Energy conservation in e-e+ annihilation

[maverick280857]

Hi,

I'm trying to convince myself (mathematically) that energy is conserved at the vertex in $e^-e^+$ 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 $T_{fi}$ 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 $\omega$ ($\hbar = 1$).

From what I understand, the transition amplitude is proportional to

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

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

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

The rule is to form the matrix element,

$$\int d^{4}x\,\phi^{*}_{outgoing}V\phi_{ingoing}$$

where ingoing and outgoing always refer to the arrows on the particle (electron) lines.

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

Thanks in advance.

 

[maverick280857]

Anyone?

 

[Meir Achuz]

The signs in your exponents are wrong. It should be
$$e^{i\omega}e^{-i (E+E')}$$, corresponding to E+E' in the initial state and omega in the final state.

 

[maverick280857]

The signs in your exponents are wrong. It should be
$$e^{i\omega}e^{-i (E+E')}$$, corresponding to E+E' in the initial state and omega in the final state.

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.