From what I understand of quantum optics, the creation and annihilation of photons is modeled by a quantum harmonic oscillator. The latter is obtained by applying the quantization "postulates"

$x(t) \rightarrow \hat{x} = x\cdot$


$p(t) \rightarrow \hat{p} = -i\hbar \frac{\partial }{\partial t}$

to the classical expression for energy obtained from Maxwell's equations. This is what produces in equally-spaced quantization of the energy corresponding to the (discrete) number of field excitations (i.e., photons).


Can one conceive of hypothetical physical field for which the energy isn't modeled by the harmonic oscillator but, say, by a infinite square well? The resulting excitations, i.e., particles, will then have energies that add up quadratically and therefore won't be equally spaced (e.g., two particles will be four times more energetic than a single particle, etc.).

Are all physical fields underpinned by a quadratic potential---i.e., by a harmonic oscillator---, and if so, is there a reason for it?

PS: Please keep the answers motivated from first principles rather than by referring to elaborate field theory constructs. I'm just trying to build an intuition for how quantization and particle creation arises from certain field potentials.


2 Answers 2


Can one conceive of hypothetical physical field for which the energy isn't modeled by the harmonic oscillator but, say, by a infinite square well?

Yes, absolutely!

Are all physical fields underpinned by a quadratic potential...?

No. The quadratic-potential case gets lots of attention in textbooks, partly because it's exactly solvable (exactly-solvable examples are rare!), and partly because it's the starting point for a useful approximation method, the method that Feynman diagrams represent.

Why non-quadratic terms are essential

One of the most familiar applications of quantum field theory (QFT) is to scattering experiments. In a scattering experiment, we start with a state of widely separated particles, each of which corresponds to the lowest-energy excitation of some field for the given momentum. The lowest-energy excitation doesn't care about the shape of the potential, as long as the potential admits at least one discrete energy eigenstate above the ground state (otherwise the field would not have any corresponding particle-like excitations). If the potential were quadratic, then "scattering" would be boring: the particles would pass through each other unaffected, completely oblivious to each other's existence. To get interactions between the particles, we need to use a non-harmonic potential — or, more generally, non-quadratic products of two or more different fields. Some intuition behind this is given in J. Murray's answer.

The non-quadratic terms in the Standard Model are what make the Standard Model interesting, and their importance isn't limited to scattering experiments. Life wouldn't be possible without them!

Fermion fields raise another type of exception: the concept of a "potential" doesn't really apply to them, because they are Grassmann-valued fields, but we can still talk about quadratic and non-quadratic terms in the lagrangian, and then the preceding comments still apply.

Can we use a square-well potential?

Consider scalar fields. For any potential $V$ with a finite lower bound, we can construct a quantum field theory of a single scalar field $\phi(x,t)$ whose lagrangian density is $$ \big(\partial^\mu\phi(x)\big)\big(\partial_\mu\phi(x)\big) - V\big(\phi(x)\big). $$ If we take $V(\phi)\propto \phi^2$, then we have a free field, which is the harmonic-oscillator case. In that case, the particles don't interact with each other: the theory is boring.

One way to make the theory interesting is to use a non-quadratic potential $V$. For example, the choice $V(\phi)\propto a\phi^2+b\phi^4$ gives what is usually called the "$\phi^4$ model." By tuning the coefficients $a,b$, we can adjust both the single-particle mass and the strength of the interaction between particles. We can also use this model to illustrate spontaneous symmetry breaking.

Yes, we can also take $V(\phi)$ to be a square-well potential, but at sufficiently low energies $V$ might as well be a low-order polynomial, at least if spacetime is four-dimensional. (The story is richer in lower-dimensional spacetime, but I won't go there.) That's because the condition "sufficiently low energy" basically means that only the few lowest-energy modes are excited, even in interactions, and we can tune the coefficients of a low-order polynomial $V$ to reproduce those same lowest-energy modes. For more about this, look up Wilson renormalization.

Using a non-quadratic potential isn't the only way to make the theory interesting, though. Non-linear sigma models use scalar fields that aren't real-valued: they take values in some other manifold instead, such as a circle or some higher-dimensional manifold with non-trivial topology. The topology of the target space (the space in which the fields take their values) makes these theories interesting, even if there is no "potential" at all. The Wikipedia article about chiral perturbation theory introduces an application of this type of model to the low-energy physics of quantum chromodynamics.

  • $\begingroup$ Do we know of any elementary particles that do not arise from quadratic Hamiltonians, though? $\endgroup$
    – Tfovid
    Commented May 3, 2020 at 15:47
  • $\begingroup$ @Tfovid Depends on what you mean by "arise from quadratic Hamiltonians." The Hamiltonian of the Standard Model is not quadratic, and the non-massless particles in that model all get their masses as a result of non-quadratic terms. Or, if we ignore interactions between particles, then as explained in the answer, we can invent a quadratic model with the same spectrum of stable particles (electron/positron, photon, and proton), but then they won't interact with each other, and the particles won't have the right magnetic moments. So it really does depend on what you mean. $\endgroup$ Commented May 3, 2020 at 17:29

I would put slightly less emphasis on quantum harmonic oscillator and more emphasis on quadratic Hamiltonian and linear equations of motion.

This might be a semantic point, but it's not a matter of loving the quantum harmonic oscillator so much that we use it to model all quantum fields. Instead, it's noting that, at least from a canonical quantization point of view, we start by considering non-interacting (free) fields. If the energy spectrum of your elementary excitations (free particles) is square well-like, then adding a particle corresponds to an energy increase which depends on how many particles you already have. In other words, the particles have an effective (and non-perturbatively strong) interaction with one another.


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