David Tong SFT Sheet 3 Solutions: Renormalisation Group

Source: David Tong, Statistical Field Theory, Example Sheet 3; 6 worked problems.

These prompts paraphrase David Tong's Statistical Field Theory, Example Sheet 3 (November 2018). Problem numbers, starred status, equations, and guidance follow the official sheet.

1 Sheet 3, Problem 1: Critical Exponents of the $O(N)$ Model

For $N$ real scalars in $d=4-\epsilon$, take \[F_1[\phi]=\int d^dx\left[ \frac12(\nabla\phi)^2 +\frac{\mu^2}{2}(\phi\cdot\phi) +g(\phi\cdot\phi)^2\right].\tag{1.1}\] The supplied flows for $\tilde g=\Lambda^{-\epsilon}g$ are \[\frac{d\mu^2}{ds}=2\mu^2+ \frac{N+2}{2\pi^2}\frac{\Lambda^4}{\Lambda^2+\mu^2}\tilde g, \qquad \frac{d\tilde g}{ds}=\epsilon\tilde g- \frac{N+8}{2\pi^2} \frac{\Lambda^4}{(\Lambda^2+\mu^2)^2}\tilde g^{2}.\tag{1.2}\] Find $\nu$ at the Wilson--Fisher fixed point. Assuming $\eta=O(\epsilon^2)$, determine $\alpha,\beta,\gamma,\delta$ to leading order in $\epsilon$.

To first order in $\epsilon$, the Wilson-Fisher fixed point is given by \[\mu_*^2 = -\frac12\frac{N+2}{N+8}\epsilon \Lambda^2, \quad \tilde g_* = \frac{2\pi^2}{N+8}\epsilon.\tag{1.3}\] Let us consider the RG flow near the W-F fixed point, which is \[\frac{d}{ds}\begin{pmatrix} \delta\mu^2 \\ \delta \tilde g \end{pmatrix} = \begin{pmatrix} 2-\frac{N+2}{N+8}\epsilon & \frac{N+2}{2\pi^2}\left(1+\frac{1}{2}\frac{N+2}{N+8}\epsilon\right)\Lambda^2\\ 0 & -\epsilon \end{pmatrix}\begin{pmatrix} \delta \mu^2 \\ \delta \tilde g \end{pmatrix},\tag{1.4}\] which has eigenvalues $\lambda_1 = 2-\frac{N+2}{N+8}\epsilon$ and $\lambda_2 = -\epsilon$. Since $\delta\mu^2 \sim t$, the first eigenvalue is precisely $\Delta_t =\lambda_1$. Then, by \[\xi \to \zeta^{-1}\xi \sim\zeta^{-1}t^{-\nu}, \quad t \to \zeta^{\Delta_t}t,\tag{1.5}\] we find \[\nu = \frac{1}{\Delta_t} = \frac{1}{2} + \frac{1}{4}\frac{N+2}{N+8}\epsilon + \mathcal{O}(\epsilon^2).\tag{1.6}\] From this, we can evaluate the remaining critical exponents. \[c \sim \frac{\partial^2 f}{\partial t^2},\tag{1.7}\] while $F = \int d^d f$, and hence $[f(t)] = -d$. Therefore, $f \sim \xi^{-d} \sim t^{\nu d}$, which implies $c\sim t^{\nu d -2}$, namely \[\alpha = 2-\nu d = \frac{4-N}{2(N+8)}\epsilon + \mathcal{O}(\epsilon^2).\tag{1.8}\] Then, consider \[\phi \sim \xi^{-\Delta_\phi} \sim t^{\nu \Delta_\phi} \implies \beta = \nu \Delta_\phi = \frac12-\frac{3}{2(N+8)}\epsilon +O(\epsilon^2),\tag{1.9}\] where we have used the fact that $\eta = \mathcal{O}(\epsilon^2)$. By the term $\int d^dx B\phi$, we now that $\Delta_\phi + \Delta_B = d$, and \[\chi = \frac{\partial \phi}{\partial B} \sim \xi^{-\Delta_\phi + \Delta_B}\sim t^{\nu(\Delta_\phi -\Delta_B)},\tag{1.10}\] which implies \[\gamma = -\nu(\Delta_\phi -\Delta_B) = -\nu(2\Delta_\phi-d) = 1+\frac{N+2}{2(N+8)}\epsilon +O(\epsilon^2) .\tag{1.11}\] Finally, \[\phi \sim \xi^{-\Delta_\phi} \sim B^{\Delta_\phi/\Delta_B} \implies \delta = \frac{\Delta_B}{\Delta_\phi} = 3+\epsilon+O(\epsilon^2).\tag{1.12}\]

