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Wick theorem for particle-antiparticle annihilation

  1. May 6, 2013 #1
    I am trying to write down the LSZ formula for e- e+ -> 2 gamma:

    $$ S_{fi} \propto \langle 0| T \{\bar{\Psi}(x_1) \vec{A}(x_2) \Psi (x_4) \vec{A}(x_3) \left\lbrack \frac{(-i)^2}{2} \int d^4x A_\mu (x) \bar{\Psi}(x) \gamma^\mu \Psi (x) \int d^4y A_\nu (y) \bar{\Psi}(y) \gamma^\nu \Psi (y) \right\rbrack \} |0\rangle $$

    So I have electron at ##x_1## and positron ##x_4##.

    But now if i try to contract I can contract ##\bar{\Psi}(x_1)## with one at x or at y, and same with ##\vec{A}(x_2)## and this would produce a factor of 4 which would cancel with 1/2 from exponential expantion giving a factor 2, but this is wrong, there should not be a factor of 2.

    Any ideas? thanks.
     
  2. jcsd
  3. May 7, 2013 #2
    Sfi for two photon annihilation is probably
    (-e)2∫d4x1t2<t1d4x2<2γ|ψ-(x1μψ(x1)Aμψ-(x2vψ(x2)Av(x2)|e-e+>,then you will have to introduce the decompositon of fermionic field into two parts and also use the creation operator on vacuum states to obtain the |e-e+> and |2γ> states.
     
  4. May 7, 2013 #3
    No, i used LSZ, but i got it, the factor of 2 i am getting is correct. Thanks.
     
  5. May 8, 2013 #4
    In any case,the final state represents two photons and initial one is |e-e+>,while your is only having sandwiched between vacuum states.so try to interpret it.
     
  6. May 8, 2013 #5
    The leptons and photons are generated by the first four operators. I just used LSZ reduction formula.
     
  7. May 8, 2013 #6
    you might like to give a reference for spinor electrodynamics lsz reduction. the term in square bracket contains the full interaction term of qed which is somewhat uneasy to me.
     
  8. May 23, 2013 #7
    Greiner "Field quantization", chapter on LSZ for fermions.
     
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