Berezin Integration and the Gross-Neveu Model

Source: Perimeter Scholars International, Quantum Field Theory II, Tutorial 7; 10 worked parts.

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

1 Grassmann Variables

Let $\theta_a$ and $\bar\theta_a$, with $a=1,\ldots,N$, be $N$ conjugate pairs of Grassmann variables, and let $A$ be an $N\times N$ matrix. Begin from the identity \[\int \prod_{a=1}^{N} d\bar\theta_a\,d\theta_a\, \exp\!\left[-\sum_{c,d}\bar\theta_c A_{cd}\theta_d\right] =\det(A).\tag{1.1}\]

Introduce independent conjugate Grassmann sources $\bar\eta_c$ and $\eta_c$, which anticommute with one another and with every $\theta_a$ and $\bar\theta_a$. Show that \[\begin{aligned} Z[\bar\eta,\eta] &=\int \prod_{a=1}^{N}d\bar\theta_a\,d\theta_a\, \exp\!\left[ -\sum_{c,d}\bar\theta_cA_{cd}\theta_d +\sum_c\bigl(\bar\eta_c\theta_c+\bar\theta_c\eta_c\bigr) \right] \\ &=\det(A)\exp\!\left[ \sum_{c,d}\bar\eta_c(A^{-1})_{cd}\eta_d \right]. \end{aligned}\tag{1.2}\]

We shift the integral measure by \[\tilde{\theta}_a = \theta_a - A_{ab}^{-1}\eta_b.\tag{1.3}\] Then, the exponent becomes \[\begin{aligned} -(\bar{\tilde{\theta}}_c + \bar{\eta}_bA_{bc}^{-1})A_{cd}(\tilde{\theta}_d + A_{da}^{-1}\eta_a) + \bar{\eta}_c(\tilde{\theta}_c + A_{ca}^{-1}\eta_a) + (\bar{\tilde{\theta}}_c + \bar{\eta}_bA_{bc}^{-1})\eta_c = -\bar{\tilde{\theta}}_cA_{cd}\tilde{\theta}_d+\bar{\eta}_cA_{cd}^{-1}\eta_d, \end{aligned}\tag{1.4}\] which follows \[Z[\bar{\eta},\eta] = \det A \exp(\bar{\eta}_cA_{cd}^{-1}\eta_d).\tag{1.5}\]
Use $Z[\bar\eta,\eta]$ and the differentiation rules for Grassmann variables to evaluate \[\left\langle\theta_a\bar\theta_b\right\rangle =\frac{ \displaystyle \int\prod_c d\bar\theta_c\,d\theta_c\, e^{-\bar\theta\cdot A\cdot\theta}\, \theta_a\bar\theta_b }{ \displaystyle \int\prod_d d\bar\theta_d\,d\theta_d\, e^{-\bar\theta\cdot A\cdot\theta} }, \qquad \left\langle\theta_a\theta_b\right\rangle.\tag{1.6}\]

Using \[\frac{\partial Z[\bar{\eta},\eta]}{\partial \bar{\eta}_a} = \int \prod_{a=1}^{N}d\bar\theta_a\,d\theta_a\,\theta_a \exp\!\left[ -\sum_{c,d}\bar\theta_cA_{cd}\theta_d +\sum_c\bigl(\bar\eta_c\theta_c+\bar\theta_c\eta_c\bigr) \right],\tag{1.7}\] and \[-\frac{\partial Z[\bar{\eta},\eta]}{\partial \eta_b} = \int \prod_{a=1}^{N}d\bar\theta_a\,d\theta_a\,\bar{\theta}_b \exp\!\left[ -\sum_{c,d}\bar\theta_cA_{cd}\theta_d +\sum_c\bigl(\bar\eta_c\theta_c+\bar\theta_c\eta_c\bigr) \right].\tag{1.8}\] We can write \[\begin{aligned} \langle\theta_a\bar{\theta_b}\rangle &= -\frac{1}{Z[0,0]}\frac{\partial^2}{\partial\bar\eta_a \partial\eta_b}Z[\bar{\eta}, \eta]\bigg|_{\eta=0}\\ &=-\frac{\partial^2}{\partial\bar\eta_a \partial\eta_b}\exp(\bar{\eta}_cA_{cd}^{-1}\eta_d)\bigg|_{\eta=0}\\ &=\frac{\partial}{\partial \bar \eta_a}\left[\bar \eta_c A_{cd}^{-1}\exp(\bar{\eta}_cA_{cd}^{-1}\eta_d)\right]\bigg|_{\eta=0}\\ &= A_{cd}^{-1}. \end{aligned}\tag{1.9}\]
Establish \[\left\langle \theta_a\bar\theta_b\theta_c\bar\theta_d \right\rangle =(A^{-1})_{ab}(A^{-1})_{cd} -(A^{-1})_{ad}(A^{-1})_{cb},\tag{1.10}\] and explain how the two pairings and their relative sign realize Wick's theorem for fermions.