2 Sheet 3, Problem 2: Flow Between Gaussian and Wilson--Fisher Fixed Points

Along the flow of Problem 1, $\mu^2=O(\epsilon)$. At leading order, write \[\frac{d\tilde g}{ds} \simeq\epsilon\tilde g-\frac{N+8}{2\pi^2}\tilde g^{2} =-A\tilde g(\tilde g-g_\star).\tag{2.1}\] Determine $A,g_\star$, then solve the flow from $g_0$ at $s=0$ to obtain \[\tilde g(s)=\frac{g_\star} {1-(1-g_\star/g_0)e^{-\epsilon s}}.\tag{2.2}\]

3 Sheet 3, Problem 3*: Cubic Anisotropy and Fixed-Point Stability

Add an $O(N)$-breaking interaction to Problem 1: \[F_2[\phi]=F_1[\phi] +\int d^dx\,\lambda\sum_{a=1}^N\phi_a^4.\tag{3.1}\] Establish the necessary positive definite condition $g+\lambda>0$. For $\tilde g=\Lambda^{-\epsilon}g$ and $\tilde\lambda=\Lambda^{-\epsilon}\lambda$, use the given flows \[\frac{d\tilde g}{ds}=\epsilon\tilde g -\frac{N+8}{2\pi^2}\tilde g^{2} -\frac3{\pi^2}\tilde g\tilde\lambda, \qquad \frac{d\tilde\lambda}{ds}=\epsilon\tilde\lambda -\frac6{\pi^2}\tilde g\tilde\lambda -\frac9{2\pi^2}\tilde\lambda^{2}.\tag{3.2}\] Find the Gaussian, Heisenberg ($\tilde\lambda=0$), Ising ($\tilde g=0$), and cubic ($\tilde g\tilde\lambda\ne0$) fixed points. Check the cubic point against the positive definite condition. Determine the stability of every point in the $(g,\lambda)$ plane, and plot the RG trajectories separately for $N<4$ and $N>4$. Hint: The determinant and trace determine the eigenvalue signs of a $2\times2$ stability matrix.

For the free energy to be stable, we require the quartic part to be positive definite. Therefore, we must have \[g \left(\sum_{a=1}^N\phi_a^2\right)^2 + \lambda\sum_{a=1}^N\phi_a^4 > 0\tag{3.3}\] for all nonzero field configurations. Taking only one component to be nonzero implies $g+\lambda>0$. By solving \[\tilde g \left(\epsilon -\frac{N+8}{2\pi^2}\tilde g -\frac3{\pi^2}\tilde\lambda\right)=0,\tag{3.4}\] \[\tilde \lambda\left(\epsilon -\frac6{\pi^2}\tilde g -\frac9{2\pi^2}\tilde\lambda\right) = 0,\tag{3.5}\] we find four fixed points: \[\mathrm{Gaussian:}\quad\tilde \lambda = 0, \tilde g = 0,\tag{3.6}\] \[\mathrm{Heisenberg:}\quad \tilde \lambda = 0 ,\tilde g = \frac{2\pi^2}{N+8}\epsilon,\tag{3.7}\] \[\mathrm{Ising:}\quad \tilde \lambda = \frac{2\pi^2}{9}\epsilon ,\tilde g = 0,\tag{3.8}\] and \[\mathrm{Cubic:}\quad \tilde \lambda = \frac{2\pi^2(N-4)}{9N}\epsilon ,\tilde g = \frac{2\pi^2}{3N}\epsilon.\tag{3.9}\] The cubic point satisfies this necessary condition for $N>1$, since $\tilde g+\tilde\lambda=2\pi^2(N-1)\epsilon/(9N)>0$.

