Working in momentum space: a collision of an electron and a photon
The "momentum space", which is used in Feynman diagrams, can be understood as working with plane wave functions inside, say, a 1 cubic meter area. We do not assume anything about the location of each particle, except that they are in such a large volume in space. We assume or calculate the energy and the momentum of each particle. The way to represent such a particle is as a plane wave.
t
^ photon
| / / /
| ____________________ e-
| / / /
| ____________________
| / / /
|
-------------------------------> x
In 1 space dimension and the time dimension we work with plane wave diagrams, like the one above.
In the diagram, the horizontal lines are an electron with 511 keV of energy, but zero momentum. That is, we have an electron which sits still somewhere in the cubic meter. The lines mark the "crests" of its wave function.
The photon moves right, and the crests of its wave function are lines that have a 45 degree slant to the right. The photon holds energy E and momentum p.
The electric potential V (or the magnetic vector potential) of the photon field disturbs the free propagation of the electron wave function.
(Note: it is the classical electromagnetic wave which disturbs the classical Dirac wave of the electron. What is the relationship between the quantum mechanical probability amplitude wave of each particle and the classical wave? We need to think about this.)
The effect of the disturbance is a to act as a "source" to a small wave, the scattered wave. Klein and Nishina, and many others, have calculated that the scattered wave is a wave where the photon has been "absorbed" to the electron. That is, the energy of the scattered electron wave is 511 keV + E, and the momentum is p.
In the scattered wave the electron swallowed the payload of the photon. The scattered wave has the energy and the momentum incompatible for a free electron, that is:
E^2 != p^2 + m^2.
Our description of "virtual" particles and waves in a blog post last week is relevant here. The electron is interacting with the photon, and the electron is off-shell or virtual. The corresponding wave is malformed. It could not exist as a free wave. It is like the transient wave which we make in water when we move an oar.
The scattered electron in our diagram has to get rid of the extra energy it is holding. It immediately "emits" a second photon which carries away just the right amount of energy and momentum, so that the electron and the emitted photon are on-shell or real. The process was a collision of an electron and a photon.
We could draw the scattered electron plane wave, as well as the emitted photon plane wave in the diagram. Also, we could add the electron wave of the bounced electron.
Unitarity of the collision: a paradox?
We can flip the direction of time in our diagram. Then it is the emitted photon which scatters the bounced electron. But the amplitudes would be wrong: the emitted photon totally scatters the bounced electron. This is in contrast to the original incoming photon which only scattered a little part of the incoming electron wave.
Quantum mechanics should be unitary. We should always be able to calculate the earlier wave function state A from a later wave function state B. But in the example, how can we know that the emitted photon will totally scatter the bounced electron? What is wrong?
The answer: we do not know the later state B completely. We used a crude perturbation method to calculate B approximately. We cannot demand unitarity from crude approximations which we calculate.
No one knows how to calculate the later wave function state B precisely.
Pair production
If we flip the time in an annihilation diagram, we get a pair production diagram.
photon 2 ~~~~~~~ ---------------- e-
/
/
photon 1 ~~~~~ ------------------ e+
We assume that a plane wave which describes the positron e+ with on-shell E and p exists. The photon 1 disturbs it and causes a scattered electron wave (the slanted line up). The photon 2 then scatters that wave further, into an on-shell electron e-.
Why can we assume that a positron exists? Maybe we could answer that if we knew the precise wave function. The above diagram is about a crude perturbation approximation.
Another way to explain pair production is to assume that there is an electron in a hole, in the -511 keV energy state. The hole might actually be a positron in the +511 keV energy state. The two photons excite the electron out of the hole, and in the process the positron, too, gets a kick.
A vacuum polarization pair
e- ---------------------------------------
|
O vacuum polarization
| p virtual photon
Z+ ---------------------------------------
Let us consider the scattering of an electron from an atomic nucleus Z+. A nucleus is so heavy that we can treat it as an almost classical object whose position we know precisely.
The vacuum polarization loop contains an electron and a positron whose combined energy is zero and combined momentum is zero.
In the plane wave diagram, the electron might look like the electron in the diagram which we drew in a previous section.
The photon only carries momentum, no energy. The photon is a component of the Fourier decomposition of the Coulomb potential of the nucleus. All the component are time-independent, which means that they only carry momentum, no energy.
t
^ Fourier component
| | | | |
| | | | |
| | | | |
------------------------------------------> x
Z+
In the diagram above, we have one Fourier component of the 1 / r electric potential of the nucleus. The component is a sine wave potential dependent on x. The lines mark maximum values of the potential. The nucleus Z+ resides at the center of two lines. The Fourier decomposition has a valley of each component located at Z+.
There is no magnetic field. The "virtual photon" is a static electric field described above. The virtual photon carries a momentum p to either direction of the x axis.
Let us superimpose an electron wave on the photon wave above. There is a disturbance in the electron wave which looks like a scattered electron that absorbed the momentum p either left or right.
Let us assume that the electron in our virtual loop absorbs that momentum p, that is, the nucleus Z+ pulls the electron with that momentum.
The virtual electron must get rid of its extra momentum p before annihilating with the virtual positron. The virtual electron can (with some probability amplitude) accomplish that by pushing the real electron which is in our Feynman diagram above.
We assume that the virtual electron has certain energy E and momentum q. We describe it with a plane wave that has those characteristics. We can choose the phase of that plane wave as we like.
Now it is clear that there is no destructive interference whatsoever associated with our virtual electron (E, q). The scattered virtual electron wave is simply (E, q + p).
Why no destructive interference? Because we treat the virtual electron like an input particle into a Feynman diagram. The virtual electron is not created by the virtual photon p. Rather, the photon p "hitches a ride" on a pre-existing electron.
Thus, our assumption that we can describe the virtual electron (E, q) with a fixed plane wave removes any possibility of significant destructive interference.
It is like our experiment would happen inside a medium which contains an infinite number of virtual electrons zigzagging with any value of E and q.
Can we give an alternative description where the photon p creates the virtual pair? What would be the phase of the virtual electron then?
What about interference between virtual electrons with various values of (E, q)?
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