Long-ranged Wilson-Fisher Fixed Points via Functional RG
Final project for Course FY828 — Videregående statistisk fysik
| Course: FY828 — Videregående statistisk fysik | Author: Valentin Weitemeyer |
Read up on what is the Wetterich(–Morris–Ellwanger) equation by, e.g., reading Ray’s lecture notes, especially Eq. (*) in Sec. 3.2 and the discussion surrounding it. Understand the derivation of the Wilson–Fisher fixed point by following the calculations and discussions in Sec. 4.2 and Sec. 6. Now calculate the flow equations for the system whose energy is given by
\[E(\phi) = \int_{\mathbb{R}^d} \left( - \frac{1}{2} \phi(x) \, \Delta^{\sigma/2} \phi(x) + \frac{1}{2} r \phi(x)^2 + \frac{1}{24} u \phi(x)^4 \right) d^d x\]with the approximation (ansatz)
\[F(\phi; k) = \int_{\mathbb{R}^d} \left( -\frac{1}{2} \phi(x) \, \Delta^{\sigma/2} \phi(x) + \frac{1}{2} r(k) \phi(x)^2 + \frac{1}{24} u(k) \phi(x)^4 \right) d^d x.\]Compute the correlation length exponent \(\nu\) at the Wilson–Fisher fixed point for \(0 < \sigma < 2\) and \(d < 2\sigma\) (why?) to first non-trivial order in \(\epsilon := 2\sigma - d\).
Note: For the derivation of the Wetterich equation I follow H. Gies, Introduction to the Functional RG and Applications to Gauge Theories, in: Renormalization Group and Effective Field Theory Approaches to Many-Body Systems, Springer (2012), pp. 289–294.
Effective Free Energy Functional
With the path integral formalism, one can define the generating functional for connected correlations \(F(J)\), also called free energy functional, as
\[e^{-F(J)} = \int \mathcal{D}\varphi \, e^{-E(\varphi) + \int J \varphi }\]where \(\int J \varphi = \int_{\mathbb{R}^d} J(x) \varphi(x) \, d^d x\) for a source field \(J(x)\).
One can easily show:
\[\frac{\delta^2 F(J)}{\delta J(x) \delta J(y)} \; = \; \langle \varphi(x) \varphi(y) \rangle - \langle \varphi(x) \rangle \langle \varphi(y) \rangle \; = \; G_c^{(2)}(x, y)\]where \(G_c^{(2)}(x, y)\) is the connected two-point correlation function.
By a Legendre transformation one can define the effective free energy functional \(F(\phi)\) as
\[F(\phi) \; = \; \sup_J \left( \int J \phi - F(J) \right)\]With suitable assumptions on \(F(J)\), one can show that for any field configuration \(\phi\) there exists a unique source field \(J_\phi = J_{\sup}\).
Because of the supremum condition one has
\[\begin{align} 0 \; &\overset{!}{=} \; \frac{\delta}{\delta J(x)} \left( \int J \phi - F(J) \right) \Bigg|_{J = J_\phi} \\[6pt] \Rightarrow \qquad \phi(x) &= \; \frac{\delta F(J)}{\delta J(x)} \Bigg|_{J = J_\phi} = \langle \varphi(x) \rangle_{J_\phi} \end{align}\]Taking the derivative of the effective free energy functional with respect to \(\phi\) gives
\[\frac{\delta F(\phi)}{\delta \phi(x)} = - \int_y \underbrace{\frac{\delta F(J)}{\delta J(y)}}_{= \phi(y)} \frac{\delta J(y)}{\delta \phi(x)} + J_\phi(x) + \int_y \phi(y) \frac{\delta J_\phi(y)}{\delta \phi(x)} = J_\phi(x)\]Thus maximizing the effective free energy functional w.r.t. \(\phi\) is equivalent to setting the source field to zero:
\[\frac{\delta F(\phi)}{\delta \phi(x)} = 0 \quad \leftrightarrow \quad J_\phi(x) = 0\]With this connection between the source field and background field \(\phi\), one can rewrite the effective free energy functional as
\[e^{-F(\phi)} = e^{-F(J_\phi) + \int J_\phi(x) \phi(x) d^d x} = \int \mathcal{D}\psi \, e^{-E(\phi + \psi) + \int J_\phi(x) \psi(x) d^d x}\]where we substituted \(\phi \to \phi + \psi\) and used the shift invariance of the path integral measure. This gives us a highly nonlinear differential equation of first order for the effective free energy functional \(F(\phi)\).
