On the Ward Identity in QED I am reading P&S, particularly Chapter 5.5. The authors are trying to derive an expression for the Ward identity (not formally, but still). They claim that the amplitude describing a photon emission (stripped off from the polarization vector of the respective photon) in any QED process can be written as
$$\mathcal{M}^{\mu}(k)=\int d^4x e^{ik\cdot x}\langle f|j^{\mu}(x)|i\rangle$$
and then they dot product $\mathcal{M}^{\mu}(k)$ with the photon momentum $k_{\mu}$ to demonstrate that the Ward identity holds,
$$k_{\mu}\mathcal{M}^{\mu}(k)=0$$
exploitting the fact that the current $j^{\mu}(x)$ is conserved (provided that classical equations of motion hold at the quantum level). My questions are the following:

*

*Is $\mathcal{M}^{\mu}(k)$ a Feynman amplitude (i.e. the scattering amplitude with the momentum conserving $\delta$ function being stripped off) or a scattering amplitude? (My guess is that $\mathcal{M}^{\mu}(k)$ is a scattering amplitude! )


*In deriving the Ward identity, we basically had to substitute $k$ with $-i\partial$ and then to exploit the fact that the four-divergence of the expression $e^{ik\cdot x} \langle f|j^{\mu}(x)|i\rangle$ vanishes inside the spacetime integral (otherwise we can not reduce $k_{\mu}\mathcal{M}^{\mu}(k)$ to a single integral $\int d^4x e^{ik\cdot x}\langle f|\partial_{\mu}j^{\mu}(x)|i\rangle$, that vanishes due to the divergence of the current being zero. Does this imply that the spinor fields vanish at the asymptotic regions in space? Or does the vanishing of the four-divergence occur because of something entirely different?
Any help will be appreciated.
 A: *

*Using translation operators the position space amplitude can be written $$\mathcal{M}^\mu(x)\equiv \langle f|j^{\mu}(x)|i\rangle=\langle f|j^{\mu}(0)|i\rangle e^{i(\sum p_f-p_i)x}$$
After Fourier transforming the exponential you indeed get a delta function ensuring $k$ is compatible with momentum conservation. This is not a problem because the delta function is not setting $k=0$, it is setting $k$ equal to the difference between the initial and final on-shell momentum.


*The vanishing of the four-divergence is a manifestation of charge conservation. Classically $\partial_\mu j^\mu = 0$ too. This is the whole point of the Ward identity, it is the analogue of classical conservation equations but for quantum mechanical correlation functions.

Here is the Fourier transform in (2) in more detail
$$G(k) \equiv k_\mu \langle f|j^{\mu}(k)|i\rangle$$
Fourier transform $G(k)$,
$$G(x)= \int \frac{d^dk}{(2\pi)^d} e^{-ikx} k_\mu \langle f|j^{\mu}(k)|i\rangle$$
$$=i\partial_\mu\int \frac{d^dk}{(2\pi)^d} e^{-ikx} \langle f|j^{\mu}(k)|i\rangle=i\partial_\mu \langle f|j^{\mu}(x)|i\rangle$$
Now we say that the final equality vanishes since $\partial_\mu j^\mu=0$ as a classical conservation law (this is the non-trivial part). So $G(x)=0$ and thus its Fourier transform $G(k)=0$.
