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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.

A. P.
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  • Related: What is the orbital angular momentum (OAM) of individual photons? (though note that the non-accepted answer there is simply wrong). – Emilio Pisanty Feb 14 '21 at 12:54
  • @J... I don't mean the wave-particle duality. This is already part of quantum mechanics. Instead I'm trying to figure out where the standard model goes beyond quantum field theory. – A. P. Feb 14 '21 at 18:02
  • @EmilioPisanty Thank you for that link. Just to make sure we're on the same page: I also didn't want to focus on the wavelength property in particular, since in the general case a mode doesn't have a sharply defined wavelength. I was under the impression that particle physics can explain quantum theory as a special case, without the need for wave-like properties. – A. P. Feb 14 '21 at 18:06

2 Answers2

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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.

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.

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.

Emilio Pisanty
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  • Thank you for this answer. Do I get it right, that it's the superposition principle which might lead to the idea that photons are composite rather than elementary particles? – A. P. Feb 14 '21 at 18:08
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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

singlepho

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.

anna v
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    There is no semantic confusion. The definitions of "photon" in quantum optics and QED / QFT are identical. The confusion comes from thinking that QFT requires the single-photon states to be based on modes with well-defined momentum. That is a convenient basis in which to develop the theory, but there is no requirement coming from QFT to dictate its use. Photons $-$ as understood from QFT $-$ need not have a well-defined wavelength / momentum / frequency / OAM. – Emilio Pisanty Feb 14 '21 at 13:06
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    "It will not work with non-plane-wave fields for its creation and annihilation operators" is an exceptionally strong claim. (For full clarity, it is wrong.) Have you got any references that contain a proof of such a statement? (For full clarity, without such references the claim is entirely hollow.) It is certainly true that QED will be much less convenient in any other basis, but it is a mistake to think that just because it's inconvenient it "won't work". – Emilio Pisanty Feb 14 '21 at 13:16
  • As for "There can be an innumerable number of field theories, based on different quantum mechanical wave equation solutions of the same equation", it's so wrong it's barely worth discussing, but for full clarity: the field theory is completely and exclusively determined by the Lagrangian. Same Lagrangian, same field theory. Period. – Emilio Pisanty Feb 14 '21 at 13:17
  • @EmilioPisanty There can be an innumerable number of field theories, based on different quantum mechanical wave equation solutions of the same equation. Particle physics standard model QED is based on the zero potential plane wave solutions of the basic quantume mechanical equation, and that is the difference, it is not convenience, it is the field theory which describes and predicts data, and the standard model is continuously validated. It will not work/calculate-correctly with non-plane-wave fields for its creation and annihilation operators. – anna v Feb 14 '21 at 13:21
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    As any QFT textbook will tell you, the field theory is entirely determined by the Lagrangian, and not by the specific solutions of the Lagrangian's wave equation that are chosen to build the modes and the creation/annihilation operators. This is elementary material in the construction of the rigorous framework that underlies particle physics. If you're not willing to work within correct, rigorous QFT, there is really no point in further discussion. – Emilio Pisanty Feb 14 '21 at 13:31
  • You can describe any physically possible wave in the plane wave basis, without a potential. You can find some examples, like a Gaussian beam here. But do I get it right, that in your oppinion an elementary photon is given by a single plane wave, while linear combinations of plane waves would be occupied by composite photons? – A. P. Feb 14 '21 at 18:21
  • Yes, that is the way I have learned to calculate elementary particle interactions. On the wave function of a plane wave imposing a creation operator and following the feynman diagrams calculation rules.. Of course any complicated wave can be expanded into plane waves, and that is why I consider the excitations within matter , which by necessity will have potentials involved, are composites of elementary photons. – anna v Feb 14 '21 at 18:40
  • As you see though experts in quantum optics have a different opinion, basing it on generic field theory arguments, ignoring the particular field theory used in the standard model., and also ignoring that the standard model aims to be the basic description of nature from which all other physics disciplines emerge. – anna v Feb 14 '21 at 18:41
  • @annav Ah, excitations within matter, that is of course something different than a purely photonic state. If this is the composite state you meant, I agree. Concerning the hierarchy of theories, I'm aware that the standard model is more fundamental than physics sub-disciplines. That's why I was asking about if the standard model makes any beyond-QFT predictions for photons. But Emilio Pisanty's answer clearly says that at least for photons this is not the case. – A. P. Feb 14 '21 at 19:33
  • Another thing which just strikes me is: How is it possible for a pure photon to be described by a plane wave (sharp momentum), but at the same time being treated as a point particle (sharp position)? – A. P. Feb 14 '21 at 22:59
  • The photon's probability of interaction is treated as a plane wave, which gives it a probability to be all over spacetime, and infinitessimaly small at a point.That is why Feynman diagrams calculations work in the momentum energy space where the momentum is well defined. A free photon probability in space has to be described by a wavepacket http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html . – anna v Feb 15 '21 at 05:06
  • The standard model is a specific field theory . New theories are needed , as string theories. or supersymmetry to make predictions beyond the standard model, and experiments to see if there are deviations . Optics trained experimentalists have their own bias on what the photon is, and I always get negative votes from them. Inmy opinion they are not correct in how QFT rules are interpreted, not taking into account that the functions defining the field on which creation and annihilation operators act defines the specific QFT. – anna v Feb 15 '21 at 05:12
  • They call it generically QED because their QFT uses functions of the quantized maxwell equations for its field definition. The two , standard model and optics QFTs coincide only in vacuum in my opinion. – anna v Feb 15 '21 at 05:26