Functional Renormalization Group

Abstract

The functional renormalization group (FRG) is a particular implementation of the renormalization group concept which combines the functional methods of quantum field theory with the renormalization group idea of Kenneth Wilson. It interpolates smoothly between the known microscopic laws and the complex macroscopic phenomena in physical systems. It is a momentum-space implementation of the renormalization group idea and can be formulated directly for a continuum field theory—no lattice regularization is required. In most approaches one uses a scale-dependent Schwinger functional or scale-dependent effective action. The scale parameter acts similarly as an adjustable screw of a microscope. For large values of a momentum scale k or equivalently for a high resolution of the microscope, one starts with the known microscopic laws. With decreasing scale k or equivalently with decreasing resolution of the microscope, one moves to a coarse-grained picture adequate for macroscopic phenomena. The flow from microscopic to macroscopic scales is given by a conceptionally simple but technically demanding flow equation for scale-dependent functionals. A priori the method is non-perturbative and does not rely on an expansion in a small coupling constant.

Appendices

12.8 Programs for Chap. 12

The octave program in Listing 12.1 calculates the flow of the couplings $E_k,\,\omega _k^2$, and λ _k in the LPA approximation and the truncation with a ₆ = 0. The flow begins at k = 10⁵, but only the values of the couplings for k = 2 down to k = 0 are plotted.

Listing 12.1 Flow of couplings in polynomial LPA

1 function x=truncflowanho_lambda

2  # calculates the flow for even potentials projected onto

3  # quartic polynoms. At the cutoff E=0. The program asks

4  # for lambda at the cutoff in units of omega. The running

5  # of the effective couplings  in the infrared is plotted.

6  # couplLambda=[0;-1;0];        # for 4th order polynomial

7 couplLambda=[0;-1;0;0;0;0;0];   # for 12th order polynomial

8 Nk=100000;disp=20;

9 couplLambda(3)=lambda;

10 k=linspace(10000,0,Nk);

11  #[coupl]=lsode("flowOpt4",couplLambda,k);

12  #[coupl]=lsode("flowCallan4",couplLambda,k);

13 [coupl]=lsode( "flowOpt12",couplLambda,k);

14 xh=coupl(Nk-disp:Nk,:,:);

15 kh=k(Nk-disp:Nk);

16 plot(kh,xh(:,1),kh,xh(:,2),kh,xh(:,3));

17 legend( ’E’, ’omega**2’, ’lambda’);

18 printf( "E0 = \t %4.4f\n",coupl(Nk,1));

19 printf( "E1 = \t %4.4f\n",coupl(Nk,1)+sqrt(coupl(Nk,2)));

20 endfunction

Listing 12.2 Called by program 12.1: optimal regulator, fourth-order polynomial

1 function coupldot=flowOpt4(coupl,k)

2 ksquare=k*k;

3 P=1/(ksquare+coupl(2));xh=ksquare*P/pi;

4 coupldot(1)=xh-1/pi;

5 coupldot(2)=-xh*P*coupl(3);

6 coupldot(3)=-6*coupldot(2)*P*coupl(3);

7 endfunction

Listing 12.3 Called by program 12.1: Callan-Symanzik regulator and fourth-order polynomial

1 function coupldot=flowCallan4(coupl,k)

2 P=1/(k*k+coupl(2));

3 rootP=k*sqrt(P);

4 coupldot(1)=0.5*(rootP-1);

5 coupldot(2)=-0.25*rootP*P*coupl(3);

6 coupldot(3)=-4.5*coupldot(2)*P*coupl(3);

7 endfunction

The functions defined in Listings 12.2, 12.3, and 12.4 are called by program 12.1 to calculate the flow for different regulators and truncation orders.

Listing 12.4 Called by program 12.1: optimal regulator and 12th-order polynomial

1 function coupldot=flowOpt12(coupl,k)

2 a0=coupl(1);a2=coupl(2);a4=coupl(3);a6=coupl(4);

3 a8=coupl(5);a10=coupl(6);a12=coupl(7);

4 ks=k*k;numflow1

5 P=1/(ks+a2);P2=P*P;P2pi=ks*P2/pi;P3=P*P2;P4=P*P3;

6 a4s=a4*a4;a4q=a4s*a4s;a6s=a6*a6;

7 coupldot(1)=-a2*P/pi;

8 coupldot(2)=-a4;

9 coupldot(3)=-a6+6*a4s*P;

10 coupldot(4)=-a8+30*a4*a6*P-90*a4s*a4*P2;

11 coupldot(5)=-a10+(56*a4*a8+70*a6s)*P\

12 -1260*a6*a4s*P2+2520*a4q*P3;

