Fermi's golden rule relies on the concept of density of states. Wikipedia offers several derivations for the rule. For a mathematical theorem one proof is enough. Why do we need several proofs? Something is fuzzy in the theorem.
The cross section if the "hole" is at the center of a spherically symmetric potential
Let us consider a transition where a classical particle has to come to within some fixed distance r of a spherically symmetric potential in an R^3 space 1. If it gets that close, it moves to another R^3 space 2 where the potential is constant. The particle is non-relativistic.
Let the potential be larger in the space 2.
The hole appears as a sphere in both spaces. The cross section is
σ_2 = π r^2
from the space 2 to the space 1.
Angular momentum is conserved in a spherical potential. The maximum angular momentum that the particle can possess when it falls down from the space 2 is
J = r m|v_2|,
where v_2 is the velocity in the space 2 and m the particle mass.
Let the velocity of the particle be v_1 in the space 1. Since angular momentum is conserved, the maximum possible impact factor b for the downfalling particle in the space 1 is
b = r v_2 / v_1.
It is obvious that if we shoot a particle in the space 1 with an impact factor smaller than b, then it will climb to a distance < r in the central potential and go to the space 2.
Any state transition path 1 -> 2 or 2 -> 1 can be traversed to the opposite direction. The laws of nature are time symmetric.
The cross section for the transition 1 -> 2 is
σ_1 = π b^2 = π r^2 (v_2 / v_1)^2.
We see that the ratio of the cross sections to opposite directions is the same as the ratio of the "areas" that the momentum vector in each space can span. In the space 1, the area is 4 π (m v_1)^2 and in the space 2 it is 4 π (m v_2)^2.
In this case the cross section to the opposite direction can be derived from the cross section to the other direction.
Is there a general rule for obtaining the cross section to the opposite direction?
In the case of a spherically symmetric potential, we were lucky to be able to use conservation of angular momentum.
If the potential is another shape, then we cannot use the conservation law.
What about more complex processes, like pair production / annihilation? They probably are, in some sense, spherically symmetric.
What is time symmetry of natural laws in a probabilistic universe?
We need to define time symmetry of natural laws in a probabilistic universe. If a state A changes to B with a 50% probability, and otherwise to C, what does it mean to say that laws are time symmetric?
Suppose that we shoot a plane wave toward a particle. The output is a complex combination of plane waves.
We would like to know what happens to a single incoming plane wave if we reverse time. Knowing that a complex combination will produce a single plane wave is not interesting because we cannot produce such a complex combination.
We can reverse time in a Feynman diagram. That might be what we want: the probability amplitude of an incoming plane wave A producing a plane wave B is the same (or actually, the complex conjugate) as to the reverse direction. In a deterministic universe, the probability is always 1 or 0.
What about wave mechanics in classical physics? Can we calculate scattered amplitudes using the symmetry rule we sketched?
The Afshar experiment shows that there is no simple way to calculate the probabilities for a reverse reaction
We wrote about the Afshar experiment on December 23, 2020. In that experiment, two plane waves "help" each other to avoid a grid. Time reversal really requires both plane waves to be present.
The Afshar experiment shows that there is no general rule that allows us to calculate the response to a plane wave which is sent to the reverse direction of the process. But there might be such rules in special cases?
The concept of unitarity of the scattering matrix S is relevant here. The inverse of a unitary S has to be its conjugate transpose:
S* S = S S* = I.
But if the output has more states than the input? Then S is not a square matrix and unitarity is not defined.
In the Afshar experiment, the photon interacts with external macroscopic equipment. In principle, the photon & the equipment might develop in a unitary way, but we have no means of specifying the state of the macroscopic system. Therefore, we cannot utilize the concept of unitarity.
Interaction with a macroscopic system can be seen as a "measurement". A measurement destroys unitarity.
A unitary scattering matrix S
In quantum field theory, if a collision of two particles can produce a third particle (e.g., bremsstrahlung), then the output state is 3-dimensional, though the input state was 2-dimensional. Can we apply the concept of unitarity then?
If the collision does not create new particles, then unitarity might apply? The state spaces are continuous in most cases.
Suppose that the input has a finite number N of possible states and the output the same number N of possible states. Then the scattering matrix S tells the probability amplitude P for each transition
n -> m: P = S(n, m)
Unitarity of S then says that the probability amplitude of the reverse transition is the complex conjugate of P.
Reversing time in a Feynman diagram
The Feynman method of keeping the probability amplitude the same (or actually, taking the complex conjugate) when reversing time looks like a good solution.
Suppose that we have particles with momenta p and -p colliding. Let the angle of p be φ relative to the x axis.
Let us first set φ = 0.
We assume that the output is two particles with momenta q and -q. Energy conservation may fix |q|. Let θ be the angle of q relative to the x axis.
The scattering function (matrix) is then a complex-valued
S(θ).
How to get the reverse function? The relative angular distribution might be given by the same function S. But the cross section of the reverse reaction may, for example, be a lot smaller than the reaction.
p q
e- ----------- ~~~~~~~~~~ photon
| p - q
| virtual electron
e+ ---------- ~~~~~~~~~~ photon
-p -q
Let us look at the Feynman diagram of annihilation / pair production. Above, p and q are spatial momenta, not 4-momenta.
Let us fix p. The Feynman integral is over the possible values of q then. To the other direction, q is fixed and the integral is over possible p.
The Feynman propagator for a virtual massive Klein-Gordon "pseudoelectron" is
G_F(p - q) = 1 / (-(p - q)^2 - m^2 + i ε).
In the 4-momentum of the virtual "pseudoelectron", the energy E = 0. That is why we just have the negative square of the spatial momentum in the denominator.
p
^
|
|
----------------------> q
The value of |p - q| only depends on the angle θ between the vectors. If we let p or q vary, then G_F(p - q) has the exact same shape in both cases because it is a function of the angle θ. The only difference is in the area of the sphere which the vector p or q draws.
The ratio of the integrals over G_F(p - q) in each case is proportional to the integration area.
We see that the integrals are proportional to the area which the momentum vector can draw on each side of the reaction.
Annihilation at low energies is a complicated process - the spins play a role
Annihilation at low energies seems to be a complicated process. The spin of the particles affects things, and a positronium atom can form.
In a blog post about the Pauli exclusion principle we calculated that the magnetic repulsion or attraction from the spins dominates when electrons come very close to each other. That fact should show up in precision measurements of scattering processes. Feynman diagrams only take into account the Coulomb force.
The repulsion of the spins of the electrons, when the spins are parallel, is extremely strong at short distances. We suggested that this is the reason for the Pauli exclusion principle: the wave function must be essentially zero for short distances if the spins are parallel.
Photon-photon scattering
In photon-photon scattering, a virtual pair is created first and then it annihilates.
In the photograph model of quantum field theory, any classical orbit of the pair contributes to scattering. The situation is complicated, as the pair can have a hyperbolic orbit and be located anywhere in that orbit. Electromagnetic radiation quickly drains away all the energy of the pair.
We ignore the spins for now.
How to assign weights to various possible paths?
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