Renormalization-Group Flow in Three Dimensions

Source: Perimeter Scholars International, Quantum Field Theory II, Tutorial 6; 7 worked parts.

1 Renormalization

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

1.1 RG Flow in Three Dimensions

Consider massless Euclidean $\phi^4$ theory in three spacetime dimensions, \[S_E[\phi] =\int d^3x\, \left[ \frac{1}{2}(\partial\phi)^2 +\frac{g}{4!}\phi^4 \right].\tag{1.1}\]

Determine the mass dimension of the coupling $g$.

$g$ has mass dimension $4-d = 1$.

For two-to-two scattering with incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, define the Mandelstam variables \[s=(p_1+p_2)^2, \qquad t=(p_1-p_3)^2, \qquad u=(p_1-p_4)^2.\tag{1.2}\] In the conventions of the sheet, \[s+t+u=-4m^2=0,\tag{1.3}\] where the final equality uses $m=0$.
Write the four-point irreducible Green function $\overline\Gamma_{(4)}$ through one loop in terms of $p_1,p_2,p_3,p_4$. Set up the loop integrals, but do not evaluate them yet.

\[\begin{aligned} -\overline\Gamma_{(4)} &= -g + \frac{(-g)^2}{2}\int d^dk \frac{1}{k^2(k+p_1+p_2)^2} + (p_2 \leftrightarrow -p_3) + (p_2 \leftrightarrow -p_4) + \cdots. \end{aligned}\tag{1.4}\]
In four dimensions the corresponding integrals are ultraviolet divergent. Show, by power counting or direct analysis, that the three-dimensional loop integrals are ultraviolet finite. No UV regulator is needed here.

For $d=4$, the integral is logarithmic divergent. For $d=3$, it remains finite.
Show that \[\overline\Gamma_{(4)}(s,t,u) =g-\frac{g^2C}{2} \left[ \sqrt{s^{-1}}+\sqrt{t^{-1}}+\sqrt{u^{-1}} \right], \qquad C>0.\tag{1.5}\] Determining the precise value of $C$ using Feynman parameters is optional.

\[\begin{aligned} \overline\Gamma_{(4)} &= g - \frac{g^2}{2}\left( 2\pi\int ^1_{-1}d\cos\theta \int^\infty_0 dk \frac{1}{k^2 + s +2\sqrt{s}\cos\theta} +(s\leftrightarrow t) + (s\leftrightarrow u)\right)\\ &=g - \frac{g^2}{2}\left( \frac{\pi}{\sqrt{s}} \int^\infty_0\frac{dk}{k} \log\left(\frac{k^2 + s+2\sqrt{s}}{k^2 + s-2\sqrt{s}}\right) +(s\leftrightarrow t) + (s\leftrightarrow u)\right)\\ &=g - \frac{g^2}{2}\left( \frac{2\pi}{\sqrt{s}} \int^\infty_0\frac{dk}{k} \log\left(\frac{k + \sqrt{s}}{k-\sqrt{s}}\right) +(s\leftrightarrow t) + (s\leftrightarrow u)\right)\\ &=g - \frac{g^2}{2}\left( \frac{2\pi}{\sqrt{s}} \int^\infty_0\frac{dk}{k} \log\left(\frac{k + 1}{k-1}\right) +(s\leftrightarrow t) + (s\leftrightarrow u)\right)\\ &=g - \frac{g^2}{2}2\pi\int^\infty_0\frac{dk}{k} \log\left(\frac{k + 1}{k-1}\right)(\sqrt{s^{-1}}+\sqrt{t^{-1}}+\sqrt{u^{-1}}), \end{aligned}\tag{1.6}\] where we have rescaled $k \to k/\sqrt{s}$ in the second last equality.

Although the symmetric choice $s=t=u=\mu^2$ is incompatible with the preceding Mandelstam relation, adopt it as the simplified subtraction point used in the sheet and define \[h_R =\mu^{-1}\overline\Gamma_{(4)} (s=t=u=\mu^2).\tag{1.7}\] Hence, \[h_R =\mu^{-1} \left[ g-\frac{3}{2}g^2\mu^{-1}C \right].\tag{1.8}\]
At an arbitrary energy $E$, define \[h_{\mathrm{eff}}(E) =E^{-1}\overline\Gamma_{(4)}(s=t=u=E^2),\tag{1.9}\] together with \[\beta(h_{\mathrm{eff}}) =E\frac{dh_{\mathrm{eff}}}{dE}.\tag{1.10}\] Show, through order $g^2$, that \[\beta(h) =-h+\frac{3}{2}Ch^2+\text{higher-order terms}.\tag{1.11}\]

\[\beta(h) = E\left(-\frac{1}{E^2}\left(g - \frac32g^2E^{-1}C\right) + \frac{1}{E}\cdot \frac32 g^2\frac{1}{E^2}C\right) = -h + \frac32 Ch^2 + \mathcal{O}(g^3).\tag{1.12}\]
Plot $\beta(h)$ as a function of $h$.
Suppose $h$ is small and positive at some finite energy. Show that $h\to0$ as $E\to\infty$, and identify this zero as the ultraviolet fixed point of the RG flow.

Solving the RG equation, we find \[h(E) = \frac{1}{\frac{3}{2}C-\frac{h_0}{\frac32Ch_0-1}\frac{E}{E_0}}.\tag{1.13}\] Clearly, as $E\to \infty$, $h \to 0$. Note that $\beta(0) = 0$, and hence this is a UV fixed point.
Starting again from small positive $h$, show that the flow approaches a finite value $h_0$ as $E\to0$, where $h_0$ is the nonzero root of $\beta(h)$. Interpret $h_0$ as an infrared fixed point.

As $E \to 0$, $h \to \frac{2}{3C}$. Note that $\beta(\frac{2}{3C}) =0$, and hence this is an IR fixed point. This suggests the existence of the Wilson-Fisher fixed point. Note that at this point, the theory become strongly coupled, since all higher order terms become relevant.