I'm having trouble grasping an idea that I think that is a very basic part of  quantum field theory.

Many introductory QFT resources I have consulted often pose the following question:

What is the amplitude for particle to travel outside its forward light-cone?

According to my book Quantum Field Theory for the Gifted Amateur,

If the amplitude is nonzero then there will be a nonzero probability for a particle to be found outside its forward light-cone. This is unacceptable and would spell the death of quantum theory as we've known it so far.

My Questions:

  1. according to the principles of special relativity, nothing travels faster than the speed of light. So is the problem then that a particle with nonzero probability outside its forward light-cone would violate those principles?

  2. who ever said a particle could have a nonzero amplitude outside the forward light-cone? . Why would I come to think this? Like I recognize that the idea is important to quantum field theory. But suppose I'd never heard of quantum field theory. What event or experiment would cause you to think that a particle's amplitude would behave that way? Obviously special relativity has been very successful, so why would we ever think that a particle would have a nonzero probability outside its forward light-cone? It just seems like an idea that directly contradicts special relativity.

EDIT: I got the impression that somehow this property is important to understanding antiparticles. Perhaps an answerer could mention how that fits in as well.

  • In this chapter, the authors calculated the probability of the particle being outside of its forward light cone and found it to be nonzero. This cannot be - it violates special relativity - and so the authors use this as a reason why the formalism used to do the calculation (single-particle quantum mechanics, albeit with a relativistic energy dispersion) is wrong! The authors aren't claiming a particle can be found outside of its forward light cone, they are claiming just the opposite. – Physics_Plasma Jun 15 '15 at 23:36
up vote 3 down vote accepted
  1. In special relativity, the light cone defines the set of points that can be reached by null geodesics originating from a point$^1$. It is essentially the boundary of the set of points that can be reached by timelike curves. We call a curve timelike if its tangent vector $u^\mu$ is normalized as follows$^{2}$: $u^\mu u^\nu\eta_{\mu\nu}>0$. Three basic tenets of special relativity are
  1. Massless particles travel on null curves.

  2. Massive particles travel on timelike curves.

  3. Tachyons travel on spacelike curves and are unphysical.

The light cone of special relativity thus defines the set of points a real particle can occupy in the future. If the amplitude is nonzero outside of this cone, then there is a probability that the particle will propagate along a curve with a spacelike tangent vector, thus violating special relativity.

  1. Perhaps your quote is a little misleading. I propose the following re-write:

If the amplitude is nonzero then there will be a nonzero probability for a real particle to be found outside its forward light-cone. This is unacceptable and would spell the death of quantum theory as we've known it so far.

If you had never heard of quantum field theory, but had heard of a little inequality $$\Delta x\Delta p\ge\frac{\hbar}{2}$$ you could reason that trajectories are a little "fuzzy" and perhaps violate special relativity.

Suppose we had heard of QFT though. Let's show the amplitude of a virtual particle is nonzero outside of the light cone. For this we have to consider the propagator of an internal line of a Feynman diagram. The canonical example here is the Feynman propagator of a real scalar field. We solve the equation $$-(\Box+m^2)D(x)=\delta^4(x)$$ by the method of Green's functions and obtain$^3$ $$D(x)=\int\frac{d^4k}{(2\pi)^2}\frac{e^{ikx}}{k^2-m^2+i\epsilon}$$ A standard calculation by the method of residues leads to$^4$ $$D(x)=-i\int\frac{d^3k}{(2\pi)^32\omega_k}\left[e^{-i(\omega_kt-\mathbf{k}\cdot\mathbf{x})}\theta(t)+e^{i(\omega_kt-\mathbf{k}\cdot\mathbf{x})}\theta(-t)\right]$$ where $\omega_k=\sqrt{\mathbf{k}^2+m^2}$ is the on-shell energy. The physical interpretation of $D(x)$ is that it describes the amplitude for a particle to travel from the origin to the point $x$. One finds that $^5$ $$D(0,\mathbf{x})\simeq ce^{-mr}$$ where $c$ is an irrelevant constant.

So virtual particles violate special relativity! So what? They're not real and special relativity only puts restrictions on real particles. This property of virtual particles is explained away by the uncertainty principle.

So how is causality, special relativity and Lorentz invariance respected in field theory? The answer is probably important enough to be called a theorem$^6$. Let $\mathcal{H}(x)$ be the interaction Hamiltonian density. Then the $S$-matrix can be written as the Dyson series $$S=\mathcal{T}\exp\left(-i\int d^4x\,\mathcal{H}(x)\right)$$ where $\mathcal{T}$ denotes time-ordering. Using the cluster decomposition principle, we can write the interaction Hamiltonian in terms of quantum fields.

Theorem. All quantum fields obey $$[\psi_\ell(x),\psi_{\ell'}(y)]_\mp=[\psi_\ell(x),\psi^\dagger_{\ell'}(y)]_\mp=0$$ for $(x-y)$ spacelike. The $-$ holds for bosons and $+$ for fermions.