RG Flow Equation
The RG flow describes how the effective free energy functional \(F(\phi; k)\) changes with a change of the scale \(k\). The idea is to interpolate between the microscopic energy functional \(E(\phi)\) and the full effective free energy functional \(F(\phi)\):
\[\lim_{k \to \infty} F(\phi; k) = E(\phi) \qquad\qquad \lim_{k \to 0} F(\phi; k) = F(\phi)\]For the free energy functional we introduce a scale dependent regularization function \(R(k)\) to modify the free energy functional:
\[F(J, k) = - \ln \int \mathcal{D}\varphi \, e^{-E(\varphi) + \int J \varphi - \frac{1}{2} \int \varphi R_k ( \varphi)}\]with the short notation \(\int \phi R_k(\phi) = \int_{\mathbb{R}^d} \phi(x) \int_{\mathbb{R}^d} R(k; x, y) \phi(y) \, d^d y \, d^d x\). To keep notation short, we define the functional:
\[\Delta E_k(\phi) := \frac{1}{2} \int \phi R_k(\phi)\]One can now define the effective free energy functional at scale \(k\) as
\[F(\phi; k) = \sup_J \left( \int J \phi - F(J, k) \right) \; - \; \Delta E_k(\phi)\]where the last term is subtracted to recover the correct limit for \(k \to \infty\). Following the same steps as before, one gets the relations:
\[\begin{align} \phi(x) &= \; \frac{\delta F(J, k)}{\delta J(x)} \Bigg|_{J = J_\phi(k)} = \langle \phi(x) \rangle_{J_\phi(k)} \\[6pt] J_\phi(x; k) &= \frac{\delta F(\phi; k)}{\delta \phi(x)} + R_k(\phi)(x) \end{align}\]One can now show:
\[G_c^{(2)}(x, y; k) = \frac{\delta^2 F(J, k)}{\delta J(x) \delta J(y)} \Bigg|_{J = J_\phi(k)} = \left( \frac{\delta^2 F(\phi; k)}{\delta \phi(x) \delta \phi(y)} + R_k(x, y) \right)^{-1}\]Proof: We use the chain rule and formulas above.
\[\begin{align} \delta(x - y) &= \frac{\delta \phi(x)}{\delta \phi(y)} = \int_z \frac{\delta J_\phi(z; k)}{\delta \phi(y)} \frac{\delta \phi(x)}{\delta J_\phi(z; k)} \\[6pt] &= \int_z \left( \frac{\delta^2 F(\phi; k)}{\delta \phi(y) \delta \phi(z)} + R_k(y, z) \right) \cdot \frac{\delta^2 F(J, k)}{\delta J_\phi(x; k) \delta J_\phi(z; k)} \end{align}\]This implies the desired result. \(\blacksquare\)
Theorem (Wetterich–Morris–Ellwanger equation)
The RG flow of the effective free energy functional \(F(\phi; k)\) is given by
\[\dot{F}(\phi; k) = \frac{1}{2} \mathrm{Tr} \left[ \left( F^{(2)}(\phi; k) + R_k \right)^{-1} \dot{R}_k \right]\]where \(\dot{F} = k \frac{\partial F}{\partial k}\) and \(F^{(2)}(\phi, k) = \frac{\delta^2 F(\phi; k)}{\delta \phi(x) \delta \phi(y)}\).
