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Energy conservation in e-e+ annihilation

  1. Mar 12, 2010 #1
    Hi,

    I'm trying to convince myself (mathematically) that energy is conserved at the vertex in [itex]e^-e^+[/itex] annihilation, while solving Exercise 3.5(b) of Halzen and Martin's book (page 83).

    I am looking at the time dependent term in the scattering amplitude [itex]T_{fi}[/itex] alone, to recover the delta function term which asserts the energy conservation. I know that I can look at the process in forward time or backward time, with the roles of the e+ and e- reversed.

    Suppose the energies of the incoming electron and positron are E and E' respectively, whereas the energy of the outgoing photon is [itex]\omega[/itex] ([itex]\hbar = 1[/itex]).

    From what I understand, the transition amplitude is proportional to

    [tex]\int dt\,(e^{-i(-E')t})^{*}e^{-i\omega t}e^{-iEt} = 2\pi\delta(E+E'+\omega)[/tex]

    which gives the wrong answer of course, since the argument of the delta function should be [itex]E+E'-\omega[/itex].

    I believe I am making a mistake in interpreting their "rule":

    What does this mean? I'm a bit confused with the convention.

    Thanks in advance.
     
    Last edited: Mar 12, 2010
  2. jcsd
  3. Mar 12, 2010 #2
  4. Mar 15, 2010 #3

    clem

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    Science Advisor

    The signs in your exponents are wrong. It should be
    [tex]e^{i\omega}e^{-i (E+E')}[/tex], corresponding to E+E' in the initial state and omega in the final state.
     
  5. Mar 15, 2010 #4
    Thanks clem. I was earlier thinking that the photon contribution will be through the potential term which is sandwiched between the final and initial states, which (I thought) referred to particles alone.
     
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