# Generating functional in $\phi^4$ theory calculation upto 1st order

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?

$$\begin{eqnarray*} \frac{\delta}{\delta J(x)} W_0[J] &=& \frac{\delta}{\delta J(x)} \exp \frac{1}{2} \int d^4 y \int d^4 z J(y) \Delta(y, z) J(z) \\ &=& \frac{1}{2} W_0[J] \frac{\delta}{\delta J(x)} \int d^4 y \int d^4 z J(y) \Delta(y, z) J(z) \\ &=& \frac{1}{2} W_0[J] \int d^4 y \int d^4 z \left( \frac{\delta J(y)}{\delta J(x)} \Delta(y, z) J(z) + J(y) \Delta(y, z) \frac{\delta J(z)}{\delta J(x)} \right) \\ &=& \frac{1}{2} W_0[J] \int d^4 y \int d^4 z \left( \delta^{(4)}(y-x)\Delta(y, z) J(z) + J(y) \Delta(y, z) \delta^{(4)}(z-x)\right) \\ &=& W_0[J] \int d^4 w_1 \Delta(x,w_1) J(w_1) \end{eqnarray*}$$ 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.

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) $$$$Z[J] = \exp\left[i\int d^4 x \left(-\frac{\lambda}{4!} \frac{\delta}{\delta J(x)}\right)^4\right] Z_0[J]$$$$

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.

• 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. Commented Mar 23, 2022 at 18:07
• I also agree that in the $Z[J]$ formula $\phi^4$ shouldn't be there. It was given wrong in the book. Commented Mar 23, 2022 at 18:12
• There shouldn't be $exp$ in your 2nd aligned equation right? Commented Mar 23, 2022 at 18:14
• @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 $Z[J]$. Commented Mar 23, 2022 at 21:39
• @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. Commented Mar 24, 2022 at 19:27