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I understand that technology makes it now possible to shoot single photons.

I have read this question, and the answers did not cover one thing.

Single photons: Is there a 90° offset of the electric to the magnetic component in the direction of propagation?

We know that in the answers it is correctly stated that the phenomenon can be successfully described in QM.

Although it is worth talking about this on the individual photon level, since it is now possible to shoot a single photon (it is possible now as we know). I know that that is for the double slit exp. but the single photon shoot is the technology we have and that is what it is all about.

Double Slit Experiment: How do scientists ensure that there's only one photon?

Question: If we shoot a single photon will that single photon at a certain position of the propagation axis be the excitation of both E and M fields, (and so the photon itself would have two fields coupled) or is it just the excitation for one of them and then 'changes' to being the excitation of the other one as it propagates?

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Árpád Szendrei
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    You've been told this before, but I'll repeat it anyway: you cannot use, in the same context, the concepts of photon and electromagnetic field. These concepts are mutually exclusive. – AccidentalFourierTransform Nov 25 '16 at 19:16
  • OK please don't downvote. can you please tell me a little bit about it more, so on individual photon level we cannot talk about EM fields? Can you please explain this a little bit more? – Árpád Szendrei Nov 25 '16 at 19:20
  • and can you explain this please: http://physics.stackexchange.com/questions/2889/properties-of-the-photon-electric-and-magnetic-field-components?rq=1 – Árpád Szendrei Nov 25 '16 at 19:22
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    it is exactly that: on individual photon level we cannot talk about EM fields. I couldn't have put it better myself :-) photons is quantum mechanics, where the fundamental objects are operators, and particles are eigenstates of these operators. EM fields are classical objects, with no relation to operators and to particles. You need to quantise Maxwell theory to get photons, they do not appear at the classical level. – AccidentalFourierTransform Nov 25 '16 at 19:23
  • wow. QM does not talk about fields? Then how is QM dealing with the photon's propagation? Is QField theory then totally different then QM? – Árpád Szendrei Nov 25 '16 at 19:27
  • In QFT there are fields, but these are operators as well! it is a complex issue, but the quantum fields of QFT are not fields in the classical sense (they are Schwartz distributions). It is complicated, I know... – AccidentalFourierTransform Nov 25 '16 at 19:29
  • Will I you think understand it if I learn about Schwartz distribution? – Árpád Szendrei Nov 25 '16 at 19:30
  • It is possible to infer the properties of the photon from a radio wave produced by the synchronous acceleration of surface electrons. However, the formalism of QM tempts some to deny the photon. – HolgerFiedler Sep 10 '20 at 05:13
  • @HolgerFiedler here is another one: https://physics.stackexchange.com/questions/578637/subluminal-photons/578833#578833 – Árpád Szendrei Sep 10 '20 at 16:06

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Field quantization is (at least usually) done in the basis of eigenmodes. In free space, those eigenmodes are plane electromagnetic waves, which carry both $\vec{E}$ and $\vec{B}$ fields. So the short answer is, yes both fields are attached to the photon, because your whole plane wave is attached to it.

But from reading the comments it seems important to adress another topic: If you quantize your modes, physical quantities are represented as expectation values of operators. You now can formulate a photon number operator as well as field operators for $\vec{E}$ and $\vec{B}$. The problem is, that the field and number operators do not commute, so similiar as for momentum and position operators in simple QM, there is an uncertainty relation between them. You can be in a state where the photon number as well as the Intensity $|E|^2$ is well-defined (Fock States), but the field values are uncertain (and have expectation value zero). Or you are in a state where the fields are defined (Coherent States) , but the photon number is uncertain. This is probably the reason why in the comments it was stated that "you cannot talk about fields and photons at the same time".

