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Dirac-delta term in a non-relativistic interaction

  1. Feb 16, 2010 #1
    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
     
  2. jcsd
  3. Feb 19, 2010 #2

    blechman

    User Avatar
    Science Advisor

    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!
     
  4. Feb 20, 2010 #3
    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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