13 coupldot(6)=-a12+(90*a4*a10+420*a6*a8)*P\

14  -(3780*a8*a4s+9450*a6s*a4)*P2\

15 +75600*a4s*a4*a6*P3-113400*a4q*a4*P4;

16 coupldot(7)=(132*a4*a12+924*a8*a8+990*a10*a6)*P\

17 -(8910*a10*a4s+83160*a4*a6*a8+34650*a6s*a6)*P2\

18 +(332640*a8*a4s*a4+1247400*a4s*a6s)*P3\

19 -6237000*a4q*a6*P4+7484400*a4q*a4s*(P4*P);

20 coupldot(2:7)=P2pi*coupldot(2:7);

21 endfunction

The octave function 12.5 solves the partial differential equation for the scale-dependent effective potential in (12.31). It calls the function defined in Listing 12.6.

Listing 12.5 Flow of effective potential in LPA

1 function flowpde(a4)

2  # Solves the partial differential equation for the

3  # flow of the effective potential by rewriting it

4  # as a system of coupled ode’s. Calls fa.m

5 global Nq;

6 global dqsquareinv;

7 global W;

8 Nq=151;L=5;Nk=800;

9 q=linspace(-L,L,Nq);

10 a2=-1;

11 qsquare=q.*q;

12 dq=2*L/(Nq-1);

13 dqsquareinv=1/(dq*dq);

14 k=linspace(800,0,Nk);

15 V=0.5*a2*qsquare+a4*qsquare.*qsquare/24;

16 Veff=lsode( "fa",V,k);

17 u=Veff(Nk,:);

18 plot(q,V, ’r’,q,u, ’b’)

19 legend( ’Vclass’, ’Veff’);

20 [umin,index]=min(u);

21 upp=(u(index+1)+u(index-1)-2*u(index))*dqsquareinv;

22 printf( "\nE0 = %4.4f\nE1 = %4.4f \n",umin,umin+sqrt(upp));

23 endfunction

Listing 12.6 Called by program 12.5: right-hand side of ode

1 function xdot=fa(V,x)

2 global dqsquareinv;

3 global Nq;

4 global W;

5 xs=x*x;

6 W=dqsquareinv*(shift(V,1)+shift(V,-1)-2*V);

7 xdot=(xs./(xs+W)-1)/pi;

8 xdot(1)=0;

9 xdot(Nq)=0;

10 endfunction

12.9 Problems

12.1 (Weak-Coupling Expansion for Anharmonic Oscillator)

In this exercise we solve the truncated flow equations (12.35) for weak couplings λ∕ω ³ ≪ 1. First we introduce dimensionless variables:

$$\displaystyle \begin{aligned} x=\frac{k}{\omega},\quad e_k=\frac{E_k}{\omega},\quad o_k=\frac{\omega_k}{\omega}\quad \mbox{{and}}\quad \ell_k=\frac{\lambda_k}{24\omega^3}\;. \end{aligned}$$

• (a) Show that the truncated flow equations for the dimensionless variables read

  $$\displaystyle \begin{aligned} \frac{d e_k}{dx}=-\frac{o_k^2}{\pi}\frac{1}{x^2+o_k^2},\quad \frac{d o^2_k}{dx}=-\frac{\ell_k}{\pi} \frac{24 x^2}{(x^2+o_k^2)^2},\quad \frac{d \ell_k}{dx}=\frac{144\ell_k^2}{\pi}\frac{x^2}{(x^2+o_k^2)^3} \;. \end{aligned}$$

• (b) For the harmonic oscillator with vanishing λ, we have e _k = 1∕2, o _k = 1, and ℓ _k = 0. The initial conditions are e _x→∞ = 0, o _x→∞ = 1, and ℓ _x→∞ = λ∕24ω ³. Thus for weak coupling the expansion has the form

  $$\displaystyle \begin{aligned} \begin{array}{rcl} e_k& =&\displaystyle \alpha_0+\alpha_1\varepsilon+\alpha_2\varepsilon^2+\dots\\ o^2_k& =&\displaystyle \;\;1+\beta_1\varepsilon+\beta_2\varepsilon^2+\dots\\ \ell_k& =&\displaystyle \hskip10.5mm\varepsilon+\gamma_2\varepsilon^2+\dots \end{array} \end{aligned} $$

  with ε = λ∕24ω ³ ≪ 1 and scale-dependent coefficient functions. Show that these functions satisfy the differential equations