You can verify (laboriously) that your standard mode expansions obey this theorem.


$^1$ In general relativity, however, this is only locally true and depends furthermore on the topology of spacetime. See, e.g., Wald, General Relativity (1984).

$^2$ Here I am using the $(+--\,-)$ convention.

$^3$ See, e.g., this Phys SE post.

$^4$ See, e.g., Cahill, Physical Mathematics (2013), p. 201.

$^5$ The full calculation is found in Zee, Quantum Field Theory in a Nutshell (2010, 2nd Ed.), p. 545.

$^6$ See, e.g., Weinberg, The Quantum Theory of Fields (1995). I can't nail down a specific page because the proof is smeared out over chapters 3, 4 and 5.

  • tour de force! thanks for this. – niels nielsen Oct 1 at 4:46

You've touched on the answer with your first question: the forward light cone of a particle describes all the spacetime coordinates to which a photon emitted from the particle could travel. Note that this is a 4D cone; the 3D component is a sphere and it expands along the time dimension.

So let's say that at this exact moment there's a supernova exploding in the Andromeda Galaxy. That galaxy is 2.5 million lightyears away, which means that you and I are not in its forward light cone; the light from the supernova will eventually reach our location in space, but will shine "past" us in time, since in 2.5 million years the two of us (probably) won't be here.

Now let's look at some matter expelled by the supernova: atomic helium let's say, so you've got a nucleus with two electrons around it. The electrons as we understand them don't have a location usually, but we can describe a probability cloud of where we could find them if we measured and looked for them. That probability cloud is technically unbounded: there's no particular reason that we wouldn't look for an electron several feet away from the nucleus (an enormous distance in atomic terms) though it would be highly unlikely.

So we take one measurement near the nucleus and find the electron, and seconds later measure again a few feet away and find the electron again. Amazing! The odds were astronomically against us for that result, but it was still possible and it happened by chance. Now let's say a few seconds later our colleagues back in the Milky Way (remember, we're in Andromeda near the supernova) measure for the electron. I said before that the probability cloud for the electron is technically unbounded, so there should be a chance with a large negative exponent that our colleagues will find our electron near them, right?

The answer here is no. Whereas the probability cloud (the amplitude, in your literature) is nonzero everywhere in space, you also have to consider time: the most obvious example is that the electron can't exist at any arbitrary point in space before it exists at all; and when it does exist it moves with finite speed and thus requires time to move from one place to another. The soonest that the electron would have a non-zero probability to be found in the Milky Way would be 2.5Myr, and eventually the entire visible universe will have such a probability. Just not now.

In other words, a particle appearing outside its forward light cone is synonymous with it traveling faster than light... which is indeed why your book states that this is "unacceptable."

If you weren't doing quantum mechanics, your particle would have it's mass $m$, energy $E$, and momentum $\vec{p}$ satisfy the following condition:

$$E^2=(c\vec{p})^2+(c^2m)^2$$

and the four-vector $(E,c\vec{p})$ would point tangent to the particle motion through space time. So the fact that the particle has a real mass forces the motion to be within the forward light cone.

In quantum mechanics you don't usually model a particle as a simple particle with a trajectory, and it's virtually unheard of to do so in Quantum Field Theory. But you can still consider states with both momentum and energy. The states with the energy and momentum that satisfy

$$E^2=(c\vec{p})^2+(c^2m)^2$$

are called on-shell. And I think it's best to imagine these as the stable states, in the sense that they can move through the vacuum with minimal change (say just a phase). More rightly they are similar to pre-second-quantization states or free-particle eigenstates. Often, these are the asymptotic (in space and time) states, such as if you are doing scattering to compute S-matrix elements.

So all around, you can think of them as stable when nothing is going on, they get along well with the vacuum.

But, unlike in classical physics, we do not require that the energy and momentum associated with a non-free non-asymptotic particle be on-shell. And when we compute the probability amplitude, instead of insisting that everything be on shell instead we might compute give a factor like :

$$\frac{1}{E^2-(c\vec{p})^2-(c^2m)^2-iq}.$$

Which is largest (in magnitude) when $E^2=(c\vec{p})^2+(c^2m)^2$, but can definitely have $ E^2-(c\vec{p})^2\neq (c^2m)^2$. These "states" are off-shell but they are not free, they are not input states, they are not output states, they are not asymptotic, and they always go together they are "virtual particles" if you want to think of them as particles, and they by definition only exist inside a calculation, in there is a whole continuum of them. You wouldn't really want to take their "motion" too literally, even if you like to think of the motion of quantum particles.

The real can of worms is that if you detect something you have to interact with it, and if you interact, what you are interacting with isn't a free particle, so virtual particles are really involved in any detector. But unless it is really close to on-shell, the amplitude will be ridiculously small. And it's hard to know exactly where it went once you have a detector since the detector itself has some spread, so the vagueness saves you from it being a problem either.

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