# In what sense is a quantum field an infinite set of harmonic oscillators?

In what sense is a quantum field an infinite set of harmonic oscillators, one at each space-time point?
When is it useful to think of a quantum field this way?

The book I'm reading now, QFT by Klauber, claims its not true, which is it?

I would like to understand this analogy a little better.

For simplicity, let us talk about a scalar field $\phi : \mathbb{R}^4 \rightarrow \mathbb{R}$.

The action for a free scalar field is

$$S[\phi] = \frac{1}{2}\int_{\mathbb{R}^4} \partial_\mu\phi\partial^\mu\phi - m^2\phi^2$$

and its classical equations of motion is the Klein-Gordon equation

$$(\partial_\mu\partial^\mu + m^2) \phi = 0$$

Now that looks suspiciously like an oscillator or wave equation, doesn't it? This inspires us to do a Fourier transform to obtain the eigenfunctions $\mathrm{e}^{\mathrm{i}px}$ solving this equation. The general solution $\phi(x)$ can then be expanded as

$$\phi(x) = \int \frac{\mathrm{d}^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}}(a(\vec p)\mathrm{e}^{\mathrm{i}px} + a^\dagger(\vec p)\mathrm{e}^{-\mathrm{i}px})$$

which is precisely the expansion one could do for any other oscillator. Now, you can talk about the modes $a(\vec p)$ and $a^\dagger(\vec p)$ of the field being excited, and you can imagine $\mathrm{e}^{\mathrm{i}px}$ describing a (basic) oscillation at any point $x$, and talk about the integral representing the field being made out of such oscillators.

This is all nonsense.

This might sound strong, but it has been the source of many annoying misunderstandings in the publicization of quantum theories to laypeople.

Just because something ($\phi$) fulfills a wave/oscillator equation and has a mode expansion (as the above is called), it does not mean that anything oscillates. It's just the same type of equation you encounter in oscillator, not the same physical situation. It's a nice pretty picture to tell ourselves that we understand the quantum field, but ultimately, there is nothing there that would justify the oscillator interpretation. Nothing physical is vibrating or oscillating here.

Furthermore, the above only holds for a free, non-interacting field. When you have a field with arbitrary interactions, its equations of motion may not look at all like the wave/oscillator equation, and it has no modes, so the picture falls apart there completely.

Since the comments show that this is more controversial than I thought, I will elaborate a bit:

Electromagnetic waves also look as if there is a harmonic oscillator at every point in space, by the same logic of mode expansion. The descriptions are formally equivalent. This led people to believe that there is the luminiferous aether, because, how else could empty space carry the wave? But this turned out to be not true, there is no aether, and there is nothing carrying the wave. The formal equivalence is misleading, there is no physical object oscillating when the wave travels through the vacuum.

I am not saying that anything about the formal treatment is false, I am trying to explain why it is not a good idea to believe that the oscillator description is a good physical interpretation of the situation, which is what I believe the OP refers to when he mentions a book saying "This is not true" about the "oscillators at every point in space" idea.

