Generating Functionals for φ⁴ Theory

Source: Perimeter Scholars International, Quantum Field Theory II, Homework 1; 10 worked parts.

Question prompts paraphrased from the official Perimeter PSI Online sheet; the equations and notation follow the source.

1 Computing Correlation Functions Using Generating Functionals

Work in Euclidean signature with the interacting scalar-field action \[S[\phi]=S_0[\phi]+\int d^4x\,V(\phi), \qquad V(\phi)=\frac{g}{4!}\phi^4,\tag{1.1}\] where $S_0[\phi]$ is the free Euclidean action.

Introduce a source $J(x)$ and consider the free generating functional \[\mathcal Z_0[J] =\int\mathcal D\phi\, \exp\!\left[ -\frac{1}{\hbar} \left( S_0[\phi]-\int d^4x\,J(x)\phi(x) \right) \right].\tag{1.2}\] Evaluate this Gaussian functional integral. As a hint, first shift the integration variable by a function determined by $J$.

In the Euclidean picture, we have \[S_0[\phi] = -\frac{1}{2}\phi (\partial^2 - m^2)\phi.\tag{1.3}\] To evaluate the integral, we make the shift of the integration variable, \[\bar \phi(x) = \phi(x) + \int d^dy \Delta(x-y)J(y) \equiv \phi_x + \Delta_{xy}J_y,\tag{1.4}\] where we have used the generalised summation convention and $(\partial^2 - m^2)\Delta(x-y) = \delta^d(x-y)$. Since $\mathcal{D}\phi = \mathcal{D}\bar \phi$, we have \[\begin{aligned} Z_0[J] &= \int \mathcal{D}\bar \phi\exp\left[-\frac{1}{\hbar}\left(-\frac{1}{2}\bar \phi_x(\partial^2-m^2)_{xy}\bar \phi_y + \frac{1}{2}J_x\Delta_{xy}J_y\right)\right] \\ &=\exp\left[-\frac{1}{\hbar}\int d^dx\int d^dy \frac{1}{2}J(x)\Delta(x-y)J(y)\right]Z_0[0]. \end{aligned}\tag{1.5}\]
For an arbitrary polynomial interaction $V(\phi)$, establish the identity \[\begin{aligned}&\exp\!\left[ -\frac{1}{\hbar}\int d^4x\,V(\phi) +\frac{1}{\hbar}\int d^4x\,J(x)\phi(x) \right] \notag\\ &\qquad= \exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta J(x)}\right) \right] \exp\!\left[ \frac{1}{\hbar}\int d^4x\,J(x)\phi(x) \right].\end{aligned}\tag{1.6}\]

\[\begin{aligned} &\exp\!\left[ -\frac{1}{\hbar}\int d^4x\,V(\phi) +\frac{1}{\hbar}\int d^4x\,J(x)\phi(x) \right]\\ &= \sum_{n=0}^\infty\frac{1}{n!}\left(-\frac{1}{\hbar}\right)^n\left(\int d^dxV(\phi)\right)^n\exp\left(\frac{1}{\hbar}\int d^dx J(x) \phi(x)\right)\\ &=\sum_{n=0}^\infty\frac{1}{n!}\left(-\frac{1}{\hbar}\right)^n\left(\int d^dx V\left(\hbar \frac{\delta}{\delta J(x)}\right)\right)^n\exp\left(\frac{1}{\hbar}\int d^dx J(x) \phi(x)\right)\\ &=\exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta J(x)}\right) \right] \exp\!\left[ \frac{1}{\hbar}\int d^4x\,J(x)\phi(x) \right]. \end{aligned}\tag{1.7}\]
For the interacting theory, define \[\mathcal Z[J] =\int\mathcal D\phi\, \exp\!\left[ -\frac{1}{\hbar} \left( S[\phi]-\int d^4x\,J(x)\phi(x) \right) \right].\tag{1.8}\] Use the preceding result to express $\mathcal Z[J]$ formally as a functional differential operator acting on $\mathcal Z_0[J]$.

