How does quantum field theory account for the movement of an electron orbiting a proton from one place to another place in the next instant of time - at a distance that even light couldn't reach in the time elapsed. The quantum mechanical wavefunction allows this.
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13Non-relativistic quantum mechanics is not causal, there is non-zero amplitude for a particle to be found outside of its future lightcone. Electrons cannot travel faster than light in QFT, no. QFT is by construction a relativistic theory of quantum mechanics. – Charlie Mar 09 '21 at 13:47
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Related question: https://physics.stackexchange.com/q/571412 – Technically Natural Mar 09 '21 at 16:06
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3What do you mean by "movement from one place to another"? There is no such thing in quantum mechanics. – ACuriousMind Mar 09 '21 at 16:07
3 Answers
Electrons are not particles which orbit like planets. In the quantum field theory (QFT) that you ask about, they are distortions or excited states of the quantum field, somewhat akin to standing waves. In a hydrogen atom the electron state surrounds the proton as a bound state and does not change. The field equation describes the shape of this "orbital". It yields the probability of finding the electron at any given location, and so is sometimes referred to as a probability cloud.
Any "collapse of the wave function" due to measurement, as it is often described, is instantaneous across space. Even a photon from a distant star, whose wave stretches lightyears across space, will collapse instantaneously when it hits the astronomer's camera. But the collapse involves no "travelling" as such.

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"Any such "collapse of the wave function" due to measurement, as it is often described, is instantaneous across space. " This is one of the many perfectly good reasons to reject the notion of collapse. – my2cts Mar 09 '21 at 19:06
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"Even a photon from a distant star, whose wave stretches lightyears across space"Why you think the wavefunction stretches lightyears across space? – Deschele Schilder Mar 09 '21 at 19:26
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@DescheleSchilder because it is black body radiation, it was not emitted by a laser. Consider say Young's slits where the distance from the source is 1 m and the separation of the slits is 1 mm. Now multiply both distances by, say, 10^22. This yields the equivalent slits for a star 1 million lightyears away; they are a thousand lightyears apart, and the wave function of the photon is a good deal wider than that. – Guy Inchbald Mar 09 '21 at 19:59
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I'm not sure I follow you. I see what you mean, but doesn't black body radiation consist of localized photons? And doesn't, in contrast, laser radiation consist out of a coherent superposition of photons (and is as such suitable for a double slit)? – Deschele Schilder Mar 09 '21 at 20:05
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@DescheleSchilder In so far as a wave function can be localized, black body photons are not so localized as to pass through a single slit in Young's classic experiment; each photon passes through both slits and then interferes with itself. If you substitute a narrow laser beam and aim it at one of the slits, it will not pass through both and therefore will not interfere with itself. This is very basic stuff, it is not open to debate. If you need to know more, ask a new question. – Guy Inchbald Mar 09 '21 at 20:14
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Do you think ordinary light (as coming from stars) is used in the double-slit experiment? When one shines sunlight on the slits, will the interference pattern appear? – Deschele Schilder Mar 09 '21 at 20:18
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See this question: https://physics.stackexchange.com/questions/212899/reproducing-double-slit-experiment-with-sunlight – Deschele Schilder Mar 09 '21 at 20:21
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@DescheleSchilder The experiment was performed with starlight a couple of years back. Yes, the interference pattern appeared. The sun is a star, I don't know if it has been used as a light source, but if it were to behave differently from other stars - now that would earn you a Nobel prize! P.S. See the answers to the question you just linked to. – Guy Inchbald Mar 09 '21 at 20:21
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One more thing. You write: "In the quantum field theory (QFT) that you ask about, they [the electrons] are field perturbations" Perturbation theory can't be used when considering a bound state like an atom. As I stated below, there are no free external lines, which is a prerequisite of Feynman's perturbation approximation. – Deschele Schilder Mar 09 '21 at 20:28
