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This question is based on section $1.2$ of Gauge Theory of Elementary Particle Physics by Ta-Pei Cheng and Ling-Fong Li.

In $\phi^4$ or $-\frac{\lambda}{4!}\phi^4$ theory let $W[J]$ be the vacuum-to-vacuum transition amplitude in the presence of an external source $J(x)$ or the generating functional.

If $\lambda=0$ (i.e. free field) then let the generating functional be denoted by $W_{0}[J]$ and given by $$W_{0}[J]=\exp \left[\frac{1}{2} \int \mathrm{d}^{4} x \mathrm{~d}^{4} y J(x) \Delta(x, y) J(y)\right].$$ For $\lambda\neq 0$ it is $$ W[J]=\left[exp\int d^4x \left(-\frac{\lambda}{4!}\left(\frac{\delta }{\delta J}\right)^4\right)\right]W_{0}[J],$$ and we can expand it as $$W[J]=W_{0}[J] (1+\lambda \omega_{1}[J]+\lambda^{2} \omega_{2}[J]+\ldots.$$ Now Cheng and Li write the following equations from above: $$\omega_{1}[J]=-\frac{1}{4 !} W_{0}^{-1}[J]\left\{\left[\mathrm{d}^{4} x\left[\frac{\delta}{\delta J(x)}\right]^{4}\right\} W_{0}[J]\right.\tag{1.86}$$

$$\omega_{1}[J]=-\frac{1}{4 !}\left[\Delta\left(x, y_{1}\right) \Delta\left(x, y_{2}\right) \Delta\left(x, y_{3}\right) \Delta\left(x, y_{4}\right) J\left(y_{1}\right) J\left(y_{2}\right) J\left(y_{3}\right) J\left(y_{4}\right)\right.$$ $$\left.+3 ! \Delta\left(x, y_{1}\right) \Delta\left(x, y_{2}\right) \Delta(x, x) J\left(y_{1}\right) J\left(y_{2}\right)\right].\tag{1.87}$$

It is understood that in eq 1.87 all arguments $(x, y_i)$ are integrated over.

Questions

  1. In Eq 1.86 shouldn't there by a $\phi^4$ term before $(\frac{\delta}{\delta J(x)})^4$?
  2. In Eq 1.87 1st term how are we getting $y_1,\dots,y_4$? Why not $-\frac{1}{4 !}\Delta\left(x, y\right)^4J\left(y\right)^4$? I know the intuitive answer from Feynman diagrams. I want the answer purely from functional derviatives.
  3. In Eq 1.87 how to get the 2nd term $-\frac{1}{4 !}\left[3 ! \Delta\left(x, y_{1}\right) \Delta\left(x, y_{2}\right) \Delta(x, x) J\left(y_{1}\right) J\left(y_{2}\right)\right]$ from functional derivatives?
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2 Answers 2

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\begin{eqnarray*} \frac{\delta}{\delta J(x)} W_0[J] &=& \frac{\delta}{\delta J(x)} \exp \frac{1}{2} \int d^4 y \int d^4 z J(y) \Delta(y, z) J(z) \\ &=& \frac{1}{2} W_0[J] \frac{\delta}{\delta J(x)} \int d^4 y \int d^4 z J(y) \Delta(y, z) J(z) \\ &=& \frac{1}{2} W_0[J] \int d^4 y \int d^4 z \left( \frac{\delta J(y)}{\delta J(x)} \Delta(y, z) J(z) + J(y) \Delta(y, z) \frac{\delta J(z)}{\delta J(x)} \right) \\ &=& \frac{1}{2} W_0[J] \int d^4 y \int d^4 z \left( \delta^{(4)}(y-x)\Delta(y, z) J(z) + J(y) \Delta(y, z) \delta^{(4)}(z-x)\right) \\ &=& W_0[J] \int d^4 w_1 \Delta(x,w_1) J(w_1) \end{eqnarray*} in the above equation $x$ is not a dummy variable (but we will later integrate over it) but $w_1$ is a dummy variable. \begin{eqnarray*} \frac{\delta^2}{\delta J(x)^2} W_0[J] &=& \frac{\delta}{\delta J(x)} \left( W_0[J] \int d^4 w_1 \Delta(x,w_1) J(w_1) \right)\\ &=& W_0[J] \int d^4 w_2 \Delta(x,w_2) J(w_2)\left(\int d^4 w_1 \Delta(x,w_1) J(w_1) \right)+W_0[J]\Delta(x,x)\\ &=& W_0[J] \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)+W_0[J]\Delta(x,x) \end{eqnarray*}