At GF, we have \[\frac{d}{ds}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix} = \begin{pmatrix} \epsilon & 0 \\0 & \epsilon \end{pmatrix}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix},\tag{3.10}\] so GF is unstable in both directions. At IF, we have \[\frac{d}{ds}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix} = \begin{pmatrix} \epsilon/3 & 0 \\-4\epsilon/3 & -\epsilon \end{pmatrix}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix},\tag{3.11}\] so it is unstable in the $(1,-1)$ direction and stable along the $\tilde\lambda$ axis. At HF, we have \[\frac{d}{ds}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix} = \begin{pmatrix} -\epsilon & -6\epsilon/(N+8) \\0 & (N-4)\epsilon/(N+8) \end{pmatrix}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix}.\tag{3.12}\] The eigenvalues are $-\epsilon$ and $(N-4)\epsilon/(N+8)$. Thus HF is stable for $N<4$ and unstable in one direction for $N>4$. Finally, at CF, we have \[\frac{d}{ds}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix} = \begin{pmatrix} -(N+8)\epsilon/3N & -2\epsilon/N \\-4(N-4)\epsilon/3N & -(N-4)\epsilon/N \end{pmatrix}\begin{pmatrix} \delta \tilde g \\ \delta \tilde \lambda \end{pmatrix}.\tag{3.13}\] The matrix has eigenvectors and eigenvalues \[\lambda_1=-\epsilon,\qquad v_1= \begin{pmatrix}3\\N-4\end{pmatrix}\tag{3.14}\] \[\lambda_2=-\frac{N-4}{3N}\epsilon,\qquad v_2= \begin{pmatrix}-1\\2\end{pmatrix}.\tag{3.15}\] Therefore, for $N>4$, CF is stable in all directions, while for $1<N<4$ it becomes unstable in the $v_2$ direction. At $N=4$, HF and CF coincide and one eigenvalue vanishes.

From these we can draw the RG diagrams.

4 Sheet 3, Problem 4*: The Sine--Gordon Perturbation

In two dimensions with ultraviolet cutoff $\Lambda$, consider \[F[\phi]=\int d^2x\left[ \frac12(\nabla\phi)^2-\lambda_0\cos(\beta_0\phi)\right].\tag{4.1}\] Determine the engineering dimensions of $\beta_0$ and $\lambda_0$. Split $\phi_\mathbf{k}=\phi_\mathbf{k}^-+\phi_\mathbf{k}^+$, with the plus modes supported on $\Lambda/\zeta<|k|<\Lambda$, and define their position-space fields. Integrate out the plus modes to first order in $\lambda_0$ to derive \[F'[\phi^-]=\int d^2x\left[ \frac12(\nabla\phi^-)^2 -\lambda_0\left\langle \cos\bigl(\beta_0(\phi^-+\phi^+)\bigr) \right\rangle_+\right].\tag{4.2}\] Define $\langle\cdot\rangle_+$ and evaluate it, showing that the cosine coefficient becomes $\lambda_0\zeta^{-\beta_0^2/(4\pi)}$. Hence find the relevance threshold $\beta_0^2=8\pi$. Hint: Expand the cosine into exponentials and use Wick's identity from Sheet 2, Problem 8, together with \[\langle\phi^+(x)\phi^+(y)\rangle_+ =\int_{\Lambda/\zeta<|k|<\Lambda} \frac{d^2k}{(2\pi)^2}\frac{e^{-ik\cdot(x-y)}}{k^2}\tag{4.3}\] in the coincident-point limit.