From b), we have \[Z[J] = \int \mathcal{D}\phi\exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta J(x)}\right) \right]\exp\!\left[ -\frac{1}{\hbar} \left( S_0[\phi]-\int d^4x\,J(x)\phi(x) \right) \right].\tag{1.9}\] Then, using the same field redefinition in a), we find \[Z[J] = \exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta J(x)}\right) \right]\exp\left[-\frac{1}{\hbar}\int d^dx\int d^dy \frac{1}{2}J(x)\Delta(x-y)J(y)\right]Z_0[0],\tag{1.10}\] or equivalently, \[Z[J] = \exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta J(x)}\right) \right]Z_0[J].\tag{1.11}\]
The two-point Green function may be obtained from \[G(x_1,x_2) =\langle\phi(x_1)\phi(x_2)\rangle =\left. \frac{\hbar^2}{\mathcal Z[0]} \frac{\delta^2\mathcal Z[J]} {\delta J(x_1)\delta J(x_2)} \right|_{J=0}.\tag{1.12}\] Compute its perturbative expansion through order $g^2$ in $\phi^4$ theory. At each order, list every contribution without evaluating the position-space integrals, draw its Feynman diagram, and give the correct symmetry factor.

Using the results from above, we have \[G(x_1,x_2) = \hbar^2\frac{\delta^2}{\delta_1 \delta_2}\exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta_x}\right) \right]\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\bigg|_{J=0}\frac{Z_0[0]}{Z[0]},\tag{1.13}\] where $Z[0]$ can be written as \[Z[0] = \exp\!\left[ -\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta_x}\right) \right]\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\bigg|_{J=0}Z_0[0].\tag{1.14}\] Expanding the first exponential to order $g^2$, we have \[\begin{aligned} \exp\!\left[-\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta_x}\right)\right] ={}&1-\frac{1}{\hbar}\frac{g}{4!}\int d^dx\, \left(\hbar \frac{\delta}{\delta_x}\right)^4\\ &+\frac{1}{2\hbar^2}\left(\frac{g}{4!}\right)^2 \int d^dx\, \left(\hbar \frac{\delta}{\delta_x}\right)^4 \int d^dy\, \left(\hbar \frac{\delta}{\delta_y}\right)^4 +\cdots. \end{aligned}\tag{1.15}\] At order $g^0$, we simply have $Z[0]=Z_0[0]$, and hence \[G_{12} = -\hbar\Delta_{12} + \cdots.\tag{1.16}\] At order $g^1$, we need to evaluate \[\begin{aligned} &\left(\hbar \frac{\delta}{\delta_x}\right)^4\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\\ &=\left(\hbar \frac{\delta}{\delta_x}\right)^3\left(-\Delta_{xy}J_y\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\right)\\ &=\left(\hbar \frac{\delta}{\delta_x}\right)^2\left((-\hbar\Delta_{xx}+\Delta_{xy}\Delta_{xz}J_yJ_z)\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\right)\\ &=\left(\hbar \frac{\delta}{\delta_x}\right)\left((3\hbar\Delta_{xx}\Delta_{xy}J_y-\Delta_{xy}\Delta_{xz}\Delta_{xw}J_yJ_zJ_w)\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\right)\\ &=\left((3\hbar^2\Delta_{xx}\Delta_{xx}-6\hbar\Delta_{xy}\Delta_{xz}\Delta_{xx}J_yJ_z + \Delta_{xy}\Delta_{xz}\Delta_{xw}\Delta_{xu}J_yJ_zJ_wJ_u)\exp\left[-\frac{1}{\hbar} \frac{1}{2}J_x\Delta_{xy}J_y\right]\right). \end{aligned}\tag{1.17}\] From this, we can read out \[Z[0] = Z_0[0]\left(1-\frac{g}{8}\hbar\int d^dx\Delta_{xx}^2+\cdots\right).\tag{1.18}\] Diagrammatically, the order-$g$ vacuum contribution is

