger_t = \hat{a}^\dagger_0 e^{it}.$$ One can then interpret the $$\hat{X}_t$$ and $$\hat{P}_t$$ as the real an imaginary parts of the complex amplitude $$\hat{a}^\dagger_t$$ and this is in perfect analogy with the classical simple harmonic oscillator. Now we can rewrite our Hamiltonian using these operators in the Schrödinger picture and naturally motivate the factorization.