From the way the question was asked i assume you are looking for a qualitative description. If you would like to see the math to back up my statements, i suggest https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

Wave and Matter
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  • I see that whenever I go from classical view to QM, he math gets way heavier. I have read the link you gave me. I have two questions: how can we explain the propagation of a photon? (classical Maxwell equations suggested that the propagation is done by way of changing E field creating a changing M field etc.) But this does not work in QM I guess. So in QM how can we explain the propagation of a photon? I mean do we just accept that it has a momentum in a direction, or is there an explanation how it can propagate? – Árpád Szendrei Nov 28 '16 at 19:07
  • In 1-particle QM a propagating particle is represented by some wave packet which is created by superposition of planewave(pw) modes. The peak of the packet moves in time and the particle moves/propagates. With a single photon in a single eigenmode this is of course impossible (only the phase changes, probability amplitude and therefore expected position are constant). But you can create a moving pulse if you superimpose single-photon states from different pw-modes. The result is a mix of several wavelengths, but has photon number expectation value 1. That's your single moving photon. – Wave and Matter Dec 02 '16 at 13:43
  • thank you. I understand that you write "The peak of the packet moves in time and the particle moves/propagates." So QM gives you a math description of the propagation. But I guess from this QM math description " peak of the packet moves" it will not say anything in a classical way how it moves. So I guess I will just have to accept that QM gives a successful math description, but does not explain in any classically understandable way HOW correct? – Árpád Szendrei Dec 02 '16 at 18:25
  • It does not describe what the particle actually does (only probabilities). But there is still something intuitive: If you take the expectation value of the position operator $(t)$ you get the same $x(t)$ as for a classical particle of the same velocity (Ehrenfest theorem). So the point of highest probability to find the particle moves like you would expect classically. But this only works for states which are appropriately constructed (for example wave packets), because as i said for example for an eigenmode, the probability distribution does not change. – Wave and Matter Dec 02 '16 at 21:44
  • you seem to know QM. How can I learn or do you have a link where to learn EPR and QM's explanation? I mean where QM explains that the two eigenstates cannot be defined at the same time (like for a particle's x and y axis spin at the same time cannot be, Heisenberg uncert.) and that the two matrices cannot have the same eigenvektor? – Árpád Szendrei Dec 02 '16 at 23:28
  • If you google for quantum mechanics lectures you should find plenty of pdf files or maybe even movies. You should follow them from the beginning and ask new questions (not cluttering this one) when they appear. You should also try to do some of the exercises. If You know linear algebra and fourier transforms it should be doable. Without following the math you will always have only a qualitative understanding and some questions (like the last one) cannot be answered at all. You could also try the "theoretical minimum" books by Susskind. – Wave and Matter Dec 03 '16 at 11:57
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The photon is a quantum mechanical point particle/entity in the table of the standard model of physics. Point means that its four vector is a function of (x,y,z,t) , it has no volume in space. It is one of the founding elements of the theory .

Question: If we shoot a single photon will that single photon at a certain position of the propagation axis be the excitation of both E and M fields, (and so the photon itself would have two fields coupled) or is it just the excitation for one of them and then 'changes' to being the excitation of the other one as it propagates?

In quantum mehanics, the point elementary particles are described by solutions of the corresponding quantum mechanical equations, Dirac for fermions, klein Gordon for bosons and a form of quantized maxwell's equation for photons.

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One usually does not discuss the wave function of a photon because the classical description of light emerges smoothly from the quantum mechanical (it does need the mathematics of quantum field theory to understand the link) and is very accurate and much more easy to use, then going to individual photon levels.

The classical electric and magnetic fields of the electromagnetic radiation are built up by a confluence of the wave functions of the zillions of photons that build it up. The fields exist in the wavefunction of the photon, i.e. before it is squared to give the probability of finding the photon with energy h*nu and an (x,y,z,t) point. It is the superposed wavefunctions that will build in confluence the classical light wave with its E and B fields.

A single photon, when detected (interacting with another particle) will only enter with the complex conjugate squared of the wave function, so the information of the electric and magnetic field is irrelevant other than its effect on the probability of the interaction happening.

anna v
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