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \dot\alpha_0& =&\displaystyle -\frac{P}{\pi},\quad \dot\alpha_1=-\frac{x^2P^2}{\pi} \beta_1,\quad \dot\alpha_2=\frac{x^2P^2}{\pi}\left(\beta_1^2P-\beta_2\right)\\ \dot\beta_1& =&\displaystyle -\frac{24x^2P^2}{\pi},\qquad \dot \beta_2=\frac{24x^2P^2}{\pi}\left(2\beta_1 P-\gamma_2\right),\quad \dot\gamma_2=144\frac{x^2P^3}{\pi}\;, \end{array} \end{aligned} $$

  where we defined P ≡ 1∕(1 + x ²). The coefficient functions vanish for x →∞, and the first differential equation has the solution $\alpha (x)=1/2-\arctan (x)/\pi$.

• (c) Calculate β ₁, γ ₂, and β ₂ with an algebraic computer program. Insert the solutions to find the functions α ₂ and α ₃.

• (d) Prove that the coefficient functions have the following small-x expansions:

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \alpha_0(x)& \sim&\displaystyle \frac{1}{2}-\frac{1}{\pi}x+\frac{1}{3\pi}x^3+\dots \\ \alpha_1(x)& \sim&\displaystyle \frac{3}{4}-2\pi x^3+\dots\\ \alpha_2(x)& \sim&\displaystyle -\frac{3}{8}\left(10-\frac{29}{\pi^2}\right) +8\left(3-\frac{4}{\pi^2}\right)x^3+\dots{}\\ \beta_1(x)& \sim&\displaystyle 6-\frac{8}{\pi}x^3+\dots\\ \beta_2(x)& \sim&\displaystyle -12\left(3-\frac{8}{\pi^2}\right)+\frac{168}{\pi}x^3+\dots\\ \gamma_2(x)& \sim&\displaystyle -9+\frac{48}{\pi}x^3+\dots \end{array} \end{aligned} $$

  (12.114)

• (e) If we make a cruder truncation with λ _k = λ of equivalently with γ ₂ = 0, how do the results change? Prove that in this case

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \alpha_2(x)& \sim&\displaystyle -\frac{3}{16}\left(8+\frac{29}{\pi^2}\right) +\frac{1}{\pi^3}(15\pi^2+16)x^3+\dots{}\\ \beta_2(x)& \sim&\displaystyle -2\left(3+\frac{16}{\pi^2}\right) +\frac{96}{\pi}x^3+\dots \end{array} \end{aligned} $$

  (12.115)

12.2 (Fixed Point Solution and Critical Exponents)

Write a program with your favorite algebraic computer system to find the fixed point solution and the critical exponents of the Z ₂ scalar field theory. You should be able to reproduce the plots in Fig. 12.5 and the critical exponents in Table 12.4.

Appendix: A Momentum Integral

In this appendix we calculate the O(p ²) contribution to the integral

$$\displaystyle \begin{aligned} F(p)=\int \mathrm{d}^d q\,\partial_k\left\{Z_kR_k(q)\right\} \, G^2_0(q)\,G_0(p+q){} \end{aligned} $$

(12.116)

for the optimized regulator function (12.10). The integrand is only nonzero for q ² ≤ k ² and in this region

$$\displaystyle \begin{aligned} \partial_k\left\{Z_kR_k(q)\right\}\,G^2_0(q)=\left(\left(k^2-q^2\right)\partial_k Z_k+2kZ_k\right) \varDelta_0^2\;.{} \end{aligned} $$

(12.117)

To proceed we need to consider two cases: |p + q|≤ k and |p + q| > k separately.

The Case |p + q| < k

This is the region located inside of both spheres in Fig. 12.11 where the Green function G ₀(q) = G ₀(p + q) = Δ ₀ is independent of the integration variable q. Let us decompose this variable as q = q _∥ + q _⊥, where q _∥ is parallel and q _⊥ perpendicular to the fixed momentum p. Then the integral has the form

$$\displaystyle \begin{aligned} I_1=\mbox{Vol}\left(S_{d-2}\right)\varDelta_0^3 \int \mathrm{d} q_\parallel \int \mathrm{d}\vert\, q_\perp\vert\, \vert q_\perp\vert^{d-2} \left(\left(k^2-q^2\right)\partial_k Z_k+2kZ_k\right)\;,{} \end{aligned} $$

(12.118)

