Week 7, Thursday

The Coleman-Weinberg Effective Potential

\begin{equation} V_\text{eff}(\varphi) = V(\varphi) – \frac{i\hbar}{2} \int \frac{d^4k}{(2\pi)^4} log \left( \frac{k^2 – V”(\varphi)}{k^2} \right) + O(\hbar^2) \end{equation}

TODO: type my lecture notes

Week 7, Wednesday

1PI Effective Action

Let us start by taking a step back and reformulate some of the combinatorics of Feynman diagrams in terms of algebra of the generating function. First, consider a Feynman diagram \begin{equation} D = \cup_{j=1}^k n_J C_j \end{equation} consisting of connected components $C_j$ with multiplicity $n_j$. The Feynman rules for $D$ contain a symmetry factor \begin{equation} \frac{1}{|\mathop{Aut} D|} = \prod_{j=1}^k \frac{1}{n_j! ~ |\mathop{Aut} C_j|} \end{equation} To evaluate the path integral we have to sum over all diagrams, connected or not. Writing $C_j$ for both the Feynman diagram and the expression obtained by applying the Feynman rules, we find \begin{equation} Z = \int \mathcal{D}\phi e^{iS} = \sum_k D_k = \prod_j \sum_{n_j} \frac{1}{n_j!} C_j = \prod_j e^{C_j} = e^{\sum_j C_j}. \end{equation} This suggests to define a functional $W[J]$ as \begin{equation} Z[J] = e^{i W[J]} ,\quad Z_E[J] = e^{W_E[J]} \end{equation} in Minkowski and Euclidean signature, respectively. The $W[J]$ then has an expansion into a sum over connected Feynman diagrams. We can define “connected” correlation functions \begin{equation} \langle \phi(x_1) \cdots \phi(x_n) \rangle_C = \tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \; \cdots \tfrac{1}{i} \frac{\delta}{\delta J(x_n)} \; W[J] \Big|_{J=0}, \end{equation} and, in particular, \begin{equation} \langle \phi(x)\rangle_C = \tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \; W[J] \Big|_{J=0} = \frac{1}{i^2} \frac{1}{Z} \frac{\delta Z}{\delta J(x)} \Big|_{J=0} = \frac{1}{i^2} \langle 0|\phi(x) |0\rangle. \end{equation} The vev of $\phi$ is, of course, $0$ in $\phi^4$-theory because of the $\phi\mapsto -\phi$ symmetry. To get something non-trivial, we should not set $J=0$, this amounts to measuring the response of the field to an external source. Hence, we define the mean field as \begin{equation} \label{eq:meanfield} \varphi(J) = \frac{\delta W}{\delta J} \end{equation} It is perhaps suggestive to think of $W$ in \begin{equation} e^{iW[J]} = \int\mathcal{D}\phi e^{i\int d^4x (\mathcal{L}+J\phi)} \end{equation} as an “effective action” which incorporates all the quantum effects from the path integral. However, it depends on the source $J$ and not on the field. Hence, to define something that deserves the name, we have to replace the $J$-dependence with a dependence on the mean field $\varphi$. The right way to do this is to use the Legendre transform, just like how we change between $(q, \dot q)$ and $(q,p)$-dependence in classical mechanics. Hence, we invert $\varphi=\varphi(J)$ to obtain $J=J(\varphi)$ and define the 1PI effective action} as \begin{equation} \Gamma[\varphi] = W[J(\varphi)] – \int d^dx \; \varphi J(\varphi). \end{equation} Functionally differentiating with respect to the mean field yields \begin{equation} \frac{\delta \Gamma[\varphi]}{\delta \varphi} + J = 0 \quad \Leftrightarrow \quad J = -\frac{\delta \Gamma[\varphi]}{\delta \varphi} \end{equation} We interpret this in the following way: The stationary point of the effective action $\Gamma[\varphi]$ is where the mean field satisfies $J(\varphi)=0$. Hence \begin{equation} \frac{\delta \Gamma[\varphi]}{\delta \varphi} \quad\Rightarrow\quad J=0 \quad\Rightarrow\quad \varphi = \frac{\delta W}{\delta J} \; \Big|_{J=0} = \tfrac{1}{i} \langle 0|\phi(x) |0\rangle, \end{equation} that is, the stationary points of the effective action determine the quantum one-point function. A similar calculation for higher derivatives shows that all correlation functions are determined by the higher derivatives of $\Gamma[\phi]$.

By this argument, the classical solutions of the effective action $\Gamma$ are equivalent to full path integral with the action $S=\int d^4x \mathcal{L}$. In terms of Feynman diagrams, this means that the sum over all tree level $\Gamma$-diagrams equals the sum over all connected $S$-diagrams. It remains to explain why $\Gamma$ is called the one-particle irreducible (1PI) effective action. This is because it can be expanded \begin{equation} \Gamma[\varphi] = S[\varphi] + \sum_{1PI} A(G) \end{equation} where the sum runs over all 1PI graphs $G$ and $A(G)$ is the corresponding amputated diagram. A mathematical proof is quite involved, but we can understand it by the following heuristic argument. According to this equation, the $\Gamma$-Feynman rules have a different type of vertex for each 1PI $S$-Feynman graph. The correspondence between all $\Gamma$ and $S$-Feynman graphs is obtained by shrinking each 1PI $S$-subgraph to the corresponding $\Gamma$-vertex.

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{equation} \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} \end{equation} where \begin{equation} T = \int \frac{d^2p}{(2\pi)^2} \frac{1}{p^2+m^2} \end{equation} 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 \begin{equation} T_L = \frac{1}{N^2} \sum_{p_L^2} \frac{1}{p_L^2 + m_L^2} \end{equation} 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 \begin{equation} m^2_L = m^2 a^2 ,\quad \lambda_L = \lambda^2 a^2 . \end{equation} 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 \begin{equation} \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). \end{equation} 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 \begin{equation} p_j^2 = 4 \sin^2\left(\tfrac{\pi j}{N}\right) ,\quad j \in \{0, \dots, N-1\}. \end{equation}

\subsubsection{Lattice Laplacian in 2D}

The 2-dimensional eigenfunctions on the $N\times N$ square lattice are \begin{equation} \phi_{ij}(x,y) = \phi_i(x) \phi_j(y) \end{equation} so there are $N^2$ eigenvalues \begin{equation} 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\}. \end{equation} Hence, the lattice approximation to the divergent tadpole integral is \begin{equation} 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) } \end{equation} A bit of complex analysis shows that the large volume limit $N\to \infty$ exists. With some more work one can show that \begin{equation} \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) \end{equation} 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, \begin{equation} \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} \end{equation} 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 \begin{equation} m^2_{L,\text{bare}} \approx -1.27 ,\quad \lambda_{L, \text{bare}} \approx 6.0. \end{equation} To get the physical parameters, we have to numerically invert eq.~\eqref{eq:2dphi4bare}, which yields \begin{equation} m_L^2 \approx 0.0980 ,\quad \lambda_L \approx 6.0 . \end{equation} 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 \begin{equation} f = \lim_{a=0} \frac{\lambda_L}{6 m^2_L} \approx 10.8 \end{equation}