Dirac-delta term in a non-relativistic interaction

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SUMMARY

The discussion centers on the emergence of a Dirac-delta term in the context of electron-electron scattering, specifically when analyzing the invariant transition amplitude derived from the Dirac equation. The participants clarify that in the nonrelativistic limit, the Fourier transform of the potential leads to a delta-function potential, which is a constant term. The discussion references the "Born Approximation" and suggests consulting Peskin and Schroeder for detailed explanations of the limits involved in this derivation. The final calculations confirm the presence of a constant term that corresponds to the delta-function potential.

PREREQUISITES
  • Understanding of the Dirac equation and its implications in quantum electrodynamics (QED).
  • Familiarity with invariant transition amplitudes and their calculations.
  • Knowledge of the Born Approximation in quantum scattering theory.
  • Proficiency in Fourier transforms and their applications in quantum mechanics.
NEXT STEPS
  • Study the derivation of the invariant transition amplitude from the Dirac equation.
  • Learn about the Born Approximation and its significance in scattering theory.
  • Review Peskin and Schroeder's textbook for detailed explanations of nonrelativistic limits in QED.
  • Explore the implications of delta-function potentials in quantum mechanics and their physical interpretations.
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Physicists, particularly those specializing in quantum mechanics and quantum field theory, as well as graduate students studying quantum electrodynamics and scattering processes.

Montejo
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Hi everyone,
I'm studying electron-electron scattering, starting with the Dirac equation it ends up calculating the invariant transition amplitude, defined as:
[tex]-iM=(ie{\overline{u}^f}_A}\gamma^\mu u^i}_A) \frac{-ig_{\mu\nu}}{q^2}(ie{\overline{u}^f}_B}\gamma^\nu u^i}_B)[/tex]

With [tex]u_A[/tex] and [tex]u_B[/tex] the electron spinors (initial and final)

After this it says that in the nonrelativistic limit a Dirac-delta spin-spin term arises in the corresponding potential. How is that?

Could anyone explain where does this dirac-delta come from? (And btw a better book to study QED)

Thanks
 
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this tree-level M corresponds to the Fourier-transform of the potential in the nonrelativistic limit (aka "Born Approximation"). So you have to take this limit, both in expanding q^2 and in the spinors. See Peskin-Schroder to see how these limits are taken, for example.

In the end, you will get (among other things) a constant term, and that corresponds to a delta-function potential (the Fourier transform of a delta function is a constant).

Hope that helps!
 
Thank you very much, I think I got the point. However, I still have a further question. Here's what I've done:

[tex] -iM=(ie{\overline{u}^f}_A}\gamma^\mu u^i}_A) \frac{-ig_{\mu\nu}}{q^2}(ie{\overline{u}^f}_B}\gamma^\nu u^i}_B)=[/tex]
[tex] = \frac{1}{q^2} \left( 1 + \frac{{p^f}_A {p^i}_A + i \sigma ({p^f}_A \times {p^i}_A) }{4m^2} + \frac{{p^f}_B {p^i}_B + i \sigma ({p^f}_B \times {p^i}_B) }{4m^2} + \O (\frac{p}{m})^4 \right)[/tex]

Now, working in CM [tex]{p^i}_A= p, {p^i}_B= -p, {p^f}_A= p', {p^f}_B=-p', q=p'-p[/tex]

[tex] = \frac{1}{q^2} \left( 1 - q^2 + {p'}^2+{p}^2+ \frac{ i ({\sigma}_A+{\sigma}_B) (p' \times p) }{4m^2} + \O (\frac{p}{m})^4 \right) =[/tex]
[tex] = -1 +\frac{1}{q^2} \left( 1 + {p'}^2+{p}^2+ \frac{ i ({\sigma}_A+{\sigma}_B) (p' \times p) }{4m^2} + \O (\frac{p}{m})^4 \right) =[/tex]

Ok, I've got a constant term which will yield a delta-function, but it has no spin-spin dependence, have I done something wrong?

Thanks
 

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