Renormalization in Massive φ⁴ Theory

Source: Perimeter Scholars International, Quantum Field Theory II, Homework 2; 15 worked parts.

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

1 Renormalization for Massive $\phi^4$ Theory

Work in Euclidean signature and retain terms through one loop. Define the renormalized action by \[S_R[\phi] =S[\phi]+\hbar\Delta_1S[\phi]+O(\hbar^2),\tag{1.1}\] where \[S[\phi] =\int d^4x\, \left[ \frac{1}{2}(\partial\phi)^2 +\frac{m_R^2}{2}\phi^2 +\frac{g_R}{4!}\phi^4 \right]\tag{1.2}\] and \[\Delta_1S[\phi] =\int d^4x\, \left[ \frac{B_1}{2}\phi^2 +\frac{C_1}{4!}\phi^4 \right].\tag{1.3}\] The cutoff-dependent coefficients $B_1$ and $C_1$ are chosen so that physical quantities expressed in terms of $m_R$ and $g_R$ remain finite as $\Lambda\to\infty$.

The one-loop four-point 1PI function is \[\begin{aligned}\widehat\Gamma_R^{(4)}(p_i) &={} g_R -\frac{\hbar g_R^2}{2} \Bigl[ I(p_1+p_2,m_R) +I(p_1+p_3,m_R) +I(p_1+p_4,m_R) \Bigr] \notag\\ &\quad+\hbar C_1+O(\hbar^2).\end{aligned}\tag{1.4}\] Using a hollow square for the quartic counterterm, these contributions have the following topologies:

In four-dimensional Euclidean spacetime, write the sharp-cutoff integral corresponding to \[I(p_1+p_2,m_R;\Lambda),\tag{1.5}\] where $\Lambda$ regulates the ultraviolet divergence.

\[\begin{aligned} I(p_1+p_2,m_R;\Lambda) &= \int \frac{d^4k}{(2\pi)^4}\frac{1}{(k^2+m_R^2)[(k+p_1+p_2)^2+m_R^2]}\\ &=\frac{4\pi}{(2\pi)^4}\int_0^\pi d\chi\sin^2 \chi\int^\Lambda_0dk \frac{k^3}{(k^2+m_R^2)(k^2+ s + 2k\sqrt{s}\cos\chi +m_R^2)}. \end{aligned}\tag{1.6}\]
Set $m_R^2=0$ and calculate the leading large-$\Lambda$ behavior of $I(p,0;\Lambda)$.

\[\begin{aligned} I(p_1+p_2,0;\Lambda) &= \frac{1}{4\pi^3}\int_0^\Lambda dk\, k \int_{0}^\pi d\chi\frac{\sin^2\chi}{k^2 +s +2k\sqrt{s}\cos\chi}\\ &= \frac{1}{4\pi^3}\int_0^\Lambda dk\, k \frac{\pi}{4k^2s}\left(k^2 +s - \sqrt{(k^2+s)^2 -4k^2s}\right)\\ &=\frac{1}{16\pi^2s}\int ^\Lambda_0\frac{1}{k}(k^2+s - |k^2-s|)\\ &=\frac{1}{16\pi^2s}\left(\int^\Lambda_{\sqrt{s}}dk \frac{2s}{k} + \int_0^{\sqrt{s}}dk \frac{2k^2}{k}\right)\\ &=\frac{1}{16\pi^2s}\left(1 + \log\frac{\Lambda^2}{s}\right). \end{aligned}\tag{1.7}\]
By differentiating under the integral, show that \[\frac{\partial}{\partial(m_R^2)}I(p,m_R;\Lambda)\tag{1.8}\] is finite as $\Lambda\to\infty$. Conclude that the divergent part in the massive theory is the same as the one identified in the massless theory.

By differentiating with respect to $m_R^2$, the integral now looks like \[\begin{aligned} \sim\int^\Lambda_0 dk \frac{k^3}{k^6}, \end{aligned}\tag{1.9}\] which is finite. Since the rate of change of the integral remains finite, the divergent part must be the same for massive and massless theory.

For parts d)--h), work in massless $\phi^4$ theory.
Determine the divergent part of $C_1$ required to make $\widehat\Gamma_R^{(4)}(p_i)$ finite and impose \[\widehat\Gamma_R^{(4)}(p_i^{\mathrm{ref}})=g_R, \qquad |p_1^{\mathrm{ref}}+p_2^{\mathrm{ref}}| =|p_1^{\mathrm{ref}}+p_3^{\mathrm{ref}}| =|p_1^{\mathrm{ref}}+p_4^{\mathrm{ref}}| =\mu.\tag{1.10}\] Only the terms in the counterterm that diverge as $\Lambda\to\infty$ need be given explicitly.

