Week 7, Monday

Lattice Renormalization

Let us go back to the 2-dimensional $\phi^4$ theory on the lattice. In perturbation theory and after Wick rotation, we have the two-point function $$\begin{split} \Gamma^{(2)} =&\; \frac{1}{k^2+m^2} + \frac{1}{(k^2+m^2)^2} \left( -\frac{\lambda}{2} T + C \right) + O(\lambda^2) \\ \frac{1}{\Gamma^{(2)}} =&\; k^2 + m^2 + \frac{\lambda}{2}T – C + O(\lambda^2) \end{split}$$ where $$T = \int \frac{d^2p}{(2\pi)^2} \frac{1}{p^2+m^2}$$ is the divergent tadpole integral and $C$ is a counterterm which cancels the divergence. On the lattice every quantity is a sum over the finitely many lattice points, so is necessarily finite. However, in the continuum limit we have to reproduce the divergence. In renormalization terms, the lattice discretion acts as a regularization just like dimensional regularization. The lattice regularized version of the divergent integral $T$ is $$T_L = \frac{1}{N^2} \sum_{p_L^2} \frac{1}{p_L^2 + m_L^2}$$ where $p_L$ are the lattice momenta and $N$ is the number of lattice points in each of the 2 directions. Also, recall the dimensionless lattice parameters (in 2d) are $$m^2_L = m^2 a^2 ,\quad \lambda_L = \lambda^2 a^2 .$$ By definition, the $p_L^2$ are the eigenvalues of the lattice Laplacian $\Delta_L$, that is, the discretized Laplacian.

Lattice Laplacian in 1D

In one dimension we can think of the discretized function as a vector with $N$ entries. The Laplacian, with periodic boundary conditions, is then the matrix $$\Delta_L = \left( \begin{smallmatrix} 2 & -1 & 0 & 0 & \cdots & 0 & -1 \\ -1 & 2 & -1 & 0 & \cdots & 0 & 0 \\ 0 & -1 & 2 & -1 & \cdots & 0 & 0 \\ 0 & 0 & -1 & 2 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 2 & -1 \\ -1 & 0 & 0 & 0 & \cdots & -1 & 2 \end{smallmatrix} \right).$$ The zero mode is the constant vector $\phi_0 = (1, \dots, 1)$ and the highest frequency mode is the alternating vector $\phi_4 = (1, -1, 1, -1, \dots, -1)$ whose eigenvector is $4$. Hence there are $N$ eigenvalues in the range $p_j^2 \in [0,4]$. A computation yields $$p_j^2 = 4 \sin^2\left(\tfrac{\pi j}{N}\right) ,\quad j \in \{0, \dots, N-1\}.$$

\subsubsection{Lattice Laplacian in 2D}

The 2-dimensional eigenfunctions on the $N\times N$ square lattice are $$\phi_{ij}(x,y) = \phi_i(x) \phi_j(y)$$ so there are $N^2$ eigenvalues $$p^2_{i} = 4 \sin\left(\tfrac{\pi i}{N}\right) + 4 \sin\left(\tfrac{\pi j}{N}\right) ,\quad i, j \in \{0, \dots, N-1\}.$$ Hence, the lattice approximation to the divergent tadpole integral is $$T_L = \frac{1}{N^2} \sum_{i,j = 0}^{N-1} \frac{1}{ m_L^2 + 4 \sin\left(\tfrac{\pi i}{N}\right) + 4 \sin\left(\tfrac{\pi j}{N}\right) }$$ A bit of complex analysis shows that the large volume limit $N\to \infty$ exists. With some more work one can show that $$\lim_{N\to \infty} T_L = \int_0^\infty e^{-m_L^2 t} \left( e^{-2t} I_0(2t) \right)^2 = \frac{2}{(4+m_L^2)\pi} K\big(\tfrac{4}{4+m_L^2}\big)$$ where $I_0$ is the Bessel function and $K$ is the elliptic integral of the first kind. In the lattice literature you can usually find only the semidefinite integral, though it is not as convenient as the closed form for actually evaluating $T_L$. Also, note that we haven’t done the continuum limit $a\to 0$ yet. In that limit, $m_L^2 = m^2 a^2 \to 0$ which diverges because $K$ has a simple pole at $K(1)$. This is of course necessary to reproduce the divergence in the field theory. In other words, this is the lattice analog to the divergence of $T$ in the dimension $d\to 2$ limit in dimensional regularization.

Renormalization Prescription

To renormalize the lattice values we need to pick a suitable renormalization prescription. We also want it to satisfy two physical requirements:

1. The renormalization prescription should distinguish the two phases, because that is what we are interested in. But just looking at the $1/\Gamma^{(2)}$ limit as $k\to 0$ does not, in the zero momentum limit the field just sits around the minimum. Both the symmetry-preserving and symmetry-breaking mimimum look locally the same, so our renormalization prescription should include non-zero momenta.

2. The renormalization prescription should be convenient for lattice calculations.

The solution to both of these requirements is to take $T_L$ as the counterterm, $$\label{eq:2dphi4bare} \begin{split} m^2_{L, \text{bare}} =&\; m_L^2 – \frac{\lambda_L}{2} T_L(m^2_L) \\ \lambda_{L, \text{bare}} =&\; \lambda_L \end{split}$$ In the lattice simulation, we necessarily picked bare values for the parameters by hand. Also, note that the lattice code used $\frac{1}{4}\lambda \phi^4$ instead of $\frac{1}{4!}\lambda \phi^4$ as interaction term, this leads to extra factors of $6$ below. Using the $\frac{}{4!}$ convention, the transition line was numerically around $$m^2_{L,\text{bare}} \approx -1.27 ,\quad \lambda_{L, \text{bare}} \approx 6.0.$$ To get the physical parameters, we have to numerically invert eq.~\eqref{eq:2dphi4bare}, which yields $$m_L^2 \approx 0.0980 ,\quad \lambda_L \approx 6.0 .$$ We now take the continuum limit while staying on the transition line. In the $m_L^2$ vs. $\lambda_L$ plane, the transition is almost on a line. The physically interesting values is the slope at the origin, which defines the critical value $$f = \lim_{a=0} \frac{\lambda_L}{6 m^2_L} \approx 10.8$$