Proof: Taking the derivative of \(F(\phi; k)\) w.r.t. \(k\) gives
\[\begin{align} \dot{F}(\phi; k) &= - \frac{\partial F(J, k)}{\partial k} \Bigg|_{\phi} \; + \; \int \frac{\partial J}{\partial k} \underbrace{\phi}_{= \frac{\delta F(J, k)}{\delta J}} \; - \; \frac{\partial \Delta E_k(\phi)}{\partial k} \\[6pt] &= - \frac{\partial F(J, k)}{\partial k} \Bigg|_{J = J_\phi(k)} \; - \; \frac{\partial \Delta E_k(\phi)}{\partial k} \\[6pt] &= \frac{1}{2} \int_{\mathbb{R}^d} \left\langle \varphi(x) \int_{\mathbb{R}^d} \dot{R}_k(x, y) \varphi(y) \, d^d y \right\rangle_{J_\phi(k)} d^d x \\ &\quad - \frac{1}{2} \int_{\mathbb{R}^d} \phi(x) \int_{\mathbb{R}^d} \dot{R}_k(x, y) \phi(y) \, d^d y \, d^d x \\[6pt] &= \frac{1}{2} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \dot{R}_k(x, y) \left( \langle \varphi(x) \varphi(y) \rangle_{J_\phi(k)} - \langle \varphi(x) \rangle_{J_\phi(k)} \langle \varphi(y) \rangle_{J_\phi(k)} \right) d^d x \, d^d y \\[6pt] &= \frac{1}{2} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \dot{R}_k(x, y) G_c^{(2)}(x, y; k) \, d^d x \, d^d y \\[6pt] &= \frac{1}{2} \mathrm{Tr} \left[ \left( F^{(2)}(\phi; k) + R_k \right)^{-1} \dot{R}_k \right] \end{align}\]where we used the definition of the connected two-point correlation function in the fourth step, \(\phi = \langle\varphi\rangle_{J_\phi(k)}\), and the previous identity for \(G_c^{(2)}\) in the last step. \(\blacksquare\)
The Trace of an operator \(A(x, y)\) is defined as
\[\mathrm{Tr}[A] = \int_{\mathbb{R}^d} A(x, x) \, d^d x\]Note: For the derivation of the RG flow equations I follow the lecture notes of Ray, especially Sec. 4.2 and Sec. 6.
Application to Long-ranged Wilson–Fisher Fixed Points
We consider the energy functional
\[E(\phi) = \int_{\mathbb{R}^d} \left( - \frac{1}{2} \phi(x) \, \Delta^{\sigma/2} \phi(x) + \frac{1}{2} r \phi(x)^2 + \frac{1}{24} u \phi(x)^4 \right) d^d x\]with \(0 < \sigma < 2\) and \(d < 2\sigma\). The fractional Laplacian \(\Delta^{\sigma/2}\) is defined via its action in momentum space.
We use the approximative ansatz for the free energy functional
\[F(\phi; k) = \int_{\mathbb{R}^d} \left( -\frac{1}{2} \phi(x) \, \Delta^{\sigma/2} \phi(x) + \frac{1}{2} r(k) \phi(x)^2 + \frac{1}{24} u(k) \phi(x)^4 \right) d^d x.\]Our goal is to compute the RG flow equations for the couplings \(r(k)\) and \(u(k)\).
With the Wetterich equation we can calculate the RG flow of the free energy functional \(F(\phi; k)\):
\[\dot{F}(\phi; k) = \int_{\mathbb{R}^d} \left( \frac{1}{2} \dot{r}(k) \phi(x)^2 + \frac{1}{24} \dot{u}(k) \phi(x)^4 \right) d^d x.\]To extract the flow equations for \(r(k)\) and \(u(k)\) we assume a constant field configuration \(\phi(x) = \phi\) and compute the second and fourth derivative of \(\dot{F}(\phi; k)\) at \(\phi = 0\):
\[\begin{align} \dot{F}(\phi; k) &= \frac{1}{2} \dot{r}(k) \phi^2 \cdot V_{\mathbb{R}^d} + \frac{1}{24} \dot{u}(k) \phi^4 \cdot V_{\mathbb{R}^d} \\[6pt] \dot{r} \cdot V_{\mathbb{R}^d} &= \left. \frac{\partial^2 \dot{F}}{\partial \phi^2} \right|_{\phi = 0}, \qquad \dot{u} \cdot V_{\mathbb{R}^d} = \left. \frac{\partial^4 \dot{F}}{\partial \phi^4} \right|_{\phi = 0} \end{align}\]For later use, it is important to note:
\[V_{\mathbb{R}^d} \; = \; \int_{\mathbb{R}^d} e^{i(0)x} \, d^d x \; = \; (2\pi)^d \delta^{(d)}(0)\]as \(\delta(p) = \mathcal{F}(1)(p)\) is the Fourier transform of the constant function \(1\).
To compute the right-hand side of the Wetterich equation, we need the second functional derivative of the free energy functional:
\[F''(\phi; k; x, y) = \frac{\delta^2 F(\phi; k)}{\delta \phi(x) \delta \phi(y)} = \left( -\Delta^{\sigma/2}_x + r(k) + \frac{1}{2} u(k) \phi(x)^2 \right) \delta(x - y)\]Acting on a plane wave basis \(e_p\) we get
\[\begin{align} (F''(\phi; k) e_p)(x) &= \left( |p|^\sigma + r(k) + \frac{1}{2} u(k) \phi(x)^2 \right) e_p(x) \\ &= \left( |p|^\sigma + r(k) + \frac{1}{2} u(k) \phi^2 \right) e_p(x) \end{align}\]to extract the flow equations for \(r(k)\) and \(u(k)\) we only need solutions for a constant field \(\phi(x) = \phi\).
