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 \begin{equation} \psi(x) \to(1+ie\alpha(x))\psi(x). \tag{9.100} \end{equation} The QED Lagrangian density transforms to \begin{equation} \mathscr{L}\to \mathscr{L}-e\partial_\mu\alpha\overline{\psi}\gamma^\mu\psi. \tag{9.101} \end{equation} 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 \begin{equation} \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} \end{equation} Then the above amplitudes converted as \begin{equation} -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} \end{equation} 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.

  • 1
    $\begingroup$ Do you understand the derivation of the Schwinger-Dyson equations? $\endgroup$
    – Qmechanic
    Aug 9, 2020 at 11:17
  • $\begingroup$ I read It from Peskin and Shroeder , I understand it with some scratches somewhere, I will come with doubt on that later@Qmechanic $\endgroup$
    Aug 10, 2020 at 19:04

1 Answer 1

  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?

  • $\begingroup$ 1)This change in lagrangian does not affect the generating functional and hence correlation functions , this happens why? @Mauro Giliberti $\endgroup$
    Aug 9, 2020 at 17:45
  • $\begingroup$ @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. $\endgroup$ Aug 9, 2020 at 18:16
  • $\begingroup$ Does this true for non-infinitisimal transformations?@ Mauro Giliberti $\endgroup$
    Aug 9, 2020 at 18:40
  • $\begingroup$ Why the author takes non-interacting vacuum $(|0⟩)$ instead of interacting vacuum$(|\Omega⟩)$@Mauro Giliberti $\endgroup$
    Aug 9, 2020 at 18:44
  • $\begingroup$ @ROBINRAJ it is true for every gauge transformation, it is the definition of gauge transformation itself. It talks about non-interacting vacuum states because that's how the generating function is defined in terms of the correlation function. Also, comments aren't for extended discussion: if you have more questions you should write them as separate posts. $\endgroup$ Aug 9, 2020 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.