UPDATE February 4, 2021: the analysis in this blog post has been improved. See our new blog post.
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Photon-photon scattering is sometimes grouped under the name Delbrück scattering. Max Delbrück in the 1930s studied single photon scattering in the static Coulomb field of a nucleus. But in this blog post we are interested in a collision of two real photons.
A collision of photons can produce a real pair. Real pair production is a tree-level Feynman diagram, and consequently, does not involve diverging integrals.
loop
~~~~ ---------- ~~~~
| |
| |
~~~~ ---------- ~~~~
2 photons 2 photons
total energy < 1.022 MeV
We may interpret the diagram in this way: in the middle is a pair whose total energy is less than 1.022 MeV. On the right the pair annihilates and produces two photons. On the left is the reverse reaction to annihilation: pair production - but the produced pair never gains enough energy so that the electron and the positron could escape from each other.
Yi Liang and Andrzej Czarnecki (2011) have written a tutorial on photon-photon scattering.
They write that there are three essentially different diagrams for scattering. Each diagram individually diverges, but when all are added together, divergences cancel. However, the sum is different for dimensional regularization and for Pauli-Villars regularization.
Pair production or scattering from the collision of two photons has not been observed experimentally.
V
^
|
| ______ ------------ 1.022 MeV
| /
| ------------------------------------------ 0 MeV
| /
| /
| |
| |
---------------------------------------------------> r
if V = 0 then r = 1.4 * 10^-15 m
high
momenta?
In the diagram, V is the total energy of a static electron-positron pair. The separation between the particles is r.
The diagram is in very poor character graphics, but the reader of this blog probably gets the idea.
Our previous blog posting suggested that the arbitrarily high 4-momenta in a diverging Feynman loop integral may be (classical) electron-positron pairs where the separation is r < 1.4 * 10^-15.
Annihilation of a pair
How fast does an electron radiate energy if it starts static at a distance r_0 = 1.4 * 10^-15 m from the positron = half of the electron classical radius.
The Coulomb force is
F = k e^2 / r_0^2 = 120 N.
The acceleration of the electron is huge, 1.3 * 10^32 m/s^2.
The Larmor formula for the radiated power is
P = e^2 a^2 / (6 π ε_0 c^3)
= 10^11 W.
The mass-energy of the electron is 10^-13 J.
The electron would lose all its mass-energy in 10^-24 s, which at the speed of light corresponds to r = 3 * 10^-16 m. That is just 1 / 5 of r_0.
Let us calculate r symbolically, using the formulas for P, a = F / m (where m is the electron mass), and
r_0 = 1 / 2 * 1 / (4 π ε_0) * e^2 / (m c^2),
k = 1 / (4 π ε_0).
We get
r = c * (m c^2 / P)
= m c^3 * 6 π ε_0 c^3
* m^2
* e^8
* 4^2 π^2 ε_0^2
/ (e^2
e^4
8^4 π^4 ε_0^4 m^4 c^8)
= 6 * 16 e^2
/ (8^3 m c^2 8 π ε_0)
= 96 / 512 r_0.
If we cut the distance to half, the acceleration is 4X and the power 16X.
If we double the distance, the power is just 1 / 16 X.
Classically, the speed of the electron will be significantly lower than c until it is close to the distance 1.4 * 10^-15 m. This is simply because the negative potential goes as 1 / r. Thus, the Larmor formula may be roughly right at larger distances.
Hypothesis: classically, in annihilation, the electron will radiate its mass-energy away at a distance 1.4 * 10^-15 m from the positron, in a short burst.
It is interesting that the classical analysis agrees this well with the classical electron radius. Suppose that the particles would radiate their energy away much slower. Then, in principle, we could have a pair which would still possess positive energy even though the particles are static and closer than 1.4 * 10^-15 m. But such a pair has negative total energy. This would be a contradiction.
We have the length scale problem here. The energy is radiated in a very sharp electromagnetic wave whose length scale is 1.4 * 10^-15 m. But the wavelength of the photons is 2 * 10^-12 m. Our earlier blog posting suggested that we must sum the electromagnetic waves over many spacetime points, over a volume 2 * 10^-12 m (uncertainty of the position of the pair), to get the wave observed by an external observer. Then the electromagnetic wave's wavelength is of the right order of magnitude.
Creation of a real pair from 1.022 MeV photons versus a virtual pair
~~~~ ---------------------- e+
| virtual
| electron
~~~~ ---------------------- e-
2 photons
Production of a real pair from a photon-photon collision is called the Breit-Wheeler process.
Let the photons have a combined energy of just over two times the electron mass.
The Feynman integral has one propagator, the virtual electron in the middle.