Using the same method, we have \[\begin{aligned} \langle \theta_a \bar \theta_b \theta_c \bar \theta_d\rangle &= -\frac{\partial^2}{\partial\bar \eta_a\partial\eta_b}\left[A_{cd}^{-1} - \bar \eta_e A^{-1}_{ed}A_{cf}^{-1}\eta_f\exp(\bar{\eta}_cA_{cd}^{-1}\eta_d)\right]\bigg|_{\eta=0}\\ &=A_{ab}^{-1}A_{cd}^{-1} - A_{ad}^{-1} A_{cb}^{-1}. \end{aligned}\tag{1.11}\] In terms of Wick's theorem, this is

Now take the continuum limit $\theta_a\to\psi_\alpha(x)$, where $\alpha=1,2,3,4$ is a suppressed Dirac index, and replace $A_{cd}$ by the integral kernel $A(x,y)$. Use \[\{\psi(x),\psi(y)\}=0, \qquad \frac{\delta\psi(x)}{\delta\psi(y)}=\delta^4(x-y).\tag{1.13}\]

In the finite-dimensional case, prove \[\left\{\frac{d}{d\theta_a},\theta_b\right\}=\delta_{ab}, \qquad \left\{\frac{d}{d\theta_a},\frac{d}{d\theta_b}\right\}=0,\tag{1.14}\] and write the corresponding functional relations for the Grassmann fields $\psi(x)$.

\[\left\{\frac{d}{d\theta_a},\theta_b\right\}f(\theta) = \delta_{ab}f(\theta) - \theta_b \frac{df}{d\theta_a} + \theta_b \frac{df}{d\theta_a} = \delta_{ab}f(\theta) \quad \forall f(\theta).\tag{1.15}\] \[\left\{\frac{d}{d\theta_a},\frac{d}{d\theta_b}\right\} = \frac{d}{d\theta_a}\frac{d}{d\theta_b} + \frac{d}{d\theta_b}\frac{d}{d\theta_a} = 0.\tag{1.16}\]
Complete the continuum Gaussian identity \[\begin{aligned} &\int\mathcal{D}\bar\psi\,\mathcal{D}\psi\, \exp\Bigg[ -\int d^4x'\,d^4x\, \bar\psi(x')A(x',x)\psi(x) \\ & +\int d^4x\, \bigl(\bar\psi(x)\eta(x)+\bar\eta(x)\psi(x)\bigr) \Bigg] =\cdots . \end{aligned}\tag{1.17}\]

We can read this off directly from the finite version, \[Z[\bar \eta,\eta ] = \det A'(x,x') \exp\left(\int d^dx \int d^dx'\bar \eta(x) A'^{-1}(x,x')\eta(x')\right).\tag{1.18}\]
Determine the two-point correlation function when \[A=i\gamma\cdot\partial-m.\tag{1.19}\]

\[\begin{aligned} \langle \psi(x) \bar \psi(y)\rangle &= A'^{-1}(x,y) =\int \frac{d^dk}{(2\pi)^d}\frac{-e^{ik\cdot(x-x')}}{\slashed{k} + m}. \end{aligned}\tag{1.20}\]

2 Gross--Neveu Model

Consider the two-dimensional Gross--Neveu model \[\mathcal{L}_{\mathrm{GN}} =\bar\psi_a i\slashed{\partial}\psi^a +\frac{g^2}{2N}\bigl(\bar\psi_a\psi^a\bigr)^2,\tag{2.1}\] where $a=1,\ldots,N$ is a colour index and each suppressed Dirac index has two values.

