La identidad Ward-Takahashi en Peskin y Schroeder (página 311)

Estoy trabajando en la identidad Ward-Takahashi en Peskin (página 311), pero no puedo obtener la Ec. (9.105) de la Ec. (9.103)

Según la Ec.(9.103)

$$\begin{align} &i \partial_{\mu}\left\langle 0\left|T j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle=-i e \delta\left(x-x_{1}\right)\left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle +i e \delta\left(x-x_{2}\right)\left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle \\ \Rightarrow&\int d^{4} x e^{-i k \cdot x} \int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}}i \partial_{\mu}\left\langle 0\left|T j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle\\ &=-\int d^{4} x e^{-i k \cdot x} \int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}}i e \delta\left(x-x_{1}\right)\left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle\\&+\int d^{4} x e^{-i k \cdot x} \int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}}i e \delta\left(x-x_{2}\right)\left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle\\ \Rightarrow&k_{\mu}\int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}}\left\langle 0\left|T j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle\\ &=-ie\int d^{4} x_{1} e^{+i (q-k) \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}} \left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle+ie\int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i (p+k) \cdot x_{2}}\left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle \end{align}$$ Considerar $$\begin{align} \int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}} \left\langle 0\left|T \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle = \mathcal{M}(0 ; p,q) \end{align}$$ Tenemos $$\begin{align} k_{\mu}\int d^{4} x_{1} e^{+i q \cdot x_{1}} \int d^{4} x_{2} e^{-i p \cdot x_{2}}\left\langle 0\left|T j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle=-i e \mathcal{M} (0 ; p , q-k)+i e \mathcal{M}(0 ; p+k , q) \end{align}$$ Compare con la ecuación (9.105) $$\begin{align} -i k_{\mu} \mathcal{M}^{\mu}(k ; p ; q)=-i e \mathcal{M}_{0}(p ; q-k)+i e \mathcal{M}_{0}(p+k ; q)\tag{9.105} \end{align}$$ El lado derecho es consistente, pero ¿cómo se deriva el lado izquierdo?