# Does the uncertainty principle imply the existence of particles that exceed the speed of light?

The uncertainty principle allows for the creation of virtual particles (with non-zero mass) that exist for very short durations. This allows empty space to have particle pairs that pop into existence for very short periods and then get destroyed when recombine. This appears to me as a violation of the energy conservation law over very short time durations.

I am wondering if the uncertainty principle, $\Delta p \Delta x \geq \frac{h}{4\pi}$, allows the existence of virtual (or real) massive particles that travel faster than the speed of light for very short distances.

I am aware of the discovery that some particles found to break the speed of light.

• It would be interesting if someone plugs in the diameter of the hydrogen atom for $\Delta x$ and calculate an estimate of $\Delta v$ (using electron's mass). – Mohammad Al-Turkistany Oct 31 '15 at 15:48

1. The link about superluminal neutrions you cite is missing the fact that later on an error was discovered, and neutrinos do not, in fact, travel faster than light (see e.g. the Wikipedia article). To date, nothing that travels faster than light is known.

2. The uncertainty principle does not "allow for the creation of virtual particles". The idea of such pairs coming into existence and annihilating again relies on a misinterpretation of a particular kind of Feynman diagram, called the "vacuum bubbles", which are just the diagrams without external legs which contribute to the vacuum energy. QFT does not assign actual particle states to lines which are not external, so there is no rigorous basis for talking about the "creation" of virtual particles.

3. The uncertainty principle does also not allow faster than light speeds. That the standard deviation $\Delta p$ of momentum means only that - if you do many momentum measurements on identically prepared states, your data set will have this as its standard deviation. But, in general, the "speed" of a quantum object is not well-defined - it is typically not localized at a point, so it does not have a speed in the classical sense, and so it is not clear how its momentum would relate to its speed, or how a large standard deviation in momentum would lead to faster than light speeds.

4. You also cannot use the uncertainty principle for $p$ and $x$ in relativistic settings because there is no straightforward relativistic position operator (the closest you get to one are the so-called Newton-Wigner operators).

• As always, thank you so much for these wonderful answers. 3 questions if I may: a) in point 3, so the Heisenberg's uncertainty bound for conjugate variables does not make much sense for single measurements? Is this related to the fact that often in weak measurments it is claimed that the heisenberg bounds are violated? b) when people speak of undefinedness of properties (speed here) of quantum objects, i really get lost...what do we mean? c) could you intuitively explain the issue with the position operator in relativistic settings? Thanks a lot – user929304 Nov 14 '15 at 20:58
• @user929304: a) Indeed. For a single measurement, it's unclear what the uncertainty - which really is a standard deviation - could mean. Much more relevant for thinking about a single measurement is the actual probability distribution. b) Often, we mean simply that the objects is not in an eigenstate of the observable for that property. Speed is different: There is no quantum analog of the classical path $x(t)$ a particle takes, so speed, which is classically $\dot{x}(t)$, has no observable associated to it. You simply can't meaningfully speak of it. – ACuriousMind Nov 14 '15 at 21:06
• @user929304: c) One component of a proper relativistic position operator would have to be time - since time is one component of the four-position, and Lorentz transformation mix time and space. But time cannot be an operator in QM since it would have to be related to energy just as position is to momentum - and that would mean that since time is continuous and unbounded, energy would have to be, too, but energy is often discrete and/or bounded, so a time operator as we might naively imagine it does not exist. – ACuriousMind Nov 14 '15 at 21:09
• thanks for your prompt reply! About the speed matter: so shocking to realise that velocity which is a fundamental part of classical lagrange mechanics has no place in QM. Out of curiosity, how about the path integral formalism, surely there one speaks of all the multiple possible paths parametrised by time, can there one define the derivative of x eigenvectors? About c) very nice point. So can i say the following: energy and time can never be fourier transform of one another, because that would imply that when energy is undefined, time has compact support, which cannot be the case? – user929304 Nov 14 '15 at 21:29
• @user929304: Well, you can define the velocity along a path, of course. But the path integral simply integrates over all paths, there is no single path associated to a quantum state, so you still don't get a "speed" for states. I would not speak of the "Fourier transform" because the classical variables energy and time, which is what one Fourier transforms, are indeed transforms of each other. It's just that energy is a property of a state, time is not. You might also see this classically in the Hamiltonian view: $H$ is a classical function on the phase space, but $t$ is not. – ACuriousMind Nov 15 '15 at 2:08

In Peskin's introduction to quantum field theory, he talks about this in chapter 2, section 4 in a subsection titled causality, which is on page 27 in my book.

He explains that yes, there is some non-zero probability that a particle will move faster than the speed of light. However, he shows any two local operators that are separated by a spacelike interval commute. This follows from lorentz invariance.

Then he argues that since these operators commute, two measurements separated by a space-like interval cannot affect each other, and so causality is preserved.

This last step seems vague to me, and this isn't really my area of expertise, but I hope I have given you a better idea of what is going on or pointed you in the right direction at least.

• If some one wants to expand on this answer or make their own better answer, I would be interested in learning about the nuances of this argument. – Brian Moths Oct 31 '15 at 16:41
• Commuting quantum mechanical operators possess simultaneous eigenstates. See bit.ly/3leOTSK . Thus, in principle you can always do an experiment that will simultaneously reveal the two observables corresponding to the two operators. And you can adjust one of those observables without affecting the other. If you can adjust these two observables independently from one another, then the value of one observable did not cause the value of the other. – David C. Aug 25 '20 at 1:35