For simplicity we choose a regularization function of the form
\[R(k; q) = k^\sigma \qquad\qquad \dot{R}(k,q) = \sigma k^\sigma\]acting as a mass term to counter IR divergencies at the Gaussian fixed point. The power of \(k\) has to match the power of momentum in the kinetic term of the energy functional due to dimensional reasons. Therefore the regularization function is diagonal in the plane wave basis as well.
Inserting this into the Wetterich equation and taking the trace:
\[\begin{align} \dot{F}(\phi; k) &= \frac{1}{2} \mathrm{Tr} \left[ \left( F''(\phi; k) + R(k) \right)^{-1} \dot{R}(k) \right] \\[6pt] &= \frac{1}{2} \int_{\mathbb{R}^d} d^d p \; \langle e_p, \left( F''(\phi; k) + R(k) \right)^{-1} \dot{R}(k) e_p \rangle \end{align}\]Using the calculations in the momentum basis above:
\[\begin{align} &= \frac{1}{2} \int_{\mathbb{R}^d} d^d p \; \frac{\sigma k^\sigma}{|p|^\sigma + r(k) + \frac{1}{2} u(k) \phi^2 + k^\sigma} \underbrace{\langle e_p, e_p \rangle}_{= \delta^d(0)} \\[6pt] &= \frac{V_{\mathbb{R}^d}}{2} \int_{\mathbb{R}^d} \frac{d^d p}{(2\pi)^d} \; \frac{\sigma k^\sigma}{|p|^\sigma + k^\sigma + r(k) + \frac{1}{2} u(k) \phi^2} \end{align}\]Now we can extract the flow equations for \(r(k)\) and \(u(k)\) by taking the second and fourth derivative w.r.t. \(\phi\) at \(\phi = 0\):
\[\begin{align} \dot{r} \cdot V_{\mathbb{R}^d} &= \left. \frac{\partial^2 \dot{F}}{\partial \phi^2} \right|_{\phi = 0} = -\frac{V_{\mathbb{R}^d}}{2} u(k) \sigma k^\sigma \int_{\mathbb{R}^d} \frac{d^d p}{(2\pi)^d} \frac{1}{\left( |p|^\sigma + r(k) + k^\sigma \right)^2} \\[6pt] \dot{u} \cdot V_{\mathbb{R}^d} &= \left. \frac{\partial^4 \dot{F}}{\partial \phi^4} \right|_{\phi = 0} = V_{\mathbb{R}^d} \cdot 3 u(k)^2 \sigma k^\sigma \int_{\mathbb{R}^d} \frac{d^d p}{(2\pi)^d} \frac{1}{\left( |p|^\sigma + r(k) + k^\sigma \right)^3} \end{align}\]By cancelling \(k^\sigma\) out of the denominators and substituting \(p = k q\):
\[\begin{align} \dot{r} \; &= \; -\frac{1}{2} u(k) \sigma k^{d-\sigma} \int_{\mathbb{R}^d} \frac{d^d q}{(2\pi)^d} \frac{1}{\left( |q|^\sigma + 1 + r(k)/k^\sigma \right)^2} \; = \; - \frac{1}{2} u(k) \sigma k^{d-\sigma} I_2(r;k) \\[6pt] \dot{u} \; &= \; 3 u(k)^2 \sigma k^{d-2\sigma} \int_{\mathbb{R}^d} \frac{d^d q}{(2\pi)^d} \frac{1}{\left( |q|^\sigma + 1 + r(k)/k^\sigma \right)^3} \; = \; 3 u(k)^2 \sigma k^{d-2\sigma} I_3(r;k) \end{align}\]where we defined the integrals
\[I_n(r; k) := \int_{\mathbb{R}^d} \frac{d^d q}{(2\pi)^d} \frac{1}{\left( |q|^\sigma + 1 + r(k)/k^\sigma \right)^n} \quad , \quad n = 2, 3\]To find fixed points of the RG flow, we introduce the dimensionless couplings.
Dimensional Analysis
To introduce dimensionless couplings, we need to know the dimensions of the fields and couplings. We define the dimension of a quantity \(X\) as \([X] = \ell^{d_X}\) where \(d_X\) is the scaling dimension of \(X\).
