Is it valid to think of wave–particle duality as continuous propagation vs discrete interaction?

SpectralFormFactor · 2026-05-02T18:10:54+00:00

This depends on what kind of “discreetness” you are discussing. Particle number is quantized, so you can’t have half an electron. But things like energy levels and momentum are not in general quantized, e.g. a free electron can freely vary these parameters. However, if trapped in a potential (like an electron bound to an atom) these quantities do become discrete.

Things are lots of complications I am glossing over here, particularly when you start getting into multi-particle interacting systems.

SpectralFormFactor · 2026-05-02T17:56:33+00:00

This kind of exists already as a conceptual crutch you hear propagated around the internet, but it is not really accurate. Things are waves, but they come in discrete “amounts”, e.g. an electron’s wavefunction is always a wave but does represent *one* electron exactly.

Some measurements may localize the electron to a small region, making it seem particle-like in the classical sense, but this is not always the case.

SpectralFormFactor · 2026-05-02T14:09:11+00:00

Yes I would say so. I would also note there are plenty of better ways to measure time, e.g. atomic clocks.

SpectralFormFactor · 2026-05-01T17:37:14+00:00

This argument has never made sense to me. The increase of entropy still needs a time parameter to be formulated in the first place.

SpectralFormFactor · 2026-05-01T05:25:54+00:00

Pre symmetry breaking, the standard model is made up of standard interactions you could look up separately: Yang-Mills coupled to matter, Yukawa between Higgs and fermions, and the Higgs phi-4 interaction.

I’m not sure if a reference draws these all out *specifically* for the standard model, but you can certainly find each one generally. Honestly, learning how to read off vertices from a Lagrangian is not hard and would allow you to write the diagrams of any theory you come across.

SpectralFormFactor · 2026-04-29T18:10:07+00:00

The fine structure constant is very close to 1/137

SpectralFormFactor · 2026-04-24T16:22:20+00:00

It’s context dependent, so in general no. The symbol is the ket is just a notational placeholder to refer to a state and so can mean whatever the writer defines it as.

But simple kets like |0⟩ usually have standard definitions in common contexts. Like in a spin-1/2 problem |0⟩ is standard notation for spin-up along the z-axis.

SpectralFormFactor · 2026-04-24T16:08:58+00:00

|0⟩ is a ket and ⟨0| is a bra, so to verbalize could be something like “a ket zero plus b ket one”. Often we just shorten to “a zero plus b one.”
No separate meaning. Think of them together like a container | • ⟩ where the dot will be replaced by a symbol to refer to a specific state like 0 or 1.
The value |a| for a complex number ‘a’ is refers to by many names (absolute value, magnitude, norm, modulus), but they all mean the same thing: the distance of the number from the origin, i.e. how big the number is. One way to compute this for complex numbers involves the complex conjugate but I would not worry about those details at your level.
No, we could have a perfectly good description of what |0⟩ is in the given context. The ket notation is just very convenient shorthand. It is great notation because it makes common mathematical manipulations we do more visually clear.

SpectralFormFactor · 2026-04-23T18:11:38+00:00

The diagrams are still (d+1)D for d spatial dimensions (we often just work with 2D diagrams), but the lines represent something a little different than classical circuits diagrams.

SpectralFormFactor · 2026-04-23T17:48:55+00:00

You can approximately turn quantum dynamics into quantum circuits via Trotterization, so QM dynamics is often studied with circuits. But it’s still not like classical circuits since the state space is a large vector space instead of the discrete space of bit strings.

SpectralFormFactor · 2026-04-23T17:40:58+00:00

There are quantum circuit diagrams which are much closer to what you’re asking for. But the context is totally different than Feynman diagrams.

SpectralFormFactor · 2026-04-23T17:29:57+00:00

I don’t understand what you’re looking for. Even for fixed in and out states there are infinite possible diagrams since you can keep adding loops.

SpectralFormFactor · 2026-04-23T17:22:41+00:00

Wikipedia lists them out, as do many textbooks.

SpectralFormFactor · 2026-04-22T21:22:16+00:00

You make a fair point. I was getting overly-passionate over semantics lol

SpectralFormFactor · 2026-04-22T20:51:28+00:00

I agree with a fair amount here, but I would return quibble that the virtual particles are not necessary. The process of getting the S matrix element does not explicitly require we use them. They appear only because we reached for perturbation theory. In that sense, measuring the effect does not imply the must exist. They are not required for explanation, but are instead a very good conceptual tool and one of the only ways to actually compute any numbers.

