k = (δ, δ)
~~~~~~~~~~~~~~~ real photon
p' /
e- ------------------------------------ p = p' + q - k
|
| q
|
Z+ ------------------------------------
In the diagram, we have set c = 1. Then the absolute value |δ| of the spatial momentum of the real photon k is the same as its energy δ.
The electron propagator measures "how far" is the electron from being on-shell. For an on-shell particle,
E² = P² + m²,
where E is its total energy and P is its spatial momentum. In the propagator formula above p is the 4-momentum, and
p² = E² - P².
The denominator above is zero for an on-shell particle. That is, there is a pole. The pole is formally removed by the i ε term.
The numerator is a 4 × 4 matrix. The components of p multiply gamma matrices. The term m above is actually m times the 4 × 4 identity matrix I. The numerator is not zero.
Suppose that the electron is off-shell by an energy δ. That is, it has δ "too little" energy, compared to its spatial momentum P:
p² - m² = (E - δ)² - (P - δ)² - m²
= -2 E δ + δ² - δ²
= -2 E δ,
where we have assumed that the spatial momentum of the electron P is normal to the momentum δ of the photon. Also, if the electron is not very fast, then |P| is a lot smaller than E, and we can ignore the cross term of P and δ.
We see that the probability amplitude of a real photon, |δ| << |P|, is governed by the electron propagator, and is
~ 1 / |δ|.
Can a scalar electrically charged particle exist?
The electron propagator is derived from the Dirac equation. How is it possible that the Dirac equation "knows" what kind of an electromagnetic wave will be born if the electron is pushed by the momentum q?
The Dirac equation is kind of a "square root" of the wave equation. This probably explains how it can know about the behavior of the electromagnetic field.
Would the propagator of a scalar charged particle work? The deminator looks like the one for the electron, but the numerator is very different.
The Peskin and Schroeder textbook on QFT (1995) gives the bremsstrahlung formula above for an electron. For a scalar charged particle, the numerators on the right would not contain the 4-momenta p and p'. The polarization ε* of the photon should be coupled to p and p' with a separate mechanism. The propagator of the scalar particle would control the spectrum of bremsstrahlung.
The photon can be emitted from the particle either before the particle scatters from Z+ or after. The propagator in the first case is
≈ -1 / (2 E δ)
and in the second case
≈ 1 / (2 E δ).
The probability amplitudes cancel each out almost completely. It looks like the scalar particle propagator does not "know" what kind of an electromagnetic wave is created by pushing the particle.
This may explain why there are no charged scalar particles.
The Dirac equation under an electric field: it really does not need to "know" anything
e- ~~~ --->
^
| E
| E
v
<--- ~~~ e-
Suppose that two wave packet electrons pass by each other so far that their distance is much larger than the size of the packet. Then we can approximate the electric field E of each electron at the other electron.
p → p + e A(r, t).
Since the field E disturbs the Dirac equation, the electron wave probably is off-shell. The Dirac equation understands this because the electron in it is minimally coupled to the field E.
But how is this related to the Green's function of the Dirac field?
Could it be so that we actually derive the properties of the electromagnetic field from the Dirac equation? Then there would be no mystery of how the Dirac equation "knows" the properties of the electromagnetic field.
Yes. We derive the interaction of an electron with the field from the Dirac equation. The macroscopic interaction of a charge and the electromagnetic field is not given to us beforehand. We derive the form of the bremsstrahlung wave from the Dirac equation, using quantum field theory.
Quantum field theory is primary. From it we derive macroscopic Maxwell's equations.
However, this is not entirely satisfactory. We showed on November 25, 2025 that Feynman diagrams miscalculate several classical limits. The classical equation is more robust.
Quantum gravity and gravitational bremsstrahlung: the problem with various propagators
Above we suggested that an electrically charged scalar particle might break macroscopic Maxwell's equations. Could there be similar problems with gravitational waves?
Bremsstrahlung depends on the propagator of the particle. It would be strange if gravitational waves for different propagators would be different.
The Higgs particle is a scalar particle with a rest mass. W and Z bosons have a rest mass.
It does not sound reasonable that the gravitational wave from such a particle would be different, depending on the propagator of the particle.
*** WORK IN PROGRESS ***
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