From the kinetic term of the energy functional we get
\[\begin{align} [\phi(x) \, \Delta^{\sigma/2} \phi(x)] \; &= \; \ell^{-d} \\ [\phi(x)]^2 \cdot \ell^{-\sigma} \; &= \; \ell^{-d} \\ \Rightarrow \quad [\phi(x)] \; &= \; \ell^{-(d - \sigma)/2} \end{align}\]From the mass term we get
\[\begin{align} [r \phi(x)^2] \; &= \; \ell^{-d} \\ [r] \cdot \ell^{-(d - \sigma)} \; &= \; \ell^{-d} \\ \Rightarrow \quad [r] \; &= \; \ell^{-\sigma} \end{align}\]From the quartic coupling term we get
\[\begin{align} [u \phi(x)^4] \; &= \; \ell^{-d} \\ [u] \cdot \ell^{-2(d - \sigma)} \; &= \; \ell^{-d} \\ \Rightarrow \quad [u] \; &= \; \ell^{-(2\sigma - d)} \end{align}\]With these dimensions we can define the dimensionless couplings
\[\begin{align} \tilde{r}(k) \; &= \; r(k) / k^\sigma \\ \tilde{u}(k) \; &= \; u(k) / k^{2\sigma - d} \end{align}\]Taking the derivative w.r.t. \(k\) gives
\[\begin{align} \dot{\tilde{r}}(k) \; &= \; \frac{\dot{r}(k)}{k^\sigma} - \sigma \tilde{r}(k) \\ \dot{\tilde{u}}(k) \; &= \; \frac{\dot{u}(k)}{k^{2\sigma - d}} - (2\sigma - d) \tilde{u}(k) \end{align}\]Inserting the flow equations for \(r(k)\) and \(u(k)\) gives
\[\begin{align} \dot{\tilde{r}}(k) \; &= \; - \sigma \tilde{r}(k) \; - \; \frac{1}{2} \tilde{u}(k) \sigma \tilde{I}_2(\tilde{r}(k))\\ \dot{\tilde{u}}(k) \; &= \; (d - 2\sigma) \tilde{u}(k) \; + \; 3 \tilde{u}(k)^2 \sigma \tilde{I}_3(\tilde{r}(k)) \end{align}\]where we conveniently can absorb the factor \(k^\sigma\) into the definition of the integrals:
\[\tilde{I}_n(\tilde{r}(k)) := \int_{\mathbb{R}^d} \frac{d^d q}{(2\pi)^d} \frac{1}{\left( |q|^\sigma + 1 + \tilde{r}(k) \right)^n} \quad , \quad n = 2, 3\]Wilson–Fisher Fixed Point
One quickly sees that there is a Gaussian fixed point \(\tilde{r}^*_0 = \tilde{u}^*_0 = 0\). Calculating the stability matrix at this fixed point:
\[\frac{d}{d\ell} \begin{pmatrix} \tilde{r}(k) \\ \tilde{u}(k) \end{pmatrix} \; = \; \begin{pmatrix} -\sigma & -\dfrac{1}{2} \sigma \tilde{I}_2(0) \\[8pt] 0 & d - 2\sigma \end{pmatrix} \begin{pmatrix} \tilde{r}(k) \\ \tilde{u}(k) \end{pmatrix}\]where we use \(\frac{d}{d\ell} = k \frac{d}{dk}\). Because the stability matrix is upper triangular, the eigenvalues are given by the diagonal elements:
\[\lambda_1 = -\sigma \quad , \quad \lambda_2 = d - 2\sigma\]For \(d > 2\sigma\) we have only one relevant direction. Therefore (apart from the UV divergencies of the integrals) the Gaussian fixed point describes the critical behavior of the system. For the long-ranged system the critical dimension is \(d_c = 2\sigma\).
For \(d < 2\sigma\) we have two relevant directions, so we do not have a second-order phase transition. However, we expect a non-Gaussian fixed point — the Wilson–Fisher fixed point — which describes the critical behavior of the system.
