I was informed by this answer from John Rennie that the particle and wave behavior of light are both approximations to the more fundamental description of light as an excitation in a quantum field.

How would one, starting with the quantum field description of light, derive/predict the particle approximation in some situations and the wave approximation in others? What are the properties of a system/environment in which the underlying field could be approximated by a wave, and how are they different from the properties of a system that would be better approximated by the particle description?

  • $\begingroup$ I've deleted some unconstructive comments. $\endgroup$
    – David Z
    Aug 5 '16 at 5:40

This is at best a partial answer and I offer it partly for fear your question will be closed before an expert can answer, and partly in the hope it will prompt a better answer from an expert. Treat what follows as a Discovery Channel level answer and bear in mind that any quantum field theorist will regard it as horrifically oversimplified.

The simplest quantum field theory, and the one traditionally taught first to new students, is the scalar noninteracting field. In this theory we have a well defined vacuum state that has no particles present. To create a particle we use a function called a creation operator, $\mathbf a_k^{\,\dagger}$, which creates a particle with wave vector $k$. In effect this transfers one quantum of energy into the quantum field, and that quantum of energy creates a particle. Likewise there is an annihilation operator, $\mathbf a_k$, that removes one quantum of energy from the field and destroys a particle.

We can tell how many particles have been created using a particle number operator. Applying this to our field returns an integer that gives the net number of particles that have been created.

So far so good, but what does the creation operator actually do? Well, the thing it creates is a plane wave with wave vector $k$. So even though we created a particle what we ended up with is a wave. In fact it's very difficult to take a quantum field and point to anything that looks like a particle. You can construct localised waves by summing up plane waves, but this is really just a wavepacket of the sort that you will probably have encountered in non-relativistic quantum mechanics.

So as a general rule, when we are adding energy to the field or taking energy away from the field this interaction looks like a particle i.e. we create or destroy a particle. However when that energy is propagating it looks like a wave. This is the physics behind the notorious wave particle duality, though it is not helpful to regard this as a particle behaving like a wave or a wave behaving like a particle. It is simply the way a quantum field behaves.

Now, I started by saying we would consider a scalar non-interacting field. This is a model theory that doesn't describe anything in the real world. The reason we use it is that for interacting fields things suddenly become a lot more complicated. For example the number of particles in the field is no longer well defined i.e. the particle number operator will not always return the same value. Even in the vacuum state the number of particles isn't well defined due to the notorious vacuum fluctuations.

Furthermore, if you consider the quantum field theory describing photons (quantum electrodynamics) then the plane wave the creation operator creates is not an electromagnetic/light wave. A light wave, with it's associated electric and magnetic fields is constructed from the theory in a more complicated way that I frankly admit I do not understand. Describing this falls into the subject of quantum optics.

And that is as far as I can take this answer. I fear it may have been more confusing than helpful, but I think this is the best that can be done without you actually learning quantum field theory.

  • $\begingroup$ Wow THANK YOU! This is really great, just what I had In mind. Never imagined I hear from you personally. $\endgroup$
    – D. W.
    Aug 5 '16 at 14:40
  • $\begingroup$ I also thank you very much your answer, I hunted exactly for this on the google. $\endgroup$
    – peterh
    Aug 5 '16 at 15:42

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