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The Optical Theorem (Total cross section?)

  1. May 10, 2015 #1
    I am going over the optical theorem (specifically from Sakurai) and I just have a simple couple of questions. The optical theorem says

    [tex]\sigma_{total}=\frac{4\pi}{k}\text{Im}(f(0))[/tex]

    Where ##f(0)## is the scattering amplitude in the forward direction. (Which I am assuming means the direction parallel to the incident direction. Or does it mean in the direction back towards the source?)

    I just want to verify what is meant by the total cross section. To me the total cross section means the amount the incoming wave function is scattered integrated in all directions over a sphere around the scatterer. Is this the correct way of thinking?
     
  2. jcsd
  3. May 11, 2015 #2

    blue_leaf77

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    The parallel one, not the antiparallel.
    Differential cross-section is defined as the fraction of the scattered probability current at certain direction, fraction means relative to the incoming probability current. The total cross section is the sum of differential cross section over all directions.
     
  4. May 14, 2015 #3
    Thank you! (Sorry for the late reply!)
     
  5. May 15, 2015 #4

    vanhees71

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    It's worth mentioning that on the left-hand side stands the total cross section, including both elastic and inelastic channels, while on the right-hand side there's the elastic scattering amplitude.

    The optical theorem can be generalized to more general scattering processes than just single-particle scattering on a potential. It reflects the very fundamental principles of S-matrix theory, causality and unitarity and is closely linked to detailed balance in kinetic theory.
     
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