But if we reduce the combined energy of the photons just under the mass of two electrons, the produced pair will inevitably annihilate. The Feynman integral then has four propagators. If we assume that the electron and the positron are almost on-shell, then the propagators for them have very large absolute values around the pole which corresponds to an on-shell electron.
We need to check what happens if we increase the mass of the electron so that it becomes a classical particle. Do the large propagator values around the pole make sense then?
What is the correct Feynman diagram for photon-photon scattering?
If the photons have combined energy < 1.022 MeV, then the produced virtual pair will certainly annihilate.
The diagram at the top of this blog post contains three "unnecessary" propagators for the electron and the positron, the ones which do not appear in the real pair production diagram. If we know that annihilation is inevitable, why should we use those three propagators?
It might be better to use the real pair production diagram, with the electron mass adjusted lower?
We know that Feynman diagrams do not handle bound states correctly. The pair with energy slightly below 1.022 MeV is kind of a bound state. The pair might even form a positronium "atom", whose lifetime is 0.12 ns for antiparallel spins, and 142 ns for parallel spins.
Numerical cross sections for pair production and scattering in a real photon-photon collision: the values differ too much?
Let us check what people have calculated for the pair production cross section at combined energies close to 1.022 MeV.
The production of virtual pairs of moderate momenta should have roughly the same cross section.
Gould and Schreder (1967) give the pair production cross section in the center of mass frame as
σ = β π r_0^2,
where the velocity of the produced electron and positron is β c, and r_0 is the classical electron radius.
If β = 0.1 c, then the cross section is 2.5 * 10^-30 m^2, or 25 millibarns.
T. Takahashi et al. (2018) have the scattering cross section in their Figure 1 as less than 10 microbarns, or less than 10^-33 m^2.
This is strange. How can the probability for a virtual pair production be less than 1 / 1,000 of the probability for real pair production at just a little higher energy?
We need to check the figures from other sources.
R. Svensson and A. Zdziarski (1990) in their introduction give a value ~ 4 * 10^-34 m^2 for the cross section of scattering of ~ 1 MeV photons. They write that scattering is a fourth order process - therefore the cross section is small. Their value is consistent with Takahashi et al.
A concrete example: combined photon energy 2 m_e +- 0.5 eV
Let us calculate a concrete example. Let β = 0.001 in the pair production formula. The combined kinetic energy of the pair is then 0.5 eV. The cross section for pair production is 10^-32 m^2.
If the pair would have the energy 2 m_e c^2 - 0.5 eV, the virtual particles could climb to a separation of 3 nanometers until falling back and annihilating. Literature above claims that the cross section is < 10^-33 m^2.
A separation of 3 nanometers is a huge distance in particle physics. The diameter of a hydrogen atom is 0.1 nm.
How can the colliding photons, that are lacking 0.5 eV, know beforehand that their energy just fails to produce a real pair, and the photons then "decide" that the cross section should be much smaller?
Or is it so that the annihilation will send the photons to the exact same directions that they arrived to the collision? Why would that be?
Feynman formulas fail to take into account the attraction between the created particles. One of the basic assumptions in Feynman diagrams is that particles fly between vertices without interaction.
What about scattering at substantially higher energies than 2 m_e? Classically, the created electron and the positron will move to opposite directions, and will never annihilate. In quantum mechanics, the positions of the particles have uncertainty. In the MeV scale, the uncertainty about the relative position is ~ 10^-12 m.
The cross section for annihilation should be roughly the same as for pair production, that is, ~ 10^-29 m^2. If the electron and the positron move through an area ~ 10^-24 m^2, the probability for annihilation is ~ 10^-5.
We conclude that the cross section for scattering if the combined energy is substantially larger than 1.022 MeV, is ~ 10^-34 m^2, or one microbarn. This figure agrees with the ones that T. Takahashi et al., and R. Svensson et al. give.
Conjecture: cross sections for ~ 1.022 MeV combined photon energy are larger both for pair production and scattering
How could the colliding photons, who have a combined energy ~ 1.022 MeV, know what to do? The formula for pair production gives a very small cross section for real pair production if the energy is only slightly over the limit 2 m_e.
The formula for scattering gives an extremely small cross section regardless of the energy.
Suppose that a real or virtual pair is produced in a photon collision.
Classically, the eventual fate of the pair would depend on its energy: if the energy is < 2 m_e, then the pair falls back and annihilates. Then it is a case of scattering.
In wave mechanics, a wave which is born from a source in a small spatial volume must go through a "hole" whose area is 4 π r^2 where r is an arbitrary distance from the source. The hole can be considered a circle whose radius is 2 r. If the de Broglie wavelength of the particle is long, it has hard time going through the hole, and is reflected back. That prevents real pair creation.