Using the result from b), we have \[g_R - \frac{\hbar g_R^2}{2}\frac{3}{16\pi^2}\left(1 + \log{\frac{\Lambda^2}{\mu^2}}\right) + \hbar C_1 = g_R \implies C_1 = \frac{3g_R^2}{32\pi^2}\left(1 + \log{\frac{\Lambda^2}{\mu^2}}\right).\tag{1.11}\]
Find $\widehat\Gamma_R^{(4)}(p_i,g_R,\mu)$ at general external momenta. Renormalization-scale independence requires \[\widehat\Gamma_R^{(4)}(p_i,g_R,\mu) =\widehat\Gamma_R^{(4)}(p_i,g_R',\mu'),\tag{1.12}\] where the primed parameters are defined at the scale $\mu'$.

\[\begin{aligned} \widehat\Gamma_R^{(4)}(p_i,g_R,\mu) &= g_R -\frac{\hbar g_R^2}{32\pi^2}\left(\log\frac{\mu^2}{s} + \log\frac{\mu^2}{t} +\log\frac{\mu^2}{u}\right)\\ &=g_R' -\frac{\hbar g_R'^2}{32\pi^2}\left(\log\frac{\mu'^2}{s} + \log\frac{\mu'^2}{t} +\log\frac{\mu'^2}{u}\right). \end{aligned}\tag{1.13}\]
Evaluate the preceding equality at a convenient momentum configuration and derive $g_R'(\mu')$ in terms of $g_R(\mu)$, $\mu$, and $\mu'$.

Take $s=u=t = \mu'^2$, we find \[g_R'(\mu') = g_R(\mu) -\frac{3\hbar g_R^2(\mu)}{32\pi^2}\log\frac{\mu^2}{\mu'^2}.\tag{1.14}\]
Define the effective coupling by \[g_{\mathrm{eff}}(E) =\begin{cases} g_R(\mu), & E=\mu,\\ g_R'(\mu'), & E=\mu', \end{cases}\tag{1.15}\] and the beta function by \[\beta(g_{\mathrm{eff}}) \equiv E\frac{d}{dE}g_{\mathrm{eff}}(E).\tag{1.16}\] Compute the one-loop beta function and plot it as a function of $g_{\mathrm{eff}}$.

The definition here is not very conventional. It is equivalent to defining \[g_\mathrm{eff}(E) = \Gamma_R^{(4)}(E,g_R,\mu),\tag{1.17}\] which immediately implies \[g_{\mathrm{eff}}(E) =\begin{cases} g_R(\mu), & E=\mu,\\ g_R'(\mu'), & E=\mu'. \end{cases}\tag{1.18}\] Then, we have \[\beta(g_\mathrm{eff}) = \frac{3\hbar g_R^2(\mu)}{16\pi^2} = \frac{3\hbar g_\mathrm{eff}^2}{16\pi^2}.\tag{1.19}\]
Explain what this beta function implies for the interaction strength at high and low energies.

Solving the beta function, we have \[\frac{1}{g_\mathrm{eff}(E)} = \frac{1}{g_0} + \frac{3\hbar}{32\pi^2}\log\frac{\Lambda^2}{E^2}.\tag{1.20}\] Therefore, the theory is free at $IR$ and becomes strongly coupled at the UV.

The four-point diagrams introduce no additional divergence in the massive theory, so the same calculation can be repeated there. For mass renormalization, consider the one-loop two-point 1PI function \[\widehat\Gamma_R^{(2)}(p) =p^2+m_R^2 +\frac{\hbar g_R}{2}T(m_R) +\hbar B +O(\hbar^2).\tag{1.21}\] Using a filled square for the quadratic counterterm, its topologies are

Write the sharp-cutoff integral $T(m_R;\Lambda)$ that regulates the UV divergences.