• @ACuriosMind Have you ever tried to write the Hamiltonian for such a free scalar theory? It is something of the form $H=\int d^3x \frac{1}{2}\left[\dot\phi^2(t,x)+\nabla_i\phi^2+m^2\phi^2(x)\right]$, which is definitely the sum of harmonic oscilators, one at every spacetime point. If your problem is with the derivative, just go to the momentum space which is even more apparent, $H\sim \int d^3k \sqrt{k^2+m^2}a^\dagger(k)a(k)$: a harmonic oscillator in every point in momentum space. It is in fact the very reason why we can speak of particles in weakly coupled field theories around the free t. Jul 18 '14 at 17:50
• @TwoB: That's again the conflation of "It looks just like the equations I know from the oscillator" and "It is a real, physical oscillator." The quantum fields are no physical objects, they are not themselves observable. Talking about oscillators at every spacetime point heavily implies that there is a manifest object oscillating there, which is, in my view, simply not true. It creates an intuition that does not really reflect the status of what the quantum field as an abstract operator is supposed to be. Jul 18 '14 at 17:55
• @ACuriosMind your critics would apply to the harmonic osciallator in plain QM too. The coordinate q is an operator as well as its conjugate momentum. There is nothing special about fields here. It's just QM. Moreover, you can take a path integral approach and get the correlation function without ever talking about operators. The nomenclature is very cleare: if the dynamical variables evolve in time according to the Hamiltonian of an harmonic oscillator we call them so, irrespectively of their quantum nature Jul 18 '14 at 18:00
• @Mark Mitchison: The quantum field that is being quantized is the gauge potential, not the electric field. And indeed, for the gauge potential, the oscillator analogy fails drastically, since there are spurious and negative norm modes (or polarization states) corresponding to excitations of the "oscillators" which have no physical meaning. (aka ghosts/BRST exact states) Jul 18 '14 at 22:29
• @ACuriousMind Not true, it is perfectly possible, although unfashionable, to quantise in terms of the electric and magnetic fields. This approach is quite common in quantum optics, where preservation of Lorentz invariance is unnecessary. My point was simply that the concept of a quantum field is much more general than what one encounters in particle physics. Jul 18 '14 at 23:15

For a free theory, say for one scalar field for simplicity, which gives a a linear differential equation for the field $$\phi$$, one can cast the Hamiltonian $$H=\frac{1}{2}\int d^3x \dot\phi^2+(\partial_i \phi)^2+ m^2\phi^2$$ in this form (basically by taking a Fourier Transform): $$H=\mathrm{const}+\int \frac{d^3 k}{(2\pi)^3} \omega(k)a^\dagger(k)a(k)\,,\qquad \omega(k)^2=k^2+m^2$$ where $$[a(k),a^\dagger(p)]=(2\pi)^3\delta^3(p-k)$$ and $$[a(k),a(p)]=0$$. You see that this is the Hamiltonian for infinitely many harmonic oscillators, one in every point of momentum space. Since the energy levels for a harmonic oscillator are evenly separated for every $$k$$, you get the particle interpretation of the free field theory given that the state of the system is identified by the occupation numbers of any such harmonic oscillator.

If the theory is weakly coupled, that is a small perturbation of the above system, you can keep thinking of the system as a collection of harmonic oscillators, although now coupled, with new phenomena such as non-trivial scattering (elastic or inelastic with particle production...) because the time evolution is deformed by one of the free oscillators/particles that we have prepared at infinity where the interactions can be neglected.

Vice versa, several QFTs, especially the strongly coupled ones, do not admit such a particle/oscillators interpretation and you should not think of them as a collection of harmonic oscillators.

Imagine a classical relativistic bosonic scalar field $\Phi(\vec x,t)$, its equation is the Klein-Gordon equation :

$$\partial_0^2 \Phi(\vec x,t) - \sum\limits_i \partial_i^2 \Phi(\vec x,t) = 0 \tag{1}$$

Now, take the spatial Fourier transform of this equation, you get an equation for $\Phi(\vec k, t)$ :

$$\partial_0^2 \Phi(\vec k,t) + \vec k^2 \Phi(\vec k,t)=0\tag{2}$$

Using the notation $\Phi_{\vec k}(t) = \Phi(\vec k,t)$, one may write :

$$\partial_0^2 \Phi_{\vec k}(t) + \vec k^2 \Phi_{\vec k}(t)=0\tag{3}$$

One sees that the $\Phi_{\vec k}(t)$ represent an collection of independent harmonic oscillators, each with a frequency $\omega_{{\vec k}} = |\vec k|$

Now, suppose that you want to quantize the field, to obtain a quantized relativistic bosonic field. This is easy, because you know how to quantize an harmonic oscillator. You just promote the classical fields $\Phi_{\vec k}(t)$ to operators. For instance, you will have :

$$[\Phi_{\vec k}(t), \partial_0 \Phi_{\vec k'}(t')]_{t=t'} = i \delta^3 (\vec k - \vec k') \tag{4}$$