Let us calculate an example. The de Broglie wavelength of an electron with 50 keV kinetic energy is 2.3X the Compton wavelength. The de Broglie wavelength of a 500 keV kinetic energy electron is 0.5X the Compton wavelength.
We also need to check if the very small cross sections in literature can be explained by destructive interference. Maybe destructive interference is very important if the output is slow electrons?
__ __
__/ \___/ \___
photon wavelength 2 * 10^-12 m
Let us assume that we collide two linearly polarized laser beams whose wavelength is the Compton wavelength of the electron.
They form a standing electromagnetic wave which has the same wavelength.
We assume that there exists a nonlinear process which causes the standing wave to induce waves in the Dirac field. It is reasonable to assume that the created waves in the Dirac field inherit their phase from the standing wave.
If we replace the Dirac field with a massive Klein-Gordon field, we observe that almost all the energy that we take from the standing wave is spent to make the harmonic oscillators to oscillate at each spatial point x. Only a very small amount of energy goes to the kinetic term d^2 ψ / dx^2 of the massive Klein-Gordon equation.
This means that the energy flow from the electromagnetic field to the Dirac field is almost unharmed by destructive interference. We conclude that destructive interference is insignificant in the process.
Pair production when a 1.1 MeV photon meets a nucleus
There exists measured data for a different process, namely, a single photon hitting the nucleus of an atom.
R. Frahm et al. (2009) measured pair production cross section at low gamma ray energies > 1.022 MeV when photons hit atoms. They calculated with the program XCOM 3.1 that the cross section for 2 MeV should be 0.2 barns per gadolinium atom and 0.0007 barns for 1.1 MeV.
Their Figure 4 shows that the measured cross section at 1.1 MeV is roughly 0.002 barns, or 3X times the calculated value.
Why is the cross section so much lower for 1.1 MeV than for 2 MeV? Bethe and Heitler apparently use Fermi's golden rule in their 1934 paper. If 1.022 MeV is spent to create the electron and the positron, then the "number of states" in the momentum universe of the final outcome is less with just 78 keV of kinetic energy.
When a nonrelativistic electron hits a sharp potential step, the transmission is roughly the ratio of kinetic energies.
Cross section for pair annihilation
The Feynman diagram formula for the probability amplitude has the same value if we turn the diagram upside down.
We know that if we put a static electron and a static positron close to each other, the probability of annihilation is 100%.
If we model the electron and the positron as de Broglie wave packets, we cannot really make them static.
σ_x σ_p >= h-bar / 2.
Let us assume that the uncertainty of the electron position is 1 meter. Then the uncertainty in its momentum is 5 * 10^-35 kg m / s, which corresponds to a velocity 5 * 10^-5 m/s.
The kinetic energy of the electron is
E = 1/2 m v^2 = 10^-39 J.
The Coulomb potential of an electron and a positron at the distance 1 m is
V = -k e^2 / r = 2 * 10^-28 J.
We see that the probability of annihilation is 100% if the electron and the positron move at a speed < 15 m/s.
A thermal electron at 300 K moves at 1.2 * 10^5 m/s.
Let us check what literature says about annihilation cross section.
Frank Rieger gives the low-energy cross section as
σ = 1 / β * π r_0^2,
where the velocity of the particles is β c and r_0 is the classical electron radius.
The annihilation cross section for a thermal 300 K electron is
2,500 * π r_0^2,
or a circle whose radius is 50 r_0 = 1.4 * 10^-13 m.
Gould and Schreder give a cross section
σ = β π r_0^2
for the inverse reaction, i.e., pair production (300 K) from the photons. It corresponds to a circle whose radius is 1/50 r_0.
The kinetic energy of a thermal 300 K electron is 0.04 eV. If the energy of a pair is 2 m_e - 0.08 eV, what is their separation? It is 2 * 10^-8 m, a huge distance in particle physics.
The Rieger formula gives a cross section in the 1 m^2 range if β = 10^-30, while our own potential calculation found that β = 5 * 10^-8 is enough to ensure a 1 m^2 cross section. The Rieger formula probably is correct for 300 K electrons, but fails for electrons whose temperature is in the microkelvin range.
Divergence of individual Feynman diagrams
Liang and Czarnecki mention that each of the scattering diagrams has a diverging Feynman integral.
The divergence is probably logarithmic, because there are four electron propagators in a diagram.
Above we argued that the diagrams are not the right way to calculate because photons cannot know in advance what exactly is the value 2 m_e.
Anyway, whatever method we use, we have to face the problem that virtual pairs with arbitrarily high 4-momenta contribute to scattering. How to get rid of divergences?
We will look next at those rogue virtual pairs.
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