Naively, the engineering dimensions are $[\phi] = 0$, $[\beta_0] = 0$, and $[\lambda_0 ] = -2$. Splitting into $\phi_\mathbf{k} = \phi_\mathbf{k}^+ + \phi_\mathbf{k}^-$, the partition function becomes \[\int_{k <\Lambda}\mathcal{D}\phi e^{-F[\phi]} = \int_{k <\Lambda/\zeta}\mathcal{D}\phi^- \int_{\Lambda/\zeta< k <\Lambda}\mathcal{D}\phi^+ e^{-F[\phi]} =\int_{k <\Lambda/\zeta}\mathcal{D}\phi^- e^{-F'[\phi^-]},\tag{4.4}\] where \[e^{-F'[\phi^-]} = e^{-F_0[\phi^-]}\int_{\Lambda/\zeta< k <\Lambda}\mathcal{D}\phi^+ e^{-F_0[\phi^+]}\left(1 + \lambda_0\cos(\beta_0(\phi^-+\phi^+)) + \mathcal{O}(\lambda_0^2)\right).\tag{4.5}\] Therefore, we find \[\begin{aligned} F'[\phi^-]&= \int d^2x\left[ \frac12(\nabla\phi^-)^2 - \log \left(1 + \lambda_0\left\langle \cos\bigl(\beta_0(\phi^-+\phi^+)\bigr) \right\rangle_+ + \mathcal{O}(\lambda_0^2)\right)\right]\\&= \int d^2x\left[ \frac12(\nabla\phi^-)^2 -\lambda_0\left\langle \cos\bigl(\beta_0(\phi^-+\phi^+)\bigr) \right\rangle_+ + \mathcal{O}(\lambda_0^2)\right]. \end{aligned}\tag{4.6}\] The correlation function can be written as \[\left\langle \cos\bigl(\beta_0(\phi^-+\phi^+)\bigr) \right\rangle_+ = \frac{1}{2}\left[\left\langle e^{i\beta_0\phi^+}\right\rangle_+e^{i\beta_0\phi^-} + \left\langle e^{-i\beta_0\phi^+}\right\rangle_+e^{-i\beta_0\phi^-}\right].\tag{4.7}\] Using the Wick identity, we find \[\left\langle e^{i\beta_0\phi^+}\right\rangle_+ = \left\langle e^{-i\beta_0\phi^+}\right\rangle_+ = \exp\left(-\frac{1}{2}\beta_0^2\left\langle \phi^+(0)\phi^+(0)\right\rangle_+\right) =\exp\left(-\frac{1}{4\pi}\beta_0^2 \log\zeta \right).\tag{4.8}\] Plugging back into the free energy, we find \[F'[\phi^-] = \int d^2x\left[\frac{1}{2}(\nabla\phi^-)^2 - \lambda_0\zeta^{-\beta_0^2/4\pi}\cos(\beta_0\phi^-)\right].\tag{4.9}\] Rescaling the UV cutoff back to $\Lambda$, namely $\mathbf{k}' = \zeta\mathbf{k}$ and $\mathbf{x}' = \mathbf{x}/\zeta$, we find \[\lambda(\zeta) = \zeta^{-\beta_0^2/4\pi +2}\lambda_0.\tag{4.10}\] Therefore, the potential it relevant when $\beta_0^2 < 8\pi$ and irrelevant when $\beta_0^2 > 8\pi$.

