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Electromagnetic waves are produced by oscillating electric charges. In quantum mechanics this fact is represented by photons mediating electromagnetic interactions of particles with an electric charge.

However, in quantum mechanics, any particle can be viewed as a wave. Can this wave also be viewed as caused by oscillations of some type of charges or quantum numbers?

For example, the electron can be viewed as a wave, just like the photon. So, if the electromagnetic wave that the photon represents is caused by oscillating electric charges, then oscillating of what causes the wave that the electron represents?

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Photons don't mediate electromagnetic interactions, Virtual photons do. They don't really exist like physical photons, they're just in mathematical formalisms. The photons emitted by accelerating charges are described by this.

AFAIK The 'waves' of matter in QM aren't waves in the same sense, depending on your interpretation they may not be physical at all, rather just information about what can be known about the system since the last measurement. Wavefunctions in QM just describe the different observable values and the probabilities tagged to them.

Hope this helps

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  • $\begingroup$ A beam of electrons changing the direction in a CRT or the process of light emission or absorption all certainly are examples of electromagnetic interactions involving real photons. Radio waves are produced by electrons oscillating in an antenna that has nothing to do with the breaking radiation. The photon wavefunction describes the probability as well, but also oscillations of the electromagnetic field. Oscillations of what field does the electron wavefunction represent? $\endgroup$
    – safesphere
    Aug 17, 2017 at 3:20
  • $\begingroup$ I don't think you read what I wrote. Just because something is emitted, doesn't mean it's 'mediating' an interaction between particles. 'Virtual photons' are the ones that do that. For example, if two electrons begin to approach, within the maths of QFT they don't exchange photons, but 'virtual photons'. Also, Breaking radiation is anything EM radiation emitted by decelerating charges. How could that not be exactly what you're talking about. $\endgroup$
    – user95137
    Aug 17, 2017 at 12:02
  • $\begingroup$ I didn't refer to mediation in my reply to you. Mediation is not important to my question. The emission of photons is an electromagnetic interaction, is it not? The breaking radiation certainly is not the only way photons are emitted, but this is irrelevant to my question as well. You are picking on irrelevant details and perhaps not understanding the question. It is not about the formalism, I am well aware of the formalism. The question is about the physical meaning of the formalism, the meaning that too often gets lost behind the mathematical formulas. $\endgroup$
    – safesphere
    Aug 17, 2017 at 18:51
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You are confusing the classical framework of Maxwell's equations with the quantum mechanical.

For example, the electron can be viewed as a wave, just like the photon.

Both the photon and the electron are described by a wave function whose complex conjugate squared gives the probability density for the particular observation.

So, if the electromagnetic wave that the photon represents is caused by oscillating electric charges,

The photon is not an electromagnetic wave: it has a wavefunction. The classical oscillating fields emerge from an enormous number of quantum mechanical photon fields. ( for a mathematical analysis look at this link). The reason that the photons, defined in the quantum mechanical framework can build up the classical maxwell electromagnetic field is because their wave function is the solution of the quantized Maxwell equation..

photwavf

then oscillating of what causes the wave that the electron represents?

The wavefunctions of electrons, positrons and the rest of massive particles in the standard model of particle physics, are based on solutions of the Dirac equation. There exists the Schrodinger-Newton equation studied as a limiting case but the analysis goes in a different direction, does not have the simple solutions which allow for the emergence of the classical electromagnetic field from the photon wavefunction.

Thus your wave analogy does not hold.

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  • $\begingroup$ Hi Anna, thanks so much for your detailed answer! My analogy for the photon is exactly the same as your reference to its wavefunction being a solution of the [quantized] Maxwell's equation. The probability density physically is the same as the field intensity. So my analogy does hold for the photon. The corresponding analogy for the electron is exactly the question. The conventional answer is that it is not there, but it is not a satisfying answer from the standpoint of physical meaning. What is the physical meaning of the quantum field of the electron? None? $\endgroup$
    – safesphere
    Aug 17, 2017 at 19:19
  • $\begingroup$ The consensus is that there is no physical meaning to the wavefunction. It is a mathematical model, the same that there is no physical meaning to the calculated parabola of a classical trajectory in the gravitational field of the earth. It is just mathematics. It is only the probability density that is connected with observations in quantum mechanical systems. The coincidence of the quantum and classical equation forms in electromagnetism does not exist for massive fermions. $\endgroup$
    – anna v
    Aug 18, 2017 at 3:48
  • $\begingroup$ Thanks Anna, I appreciate the explanation. I think I better understand this now. All single particles behave like waves, but a bunch of them gotta obey different distributions, either Bose or Fermi. So a swarm of bosons still look like a wave, but a pack of fermions may look more like a solid rock, etc. And the rock essentially is the sum of the wavefunction probabilities and, in this sense, the intensity of the fermion "field". Just a very different way things add up in two different distributions. Thanks again and upvote :) $\endgroup$
    – safesphere
    Aug 18, 2017 at 4:28

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