# Proof of Ward-Takahashi Identity in Peskin and Schroeder page 311

I am studying the derivation of Ward Takahashi identity using Peskin and Schroeder (Page number 311) What I understand from his statements is as follows, for a change of variables $$$$\psi(x) \to(1+ie\alpha(x))\psi(x). \tag{9.100}$$$$ The QED Lagrangian density transforms to $$$$\mathscr{L}\to \mathscr{L}-e\partial_\mu\alpha\overline{\psi}\gamma^\mu\psi. \tag{9.101}$$$$ I agreed with his statements till here. Then he says

"This transformation leads to the following identity for the functional integral over two fermion fields \begin{align} 0=\int\mathscr{D}\overline{\psi}\mathscr{D}\psi\mathscr{D}Ae^{i\int d^4x\mathscr{L}}\Bigg\{-i\int d^4x\partial_\mu\alpha(x)\Bigg[j^\mu(x)\psi(x_1)\overline{\psi}(x_2)\Bigg]\\+\bigg(ie\alpha(x_1)\psi(x_1)\bigg)\overline{\psi}(x_2)+\psi(x_1)\big(-ie\alpha(x_2)\overline{\psi}(x_2)\big)\Bigg\} \end{align} \tag{9.102} with $$j^\mu=e\overline{\psi}\gamma^\mu\psi$$. [...] Dividing this equation by $$Z$$ gives \begin{align} i\partial_\mu⟨0|Tj^\mu(x)\psi(x_1)\overline{\psi}(x_2)|0⟩=-ie\delta (x-x_1)⟩⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩ +ie\delta (x-x_2)⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩.\end{align} \tag{9.103} To put this equation into a more familiar form, compute its Fourier transform by integrating $$$$\int d^4xe^{-ik\cdot x}\int d^4x_1e^{iq\cdot x_1}\int d^4x_1e^{-ip\cdot x_2}. \tag{9.104}$$$$ Then the above amplitudes converted as $$$$-ik_\mu\mathscr{M}^\mu(k;p;q)=-ie\mathscr{M}_0(p;q-k)+ie\mathscr{M}_0(p+k;q). \tag{9.105}$$$$ This is exactly the Ward-Takahashi identity for two external fermions."

My questions are

1. How did he get \begin{align} 0=\int\mathscr{D}\overline{\psi}\mathscr{D}\psi\mathscr{D}Ae^{i\int d^4x\mathscr{L}}\Bigg\{-i\int d^4x\partial_\mu\alpha(x)\Bigg[j^\mu(x)\psi(x_1)\overline{\psi}(x_2)\Bigg]\\+\bigg(ie\alpha(x_1)\psi(x_1)\bigg)\overline{\psi}(x_2)+\psi(x_1)\big(-ie\alpha(x_2)\overline{\psi}(x_2)\big)\Bigg\}~? \end{align} \tag{9.102}

2. On dividing this equation by $$Z$$ gives \begin{align} i\partial_\mu⟨0|Tj^\mu(x)\psi(x_1)\overline{\psi}(x_2)|0⟩=-ie\delta (x-x_1)⟩⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩ +ie\delta (x-x_2)⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩,\end{align}\tag{9.103} how did he get this?

3. In taking Fourier transform, why does he take $$(-k,q,-p)$$ as the set of momentum in exponential instead of $$(-k,-q,-p)$$ ? Please help me to get an intuition.

• Do you understand the derivation of the Schwinger-Dyson equations? Aug 9, 2020 at 11:17
• I read It from Peskin and Shroeder , I understand it with some scratches somewhere, I will come with doubt on that later@Qmechanic Aug 10, 2020 at 19:04

1. You neet to take the standard functional integral $$\int \mathcal{D}[\psi, \bar{\psi}, A] e^{i\int d^4x\mathcal{L}[\psi, \bar{\psi}, A]} (\psi \bar{\psi})$$ and expand both the fields and the Lagrangian. Using the fact that the functional measure $$\mathcal{D}$$ and the integral as a whole are invariant, you'll have a term that is the same as the non-expanded one (hence the $$0=...$$) and other terms that are the ones that you can see in the book.
2. The definition of the correlation function in terms of the functional integral can be given as the integral of the fields (that is, the derivative of $$Z[J]$$ valued at $$J=0$$), divided by $$Z[0]$$. Check the previous pages of the book for examples of this.
3. I'm not sure what you mean. Why should he have chosen $$-q$$ instead?
• @ROBINRAJ because both $\psi$ and its transform $\psi '$ are solutions of the same equation of motion, and account to the same generating functional. $\int \mathcal{D}[\psi, \bar{\psi}, A] e^{i\int d^4x\mathcal{L}[\psi, \bar{\psi}, A]} (\psi \bar{\psi})=\int \mathcal{D}[\psi' , \bar{\psi'}, A] e^{i\int d^4x\mathcal{L}[\psi', \bar{\psi'}, A]} (\psi' \bar{\psi'})$ and using the fact that the functional measure doesn't change, you get the above equation. Aug 9, 2020 at 18:16
• Why the author takes non-interacting vacuum $(|0⟩)$ instead of interacting vacuum$(|\Omega⟩)$@Mauro Giliberti Aug 9, 2020 at 18:44