5 Sheet 3, Problem 5: Renormalisation of a Fluctuating Membrane

For a two-dimensional membrane in three-dimensional space, use \[F[h]=\int d^2x\left[ \frac{r_0}{2}(\nabla h)^2 +\frac{\lambda_0}{2} \bigl(1+(\nabla h)^2\bigr)^{-5/2}(\nabla^2h)^2\right].\tag{5.1}\] Expand it through quartic order as \[F[h]\simeq\int d^2x\left[ \frac{r_0}{2}(\nabla h)^2 +\frac{\lambda_0}{2}(\nabla^2h)^2 -\frac{5\lambda_0}{4}(\nabla h)^2(\nabla^2h)^2 +\cdots\right].\tag{5.2}\] Find the engineering dimensions of $h,r_0,\lambda_0$. Separate $F=F_0+F_I$ into quadratic and quartic pieces and split $h_k=h_k^-+h_k^+$ with $\Lambda/\zeta<|k|<\Lambda$ for $h_k^+$. Derive the Gaussian fast-mode covariance \[\langle h_{k_1}^+h_{k_2}^+\rangle_+ =\frac{(2\pi)^2\delta(k_1+k_2)} {r_0k_1^2+\lambda_0k_1^4}.\tag{5.3}\] Fourier transform $F_I$, isolate the term that renormalises $(\nabla^2h)^2$ while ignoring field renormalisation, and derive $d\lambda/ds\simeq-5/(4\pi)$ for $\zeta=e^s$. Does short-distance fluctuation make the membrane more or less flexible in the infrared?

The engineering dimensions are $[h] = 1$, $[r_0] = -2$, and $[\lambda_0] = 0$. We read off $F_0[h_\mathbf{k}^+]$, which is \[F_0[h_\mathbf{k}^+] = \int \frac{d^2k}{(2\pi)^2}\left(\frac{r_0}{2}k^2 + \frac{\lambda_0}{2}k^4\right)h_\mathbf{k}^+ h_{-\mathbf{k}}^+.\tag{5.4}\] This immediately gives \[\langle h_{k_1}^+h_{k_2}^+\rangle_+ =\frac{(2\pi)^2\delta(k_1+k_2)} {r_0k_1^2+\lambda_0k_1^4}.\tag{5.5}\] The interaction free energy is \[F_I[h_\mathbf{k}] = \frac{5\lambda_0}{4}\int\prod_{i=1}^4\frac{d^2k_i}{(2\pi)^2}\mathbf{k}_1 \cdot \mathbf{k}_2 k_3^2 k_4^2h_{\mathbf{k}_1}h_{\mathbf{k}_2}h_{\mathbf{k}_3}h_{\mathbf{k}_4}(2\pi)^2\delta^2\left(\sum_i\mathbf{k}_i\right).\tag{5.6}\] We shall focus only on the lowest order here. Using \[F'[h_\mathbf{k}^-] = F_0[h_\mathbf{k}^-] + \langle F_I[h_\mathbf{k}^+, h^-_\mathbf{k}]\rangle_+ + \cdots,\tag{5.7}\] the only contributing term in $\langle F_I\rangle$ at this order is \[\frac{5\lambda_0}{4}\int\prod_{i=1}^4\frac{d^2k_i}{(2\pi)^2}\mathbf{k}_1 \cdot \mathbf{k}_2 k_3^2 k_4^2\langle h_{\mathbf{k}_1}^+h_{\mathbf{k}_2}^+\rangle_+h_{\mathbf{k}_3}^-h_{\mathbf{k}_4}^-(2\pi)^2\delta^2\left(\sum_i\mathbf{k}_i\right).\tag{5.8}\] Note that there is no polynomial factors here. This gives \[-\frac{5\lambda_0}{4}\int\frac{d^2k}{(2\pi)^2}\frac{d^2q}{(2\pi)^2} k^4\frac{1}{r_0 + \lambda_0 q^2}h_{\mathbf{k}}^-h_{-\mathbf{k}}^-.\tag{5.9}\] Since $\lambda_0$ has engineering dimension zero, and the canonical term remains unchanged, we find \[\lambda(\zeta) = \lambda_0 - 2\cdot \frac{5\lambda_0}{4(2\pi)}\int^\Lambda_{\Lambda/\zeta}dq \frac{q}{r_0 + \lambda_0q^2} + \cdots .\tag{5.10}\] Recall \[\frac{d}{ds}\int^\Lambda_{\Lambda e^{-s}}dq f(q) = \Lambda f(\Lambda),\tag{5.11}\] which gives \[\frac{d\lambda}{ds} = -\frac{5\lambda_0}{4\pi}\frac{\Lambda^2}{r_0 + \lambda_0 \Lambda^2} + \cdots \approx -\frac{5}{4\pi},\tag{5.12}\] where we have assumed $\Lambda^2 \gg r_0/\lambda_0$. Therefore, by integrating out the short distance behaviour, $\lambda_0$, which measures the stiffness of the membrane, becomes smaller. That is, short-distance fluctuation makes the membrane more flexible in the IR.

