Short question: Spontaneous emission from Feynman diagrams?

[blue2script]

Hey all,
I am just wondering if one can directly calculate the Einstein coefficient in spontaneous emission of, say, two-level atoms through feynman diagrams? I searched for sources in google but could not find anything.

Thanks a lot for an answer!

Wit best regards,
blue2script

 

[blue2script]

Maybe I can render the question more precisely: Given a proton in an outer magnetic field I get two energy levels depending on the direction of spin. I can plug the outer field into the Lagrangian. After determing the feynman rules I should be prepared to calculate the decay rate of a proton in the upper energy state.

Am I right?

 

[Entropia]

Hello blue2script,

Thank you for your question. The short answer is no, Feynman diagrams cannot be used to directly calculate the Einstein coefficient in spontaneous emission. Feynman diagrams are a graphical representation of the mathematical calculations in quantum field theory, and they are primarily used to calculate scattering amplitudes and cross-sections.

The Einstein coefficient in spontaneous emission is related to the probability of an atom transitioning from an excited state to a lower energy state and emitting a photon. This process is described by quantum electrodynamics (QED), which is a different theory than quantum field theory.

However, there have been attempts to use Feynman diagrams in a modified form to calculate the Einstein coefficient. One approach is to use the Dyson-Schwinger equations, which are a set of equations that relate the Green's functions in QED to the Feynman diagrams. These equations can be solved to obtain the Einstein coefficient, but the calculations are still quite complicated.

In summary, while Feynman diagrams are a powerful tool in quantum field theory, they cannot be directly used to calculate the Einstein coefficient in spontaneous emission. Other methods, such as the Dyson-Schwinger equations, may be used, but they are still complex and challenging calculations. I hope this helps answer your question.

Best regards,