$\displaystyle \frac{Z[0]}{Z_0[0]} =1+$
$\displaystyle +\cdots, \qquad S=8.$(1.19)

Hence \[\begin{aligned} G_{12} &=\left[-\hbar\Delta_{12} +\frac{g}{8}\hbar^2\Delta_{12}\int d^dx\,\Delta_{xx}^2 +\frac{g}{2}\hbar^2\int d^dx\, \Delta_{xx}\Delta_{1x}\Delta_{2x} +\mathcal O(g^2)\right]\\ &\quad\times \left[1+\frac{g}{8}\hbar\int d^dy\,\Delta_{yy}^2 +\mathcal O(g^2)\right]\\ &=-\hbar\Delta_{12} +\frac{g}{2}\hbar^2\int d^dx\, \Delta_{xx}\Delta_{1x}\Delta_{2x} +\mathcal O(g^2). \end{aligned}\tag{1.20}\] The disconnected free propagator times the vacuum bubble has cancelled. The surviving diagrams are

$\displaystyle G_{12} =$
$\displaystyle +$
$\displaystyle +\mathcal O(g^2), \qquad S_{g^0}=1,\quad S_{g^1}=2.$(1.21)

Now, we turn to order $g^2$, where we need to evaluate \[\begin{aligned} &\left(\hbar \frac{\delta}{\delta_x}\right)^4 \left(\hbar \frac{\delta}{\delta_y}\right)^4 \exp\left[-\frac{1}{\hbar}\frac{1}{2} J_x\Delta_{xy}J_y\right]\\ &=\left(\hbar \frac{\delta}{\delta_y}\right)^4 \Bigg[\Big( 3\hbar^2\Delta_{xx}^2 -6\hbar\Delta_{xx}\Delta_{xv}\Delta_{xw}J_vJ_w +\Delta_{xv}\Delta_{xw}\Delta_{xz}\Delta_{xu} J_vJ_wJ_zJ_u \Big) \exp\left[-\frac{1}{\hbar}\frac{1}{2} J_x\Delta_{xy}J_y\right]\Bigg]\\ &=\Bigg\{ -\hbar^3\Big[ 18\Delta_{xx} \left(\Delta_{xx}\Delta_{yy}+4\Delta_{xy}^2\right) \Delta_{yr}\Delta_{ys}J_rJ_s\\ &\qquad\qquad +48\Delta_{xy} \left(3\Delta_{xx}\Delta_{yy}+2\Delta_{xy}^2\right) \Delta_{xv}\Delta_{yr}J_vJ_r\\ &\qquad\qquad +18\Delta_{yy} \left(\Delta_{xx}\Delta_{yy}+4\Delta_{xy}^2\right) \Delta_{xv}\Delta_{xw}J_vJ_w \Big]\\ &\qquad +\hbar^4\Big( 9\Delta_{xx}^2\Delta_{yy}^2 +72\Delta_{xx}\Delta_{yy}\Delta_{xy}^2 +24\Delta_{xy}^4 \Big)\Bigg\} \exp\left[-\frac{1}{\hbar}\frac{1}{2} J_x\Delta_{xy}J_y\right]\\ &\qquad +\text{terms containing more than two currents}. \end{aligned}\tag{1.22}\] We have discarded terms with more than two currents because they vanish after the two external derivatives are taken and $J$ is set to zero. From this, we find \[\begin{aligned} Z[0]=Z_0[0]\Bigg[1 &-\frac{g}{8}\hbar\int d^dx\,\Delta_{xx}^2\\ &+g^2\hbar^2\int d^dx\,d^dy\left( \frac{1}{128}\Delta_{xx}^2\Delta_{yy}^2 +\frac{1}{16}\Delta_{xx}\Delta_{yy}\Delta_{xy}^2 +\frac{1}{48}\Delta_{xy}^4 \right)+\mathcal O(g^3)\Bigg]. \end{aligned}\tag{1.23}\] The three order-$g^2$ vacuum diagrams are

