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Virtual particles even today contrary to the overwhelming evidence in my opinion that they have a real effect on normal particles (i.e. Casimir effect, $g-2$ muon Fermilab experiment) are considered by many scientists that they are really not existing but merely mathematical inventions to describe known quantum phenomena in nature.

Even so, seen as a mathematical formalism artifact included in established theories assisting explanations and descriptions for these theories like quantum field theory, my question is are there any particular cases of analysis concerning phenomena where the calculations turn out to generate virtual particles that exceed the speed of light $c$ limit in a vacuum?

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Yes.

Consider the one-loop contribution to the photon propagator, with a virtual electron/positron loop. To evaluate this Feynman diagram, one must independently integrate over all possible relativistic loop energies $E$ and relativistic loop momenta $\mathbf p$… from zero to infinitely large.

In relativistic mechanics, velocity can be determined from $E$ and $\mathbf p$ using their ratio:

$$\mathbf v = \frac{\mathbf pc^2}{E}.$$

Thus if one really wants to ascribe velocities to the virtual particles in the loop, these velocities must also range from zero to infinitely large.

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    $\begingroup$ That looks like a good reason not to take the semi-classical particle picture seriously. It doesn't even make sense semi-classically. $\endgroup$ Commented Jun 19, 2023 at 20:50
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    $\begingroup$ My take on it is that people who for whatever reason want to think that virtual particles are particles need to accept that if they are particles, they are nothing at all like any particles we have ever observed or ever will observe. Perhaps they should be called “varticles”. :) $\endgroup$
    – Ghoster
    Commented Jun 19, 2023 at 20:54
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    $\begingroup$ @FlatterMann But the OP does not pine for taking semiclassical thinking seriously. He asks: "Even so, seen as a mathematical formalism artifact included in established theories assisting explanations and descriptions for these theories like quantum field theory", "...phenomena where the calculations [involve] virtual particles that exceed the speed of light 𝑐" ... $\endgroup$ Commented Jun 19, 2023 at 20:54
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    $\begingroup$ I wish I could give more than one thumbs up for that comment! $\endgroup$ Commented Jun 19, 2023 at 20:55
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    $\begingroup$ @FlatterMann I fear I missed any assignment of "actual" to virtual particles, internal lines for most of us... There are dozens of questions on this site shadow-boxing with bad science reporters... $\endgroup$ Commented Jun 19, 2023 at 21:06
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Not that I'm aware of.

If you look at the real space representation of the computations, they're based on the Green's function of the Klein-Gordon equation (and its first derivative, if you're using fermions) which is nice and confined to within the light cones (forward and backward). Usually, we use the Feynman propagator, which includes the Green's function in its real part and the vacuum fluctuations in its imaginary part. The vacuum fluctuations have correlations outside of the light cone, but those aren't propagating signals, they're coincidences enforced by the fact that the spectrum of ground state fluctuations is fixed by the Hamiltonian. The formula for the Feynman propagator is \begin{align} \Delta_F(s) = -\frac{1}{4\pi} \delta(s) - \frac{imc}{4\pi^2 \hbar\sqrt{s}} K_1\left(\frac{mc\sqrt{s}}{\hbar}\right), \end{align} where $K_1$ is a modified Bessel function of the second kind, $m$ is the mass of the particle, and if the propagator is between the spacetime points $x$ and $y$ then $s = (\mathbf{x}-\mathbf{y})^2 - (x^0 - y^0)^2$. The important thing is that the imaginary part of $\Delta_F$ comes from vacuum correlations, and the real part is what deals with the creation and annihilation of virtual particles.

The idea that virtual particles can travel faster than light comes from looking at the Fourier space representation of the Feynman propagator. If we take the full 4-$d$ Fourier transform of $\Delta_F$ using the unitary convention we get that \begin{align} \Delta_F(p) &=\frac{\hbar^2}{\left(2\pi\right)^2\left[-(p^0)^2 c^{-2} + \mathbf{p}^2 + m^2c^2\right]} \nonumber \\ &\hphantom{=}+ \frac{2\pi i}{(2\pi)^2} \delta\left(\frac{-(p^0)^2 c^{-2} + \mathbf{p}^2 + m^2c^2}{\hbar^2}\right). \end{align} Note that the usual formulation only includes the first term, and the second term is produced by a choice of how to dodge the pole in the propagator using a complex contour integral. In the above formulation, the pole is assumed to be integrated using the Cauchy principle value, and the mass shell term is included explicitly. Because there is now no contour integral, you can see that the part of the propagator that is confined to the mass shell (the dispersion relation) produces the real-space vacuum fluctuations. Those Fourier modes travel at $c$ (or less) because they obey a dispersion relation \begin{align} \omega^2 = \frac{m^2 c^4}{\hbar^2} + \mathbf{k}^2c^2. \end{align}

Does that mean that the propagator requires particles that travel faster than light to get the real part of the real space propagator? Not really. There is a formalism known as "on shell recursion." I haven't studied it in detail, so I cannot comment on how they do it, but the simple implication is that they don't use off-shell virtual particles. What I can comment on is that the on-shell part of the Feynman propagator is $p^0$-even. The $p^0$-odd version has the Fourier space form \begin{align} \frac{2\pi i}{(2\pi)^2} \operatorname{sgn}(p^0) \ \delta\left(\frac{-(p^0)^2 c^{-2} + \mathbf{p}^2 + m^2c^2}{\hbar^2}\right). \end{align} If we Fourier transform that back to real space, we get \begin{align} \operatorname{sgn}(t)\,\Theta([ct]^2 - r^2) \frac{mc J_1\left(\frac{mc\sqrt{(ct)^2 - r^2}}{\hbar}\right)}{4\pi \hbar \sqrt{(ct)^2 - r^2}}. \end{align} Note that, up to the sign function, that's identical to the real part of the Feynman propagator. In principle, then, it should be possible to work with only on mass shell modes in the Fourier transform of the propagator, if you're careful enough.

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  • $\begingroup$ Great canonical expert answer! Thank you for the time taken and effort for answering this question. $\endgroup$
    – Markoul11
    Commented Jun 20, 2023 at 11:07

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