Expansion of the solution if interacting KG equation

  • Context: Graduate 
  • Thread starter Thread starter ShayanJ
  • Start date Start date
  • Tags Tags
    Expansion
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 replies · 967 views
Messages
2,802
Reaction score
605
The exponentials ##\phi_p(x)=e^{ipx} ##(where ## px=p_\mu x^\mu##), are solutions of the free KG equation ## (\partial_\mu \partial^\mu+m^2) \phi =0##. I can expand the solutions of the interacting KG equation ## (\partial_\mu \partial^\mu+m^2)\psi=V\psi ## in terms of solutions of the free KG equation ## \psi(x)=\int e^{ipx} \pi(p) d^4p ##.
But because there is an interaction, it seems to me that the evolution of ## \psi(x) ## should be different from a free field and so I think the function ##\pi## should also be a function of time. But because we want to think covariantly and also because it seems a natural generalization to let the field have different expansion coefficients in different events of space-time, I think we should let ## \pi ## depend on all four coordinates, i.e. we should write ## \psi(x)=\int e^{ipx} \pi(p,x) d^4p ##
Is this the right way of thinking about it?
Thanks
 
Physics news on Phys.org
It seems an advanced level question requiring tensors.
Please make the question simple if possible!