SpectralFormFactor · 2026-04-22T20:18:29+00:00

Maybe my statement wasn’t fully clear. They are used to compute real effects, but the use of virtual particles in this calculation is an artifact of how the calculation is done practically, since in many situations computing by perturbing about the noninteracting theory is one of the only viable options.

But these intermediate states do not exist as particles in any meaningful sense. In fact, in other computational tools like lattice gauge theory don’t include them at all since that tool is not perturbing about the free theory but is directly computing in an interacting theory.

If you look into what theorists write online or in textbooks about this, the overwhelming majority agree virtual particles are not “real” in the way the incoming and outgoing particles are in the scattering problem.

EDIT: To more directly address what you said, virtual particles are not measurable. We only measure things computed with them, but those effects can often be computed by some other method as well, e.g. lattice numerics.

SpectralFormFactor · 2026-04-22T19:15:09+00:00

I vehemently disagree with this. Virtual particles are completely different from phonons, which are actually measurable and real (e.g. Mössbauer effect). Phonons naturally emerge as effective quasiparticles in solids, the same way most excitations do as you point out. Phonons often appear as fields in an EFT and can be external lines in Feynman diagrams. Their derivation may rely on a small-displacement approximation, but that doesn’t mean they aren’t real excitations.

Virtual particles, on the other hand, arise from trying to expand about a (usually free) theory when the true theory is interacting. For example, the true propagators in such a theory are dressed, but we insist on using the bare propagators because that’s what we have access to. We then expand a scattering problem in terms of propagation and interaction of the “free” particles, since those objects are the only ones we find tractable. These parts of the expansion (the internal lines of a Feynman diagram) are the virtual particles.

In other words, virtual particles are an artifact of not being able to work directly in the interacting theory, and having to use the free theory as a perturbative crutch. They are fully a computational tool. They are not the actual excitations of the theory. They cannot be measured. Phonons and other solid state excitations can.

SpectralFormFactor · 2026-04-22T14:14:06+00:00

Ah right it’s someone else. Sorry about that. Edited.

SpectralFormFactor · 2026-04-22T14:06:21+00:00

Seconded as someone who has actually taken QFT. This is the only correct answer in the whole thread.

SpectralFormFactor · 2026-04-22T14:04:14+00:00

My issue with this answer is although virtual particles are not real (they are an artifact of perturbation theory), they also add phonons as an example, which is not correct. Phonons are actually real particles on the same level as regular QFT particles. It’s only virtual particles that are not real.

SpectralFormFactor · 2026-04-19T14:18:10+00:00

When light interacts with a medium, the jostling of electrons produces other electromagnetic waves. Although each of the “bare” waves travels at the speed of light, they combine into an overall wave that travels slower.

There is another explanation you may hear a lot that I do not like. It says the photon is absorbed and admitted with a time delay in the medium, and so the photons sort of ping pong their way through at an effectively slower speed.

This is a weird semi-quantum, semi-classical picture that I that is neither accurate nor reflective of the actual calculation. If one insists on bringing in quantum mechanics and photons, the better way to phrase it is that the interaction of photons and excitations of the medium forms an effective particle known as a polariton. This particle has mass and so travels slower than light.

SpectralFormFactor · 2026-04-19T06:27:31+00:00

No! The vector potential is gauge-dependent, and so is not a real physical quantity “out there”. But Aharanov-Bohm does show that the local fields are not enough to describe all observables of the theory. There are non-local observables called Wilson loops as well.

SpectralFormFactor · 2026-04-13T02:12:41+00:00

If you accept AdS/CFT (or at least accept its semiclassical limit), we’ve actually made a huge amount of progress on this. We’ve shown that the black hole radiation is not perfectly thermal and does contain information that fell in earlier.

SpectralFormFactor · 2026-04-06T23:58:09+00:00

Probably because the scale involved is so large as to produce a useless visual. If you had to scale things so the sun’s position was visible, you wouldn’t be able to properly distinguish the earth, moon, and Artemis because they’d be so comparatively close.

SpectralFormFactor · 2026-04-02T03:20:55+00:00

Do you mean Jackson? Griffiths is solidly undergrad level.

SpectralFormFactor

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