$\displaystyle _{S=128} \quad+\quad$
$\displaystyle _{S=16} \quad+\quad$
$\displaystyle _{S=48}.$(1.24)

As at order $g$, all vacuum-disconnected terms cancel against the normalization by $Z[0]$. The connected result is therefore \[\begin{aligned} G_{12}={}&-\hbar\Delta_{12} +\frac{g}{2}\hbar^2\int d^dx\, \Delta_{1x}\Delta_{xx}\Delta_{x2}\\ &-g^2\hbar^3\int d^dx\,d^dy\Bigg( \frac{1}{4}\Delta_{1x}\Delta_{2x} \Delta_{xy}^2\Delta_{yy} +\frac{1}{4}\Delta_{1x}\Delta_{xx} \Delta_{xy}\Delta_{yy}\Delta_{y2}\\ & +\frac{1}{6}\Delta_{1x}\Delta_{xy}^3\Delta_{y2} \Bigg)+\mathcal O(g^3). \end{aligned}\tag{1.25}\] The order-$g^2$ terms are, in the same order as in the preceding equation,

$\displaystyle _{S=4} \quad+\quad$
$\displaystyle _{S=4} \quad+\quad$
$\displaystyle _{S=6}.$(1.26)
Mark the one-particle-irreducible subgraphs in the diagrams obtained in the previous part.
Repeat the preceding two parts for the four-point correlation function, retaining all contributions through order $g^2$.

Using the results from d), we have \[G_{1234} =\left.\hbar^4\frac{\delta^4}{\delta_1\delta_2\delta_3\delta_4} \exp\!\left[-\frac{1}{\hbar}\int d^4x\, V\!\left(\hbar\frac{\delta}{\delta_x}\right)\right] \exp\!\left[-\frac{1}{\hbar}\frac{1}{2} J_x\Delta_{xy}J_y\right]\right|_{J=0} \frac{Z_0[0]}{Z[0]}.\tag{1.27}\] As before, all vacuum-disconnected terms cancel against the normalization by $Z[0]$. At order $g^0$, we find \[G_{1234}^{(0)} =\hbar^2\left( \Delta_{12}\Delta_{34} +\Delta_{13}\Delta_{24} +\Delta_{14}\Delta_{23} \right).\tag{1.28}\] The remaining disconnected terms are already fixed by the two-point result. For each of \[(ij\vert kl)=(12\vert34),\ (13\vert24),\ (14\vert23),\tag{1.29}\] we obtain \[\begin{aligned} G_{ij}G_{kl}={}&\hbar^2\Delta_{ij}\Delta_{kl}\\ &-\frac{g}{2}\hbar^3\int d^dx\,\Big( \Delta_{ij}\Delta_{kx}\Delta_{xx}\Delta_{xl} +\Delta_{kl}\Delta_{ix}\Delta_{xx}\Delta_{xj} \Big)\\ &+g^2\hbar^4\int d^dx\,d^dy\,\Bigg\{ \Delta_{ij}\Bigg( \frac{1}{4}\Delta_{kx}\Delta_{lx} \Delta_{xy}^2\Delta_{yy} +\frac{1}{4}\Delta_{kx}\Delta_{xx} \Delta_{xy}\Delta_{yy}\Delta_{yl} +\frac{1}{6}\Delta_{kx}\Delta_{xy}^3\Delta_{yl} \Bigg)\\ & +\Delta_{kl}\Bigg( \frac{1}{4}\Delta_{ix}\Delta_{jx} \Delta_{xy}^2\Delta_{yy} +\frac{1}{4}\Delta_{ix}\Delta_{xx} \Delta_{xy}\Delta_{yy}\Delta_{yj} +\frac{1}{6}\Delta_{ix}\Delta_{xy}^3\Delta_{yj} \Bigg)\\ & +\frac{1}{4}\Delta_{ix}\Delta_{xx}\Delta_{xj} \Delta_{ky}\Delta_{yy}\Delta_{yl} \Bigg\}+\mathcal O(g^3). \end{aligned}\tag{1.30}\] The connected terms are \[\begin{aligned} &G_{1234}-G_{12}G_{34}-G_{13}G_{24}-G_{14}G_{23}\\ ={}&-g\hbar^3\int d^dx\, \Delta_{1x}\Delta_{2x}\Delta_{3x}\Delta_{4x}\\ &+\frac{g^2}{2}\hbar^4\int d^dx\,d^dy\,\Bigg\{ \Delta_{xx}\Delta_{xy}\Big( \Delta_{1x}\Delta_{2y}\Delta_{3y}\Delta_{4y} +\Delta_{2x}\Delta_{1y}\Delta_{3y}\Delta_{4y}\\ & +\Delta_{3x}\Delta_{1y}\Delta_{2y}\Delta_{4y} +\Delta_{4x}\Delta_{1y}\Delta_{2y}\Delta_{3y} \Big)\\ & +\Delta_{xy}^2\Big( \Delta_{1x}\Delta_{2x}\Delta_{3y}\Delta_{4y} +\Delta_{1x}\Delta_{3x}\Delta_{2y}\Delta_{4y} +\Delta_{1x}\Delta_{4x}\Delta_{2y}\Delta_{3y} \Big) \Bigg\}+\mathcal O(g^3). \end{aligned}\tag{1.31}\] These corresponds to the diagrams