6 Sheet 3, Problem 6: Integrating Out Two Coupled Order Parameters

Consider fields $\phi_1,\phi_2$ with independent sign flips and \[F[\phi_1,\phi_2]=\int d^dx\left[ \sum_{i=1}^2\left( \frac12(\nabla\phi_i)^2+\frac{\mu_i^2}{2}\phi_i^2 +g_i\phi_i^4\right)+\lambda\phi_1^2\phi_2^2\right].\tag{6.1}\] Set $\mu_i^2=0$ and integrate out short-wavelength modes in $d=4-\epsilon$. Show, for a common constant $I$, that \[\frac{dg_i}{ds}=\epsilon g_i-(36g_i^2+\lambda^2)I, \qquad \frac{d\lambda}{ds} =\epsilon\lambda-(8\lambda^2+12\lambda g_1 +12\lambda g_2)I.\tag{6.2}\] Find the fixed points and show that the stable one has enhanced $O(2)$ symmetry.

Recall \[F'[\phi_i^-] = F[\phi_i^-] + \langle F_I\rangle_+ - \frac{1}{2}[\langle F_I^2\rangle_+ - \langle F_I\rangle_+^2] + \cdots.\tag{6.3}\] We shall keep $\mu_i^2 = 0$ through out the calculation, and hence neglect the running of it as well. $\langle F_I\rangle_+$ has no contributing terms to the running of $\lambda, g_i$. The leading order contribution comes from $ \frac{1}{2}[\langle F_I^2\rangle_+ - \langle F_I\rangle_+^2]$. The bubble diagrams give coefficients $36g_1^2+\lambda^2$ for $\delta g_1$, $36g_2^2+\lambda^2$ for $\delta g_2$, and $12\lambda g_1+12\lambda g_2+8\lambda^2$ for $\delta\lambda$. The original bubble drawings are in the downloadable PDF. The factor of $36$, for example, comes from $\begin{pmatrix} 4\\2 \end{pmatrix}^2$. All of these diagrams involves the integral \[f(\mathbf{k}_1+\mathbf{k}_2) = \int^\Lambda_{\Lambda/\zeta} \frac{d^dk}{(2\pi)^d}\frac{1}{k^2(\mathbf{k} + \mathbf{k}_1 +\mathbf{k}_2)^2} =\int^\Lambda_{\Lambda/\zeta} \frac{d^dk}{(2\pi)^d} \frac{1}{k^4}(1 + \mathcal{O}(\mathbf{k}_1,\mathbf{k}_2)),\tag{6.4}\] where we can ignore terms including external momenta, since these generate terms of the form $(\phi^-)^2 \nabla^2(\phi^-)^2$. These are irrelevant terms which we are not interested in. Since the couplings has engineering dimensions $[g_i] = [\lambda] = -\epsilon$, by rescaling we find \[g_i(\zeta) = \zeta^{\epsilon}[g_i - (36g_i^2 + \lambda^2)f(0)] + \cdots,\tag{6.5}\] \[\lambda(\zeta) = \zeta^\epsilon[\lambda - (12\lambda g_1 + 12\lambda g_2 + 8\lambda^2)f(0)] + \cdots.\tag{6.6}\] Using \[\frac{df(0)}{ds} = \frac{\Omega_{d-1}}{(2\pi)^d}\Lambda^{d-4} = \frac{\Lambda^{-\epsilon}}{8\pi^2}.\tag{6.7}\] Introducing the dimensionless couplings $\tilde g_i = \Lambda^{-\epsilon} g_i$ and $\tilde \lambda = \Lambda^{-\epsilon}\lambda$ and differentiating with respect to $s$, we find \[\frac{dg_i}{ds}=\epsilon g_i-\frac{1}{8\pi^2}(36g_i^2+\lambda^2), \qquad \frac{d\lambda}{ds} =\epsilon\lambda-\frac{1}{8\pi^2}(8\lambda^2+12\lambda g_1 +12\lambda g_2),\tag{6.8}\] where we have suppressed the tilde.

