The Ward-Takahashi identity in Peskin and Schroeder (page 311)

I'm working on the Ward-Takahashi identity in Peskin (page 311), but I canʻt obtain Eq.(9.105) from Eq.(9.103)

According to Eq.(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} Consider \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} We have \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 with Eq.(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} The right-hand side is consistent, but how does the left-hand side derive?

• Could you state more specifically what is not clear? Are you worried about sign conventions? Oct 17, 2020 at 10:04
• I just want to know, how do we obtain $-i k_{\mu} \mathcal{M}^{\mu}(k ; p ; q)$ form $i \partial_{\mu}\left\langle 0\left|T j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right| 0\right\rangle$ ? @Qmechanic
– sky
Oct 17, 2020 at 21:24

Your equation means that $$$$i k^\mu \langle 0| {\bf T} \{ \widetilde{j}^\mu(k) \widetilde{\psi}(q) \bar{\widetilde{\psi}}(k) \} |0\rangle = -ie \left(\langle 0| {\bf T} \{ \widetilde{\psi}(q-k) \bar{\widetilde{\psi}}(p) \} |0\rangle + \langle 0| {\bf T} \{ \widetilde{\psi}(q) \bar{\widetilde{\psi}}(p+k) \} |0\rangle \right)$$$$ You can extract an amplitude from this by using the LSZ formalism. Note that $$j^\mu = e \bar{\psi}\gamma^\mu\psi$$ and this corresponds to a vertex coupling to an electron-positron pair, giving an external photon. This leads directly to $$$$-i k^\mu \mathcal{M}_\mu (k;p,q) = -ie \mathcal{M}_0(p,q-k) + ie \mathcal{M}_0(p+k,q)$$$$ where $$\mathcal{M}_0$$ is the process $$\mathcal{M}$$ without the external photon.

• From your homepage I found your note about Peskin, in page 374 Eq.(6.5.33), you said $$\int d^{4} x e^{-i k \cdot x} i \partial_{\mu}\left\langle 0\left|\mathbf{T}\left\{j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right\}\right| 0\right\rangle=-\int d^{4} x \partial_{\mu}\left(e^{-i k \cdot x}\right)\left\langle 0\left|\mathbf{T}\left\{j^{\mu}(x) \psi\left(x_{1}\right) \bar{\psi}\left(x_{2}\right)\right\}\right| 0\right\rangle$$, Why is that?
– sky
Oct 5, 2020 at 19:05
• partial integration Oct 7, 2020 at 8:17
• If it is partial integration, the right hand side seems to be missing an i ?@Oбжорoв
– sky
Oct 8, 2020 at 19:12
• I must have dropped an $i$ somewhere. If the answer is correct, can you accept it? Oct 9, 2020 at 7:59
• If you could write the derivation in more detail, I would accept it
– sky
Oct 9, 2020 at 8:20