\begin{eqnarray*} \Rightarrow\frac{\delta^4}{\delta J(x)^4} W_0[J] &=& \frac{\delta^2}{\delta J(x)^2} \left(W_0[J] \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)+W_0[J]\Delta(x,x) \right)\\ &=& \frac{\delta^2}{\delta J(x)^2} \left(W_0[J] \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)\right)+\Delta(x,x)\frac{\delta^2}{\delta J(x)^2}W_0[J]\\ &=& \frac{\delta^2}{\delta J(x)^2}\left(W_0[J]\right) \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)\\ &&+2\frac{\delta}{\delta J(x)}W_0[J] \frac{\delta}{\delta J(x)} \left( \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)\right)\\ &&+W_0[J] \frac{\delta^2}{\delta J(x)^2} \left( \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)\right)\\ &&+\Delta(x,x)\frac{\delta^2}{\delta J(x)^2}W_0[J]\\ &=&\left( W_0[J] \int d^4 w_3d^4 w_4 \Delta(x,w_3)\Delta(x,w_4) J(w_3)J(w_4)+W_0[J]\Delta(x,x)\right)\\ &&\times\int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)\\ &&+2(W_0[J] \int d^4 w_1 \Delta(x,w_1) J(w_1)) \left(2 \int d^4 w_3 \Delta(x,x)\Delta(x,w_3)J(w_3)\right)\\ &&+W_0[J] \left( \Delta(x,x)^2\right)\\ &&+\Delta(x,x)\frac{\delta^2}{\delta J(x)^2}W_0[J]\\ &=&W_0[J] \int d^4 w_1d^4 w_2d^4 w_3d^4 w_4 \Delta(x,w_1)\Delta(x,w_2)\Delta(x,w_3)\Delta(x,w_4) J(w_1)J(w_2)J(w_3)J(w_4)\\ &&+W_0[J]\Delta(x,x)\int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2) \\ &&+4W_0[J] \Delta(x,x) \int d^4 w_1 d^4 w_3 \Delta(x,w_1)\Delta(x,w_3) J(w_1)J(w_3)\\ &&+W_0[J] \left( \Delta(x,x)^2\right)\\ &&+\Delta(x,x)(W_0[J] \int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)+W_0[J]\Delta(x,x))\\ &=&W_0[J] \int d^4 w_1d^4 w_2d^4 w_3d^4 w_4 \Delta(x,w_1)\Delta(x,w_2)\Delta(x,w_3)\Delta(x,w_4) J(w_1)J(w_2)J(w_3)J(w_4)\\ &&+5W_0[J]\Delta(x,x)\int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2) \\ &&+W_0[J] \left( \Delta(x,x)^2\right)\\ &&+W_0[J]\Delta(x,x)\int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2)+W_0[J]\Delta(x,x)^2\\ &=&W_0[J] \int d^4 w_1d^4 w_2d^4 w_3d^4 w_4 \Delta(x,w_1)\Delta(x,w_2)\Delta(x,w_3)\Delta(x,w_4) J(w_1)J(w_2)J(w_3)J(w_4)\\ &&+6W_0[J]\Delta(x,x)\int d^4 w_1d^4 w_2 \Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2) \\ &&+2W_0[J]\Delta(x,x)^2 \end{eqnarray*} The term $6$ can now be written as $3!$. We know make the notation compact by removing the integrations over $w_i$ and neglecting the term independent of $w_i$ \begin{eqnarray*} \Rightarrow\frac{\delta^4}{\delta J(x)^4} W_0[J] &=&W_0[J] \Delta(x,w_1)\Delta(x,w_2)\Delta(x,w_3)\Delta(x,w_4) J(w_1)J(w_2)J(w_3)J(w_4)\\ &&+3!W_0[J]\Delta(x,x)\Delta(x,w_1)\Delta(x,w_2) J(w_1)J(w_2) \\ \end{eqnarray*} replacing the dummy variables $w_i\to y_i$ and substituting in the formula for $\omega_{1}[J]$ and cancelling $W_0[J]^{-1}W_0[J]$ we get