\[T(m_R;\Lambda) = \int \frac{d^4k}{(2\pi)^4}\frac{1}{k^2 +m^2} = \frac{1}{8\pi^2}\int^\Lambda_0dk \frac{k^3}{k^2+m^2.}\tag{1.22}\]
Evaluate $T(m_R;\Lambda)$ at large $\Lambda$. Identify separately the quadratic divergence, logarithmic divergence, and term that remains finite as $\Lambda\to\infty$.
Hint: Radial coordinates may be useful. The area of the unit $n$-sphere is \[\int d^n\Omega =\frac{2\pi^{n/2}}{\Gamma(n/2)}.\tag{1.23}\]

\[T(m_R;\Lambda) = \frac{1}{16\pi^2}\left(\Lambda^2 - m_R^2\log \frac{\Lambda^2 +m_R^2}{m_R^2}\right) = \frac{1}{16\pi^2}\left(\Lambda^2 - m_R^2\log \frac{\Lambda^2 }{m_R^2} + \mathcal{O}(\Lambda^{-2})\right).\tag{1.24}\]

The physical mass is determined by the pole of the propagator, equivalently by \[\widehat\Gamma_R^{(2)} (p_i^{\mathrm{ref}},m_R,g_R,\mu)=0, \qquad |p_i^{\mathrm{ref}}|=m_{\mathrm{phys}},\tag{1.25}\] for any $\mu$. For practical subtraction, impose instead \[\widehat\Gamma_R^{(2)}(0,m_R,g_R,\mu)=m_R^2, \qquad \mu=m_R,\tag{1.26}\] and take the mass counterterm to have the form \[B(m_R,g_R,\mu,\Lambda) =B_0(g_R,\mu,\Lambda) +m_R^2B_1(g_R,\mu,\Lambda).\tag{1.27}\]
Determine $B_0(g_R,\mu,\Lambda)$ and $B_1(g_R,\mu,\Lambda)$ so that the preceding renormalization condition holds.

We can rewrite \[\widehat\Gamma_R^{(2)}(p,m_R,g_R,\mu) = p^2 + m_R^2 + \hbar\left(\frac{g_R(\mu)\Lambda^2}{32\pi^2} + B_0 + m_R^2\left(-\frac{g_R(\mu)}{32\pi^2}\log\frac{\Lambda^2}{\mu^2} - \frac{g_R(\mu)}{32\pi^2}\log\frac{\mu^2}{m_R^2}+B_1\right)\right).\tag{1.28}\] The renormalisation condition implies \[B_1(g_R,\mu,\Lambda) = \frac{g_R}{32\pi^2}\log\frac{\Lambda^2}{\mu^2} , \quad B_0(g_R,\mu,\Lambda) = -\frac{g_R\Lambda^2}{32\pi^2}.\tag{1.29}\]
Check that these counterterms make $\widehat\Gamma_R^{(2)}(p_i,m_R,g_R,\mu)$ finite for arbitrary $m_R$ and $p$.

\[\widehat\Gamma_R^{(2)}(p,m_R,g_R,\mu) = p^2 + m_R(\mu)^2 -\frac{\hbar g_R(\mu)}{32\pi^2}\log\frac{\mu^2}{m_R(\mu)^2}.\tag{1.30}\]
Require $\widehat\Gamma_R^{(2)}$ to be independent of the renormalization scale, as for the four-point function, and derive an equation relating $[m_R'(\mu')]^2$ and $[m_R(\mu)]^2$. Retain terms only through order $\hbar$ and express the result in terms of $g_R$.

\[m_R^2(\mu) - \frac{\hbar g_R(\mu)m_R^2(\mu)}{32\pi^2} \log\left(\frac{\mu^2}{m_R^2(\mu)}\right) = [m_R'(\mu')]^2 - \frac{\hbar g_R'(\mu')[m_R'(\mu')]^2}{32\pi^2} \log\left(\frac{\mu'^2}{[m_R'(\mu')]^2}\right).\tag{1.31}\] This implies \[[m_R'(\mu')]^2 = m_R^2(\mu) + \frac{\hbar g_R(\mu)m_R^2(\mu)}{32\pi^2} \log\left(\frac{\mu'^2}{\mu^2}\right) +O(\hbar^2).\tag{1.32}\]
Show that when $\mu^2=m_R^2$, the renormalized mass agrees in magnitude with the physical mass: \[|m_R|=|m_{\mathrm{phys}}|.\tag{1.33}\]

\[\widehat\Gamma_R^{(2)}(|m_\mathrm{phys}|,m_R = \mu,g_R,\mu) = 0 \implies |m_\mathrm{phys}| = |m_R|.\tag{1.34}\]
Calculate the anomalous dimension of the mass-squared operator, \[\gamma_{m^2}(g_R) \equiv \mu\frac{\partial}{\partial\mu}\ln m_R^2(\mu).\tag{1.35}\]