Let's find the fixed points of the theory by solving \[\begin{aligned}0 &= \epsilon g_1-\frac{1}{8\pi^2}(36g_1^2+\lambda^2),\\ 0 &= \epsilon g_2-\frac{1}{8\pi^2}(36g_2^2+\lambda^2),\\ 0 &= \lambda\left(\epsilon-\frac{1}{8\pi^2} (8\lambda+12g_1+12g_2)\right).\end{aligned}\tag{6.9}\] Clearly, it has a trivial Gaussian fixed point $\lambda=g_i=0$.

For $\lambda=0$, we have \[g_i=0 \qquad \mathrm{or} \qquad g_i=\frac{2\pi^2}{9}\epsilon,\tag{6.10}\] giving the three additional fixed points \[(g_1,g_2,\lambda) = \left(\frac{2\pi^2}{9}\epsilon,0,0\right), \quad \left(0,\frac{2\pi^2}{9}\epsilon,0\right), \quad \left(\frac{2\pi^2}{9}\epsilon, \frac{2\pi^2}{9}\epsilon,0\right).\tag{6.11}\]

For $\lambda\neq0$, combining the first two equations gives \[(g_1-g_2)\left[ \epsilon-\frac{9}{2\pi^2}(g_1+g_2) \right]=0.\tag{6.12}\] The nontrivial solutions have $g_1=g_2=g$. The third equation then gives \[\lambda=\pi^2\epsilon-3g.\tag{6.13}\] Substituting this into the first equation gives \[45g^2-14\pi^2\epsilon g+\pi^4\epsilon^2=0,\tag{6.14}\] and hence \[g_1=g_2=\frac{\pi^2}{5}\epsilon, \qquad \lambda=\frac{2\pi^2}{5}\epsilon,\tag{6.15}\] or \[g_1=g_2=\frac{\pi^2}{9}\epsilon, \qquad \lambda=\frac{2\pi^2}{3}\epsilon.\tag{6.16}\]

Expanding about \[\lambda=\lambda_*+\delta\lambda, \qquad g_i=g_{i*}+\delta g_i,\] the linearised flow is \[\frac{d}{ds} \begin{pmatrix} \delta\lambda\\ \delta g_1\\ \delta g_2 \end{pmatrix} = M \begin{pmatrix} \delta\lambda\\ \delta g_1\\ \delta g_2 \end{pmatrix}.\tag{6.17}\] The stability matrix is \[M= \begin{pmatrix} \displaystyle \epsilon-\frac{4\lambda_*+3g_{1*}+3g_{2*}}{2\pi^2} & \displaystyle-\frac{3\lambda_*}{2\pi^2} & \displaystyle-\frac{3\lambda_*}{2\pi^2} \\[6pt] \displaystyle-\frac{\lambda_*}{4\pi^2} & \displaystyle\epsilon-\frac{9g_{1*}}{\pi^2} & 0 \\[6pt] \displaystyle-\frac{\lambda_*}{4\pi^2} & 0 & \displaystyle\epsilon-\frac{9g_{2*}}{\pi^2} \end{pmatrix}.\]

At the Gaussian fixed point, \[M_{\mathrm{G}}=\epsilon \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix},\tag{6.18}\] so all three eigenvalues are positive. Hence the Gaussian fixed point is unstable in all directions.