To find this fixed point, we simplify the integrals \(\tilde{I}_n(\tilde{r})\) in the flow equations by expanding them for small \(\tilde{r}, \tilde{u} = \mathcal{O}(2\sigma - d)\):
\[\begin{align} \dot{\tilde{r}} \; &= \; \left[ - \sigma - \frac{1}{2} \tilde{u}\sigma \tilde{I}'_2(0) \right] \tilde{r} - \frac{1}{2} \sigma I_2(0) \tilde{u} \\[6pt] \dot{\tilde{u}} \; &= \; \left[ d - 2\sigma + 3 \tilde{u} \sigma \tilde{I}_3(0) \right] \tilde{u} + 3 \tilde{u}^2 \sigma \tilde{I}'_3(0) \tilde{r} \end{align}\]By assuming \(\tilde{r}^*_{WF}, \tilde{u}^*_{WF} = \mathcal{O}(2\sigma - d)\) we get:
\[\begin{align} 0 \; &= \; \underbrace{- \sigma \tilde{r}^*_{WF} - \frac{1}{2} I_2(0) \tilde{u}^*_{WF}}_{\mathcal{O}(2\sigma - d)} \; - \; \underbrace{\frac{1}{2}I'_{2}(0)\tilde{u}^*_{WF} \tilde{r}^*_{WF}}_{\mathcal{O}((2\sigma - d)^2)} \\[8pt] 0 \; &= \; \underbrace{(d - 2\sigma) \tilde{u}^*_{WF} + 3 \sigma \tilde{I}_3(0) (\tilde{u}^*_{WF})^2}_{\mathcal{O}((2\sigma - d)^2)} \; + \; \underbrace{3 \sigma \tilde{I}'_3(0) (\tilde{u}^*_{WF})^2 \tilde{r}^*_{WF}}_{\mathcal{O}((2\sigma - d)^3)} \end{align}\]We neglect the higher order terms and solve for the fixed point values:
\[\begin{align} \tilde{u}^*_{WF} \; &= \; \frac{2\sigma - d}{3 \sigma \tilde{I}_3(0)} \\[6pt] \tilde{r}^*_{WF} \; &= \; - \frac{I_2(0)}{2\sigma} \tilde{u}^*_{WF} \; = \; - \frac{I_2(0)}{6 \sigma^2 \tilde{I}_3(0)} (2\sigma - d) \end{align}\]This is the Wilson–Fisher fixed point for the long-ranged system. Calculating the stability matrix at this fixed point:
\[\frac{d}{d\ell} \begin{pmatrix} \delta \tilde{r}(k) \\ \delta \tilde{u}(k) \end{pmatrix} \; = \; \begin{pmatrix} -\sigma - \dfrac{2\sigma - d}{6} \dfrac{\tilde{I}'_2(0)}{\tilde{I}_3(0)} & -\dfrac{1}{2} \sigma \tilde{I}_2(0) \\[12pt] \mathcal{O}((2\sigma - d)^2) & (2\sigma - d) \end{pmatrix} \begin{pmatrix} \delta \tilde{r}(k) \\ \delta \tilde{u}(k) \end{pmatrix}\]where \(\delta \tilde{r}(k) = \tilde{r}(k) - \tilde{r}^*_{WF}\) and \(\delta \tilde{u}(k) = \tilde{u}(k) - \tilde{u}^*_{WF}\). The lower-left entry is of order \(\mathcal{O}((2\sigma - d)^2)\) and is therefore neglected to first order in \(\epsilon := 2\sigma - d\). The eigenvalues are then given by the diagonal elements:
\[\lambda_1 = -\sigma - \frac{\epsilon}{6} \frac{\tilde{I}'_2(0)}{\tilde{I}_3(0)} \quad , \quad \lambda_2 = \epsilon\]A quick calculation gives \(\tilde{I}'_2(0) = -2 \tilde{I}_3(0)\). Thus the eigenvalues are:
\[\lambda_1 = -\sigma + \frac{\epsilon}{3} \quad , \quad \lambda_2 = \epsilon\]So we can identify the correlation length exponent \(\nu\) as the inverse of the (negative) relevant eigenvalue at the fixed point:
\[\boxed{ \nu \; = \; -\frac{1}{\lambda_1} \; = \; \frac{1}{\sigma\!\left(1-\frac{\epsilon}{3\sigma}\right)} \; \approx \; \frac{1}{\sigma} + \frac{2\sigma - d}{3 \sigma^2} + \mathcal{O}((2\sigma - d)^2) }\]For \(\sigma = 2\) we recover the well-known result for the short-ranged Wilson–Fisher fixed point:
\[\nu \; = \; \frac{1}{2} + \frac{1}{12} (4 - d) + \mathcal{O}((4 - d)^2)\]