Feynman paths of FTL velocity have imaginary momentum?

Question

In this Phys.SE answer it is discussed that Feynman path integrals sums amplitudes for all possible paths, including those that are not time-like. If you take the momentum-space path integrals, I would naively expect that such space-like coordinate paths would contribute an imaginary momentum in the momentum-space path integral, that would result in evanescent exponentially decaying amplitudes outside the light-cone.

Is this a correct interpretation and/or expectation to have? furthermore, In the momentum propagator integrals discussed in the lecture notes Feynman Diagrams for Beginners (PDF) by Kresimir Kumericki, the integrands are $d^3 k$, so it seems that they cover the whole 3D real momentum space, which seems to me to be restricted inside the light-cone (by nature of it being real). This confuses me because it seems to contradict the idea that all paths are included in the integral. What am I missing?

Answer 1

The momentum being real has totally and absolutely 100% nothing whatsoever to do with whether a tangent to a curve is timelike, lightlike or spacelike.

The energy-momentum vector points in the direction of the tangent to the worldline of the particle. The worldline is in a real four dimensional spacetime, so its tangent has 4 real components. Whether tha tangent is timelike, lightlike or spacelike is detemined by whether: $$(c\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2,$$ is positive, zero, or negative respectively.

To have a spacelike tangent, just have $(c\Delta t)^2<(\Delta x)^2+(\Delta y)^2+(\Delta z)^2.$ To have a timelike tangent, just have $(c\Delta t)^2>(\Delta x)^2+(\Delta y)^2+(\Delta z)^2.$ Nothing imaginary required for either case.

If you had a massive particle with a timelike tangent you can scale the "unit" tangent vector: $$u=\frac{(c\Delta t,\Delta x,\Delta y,\Delta z)}{\sqrt{(c\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2}},$$ by the mass to get the energy-momentum vector: $$mu=\frac{m(c\Delta t,\Delta x,\Delta y,\Delta z)}{\sqrt{(c\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2}}.$$

If you had a massive particle with a spacelike tangent you can scale the "unit" tangent vector: $$u=\frac{(c\Delta t,\Delta x,\Delta y,\Delta z)}{\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(c\Delta t)^2}},$$ by the mass to get the energy-momentum vector: $$mu=\frac{m(c\Delta t,\Delta x,\Delta y,\Delta z)}{\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(c\Delta t)^2}}.$$

In both cases, there is a worldline, it has a tangent with a unit magnitude, and an energy-momentum vector points in the exact same direction in spacetime, but is scaled by the mass. A lightlike curve is actually different because then an energy-momentum vector is just a vector that is tangent just there is no length associated with a particular magnitude of the energy-momentum vector.

So let's get at what is means to have different tangents. They are vectors. They can have different directions, they are in spacetime, so these different directions do correspond to different speeds, they even correspond to FTL speeds if that's the way a curve is. So what is the mass of an energy-momentum vector? It is just the length. It is nothing more, it never was. It is not the source of gravity (energy and momentum and stress are, and always were) it isn't something you add up to get a total (only when the vectors point in almost the same direction is the length of the sum almost equal to the sum of the lengths), it's nothing but a length.

Given the energy and the length, you can find out the magnitude of the momentum. Given the momentum and the length you can find out the magnitude of the energy. It's really about the balance between energy and momentum.

Timelike means more energy than momentum. Spacelike means more momentum than energy. Lightlike means equal amounts of both. Absolutely nothing deeper. So spacelike just means you have an excess of momentum for your energy. It is not about imaginary momentum, this is a geometrical thing in a real 4d manifold.

So if you want to move in a where your momentum and energy are not balanced the way they usually are, you have to move off-shell, which just means with an unusual length. This happen in path-integral methods. You just have to accept that paths can go any direction they want, and that the energy momentum vector points in the direction in spacetime that the particle goes, so FTL particle paths are just energy-momentum vectors with more momentum than energy.

To be clear, you can pick an aribtrarily small (but real momentum) and as long as you make the energy super even smaller, then you can move at arbitrarily fast speeds. Give it zero energy (so clearly off-shell) and it moves infinitely fast. We cover any speed using only real momentum.

People that want to sell you imaginary momentum probably just have some assumption or bias they packed with them.