Starting from the physical mass \[m_{\mathrm{phys}}^2 = m_R^2(\mu) \left[ 1- \frac{\hbar g_R(\mu)}{32\pi^2} \log\left(\frac{\mu^2}{m_R^2(\mu)}\right) \right] +O(\hbar^2),\tag{1.36}\] and hence \[\log m_{\mathrm{phys}}^2 = \log m_R^2(\mu) - \frac{\hbar g_R(\mu)}{32\pi^2} \log\left(\frac{\mu^2}{m_R^2(\mu)}\right) +O(\hbar^2).\tag{1.37}\] Then, \[0 = \frac{d\log m_\mathrm{phys}^2}{d\log \mu} = \gamma_{m^2}(g_R) - \frac{2\hbar g_R}{32\pi^2} + \mathcal{O}(\hbar^2),\tag{1.38}\] which implies \[\gamma_{m^2}(g_R) = \frac{\hbar g_R}{16\pi^2} +O(\hbar^2).\tag{1.39}\]

At one loop in $\phi^4$ theory, no wavefunction renormalization is required; the two-point correction is momentum independent at this order.

2 Renormalization Group Flow for the $O(N)$ Model

Let $\vec\phi=\{\phi_\alpha\}_{\alpha=1}^{N}$ be an $N$-component scalar field with $O(N)$ symmetry. In $D$-dimensional Euclidean space, take \[S_E[\vec\phi] =\int d^Dx\, \left[ \frac{1}{2}(\nabla\vec\phi)^2 +U(\vec\phi^{\,2}) \right],\tag{2.1}\] where \[X\equiv\vec\phi^{\,2} =\sum_{\alpha=1}^{N}\phi_\alpha^2,\tag{2.2}\] and $U(X)$ is regular. Restrict the Fourier support to $|k|\leq\Lambda$ and write \[Z =\int\mathcal D_\Lambda[\vec\phi]\, \exp\!\left[-\frac{1}{\hbar}S_E[\vec\phi]\right].\tag{2.3}\] Separate the field into slow and fast components, \[\vec\phi =\widetilde{\vec\phi}+\vec\phi_{\mathrm{fast}},\tag{2.4}\] where $\widetilde{\vec\phi}$ has modes $|k|\leq\Lambda/S$ and $\vec\phi_{\mathrm{fast}}$ has modes $\Lambda/S<|k|\leq\Lambda$. Integrate a thin shell using \[S=1+ds, \qquad 0<ds\ll1.\tag{2.5}\]

Hold $\widetilde{\vec\phi}$ fixed and expand $U(X)$ through second order in $\vec\phi_{\mathrm{fast}}$ (the Gaussian approximation), expressing the result in terms of $U'$ and $U''$. Keep the vector structure explicit.
In the local-potential approximation, take $\widetilde{\vec\phi}$ to be constant, so only its $k=0$ mode is nonzero. Show that the fast-mode integral separates into one longitudinal mode parallel to $\widetilde{\vec\phi}$ and $N-1$ decoupled transverse modes.
Show that the two kinds of mode renormalize the potential according to \[U\longrightarrow U_{\mathrm{eff}} =U+ds\left( \Delta U_{\parallel} +(N-1)\Delta U_{\perp} \right),\tag{2.6}\] and calculate $\Delta U_{\parallel}(X)$ and $\Delta U_{\perp}(X)$.
Show that the rescalings \[\vec\phi\longrightarrow (1-ds)^{(2-D)/2}\vec\phi, \qquad x\longrightarrow(1-ds)x\tag{2.7}\] restore the cutoff to $\Lambda$.
Determine the local-potential flow equation \[S\frac{\partial}{\partial S}U_S(X)=?\tag{2.8}\] for the $O(N)$ model. As in the lecture, set $\Lambda=1$ if desired.
Check that $N=1$ reproduces the one-component flow for $V(\phi)$, using $U(\phi^2)=V(\phi)$.
Truncate the potential at second order in $X=\vec\phi^{\,2}$, \[U(X)=\frac{t}{2}X+\frac{u}{8}X^2.\tag{2.9}\] Expand the flow equation consistently through $X^2$. Derive both the flow equations and the Wilson functions $S\,\partial g_S/\partial S$ for the renormalized couplings $t_S$ and $u_S$.