Find the mass dimension of $g$.

$[g] =0$.
Choose \[\gamma^0=\sigma^2, \qquad \gamma^1=i\sigma^1, \qquad \gamma^5=\gamma^0\gamma^1=\sigma^3.\tag{2.2}\] Show that $\mathcal{L}_{\mathrm{GN}}$ is invariant under $\psi_a\mapsto\gamma^5\psi_a$, and explain why this symmetry excludes a fermion mass term.

Under the transformation \[\bar \psi \gamma^\mu \psi \to \psi^\dagger \gamma^5\gamma^0 \gamma^\mu \gamma^5 \psi = \bar \psi\gamma^\mu\psi.\tag{2.3}\] This can be checked using explicit matrix multiplication. Meanwhile, \[\bar \psi \psi = \psi^\dagger \gamma^0 \psi \to \psi^\dagger \gamma^5 \gamma^0 \gamma^5 \psi = - \psi^\dagger \gamma^0 = -\bar \psi \psi.\tag{2.4}\] Therefore, the Dirac mass is not allowed, and \[(\bar \psi \psi)^2 \to (\bar \psi \psi)^2.\tag{2.5}\]
Show that the model is equivalent to the auxiliary-scalar theory \[\mathcal{L}_{\psi,\sigma} =\bar\psi_a(i\slashed{\partial}-\sigma)\psi^a -\frac{N}{2g^2}\sigma^2,\tag{2.6}\] where $\sigma(x)$ is a scalar field. Hint: perform the functional integral over $\sigma$.

Integrating over the scalar field, we have \[\begin{aligned} &\int \mathcal{D}\sigma\exp\left(- \int d^2x\bar\psi_a(i\slashed{\partial}-\sigma)\psi^a -\frac{N}{2g^2}\sigma^2\right)\\ &=\exp\left(-\int d^2x\bar\psi_ai\slashed{\partial}\psi\right)\int \mathcal{D}\sigma\exp\left(-\int d^2x\left(\frac{N}{2}\frac{1}{g}\sigma + \frac{g}{\sqrt{2N}}(\bar\psi _a\psi^a)\right)^2 + \frac{g^2}{2N}(\bar\psi _a\psi^a)^2\right)\\ &= \mathrm{const} \times \exp\left(-\int d^2x \mathcal{L}_{\mathrm{GN}}\right). \end{aligned}\tag{2.7}\]
Instead perform the functional integral over $\psi$ and show that the model becomes the scalar-field theory \[\mathcal{L}_\sigma =-\frac{N}{2g^2}\sigma^2 +N\operatorname{tr}\log(i\slashed{\partial}-\sigma).\tag{2.8}\]

Integrating over the fermion fields, we have \[\begin{aligned} &\int \mathcal{D}\bar \psi_a\mathcal{D}\psi_a \exp\left(- \int d^2x\bar\psi_a(i\slashed{\partial}-\sigma)\psi^a -\frac{N}{2g^2}\sigma^2\right)\\ &=\exp\left(-\int d^2x - \frac{N}{2g^2}\sigma^2\right)\left(\int \mathcal{D}\bar\psi\mathcal{D}\psi\exp\left(- \int d^2x\bar\psi(i\slashed{\partial}-\sigma)\psi \right)\right)^N\\ &=\exp\left(-\int d^2x - \frac{N}{2g^2}\sigma^2\right)\det(i\slashed{\partial}-\sigma)^N\\ &=\exp\left(-\int d^2x - \frac{N}{2g^2}\sigma^2 +N\operatorname{tr}\log(i\slashed{\partial}-\sigma)\right). \end{aligned}\tag{2.9}\]