$\mathcal O(g^0):\quad$

$\mathcal O(g^1):\quad$
Define \[G_{\mathrm{CC}}(x_1,x_2,x_3,x_4) =\left. \hbar^3 \frac{\delta}{\delta J(x_1)}\cdots \frac{\delta}{\delta J(x_4)}\mathcal W[J] \right|_{J=0}, \qquad \mathcal W[J]=\hbar\log\mathcal Z[J].\tag{1.32}\] Show that this construction produces connected Feynman diagrams only.

\[\begin{aligned} &G_\mathrm{CC}(x_1,x_2,x_3,x_4)\\ &= \hbar^4\delta_1\delta_2\delta_3\left(\frac{\delta_4 Z[J]}{Z[J]}\right)\bigg|_{J=0}\\ &=\hbar^4\left(\frac{\delta_1\delta_2\delta_3\delta_4 Z[J]}{Z} - \frac{\delta_1\delta_2 Z[J]\delta_3\delta_4 Z[J] + \delta_1\delta_4 Z[J]\delta_2\delta_4 Z[J] + \delta_1\delta_3 Z[J]\delta_2\delta_4 Z[J]}{Z^2}\right)\bigg|_{J=0}\\ &=G_{1234} - G_{12}G_{34}-G_{13}G_{24}-G_{14}G_{23} &= (\mathrm{connected 4-point diagrams}), \end{aligned}\tag{1.33}\] where we have used the fact that there are no tadpole diagrams for $\phi^4$ theory.

2 Generating Functional for One-Particle-Irreducible Diagrams

Define the background field in the presence of $J$ by \[\varphi(x) =\frac{\delta\mathcal W[J]}{\delta J(x)} =\langle\phi(x)\rangle[J],\tag{2.1}\] and introduce the effective action through the Legendre transform \[\Gamma[\varphi] =\int d^4x\,J(x)\varphi(x)-\mathcal W[J].\tag{2.2}\] Recall that the connected generating functional has the expansion \[\mathcal W[J] =\sum_{n=0}^{\infty}\frac{1}{n!} \left[ \prod_{k=1}^{n}\int d^4x_k\,J(x_k) \right] G_{\mathrm{CC}}(x_1,\ldots,x_n),\tag{2.3}\] where $G_{\mathrm{CC}}(x_1,\ldots,x_n)$ is the fully connected $n$-point function.

