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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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    $\begingroup$ Non-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. $\endgroup$ – Charlie Mar 9 at 13:47
  • $\begingroup$ Related question: physics.stackexchange.com/q/571412 $\endgroup$ – JoshuaTS Mar 9 at 16:06
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    $\begingroup$ What do you mean by "movement from one place to another"? There is no such thing in quantum mechanics. $\endgroup$ – ACuriousMind Mar 9 at 16:07
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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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  • $\begingroup$ What distorts the field ? $\endgroup$ – user291095 Mar 9 at 18:11
  • $\begingroup$ "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. $\endgroup$ – my2cts Mar 9 at 19:06
  • $\begingroup$ "Even a photon from a distant star, whose wave stretches lightyears across space"Why you think the wavefunction stretches lightyears across space? $\endgroup$ – Deschele Schilder Mar 9 at 19:26
  • $\begingroup$ @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. $\endgroup$ – Guy Inchbald Mar 9 at 19:59
  • $\begingroup$ 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)? $\endgroup$ – Deschele Schilder Mar 9 at 20:05
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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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    $\begingroup$ I 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. $\endgroup$ – Jonas Mar 9 at 20:01
  • $\begingroup$ @Jonas I had exactly the same thought before I read your comment. $\endgroup$ – Deschele Schilder Mar 9 at 20:07
  • $\begingroup$ Assuming you mean the Lorenz factor, you are referring to classical quantum mechanics, and speeds exceeding c are present there. $\endgroup$ – Deschele Schilder Mar 9 at 20:15
  • $\begingroup$ 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. $\endgroup$ – my2cts Mar 9 at 21:10
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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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  • $\begingroup$ Nevertheless, whatever you mean by $v$, it cannot exceed $c$. $\endgroup$ – my2cts Mar 9 at 19:05
  • $\begingroup$ @my2cts Isn't that what I stated? $\endgroup$ – Deschele Schilder Mar 9 at 19:09
  • $\begingroup$ That is true, but I had a more straightforward answer in mind, which I now posed. $\endgroup$ – my2cts Mar 9 at 19:58

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