Say we just created a particle (high probability of one-particle state), is the probability of a very far away detector getting triggered at the time of creation (probability of finding a particle outside of its light cone) zero according to QFT?

Since we can detect particles and make histograms of the positions where they're found using detectors, this seems like a reasonable question to ask. I hope that QFT says that detector cannot detect particles outside its lightcone because if that's not the case, we can imagine an experiment where information can be sent FTL:

Consider a ridiculous amount of hydrogen atoms/electrons near person A and a very far away B measuring the rate of particles he detects. So when A makes some movement, if the probability outside light cone changes immediately, B's rate of detection immediately changes and hence this can be used for communication.

If you say that the probability outside light cone doesn't change immediately, that leads us to a grave situation. Assume A himself has a detector, and if he sweeps/detects most of the particles (sweeps through the high probability region, peaks). It makes zero sense to say B observes a same rate of particle detection since particles are already 'used' up by A.

(Some clarification: When I say particles near A, I mean by this is that we intuitively have an idea that particles/fields must have some kind of probability distribution. It is reasonable to assume that the there is some peak in distribution of atoms/electrons of my phone in my hand and probability of electrons of phone's atoms is extremely low far away. So even if QFT doesn't have position operator or whatever, it should somehow be able to talk about this.)


3 Answers 3


There is no position operator in QFT, and you cannot localize a particle to a precise position. In QFT, physical particle states are constructed from smooth wave-packets of momentum eigenstates. Such particle has a non-zero probability of being anywhere.

You can read the full explanation in Lecture 1 of Sidney Coleman's famous QFT series. He showed that if one attempts to construct a "position operator" similar to that in non-relativistic quantum mechanics and create a localized particle, the probability that the particle is found outside the light cone is non-zero. He also offered a physical why we cannot localize particles: if a particle is "squeezed" into a small enough box, it would have enough energy(as dictated by the uncertainty principles) for pair production , i.e. we don't have a single particle anymore, but many-particle states.

Edit: To address the topic of FTL communication, one of the main conditions in constructing a physical quantum field theory is that observables at space-like separations must commute. So you can rest assured that no FTL communication is possible. QFT is built on the assumption of relativistic causality.

  • $\begingroup$ So, are you suggesting that QFT implies that my method of FTL communication is possible? I specifically used the terminology of detector and histogram to counter the arguments that position operator doesn't exist and hence no probability distribution or something. $\endgroup$
    – Razor
    Commented May 9, 2021 at 21:34
  • $\begingroup$ Btw I didn't mean to imply that A has particles localized. I said hydrogen atoms, so $\psi(r)=e^{(-r/a)}$ and when I said free electrons, I meant wavepackets. $\endgroup$
    – Razor
    Commented May 9, 2021 at 21:36
  • 2
    $\begingroup$ QFT does not imply FTL travel is possible, the point is that you would get nonsense results if you force a particle to a location (e.g. the attempt to construct a position operator). Since there is no location, there is no sensible concept of "travel distance" either. Without a precise location, the particle does not have a definite light cone. If your particle is a smooth wave packet, its probability amplitude of being found anywhere is already non-zero, no travel needed. $\endgroup$
    – JF132
    Commented May 9, 2021 at 21:50
  • $\begingroup$ I'm not sure if I follow. We can set up a array of detectors at different positions/angles and they detect particles and we can plot this histogram. So are you saying QFT cannot predict this histogram? $\endgroup$
    – Razor
    Commented May 9, 2021 at 21:54
  • 1
    $\begingroup$ Thank you. I still don't understand though. I don't know shit in qft so I don't understand why observables at spacelike seperations commuting must mean anything. Regardless, you seem to suggest that detector at B will not observe a different rate of detection of particles instantly when A moves the particles (N of.them). (By moving particles I mean that changing the location of peak of spatial probability of high probable one particle state of many particles or high probable N particle state idk). I'm using the word probability because of your statement that "probability anywhere is non-zero". $\endgroup$
    – Razor
    Commented May 9, 2021 at 22:36

The correct representation of causality in quantum field theory is that all observables whose supports are space-like separated commute: $$[\mathcal{O}_1(x_1), \mathcal{O}_2(x_2)] = 0$$ for all observables in the fields when $x_1 - x_2$ is space-like. This means that measurements at space-like separated position cannot influence each other, precisely to prevent issues like the ones in your question where two people with detectors could probabilistically transmit FTL information. Relativistic quantum field theories obey this property more or less as an axiom.

There's also a different way to look at causality more in line with what you were probably hoping for. A common folklore argument that ignores formal issues with the existence of a position operator given in another answer goes like this:

The probability amplitude for a particle to move from $x$ to $y$ is straightforward - it's the propagator $\langle \phi(x) \phi(y)\rangle$. This propagator is non-zero outside the light cone. However, there is a second process for $x,y$ at spacelike separation that could lead to the same observation: The particle could start at $y$ and end up at $x$ - since the events are at spacelike separation, there is no unique notion of one of them being "prior" to the other, and the amplitude here is of course the propagator $\langle \phi(y)\phi(x)\rangle$. These two amplitudes precisely cancel each other for spacelike separation, so there is no overall probability to detect a particle outside its light-cone.

  • $\begingroup$ So if measurements at space-like separated positions cannot influence each other, are you saying that B will observe the same rate of particles even though A used up/detected "all" the particles? I.e. so many new particles are created so that B doesn't know that A used em all up? $\endgroup$
    – Razor
    Commented May 13, 2021 at 19:57
  • 1
    $\begingroup$ @Razor If A and B are space-like separated, then neither of them is in a position to block particles from reaching the other. They cannot "use up" particles. If A is in a position to detect all the particles, then they must be at time-like separation from B. If I "detect" all particles by putting a sheet in front of the source, no one standing behind me will be able to detect them anymore (after those that were emitted before I put up the sheet have reached them). This isn't somehow a problem with causality. $\endgroup$
    – ACuriousMind
    Commented May 13, 2021 at 20:02
  • $\begingroup$ Very interesting but weird resolution. I thought A's ability to detect particles is independent of who else is detecting it. If A detects all/fraction of particles in a very small $\Delta t$ time, there are very few points in space which can be timelike with A. It's extremely weird that A's ability to detect particles is determined by an arbitrary B. (You probably understood but just to clarify, in my exp, no source and particle velocities. All particles "with" A.) $\endgroup$
    – Razor
    Commented May 13, 2021 at 20:38
  • $\begingroup$ I enjoy your folklore! -NN $\endgroup$ Commented May 16, 2021 at 23:23
  • $\begingroup$ Though I don't understand the solution, I'm giving you the bounty. Thanks. $\endgroup$
    – Razor
    Commented May 18, 2021 at 7:38

Someone actually wrote a paper on this exact problem


The last statement in the conclusion reads :

"The main point of our results seems to be that instantaneous spreading holds already under amazingly few assumptions. Neither the existence of fields nor the usual axioms of field theory are assumed -the only input is Hilbert space and positivity of the energy. Our results seem to indicate the need for a mechanism like vacuum fluctuations, clouds of virtual particles, particle-antiparticle pairs, spontaneous excitations, or something like that in order to retain Einstein causality. Our results are compatible with quantum field theory which uses much stronger assumptions and in which vacuum fluctuations etc. are present."

You may check out the references too.


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