0
$\begingroup$

For a given Feynman diagram one can calculate the matrix Element by translating the diagram into math using Feynman rules.

In these calculations one will encounter incoming and outgoing particles (and antiparticles) which are represented by spinors. These spinors are solutions of the Dirac equation.

If the things I said aren't wrong I would like to know whether these spinors are always (e.g. for Bhabha scattering) solutions of the free Dirac equation or not.

$\endgroup$

1 Answer 1

0
$\begingroup$

Feynman diagrams are a pictorial representation of a mathematical integral in Quantum Field Theory (QFT). QFTs by consruction have creation and annihilation operators operating on fields. The fields are plane wave solutions of a quantum mechanical equation, and are like a coordinate syste, occupying all space time. On the fields creation and annihilation operators operate generating or annihilating a particle represented by the field.

In particle physics, for all the particles in the axiomatic particle table of the standard model, there exists a field, an electron field , a neutrino field, etc extending through all space time, with the plane wave solution for the corresponding equation. For the spinor particles it is the plane wave solutions of the corresponding Dirac equation that represent the field, on which the creation and annihilation operators operate, for all calculations of a Feynman diagram.

Feynman diagrams are a way of using the plane wave solutions of spinors, in this case, to represent a term in the perturbative expansion of the full solution of a scattering of two spinors. If one could calculate the total solution it certainly would not be a plane wave solution of the Dirac equation, since there is an interaction.

The incoming, before the interaction, and outgoing, after the interaction, can be represented by operators operating on plane wave solutions, but not the interaction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.