Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Average fermion current

  1. Jan 24, 2013 #1

    I have a little problem.

    Consider a 4-fermion interaction (neglecting constant factors) of the form [itex] \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} [/itex] .
    I want to average this interaction over a background consisting of fermions (so it corresponds to the situation where fermions propagate in a background consisting of fermions).

    To this purpose, the left-handed current [itex] \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} [/itex] is approximated by the average value [itex] \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \rangle [/itex]

    There is the following relation for the averaged value produced by this interaction:
    \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} \to & \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \rangle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} + \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} \rangle
    & \quad + \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{d \mathrm{L}} \rangle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}} + \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{d \mathrm{L}} \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}} \rangle
    & \quad - \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \rangle \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} \rangle - \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{d \mathrm{L}} \rangle \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}} \rangle.

    But I don't really understand how to obtain this relation. The first two terms look reasonable, but I don't understand the remaining ones.

    Ok, the 3rd and 4th terms might correspond to an "exchange term", where fermion c and d are interchanged, because in the case where c is equal to d, we cannot distinguish between the propagating and the background fermion (on the other hand, if c is not equal to d, this exchange term should be 0). But why is there no additional minus sign, because of Fermi-Dirac statistics?

    Maybe it has something to do with a Fierz transformation, where the minus sign cancels out, i.e.,
    [tex] \overline{\psi}_{a \mathrm{L}} \gamma^{\lambda} \psi_{b \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} = \overline{\psi}_{a \mathrm{L}} \gamma^{\lambda} \psi_{d \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}}.[/tex]

    And how could the last two terms be interpreted?

    I hope somebody could help me understanding the relation above.
  2. jcsd
  3. Jan 26, 2013 #2
    can you provide a reference for that averaging procedure.How is it defined.Also it is not Fierz reshuffling otherwise you will also get pseudoscalar and pseudovector part(apart from identity)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Average fermion current
  1. Fermions are massless (Replies: 8)

  2. Proton as a fermion (Replies: 7)

  3. Vectorlike fermion (Replies: 5)

  4. Fermions and Bosons (Replies: 34)