$$\Rightarrow\omega_{1}[J]=-\frac{1}{4 !}\left[\Delta\left(x, y_{1}\right) \Delta\left(x, y_{2}\right) \Delta\left(x, y_{3}\right) \Delta\left(x, y_{4}\right) J\left(y_{1}\right) J\left(y_{2}\right) J\left(y_{3}\right) J\left(y_{4}\right)\right.$$ $$\left.+3 ! \Delta\left(x, y_{1}\right) \Delta\left(x, y_{2}\right) \Delta(x, x) J\left(y_{1}\right) J\left(y_{2}\right)\right].$$


This is quite possibly the most cumbersome calculation I have ever done in my life.

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  1. Your expression for $W[J]$ is not correct; also what you are calling $W$ is normally called $Z$. It should be (sorry but I'm probably getting the $i$'s and $2$'s wrong) \begin{equation} Z[J] = \exp\left[i\int d^4 x \left(-\frac{\lambda}{4!} \frac{\delta}{\delta J(x)}\right)^4\right] Z_0[J] \end{equation}

This explains why there are no $\phi$'s.

2+3. Actually doing all the functional derivatives is a painful exercise I wouldn't want to deprive you of :-) But let's write out how to do one of them. \begin{eqnarray} \frac{\delta}{\delta J(x)} Z_0[J] &=& \frac{\delta}{\delta J(x)} \exp i \int d^4 y \int d^4 z J(y) \Delta(y, z) J(z) \\ &=& i Z_0[J] \frac{\delta}{\delta J(x)} \int d^4 y \int d^4 z J(y) \Delta(y, z) J(z) \\ &=& i Z_0[J] \int d^4 y \int d^4 z \left( \frac{\delta J(y)}{\delta J(x)} \Delta(y, z) J(z) + J(y) \Delta(y, z) \frac{\delta J(z)}{\delta J(x)} \right) \\ &=& i Z_0[J] \int d^4 y \int d^4 z \left( \delta^{(4)}(y-x)\Delta(y, z) J(z) + J(y) \Delta(y, z) \delta^{(4)}(z-x)\right) \\ &=& 2i Z_0[J] \int d^4 w \Delta(x,w) J(w) \end{eqnarray} The first line is just the definition of $Z_0$, the second line is the chain rule, the third line is the product rule, the fourth line is evaluating $\delta J(y)/\delta J(x)$, the fifth line is doing the delta function integrals, relabeling the dummy integration variable to $w$, and combining like terms.

The way to do an $n$-th functional derivative is to carefully do each of the derivatives one by one, like the one shown above.

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  • $\begingroup$ I know that it is normally called $Z$. In A Zee QFT also he calls that $Z$ and $W$ is the logarithm of $Z$. But I followed Cheng and Li's unpopular convention because my question is from there. $\endgroup$ Commented Mar 23, 2022 at 18:07
  • $\begingroup$ I also agree that in the $Z[J]$ formula $\phi^4$ shouldn't be there. It was given wrong in the book. $\endgroup$ Commented Mar 23, 2022 at 18:12
  • $\begingroup$ There shouldn't be $exp$ in your 2nd aligned equation right? $\endgroup$ Commented Mar 23, 2022 at 18:14
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    $\begingroup$ @KasiReddySreemanReddy In addition to $y_1, \cdots, y_4$ being dummy variables, $x$ is also a dummy variable. Note that I didn't write the $\int d^4 x$ in my example functional derivative, but there is an integral over $x$ in the expression for $Z[J]$. $\endgroup$
    – Andrew
    Commented Mar 23, 2022 at 21:39
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    $\begingroup$ @KasiReddySreemanReddy It looks like it has the right level of complexity :-) I still remember my hand aching after writing up the problem set where I had to do this. $\endgroup$
    – Andrew
    Commented Mar 24, 2022 at 19:27

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