Fig. 12.11

Sketch of the integration regions in momentum space. Only momenta inside of the ball centered at the origin contribute to the integral. The Green functions are constant in the gray region inside of both spheres. Inside the sickle-shaped light-gray region, G ₀(p, q) is momentum dependent. The two spheres intersect at the lower-dimensional sphere defined by $\{q_\parallel ^*,q_\perp ^* \}=\{-p/2,k^2-p^2/4\}$

where $q^2=q_\parallel ^2+q_\perp ^2$. The volume of the unit sphere originates from the integration over the directions of q _⊥. Now we split the integration domain inside of both spheres, the region marked gray in Fig. 12.11, into two spherical caps:

$$\displaystyle \begin{aligned} \begin{array}{rcl} -k\leq q_\parallel<-\frac{p}{2} \quad & \mbox{and}&\displaystyle \quad 0\leq q^2_\perp\leq k^2-q_\parallel^2\\ -\frac{p}{2}\leq q_\parallel<k-p \quad & \mbox{and}&\displaystyle \quad 0\leq q^2_\perp\leq k^2-(p+q_\parallel)^2\;. \end{array} \end{aligned} $$

The domain of integration in (12.118) is just the union of these two caps. After a shift q _∥→ q _∥− p in the integral over the second cap, we are left with the one-dimensional integral

$$\displaystyle \begin{aligned} &\frac{\mbox{Vol}\left(S_{d-2}\right)\varDelta_0^3}{d-1}\int_{p/2}^k \!\!\mathrm{d} q_\parallel \left\{\left(\frac{4k^2-4q_\parallel^2}{d+1}+2q_\parallel p-p^2\right) \partial_k Z_k +4kZ_k\right\}\\ &\quad \times \left(k^2-q_\parallel^2\right)^{(d-1)/2}\;. \end{aligned} $$

Its second derivative with respect to p at p = 0 is

$$\displaystyle \begin{aligned} \frac{\mathrm{d}^2 I_1}{\mathrm{d} p^2}\Big\vert_{p=0}=-k^dV(B_d)\,\varDelta_0^3\,\partial_k Z_k\;.{} \end{aligned} $$

(12.119)

The Case |p + q| < k

This is the sickle-shaped region inside of the sphere centered at the origin but outside the displaced sphere, marked light-gray in Fig. 12.11. The integral over this region can be written as a difference of two integrals as follows:

$$\displaystyle \begin{aligned} I_2&=\varDelta_0^2\; \left(\int_{-p/2}^k \mathrm{d} q_\parallel\!\! \int_{0}^{\sqrt{k^2-q_\parallel^2}}\mathrm{d} q_\perp \,h\left(q_\parallel,q_\perp\right) \right.\\ & \quad \left.-\int_{-p/2}^{k-p} \mathrm{d} q_\parallel\!\! \int_{0}^{\sqrt{k^2-(q_\parallel+p)^2}}\mathrm{d} q_\perp \,h\left(q_\parallel,q_\perp\right)\right)\;. \end{aligned} $$

The integrand of both integrals is

$$\displaystyle \begin{aligned} h(q_\parallel,q_\perp)=\frac{\left(k^2-q_\parallel^2-q_\perp^2\right)\partial_k Z_k +2kZ_k}{Z_k(q_\parallel+p)^2+Z_k q_\perp^2+a_2}\;. \end{aligned}$$

It is convenient to shift the variable q _∥ in the second integral by − p such that

$$\displaystyle \begin{aligned} I_2&=\varDelta_0^2\,\left( \int_{-p/2}^k \mathrm{d} q_\parallel\!\! \int_{0}^{\sqrt{k^2-q_\parallel^2}}\mathrm{d} q_\perp \,h\left(q_\parallel,q_\perp\right)\right.\\ & \quad \left.-\int_{p/2}^k \mathrm{d} q_\parallel\!\! \int_{0}^{\sqrt{k^2-q_\parallel^2}}\mathrm{d} q_\perp \,h\left(q_\parallel-p,q_\perp\right)\right)\;. \end{aligned} $$

The second derivative of this integral with respect to p at p = 0 is given by

$$\displaystyle \begin{aligned} \frac{\mathrm{d}^2 I_2}{\mathrm{d} p^2}\Big\vert_{p=0}=k^dV(B_d)\,\varDelta_0^4 \left(a\partial_k Z_k+k^2 Z_k\partial_k Z_k-2kZ_k^2\right)\;.{} \end{aligned} $$

(12.120)

Adding the two results (12.119) and (12.120) leads to the simple expression,

$$\displaystyle \begin{aligned} \frac{\mathrm{d}^2 F}{\mathrm{d} p^2}\Big\vert_{p=0}=-2k^{d+1}V(B_d)\,\varDelta_0^4\, Z^2_k\;{} \end{aligned} $$

(12.121)

for the curvature of F(p) in (12.116) at the origin in momentum space.