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Matrix valued propagators

  1. Apr 9, 2009 #1
    For scalar fields the propagator is just a number that represents the amplitude for a particle to go from one space time point to another.

    For fermions, the propagator is matrix valued. What then is the amplitude for a fermion to go from one point to another? How are the elements of the matrix to be interpreted in terms of probability ampltudes?
     
  2. jcsd
  3. Apr 9, 2009 #2

    malawi_glenn

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    row-vector times matrix times column vector = number.
     
  4. Apr 9, 2009 #3
    The fermion propagator involves the dot product of a 4 vector (momentum) and the gamma matrices. Im pretty sure the result is a matrix.
     
  5. Apr 9, 2009 #4

    malawi_glenn

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    No, since the states are vectors... (4-spinors)

    Have you never worked with relativistic quantum identities? What part of my first post did you not understand?
     
  6. Apr 9, 2009 #5
    The signature :uhh:
     
  7. Apr 9, 2009 #6

    malawi_glenn

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    shall I change it? :uhh:

    But seriously, what is wrong with my explanation?
     
  8. Apr 9, 2009 #7

    blechman

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    The fermion propogator IS a matrix, Bobhawke is correct. It is an outer product of spinors, not an inner product.

    I think the answer he is looking for is [itex]S_{\alpha\beta}\equiv\langle T\psi_\alpha\bar{\psi}_\beta\rangle[/itex] is the amplitude for a fermion of polarization [itex]\alpha[/itex] to propogate to a fermion of polarization [itex]\beta[/itex].

    Up to signs and whatnot!

    Malawi_glenn: don't change your signature for anyone! :blushing:
     
  9. Apr 9, 2009 #8
    I think we agree, there is not much more to this equation than what malawi_glenn was writing in "row-vector times matrix times column vector". It is also written in "the states are vectors... (4-spinors)".
     
  10. Apr 9, 2009 #9

    malawi_glenn

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    yes, the propagator is matrix, but the amplitude is a number. And that was the question, if the propagator is a matrix, what will happen to the amplitude.
     
  11. Apr 9, 2009 #10

    blechman

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    I think that is the question I answered. The matrix elements of the propagator are the amplitudes for the polarization states to propagate.
     
  12. Apr 9, 2009 #11
    Thanks for the replies everyone!
     
  13. Apr 11, 2009 #12

    Hans de Vries

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    Actually, the propagator is generally the propagation from a source.

    So in QED the photon is propagated from the transition current (the interference
    pattern caused by an electron in two momentum states) and the electron is
    propagated from the interference term [itex]e\gamma^\mu\,A_\mu\psi[/itex]


    It is indeed the amplitude per polarization state but there are some interesting
    details about the interplay between SU(2) and U(1). For instance in a magnetic
    field B the energy will be different per polarization state:


    [tex]\binom{~~\exp(-i[E+\Delta E]t)~~}{~~\exp(-i[E-\Delta E]t)~~} ~~=~~ \binom{~~\exp(-i\Delta Et)~~}{~~\exp(+i\Delta Et)~~}~\exp(-iEt})[/tex]

    At the RHS the energy is the same for both states but the spinor represents
    a precessing spinor around the direction of the magnetic field. Note that:

    [tex] x\uparrow=\binom{1}{1}, ~~y\uparrow=\binom{1}{i},~~x\downarrow=\binom{1}{-1}, ~~y\downarrow=\binom{1}{-i}[/tex]

    (up to an overal factor of [itex]1/\sqrt{2}[/itex])


    Regards, Hans.
     
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