My question is not referring to the fact that the harmonic oscillator (HO) is employed in a lot of physics models. For example, this question asks about the universality of HO's, and the answers focus on the fact that many potentials can be approximated by a HO term in their Taylor expansions.

What I want to know is why the mathematical form of the quantum harmonic oscillator (QHO) seems to be ubiquitous in quantum mechanics. Note that the QHO term in my examples are only formal, and do not actually represent any physical restoring force or potential.

Two vastly unrelated examples I came across are the following:

1. Landau levels

For a uniform magnetic field $\mathbf{B} = B_0\mathbf{\hat{z}}$, the corresponding magnetic vector potential is $$\mathbf{A} = \frac{B_0}{2}\left(-y\mathbf{\hat{x}}+x\mathbf{\hat{y}}\right).$$

Consider a particle with charge $q$ and mass $m$, and define the following variables: $$ P \equiv \frac{1}{\sqrt{2}}\left[ \left(p_x+p_y\right) - \frac{qB_0}{2}\left(x-y\right) \right], \quad Q \equiv \frac{1}{\sqrt{2}}\left[ \frac{1}{qB_0}\left(p_x-p_y\right) + \frac{1}{2}\left(x+y\right) \right]. $$

Then it can be checked that $P$ and $Q$ satisfy the canonical commutation relation: $$ \left[Q, P\right] = i\hbar. $$ So $P$ and $Q$ are formally momentum and position. Then we find that the Hamiltonian can be written as $$ \hat{H} = \frac{1}{2m}\left(\mathbf{p}-q\mathbf{A}\right)^2 = \frac{1}{2m}P^2 + \frac{1}{2}m\omega^2Q^2, $$ where $\omega$ is equal to $\frac{qB_0}{m}$.

The system is analogous to a QHO, with energy eigenvalues $E_n = \hbar\omega\left(n+\frac{1}{2}\right).$

2. Proof that orbital angular momenta do not have half-integer values

This proof has several appearances on PSE, e.g. here. I will provide my own brief summary. First we write the non-dimensionalised positions and momenta: $$ \xi_x = \sqrt{\frac{m\omega}{\hbar}}x, \quad \xi_y = \sqrt{\frac{m\omega}{\hbar}}y, \quad \pi_x = \frac{p_x}{\sqrt{m\omega\hbar}}, \quad \pi_y = \frac{p_y}{\sqrt{m\omega\hbar}}. $$ Then define $$ P_1 \equiv \frac{\pi_x-\xi_y}{\sqrt{2}}, \quad P_2 \equiv \frac{\pi_x+\xi_y}{\sqrt{2}}, \quad Q_1 \equiv \frac{\xi_x+\pi_y}{\sqrt{2}}, \quad Q_2 \equiv \frac{\xi_x-\pi_y}{\sqrt{2}}. $$

These $\left(P_i, Q_i\right)$ pairs also satisfy the canonical commutation relation, and the orbital angular momentum $L_z = xp_y-yp_x$ turns out to be $$ \frac{\hbar}{2}\left(Q_1^2 + P_1^2\right) - \frac{\hbar}{2}\left(Q_2^2 + P_2^2\right). $$ Which is again analogous to the QHO, so we can write $L_z$ in terms of the Hamiltonians: $$L_z = \frac{1}{\omega}\left(H_1-H_2\right) = \hbar\left(n_1-n_2\right) = \hbar m. $$

What's the significance behind the fact that the mathematical form of QHO keeps appearing in seemingly unrelated phenomena?


  • Griffiths - Introduction to Quantum Mechanics
  • Ballentine - Quantum Mechanics
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    $\begingroup$ The Harmonic oscillator often appears because it can approximate more complicated potential through a Taylor series expansion. It can be seen as a good perturbative limit for many more complicated potentials. $\endgroup$
    – StarBucK
    Commented Jun 15, 2023 at 12:48
  • $\begingroup$ Possible duplicates: Why is the harmonic oscillator so important? and links therein. $\endgroup$
    – Qmechanic
    Commented Jun 15, 2023 at 13:36
  • $\begingroup$ @Qmechanic my question is asking why the same mathematical form appears in vastly different calculations. The possible duplicate you linked to is asking why many systems can be approximated by a HO. $\endgroup$ Commented Jun 15, 2023 at 15:54
  • $\begingroup$ Because this is something we can calculate very well, whereas most other systems we can not calculate well/we can't calculate at all. If one can approximate something hard to understand with something easy to understand, then that's a good strategy in physics. $\endgroup$ Commented Jun 15, 2023 at 19:16

