I'm studying chapter 6 section 2 of Peskin and Schroeder, on page 185 the unnumbered equation between 6.29 and 6.30, we are trying to compute the $S$-matrix element of the scattering of an eletron under the Hamiltonian $$\Delta H_{int}=\int d^3x\,eA_{\mu}^{cl}j^\mu$$ where $j^\mu(x)=\overline{\psi}(x)\gamma^\mu\psi(x)$ is the EM current and $A_\mu^{cl}$ is a fixed classical potential. $\tilde{A}^{cl}_\mu$ is its Fourier transform.

$$i \mathcal{M}(2\pi)\delta (p^0{'}-p^0)=-ie\overline{u}(p')\gamma^\mu u(p)\cdot\tilde{A}_\mu^{cl}(p'-p).$$

I'm confused about the left hand side of the equation. I think the right hand side of the equation is simply $$\langle 0|a_{p'}(-i)\mathcal{T}(\int d^4x\, eA_\mu^{cl}j^\mu) a_p^\dagger|0\rangle.$$ This can be simplified using Wick's theorem. However, I can't get how we can extract the $(2\pi)\delta(p^0{'}-p^0)$ factor.

Another problem is that this doesn't seem to be the same as

$$\tag{4.73}\langle p_1p_2...|iT|k_A k_B\rangle=(2\pi)^4\delta^4(k_A+k_B-\sum p_f)\cdot i\mathcal{M}(k_A,k_B\rightarrow p_f).$$


1 Answer 1


i suggest you to read Mandl and Shaw quantum field theory; in the chapter "QED processes in lowest order" they explain that the presence of delta for energy conservation $\delta (E'-E)$ is due to the presence of the classical field $A^\mu_e$. In fact we are dealing with a diagram with three external lines in a vertex; this is a possible diagram because we are considering the momentum vector $\vec k$ as injected by the classical field $A^\mu_e$, ignoring what happens to the total momentum of the source of the field. What happens here is that the photon exchanged between the vertex and the external field is a virtual photon for which the Einstein's relation ($|\vec k|^2 c^2=\omega^2_k$) does not hold anymore. What is not quite clear to me is how to obtain exactly the $\delta (E'-E)$ instead of the usual $\delta (p'-p)$ using the last statement; asking for help too on the argument.


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