Electromagnetism – How to Get Photon Probability from Electromagnetic Potentials?

electromagnetismprobabilityquantum-electrodynamics

I have a question about probabillity of photon from electromagnetic fields.

We know that electromagnetic four-potential, which can be found with QED equations

$A_{\mu}=\bigl(\begin{smallmatrix}\frac{\phi}{c} & A_x & A_y & A_z\end{smallmatrix}\bigr)$

Such that

$ \vec{E}=-\nabla\: \phi-\frac{\partial \vec{A}}{\partial t}$

$ \vec{B}=\nabla × \:\vec{A}$

How do I get photon probabillity out of this potential?

Best Answer

In quantum field theory, observables (such as decay rates and scattering cross sections) are probabilistic and often characterized in terms of correlation functions. For example:

  • The two point function $\langle 0| A(x) A(y)|0 \rangle$ is related to the probability amplitude for a photon to travel from spacetime point $x$ to spacetime point $y$.
  • The three point function $\langle 0 |\bar\psi(x) A(y) \psi(z) | 0\rangle$ (where $\psi$ is the electron field) is related to the probability amplitude for an electron and positron to annihilate into a photon.

where $|0\rangle$ is the vacuum state.

There are important subtleties related to operator ordering, Lorentz transformations, and gauge invariance that I am suppressing in this answer for simplicity; if you want the full story you will need to work through a quantum field theory text. David Tong's lecture notes are a good resource https://www.damtp.cam.ac.uk/user/tong/qft.html.

The correlation functions are computed using the Hamiltonian of the system, typically using the Dyson series.

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