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Peskin and Schroeder - page 4 - spin and cross section

  1. Jan 21, 2016 #1
    In chapter 1 of Peskin and Schroeder, the reaction ##e^{+}e^{-}\rightarrow \mu^{+}\mu^{-}## is studied in detail. The related following paragraph is taken from page 4 of Peskin and Schroeder:

    Both the electron and the muon have spin ##1/2##, so their spin orientations must be specified. It is useful to take the axis that defines the spin quantization of each particle to be in the direction of its motion - each particle can then have its spin polarized parallel or antiparallel to this axis. In practice, electron and positron beams are often unpolarized, and muon detectors are normally blind to the muon polarization. Hence one should average the cross section over electron and positron spin orientations, and sum the cross section over muon spin orientations.

    I have the following questions regarding the content of the paragraph:

    1. Why is it useful to have the spin of each particle polarized parallel or antiparallel to the direction of its motion?
    2. Why does the fact that electron and positron beams are often unpolarized in practice imply that one should average the cross section over electron and positron spin orientations?
    3. Why does the fact that muon detectors are normally blind to the muon polarization imply that one sum the cross section over muon spin orientations?


     
  2. jcsd
  3. Jan 21, 2016 #2

    mfb

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    It makes the calculation of the scattering process easier.
    What else would you do? Sometimes the collision happens with one set of polarizations, sometimes with a different - all you can do is average over them.
    What else would you do? If you get green and red apples, but cannot measure the color, just the number of apples, you have to sum over the two.
     
  4. Jan 21, 2016 #3
    In what ways is the calculation made easier?

    Would it be possible for you to provide details of the simplifications (associated with axis of spin quantisation taken parallel to direction of motion) versus complications (associated with axis of spin quantisation not taken parallel to direction of motion)?
     
  5. Jan 21, 2016 #4
    The following is taken from page 4 of Peskin and Schroeder and is valid for the annihilation reaction ##e^{+}e^{-}\rightarrow \mu^{+}\mu^{-}##:

    For a given set of spin orientations, with ##\mu^{-}## produced into a solid angle ##d\Omega##, the differential cross section is

    ##\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^{2}E^{2}_{\text{cm}}}\ \cdot{\lvert\mathcal{M}\lvert^{2}}.##


    My questions are as follows:
    1. Why does the differential cross section ##\frac{d\sigma}{d\Omega}## change with the given set of spin orientations?
    2. Does ##E_{\text{cm}}## change with the given set of spin orientations?
    3. Does ##\mathcal{M}## change with the given set of spin orientations?
    4. Why is the production of ##\mu^{-}##, and not ##\mu^{+}##, into a solid angle ##d\Omega## considered in the differential cross section ##\frac{d\sigma}{d\Omega}##?
    5. Why is ##\mathcal{M}## called the quantum-mechanical, and not the quantum-field-theoretic, amplitude for the annihilation reaction to occur?
     
  6. Jan 21, 2016 #5

    mfb

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    Look at the actual calculation, try to see how it would look like for other spin directions.
    I'm not a textbook or QFT course.

    Concerning your new questions (I merged the thread): What do you think? Some of them should be easy to answer, as they do not require any QFT knowledge.
     
  7. Jan 22, 2016 #6
    The spin orientations of the incoming electron beam and the incoming positron beam affect the numbers of ##\mu^{+}## and ##\mu^{-}## produced into each solid angle ##d\Omega##, hence the dependence of the differential cross section ##\frac{d\sigma}{d\Omega}## with the given set of spin orientations.

    No, ##E_{\text{cm}}## is solely the kinetic energy of the system in the centre-of-mass frame. The contribution to the potential energy of the system due to the given set of spin orientations is negligible.

    Yes.

    We could equally well have considered the production of ##\mu^{+}## into the solid angle ##d\Omega##. The momenta of the ##\mu^{-}## and the ##\mu^{+}## are equal and opposite, so equal numbers of ##\mu^{-}## and ##\mu^{+}## are produced into diametrically opposite solid angles ##d\Omega##.

    The general framework (Heisenberg and Schrodinger pictures, perturbation theory, Hermitian operators, etc.) of quantum mechanics is applied to formulate quantum field theory. Indeed, whilst the basic course in quantum mechanics studies the quantisation of single particle systems, quantum field theory studies the quantisation of relativistic fields.
     
  8. Jan 22, 2016 #7
    What do you think?
     
  9. Jan 22, 2016 #8

    mfb

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    Right.

    For "quantum mechanical": just historic reasons. Naming is often arbitrary.
     
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