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@DescheleSchilder Chasing logic games into the ground is not what these comments are for. I will repeat one last time, if you have any serious doubts, ask a proper new question. – Guy Inchbald Mar 09 '21 at 20:30
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What do you mean by logic games? Only physics is involved. I just asked a physical question. No logic involved. If you put the two slits 1000 kilometers from each other, will the pattern still appear? – Deschele Schilder Mar 09 '21 at 20:34
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"In the quantum field theory (QFT) that you ask about, they are field perturbations" Perturbation can't be applied to bound states in QFT. QFT is not well suited for bound states. It is for free fields that a perturbation can be made (with Feynmann diagrams which contain free external lines). – Deschele Schilder Mar 10 '21 at 09:17
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@DescheleSchilder Really? See for example https://core.ac.uk/download/pdf/216219239.pdf – Guy Inchbald Mar 10 '21 at 09:28
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Yeah, I've been looking at it. Heavy stuff! I actually used the link to pose a question, because right now I'm not sure anymore. I think that they have used the wrong basic fields and for the real basis fields the approximation procedure might not work. – Deschele Schilder Mar 10 '21 at 10:08
In relativistic quantum field theory (QFT) it is not the velocity that is fundamental but four-momentum:
$$p=\frac{mv}{\sqrt{1-v^2/c^2}}.$$
So all intermediate "trajectories" between two spacetime points (virtual trajectories, in the path integral formalism of QFT) do not have the quality of lightspeed greater than $c$. This speed is never exceeded. There is an infinite range of the vales of $p$ though.
How does quantum field theory account for the movement of an electron orbiting a proton from one place to another place in the next instant of time - at a distance that even light couldn't reach in the time elapsed?
It should be emphasized that QFT has difficulties in handling bound states. It allows us to calculate scattering amplitudes and decay rates with high precision. But in doing so external lines in the associated Feynmann diagrams refer to external lines that represent free fields. For an atom, no external lines are there. Hence the difficulty. Of course, in QFT no velocities $v$ which exceed the speed of light are present, as might be clear from what I wrote above.
You write:
How does quantum field theory account for the movement of an electron orbiting a proton from one place to another place in the next instant of time - at a distance that even light couldn't reach in the time elapsed.
First, there is no next instant of time. Instants have zero extent and putting two instants together will give the same instant. There is no next instant though it's obvious what you mean: to take the differential of time $dt$. They are defined as intervals with an extent approaching zero.
Secondly, in QFT an electron doesn't move in an orbit around a proton. So to say that in one $dt$ the electron is at a different position than the next $dt$ isn't
correct. Maybe this is the case in ordinary QM, but seeing QFT as lying beneath ordinary QM, this can't be the case.
The path integral in QFT stands in direct relation to the wavefunction in ordinary QM. But a thorough derivation of the wavefunction of an electron in interaction with a proton hasn't been derived, although the wavefunction can be derived in ordinary QM (but QFT holds the key to what's really "going on").
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That is true, but I had a more straightforward answer in mind, which I now posed. – my2cts Mar 09 '21 at 19:58
The quantity $v$ that appears in $\gamma = 1/\sqrt{1-v^2}$ is the group velocity of the wave function and it cannot exceed $c$. Hence no energy, charge or other physical quantity can travel faster than light in quantum mechanics.
To simplify notation I used units for which $c=1$.

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1I have no background whatsoever with QFT, but if this is the same $\gamma$ as the Lorentz factor (?), it should be $\gamma = 1/\sqrt{1-v^2/c^2}$. Though it could of course be the case that I am on a totally wrong track and those aren't related. – jng224 Mar 09 '21 at 20:01
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@Jonas I had exactly the same thought before I read your comment. – Deschele Schilder Mar 09 '21 at 20:07
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Assuming you mean the Lorenz factor, you are referring to classical quantum mechanics, and speeds exceeding c are present there. – Deschele Schilder Mar 09 '21 at 20:15
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Well spotted! Force of habit, I always use units with $c=1$ to avoid cluttered notation. I do indeed use $\gamma$ for the famous special relativity factor. – my2cts Mar 09 '21 at 21:10