What is the connection between quantum optical photons and particle physics' photons?

Question

There has been quite some debate amongst users with different backgrounds about the usage of the word photon.^{[1, 2]} The most apparent disagreement was on whether a photon has a wavelength. I don't want to start a discussion about which viewpoint is more correct, because quantum optics is clearly only a sub-field of the standard model. Instead I would like to understand what additional predictions about photons the standard model allows to make and how one can construct the properties of the quantum optics' photon from it.

In the quantum optics community a photon is a quantum of excitation of an electromagnetic (EM) mode. The mode is a solution of (the relativistic) Maxwell's equations. Therefore asking about the wavelength of a photon boils down to the wavelength of the EM mode. The mode doesn't need to be a plane wave.

Now the particle physics perspective – I don't know much about it, but there were some statements which confused me: Photons are point particles without a wavelength. Moreover, the entity quantum optics people term "photon" is a composite particle or quasiparticle.

I especially wonder how the absence of a wavelength does not contradict the explanation of diffraction experiments. The diffraction of a quantum optics' photon follows quite naturally from the fact that the EM mode is different in the presence of e.g. a grating compared to without the grating. But how are the wave-like properties modelled in particle physics?

Please note that I'm not asking about the wave-particle duality, but about the apparent contradiction of the mentioned statements with interference phenomena.

Answer 1

The models that describe photons used in quantum optics and in particle physics are one and the same: the Standard Model of particle physics (often replaceable with just its quantum electrodynamics component) as encased within the formalism of quantum field theory. Moreover, the definition of photons (more specifically, single-photon states of the field) are identical in both fields.

To be unambiguous: in both quantum optics and particle physics, photons need not have a well-defined wavelength.

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The definition you give,

a photon is a quantum of excitation of an electromagnetic (EM) mode; the mode is a solution of (the relativistic) Maxwell's equations,

is essentially right. Quantum electrodynamics builds on top of the classical Maxwell equations, by building creation and annihilation operators $\hat a_q$ and $\hat a_q^\dagger$ associated with each mode (indexed by $q$), and it provides us with number operators for each mode, $\hat n_q=\hat a_q^\dagger \hat a_q$, and a global number operator, $$ \hat N = \sum \!\!\!\!\!\!\!\!\!\; \int\limits_q \ \hat a_q^\dagger \hat a_q. $$ Single-photon states of the field are the eigenstates of $\hat N$ with eigenvalue $1$.

Some of those states are also eigenstates of one of the individual $\hat n_q$ for a mode $q$ which has a well-defined frequency (resp. wavelength, momentum, but also possibly e.g. orbital angular momentum), in which case the photon itself will have a well-defined frequency (resp. wavelength, momentum, OAM). If that premise is not fulfilled, then the photon does not have that well-defined quantity.

In other words, your quote

Therefore asking about the wavelength of a photon boils down to the wavelength of the EM mode

is not wrong as such, but it seems to implicitly assume that "asking about the wavelength of the EM mode" will always have a well-defined answer and (with frequency, wavelength, momentum, OAM, etc) the answer is sometimes yes but also sometimes no.

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For full clarity, the answer you linked to is dead wrong. There is no semantic confusion. What can happen is that some phenomenological particle physicists in older generations, trained in the consequences of the QFT framework but not necessarily in its full mathematical generality, can sometimes fail to appreciate the full breadth of generality of the consequences of the QFT formalism. There's a lot of value in the traditional perspectives of phenomenological particle physics, but when their conclusions are in conflict with the mathematical formalism of QFT that underlies the theory, then the conclusions drawn from those perspectives are wrong.

Answer 2

What is important to keep in mind is what is considered the wave characterizing a particle in particle physics. The wave is a probability wave, the probability function given by the solution of the quantum mechanical equation corresponding to the given point particle in the table of elementary particles.

That photons have a footprint of a particle as seen in this experiment

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

Individual photons leave a footprint of a point on the screen, they are not spread all over the space. The probability distribution is given by the $Ψ^*Ψ$, where $Ψ$ is the wavefunction, and the interference pattern is seen when the number of events accumulates.

In the field theory of the standard model, creation and annihilation operators work on the photon (electron etc) plane wave wavefunctions which are the solution of the quantized Maxwell equations( Dirac for electron, etc), on which fields, creation and annihilation operators act, allowing for the pictorial description with Feynman diagrams for the calculation of interactions. Plane wave solutions mean there are no potentials involved.

The mode is a solution of (the relativistic) Maxwell's equations.

Maxwell' equations are by construction relativistic. If you mean the quantized Maxwells equations, then the difference between particle physics definition and quantum optics is in the type of "solution ". The field theory of particle physics acts on plane wave solutions of the quantized Maxwell equations, and it is possible the quantum optics solutions may be be quite complex and general depending on the boundary conditions.

Thus the answer to:

What is the connection between quantum optical photons and particle physics' photons?

In my opinion, the particle physics photons are created on the field of plane-wave wave functions (no potentials), the quantum optics ones on solutions of the maxwell equations with some form of potentials,so cannot be the same mathematical form, except at the limit in vacuum with no potentials.

In this blog post how classical electromagnetic fields emerge from QED is shown.