Answer 2

Since your question is about special relativistic phenomena i think the best way to answer your question si in the context of quantum field theory, the propagator is defined, in the case of a scalar field theory,as: $$W(x-y)=\langle0|T[\phi(x)\phi(y)]|0\rangle$$ which can be put in relation to the path integral forumlation of QFT via the following equation: $$W(x-y)=\int D[\phi]\phi(x)\phi(y)e^{iS[\phi]}$$ that is the expectation value of the composite operator $\phi(x)\phi(y)$ on all possible field configuration weighted with $\\e^{iS[\phi]}$. In the path integral all field configurations are evaluated, that's because unphysical configuration also contribute to the probability amplitude even if they are not observable. Such field configurations have a lower weight on the result of the integration since they interfere with each other leaving the classical configuration (the one that makes the action stationary $\delta S=0$) to be the most relevant. Now i stress all that since it's important to understand that in QFT particles aren't well defined objects, since they can't be localized beyond their compton wavelenght $\lambda_c=\frac{\hbar}{mc}$ for that reason the path of a particle is not a well defined concept at every scale. That's one of the many reasons why we take the field description. The idea of sum over paths losely speaking is repleaced with the sum over all possible field configurations which creates the particle/s at a certain spacetime point. And now back to the propagator (for the sake of simplicity we will assume that $(x_0-y_0)>0$ so that $T(\phi(x)\phi(y))=\phi(x)\phi(y))$): let's take a free scalar quantum field theory as an example: $$L=\frac{1}{2}(\partial^\mu\phi)^2-\frac{1}{2}m^2\phi^2$$ since it's a free theory there will be only single particle contributions to the propagator then we can write inserting the identity $$\mathbb I=\sum|n\rangle\langle n|$$ where $|n\rangle $ is an n-particle eigenstate of the four impulse $\hat{P}^\mu$ now since only the one particle contribution matters we can write: $$W(x-y)=\int d^3p\frac{1}{2\omega_p}\langle0|\phi(x)|p\rangle\langle p|\phi(y)|0\rangle$$ that's crucial, infact since we are using the eigentates of the four-impulse operator which is real valued the sum over all possible impulse states is done over real values of it! We can use translational symmetry to write $U(x)\phi(0)U^\dagger(x)=\phi(x)$ where $$U(x)=e^{iP\centerdot x}$$ so $$W(x-y)=\int d^3p \frac{1}{2\omega_p}e^{-iP\centerdot(x-y)}|\langle0|\phi(0)|p\rangle|^2$$ it follows form lorentz invariance that $|\langle0|\phi(0)|p\rangle|^2=constant=\frac{1}{(2\pi)^{3}}$ so we have $$W(x-y)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p}e^{-iP\centerdot(x-y)}=i\Delta_F(x-y)$$ which is (for the case $(x_0-y_0)>0$) the usual feynman propagator. As you can see in this (concise) derivation no use of any immaginary momentum has been done since not even quantum mechanics requests the existence of immaginary observable quantities. Now i will demonstrate that the probability for a particle to be found outside the lightcone is non zero but that fact does not imply the existence of immaginary momentum (as i hope i already made clear) nor the fact that such thing can be observed,let us begin:

Consider $\Delta_F(x-y)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p}e^{-iP\centerdot(x-y)}$ in the case of the distance $|x-y|$ to be spacelike, there will be a frame in which $x_0=y_0$ and $\vec{x}-\vec{y}=\vec{r}$ therefore we will have $$\Delta_F(x-y)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p}e^{i \vec{p}\centerdot\vec{r}}=\frac{2\pi}{(2\pi)^3}\int_{0}^{\infty} \int_{0}^{\pi} dpd{\phi}\frac{p^2 \sin(\phi)}{2\omega_p}e^{i pr\cos(\phi) }$$ now the integral $$\int_{0}^{\pi} d{\phi} \sin(\phi) e^{i pr\cos(\phi) } =2 \frac{sin(pr)}{pr}$$ thus we will write $$\Delta_F(x-y)=\frac{-i}{(2\pi)^2} \int_{0}^{\infty} \frac{p^2}{2 \omega_P} \frac{e^{ipr}-e^{-ipr}}{pr}=\frac{-i}{2(2\pi)^2r} \int_{-\infty}^{\infty} \frac{p}{\sqrt{p^2+m^2}} e^{ipr}$$ we just have put togheter the two integrals with $e^{ipr}-e^{-ipr}$ just by doing in the second one the substitution $p \rightarrow -p$. Now in order to evaluate such integral we will use complex analysis methods, it's important to understand that this is just an intermediate step after which we will operate a limit to return in the real momentum space! No physical interpretation is or should be done of such techincal step, it's just an easy way to compute the integration. Since it's a pretty dull operation i'll tell you what is found: $$\Delta_F(r)\sim e^{-mr}$$ outside the light cone the propagation amplitude is vanishing but non zero, yet causality is preserved since no observable quantity can be related via such phenomena. (that's one of the main reasons of the theoretical prediction of existence of antimatter!). On a last note i would also stress that the familiar relation $$P^\mu=\left(mc\gamma,mv\gamma\right)$$ which would give you an immaginary momentum if $v=nc$ with $n>1$ is not respected by off-shell particles which have arbitrary energy not related to the usual classical special relativistic dispersion relation, a particle "following" (in the limited sense of quantum mechanics)a FTL path would not be observable. I hope that covers a little your question's points!

Answer 3

As it turns out, your intuition might be in the right path. As was recently noted and published by physicists of the caliber of Arkani Hamed and Freddy Cachazo, you can avoid the trouble of doing your Feynmann integrals off-shell, by replacing the vertices with on-shell particles of complex momenta and coordinates. There is a conjecture that this replacement is physically equivalent to all orders, but to my knowledge there is no proof yet.

This of course, is a calculational tool and you might do well in not giving too much weight into the interpretation of the complex extension of spacetime these integrals go over. The current interpretation is that the complex extension is a symptom of the equivalence between off-shell integrals and spacetime nonlocality