Derive the relation \[J(x)=\frac{\delta\Gamma[\varphi]}{\delta\varphi(x)}.\tag{2.4}\]

Using chain rules, we have \[\begin{aligned} \frac{\delta\Gamma[\varphi]}{\delta\varphi(x)} &= J(x) + \int d^dy\frac{\delta J(y)}{\delta\varphi(x)}\varphi(x) - \int d^dy\frac{\delta \mathcal{W}}{\delta J(y)}\frac{\delta J(y)}{\delta \varphi(x)}=J(x). \end{aligned}\tag{2.5}\]
Consider the Hessian \[\frac{\delta^2\Gamma} {\delta\varphi(x)\delta\varphi(y)}.\tag{2.6}\] Show that its functional inverse is the connected two-point function, \[G_{\mathrm{CC}}(x_1,x_2) =\left[ \frac{\delta^2\Gamma} {\delta\varphi(x_1)\delta\varphi(x_2)} \right]^{-1}, \qquad G_{\mathrm{CC}}(x_1,x_2) =\langle\phi(x_1)\phi(x_2)\rangle_{\mathrm{conn.}}.\tag{2.7}\] You may use the functional chain rule \[\frac{\delta}{\delta J(x)} =\int d^4y\, \frac{\delta\varphi(y)}{\delta J(x)} \frac{\delta}{\delta\varphi(y)}.\tag{2.8}\]

\[\begin{aligned} \frac{\delta^2\Gamma} {\delta\varphi(x_1)\delta\varphi(x_2)} &= \frac{\delta J(x_2)}{\delta \varphi(x_1)} = \left(\frac{\delta^2\mathcal{W}[J]}{\delta J(x_1)\delta J(x_2)}\right)^{-1} \implies G_\mathrm{CC}(x_1,x_2) =\left[ \frac{\delta^2\Gamma} {\delta\varphi(x_1)\delta\varphi(x_2)} \right]^{-1}. \end{aligned}\tag{2.9}\]
Fourier transform the connected two-point function according to \[G_{\mathrm{CC}}(x_1,x_2) =\int\frac{d^4k}{(2\pi)^4}\, e^{ik\cdot(x_1-x_2)}\widetilde G(k).\tag{2.10}\] Starting from its perturbative expansion, show that repeated 1PI self-energy insertions form a geometric series and hence give \[\widetilde G(k)=\frac{1}{k^2+m^2+M^2(k)},\tag{2.11}\] where $-M^2(k)$ denotes the sum of one-particle-irreducible insertions. The diagrammatic series is

Note that at tree-level $\Gamma = S$, and therefore we can write \[\Gamma_2(x,y) = G_0(x,y)^{-1} + \cdots, \quad \Gamma_2(x,y) \equiv \frac{\delta \Gamma[\varphi]}{\delta \varphi(x) \delta\varphi(y)}.\tag{2.12}\] In momentum space, this can be written as \[\Gamma_2(k) = G_0(k)^{-1} +M^2(k) = \tilde{G}(k)^{-1},\tag{2.13}\] where $M^2(k)$ is the 1PI self-energy insertion. Using \[\int d^dz\, \Gamma_2(x,z)G_\mathrm{CC}(z,y) = \delta^d (x-y), \quad \Gamma_2(x,y) \equiv \frac{\delta \Gamma[\varphi]}{\delta \varphi(x) \delta\varphi(y)}.\tag{2.14}\] This gives \[\begin{aligned} \tilde{G}(k) &= \frac{1}{k^2 + m^2 +M^2(k)}\\ &=\frac{1}{k^2+m^2}\sum_{n=0}^\infty\left(\frac{M^2(k)}{k^2+m^2}\right)^n, \end{aligned}\tag{2.15}\] which is exactly the geometric series.