For the fixed point \[(g_{1*},g_{2*},\lambda_*) = \left(\frac{2\pi^2}{9}\epsilon,0,0\right),\tag{6.19}\] we find \[M=\epsilon \begin{pmatrix} \frac23&0&0\\ 0&-1&0\\ 0&0&1 \end{pmatrix}.\tag{6.20}\] The eigenvalues and corresponding eigenvectors are \[\begin{aligned}\omega_1=\frac23\epsilon,\quad v_1&= \begin{pmatrix}1\\0\\0\end{pmatrix}, \\ \omega_2=-\epsilon,\quad v_2&= \begin{pmatrix}0\\1\\0\end{pmatrix}, \\ \omega_3=\epsilon,\quad v_3&= \begin{pmatrix}0\\0\\1\end{pmatrix}.\end{aligned}\tag{6.21}\] Thus this fixed point has one stable and two unstable directions. The fixed point \[(g_{1*},g_{2*},\lambda_*) = \left(0,\frac{2\pi^2}{9}\epsilon,0\right)\tag{6.22}\] is equivalent under $g_1\leftrightarrow g_2$ and has the same stability properties.

For the decoupled fixed point \[(g_{1*},g_{2*},\lambda_*) = \left( \frac{2\pi^2}{9}\epsilon, \frac{2\pi^2}{9}\epsilon, 0 \right),\tag{6.23}\] we obtain \[M_{\mathrm{dec}} = \epsilon \begin{pmatrix} \frac13&0&0\\ 0&-1&0\\ 0&0&-1 \end{pmatrix}.\tag{6.24}\] Hence \[\omega_1=\frac13\epsilon, \qquad \omega_2=\omega_3=-\epsilon,\tag{6.25}\] with eigenvectors along the $\lambda$, $g_1$, and $g_2$ directions respectively. Therefore, this fixed point is stable in the $g_1$ and $g_2$ directions but unstable to switching on $\lambda$.

For \[g_{1*}=g_{2*}=\frac{\pi^2}{9}\epsilon, \qquad \lambda_*=\frac{2\pi^2}{3}\epsilon,\tag{6.26}\] we find \[M = \epsilon \begin{pmatrix} -\frac23&-1&-1\\ -\frac16&0&0\\ -\frac16&0&0 \end{pmatrix}.\tag{6.27}\] Its eigenvalues are \[\omega_1=-\epsilon, \qquad \omega_2=0, \qquad \omega_3=\frac13\epsilon,\tag{6.28}\] with corresponding eigenvectors \[v_1= \begin{pmatrix}6\\1\\1\end{pmatrix}, \qquad v_2= \begin{pmatrix}0\\-1\\1\end{pmatrix}, \qquad v_3= \begin{pmatrix}-2\\1\\1\end{pmatrix}.\tag{6.29}\] This fixed point therefore has a stable direction, an unstable direction, and a marginal direction at this order, and is not IR stable.

Finally, consider \[g_{1*}=g_{2*}=\frac{\pi^2}{5}\epsilon, \qquad \lambda_*=\frac{2\pi^2}{5}\epsilon.\tag{6.30}\] The stability matrix is \[M_{O(2)} = \epsilon \begin{pmatrix} -\frac25&-\frac35&-\frac35\\ -\frac1{10}&-\frac45&0\\ -\frac1{10}&0&-\frac45 \end{pmatrix}.\tag{6.31}\] Its eigenvalues are \[\omega_1=-\epsilon, \qquad \omega_2=-\frac45\epsilon, \qquad \omega_3=-\frac15\epsilon\tag{6.32}\] with corresponding eigenvectors \[v_1= \begin{pmatrix}2\\1\\1\end{pmatrix}, \qquad v_2= \begin{pmatrix}0\\-1\\1\end{pmatrix}, \qquad v_3= \begin{pmatrix}-6\\1\\1\end{pmatrix}.\tag{6.33}\] Since all three eigenvalues are negative, this fixed point is IR stable.

Moreover, at this fixed point $\lambda_*=2g_*$, so the quartic interaction becomes \[g_*(\phi_1^4+\phi_2^4) +\lambda_*\phi_1^2\phi_2^2 = g_*(\phi_1^2+\phi_2^2)^2.\tag{6.34}\] Thus the stable fixed point has an enhanced $O(2)$ symmetry, as required.