2 Answers 2


One argument is as follows: suppose you could describe a physical system by its energy which depends on some potential $V(q)$, where $q$ is some canonical variable. Then, approximating the potential \begin{align} V(q) \approx V(q_{0}) + V'(q_{0}) (q- q_{0}) + V''(q-q_{0})^{2}+.. \end{align} From the perspective of dynamics, the first term is irrelevant as it is just a shift. Suppose further that you are interested in the special configuration $\tilde{q}_{0}$ at which the potential is extremized, then the second term vanishes. We see that the potential ,at leading order, is encoded by \begin{align} V(q) \approx k (q-q_{0})^2, \end{align} which is precisely the potential for a harmonic oscillator.

In other words, the harmonic oscillator may serve as an approximation for a whole class of systems amenable to the above assumptions, which is why it shows up everywhere.

  • $\begingroup$ While very correct, I don't think this has to do with my question. In my above exmaples, the formal position and momentum variables do not represent any physical restoring force, instead they just have the same form as a QHO, so can be solved that way. $\endgroup$ Commented Jun 15, 2023 at 15:52

A really key mathematical feature of the quantum harmonic oscillator is that it has a discrete, uniformly spaced, set of quantized energy states that can be manipulated through the use of Dirac's so-called "ladder operators." Each of the problems that you mentioned utilize this formulation in some way or another, though with varying levels of complexity and generality. In the ladder operator formalism, we can use the operators like $\hat{P}$ and $\hat{Q}$ that you defined above to construct operators $\hat{A}$ and $\hat{A}^\dagger$ that have the effect $\hat{A}^\dagger |n\rangle \propto |n+1\rangle$ and $\hat{A} |n\rangle \propto |n-1\rangle$ where $n$ refers to the $n^\text{th}$ state of the system and $\hat{A} |0\rangle = 0$. Note also that the Hamiltonian itself can be rewritten in terms of these ladder operators, which has some interesting consequences. This means that we can essentially bump the harmonic oscillator states up or down in energy level using these operators, which is quite useful in a huge number of problems in quantum mechanics. The reason that this model appears so often is that it is a well-understood and somewhat simple construct that can be massaged to fit into a huge number of apparently disparate scenarios. It is not so much that the "harmonic oscillator" is actually showing up everywhere, but that nature has a conveniently large number of phenomena that can be reasonably modeled with similar mathematics to a harmonic oscillator.

  • $\begingroup$ "...but that nature has a conveniently large number of phenomena that can be reasonably modeled with similar mathematics to a harmonic oscillator." Yes, and my question is whether the "similar mathematics" is just a coincidence, which I doubt is. $\endgroup$ Commented Jun 15, 2023 at 12:26
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    $\begingroup$ It is most certainly not. At its simplest, the underlying physical structure is that of a simple linear "restoring force," which is something that comes up in a huge number of different contexts. You can imagine electromagnetic attraction and repulsion in molecules in exactly this fashion, among countless other examples. There is likely an even more elegant connection with the algebraic structure of the spectra of operators and the ladder operators, but I do not believe I have the chops to adequately express that. $\endgroup$ Commented Jun 15, 2023 at 12:36
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    $\begingroup$ The Harmonic oscillator is often a good perturbative limit of more complicated potentials where the ground state of the potential is approximated by a parabola. Hence it can approximate a wide range of physical scenario. $\endgroup$
    – StarBucK
    Commented Jun 15, 2023 at 12:50
  • $\begingroup$ @Jono94 A simple example of something that can not be approximated this way are systems with energy gaps, e.g. superconductors. We can only approximate the states above the gap with linear models, but not the gap itself. $\endgroup$ Commented Jun 15, 2023 at 19:19

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