## Lattice Simulation Results

We now have everything prepared, all that remains is to pick any observable. We then just have to evaluate it on sample field configurations and average over the samples to approximate the path integral with that observable inserted. The simplest observable would be the average field over the $N\times N$ lattice sites: \begin{equation} \bar\phi = \frac{1}{N^2} \sum_{n\in (\mathbb{Z}/N)^2} \phi(n). \end{equation} While the distribution of samples shows some interesting patterns, the average alone is always going to be zero because of the $\phi\to -\phi$ symmetry of our theory. A more interesting observable would be, for example, \begin{equation} |\bar\phi| = \frac{1}{N^2} \sum_{n\in (\mathbb{Z}/N)^2} |\phi(n)|. \end{equation} By plotting it as a function of $m_L^2$, $\lambda_L$ we make the following observations:

- There seems to be a phase transition between an “unbroken phase” with $\langle|\bar\phi\rangle=0$ for large $m^2$ and a “broken phase”with $\langle|\bar\phi\rangle>0$ for small (large negative) $m^2$.
- The actual place where the phase transition takes place is at finite negative $m_L^2$, for example $\lambda_L\approx 1.0$, $m_L^2\approx -1.3$.
- In the broken phase the actual value of $\langle|\bar\phi\rangle>0$ depends on $m_L^2$, $\lambda_L$. \end{itemize} At least the last part is clear: the potential \begin{equation} V(\phi) = \frac{1}{2}m_L^2 \phi^2 + \frac{1}{4!} \lambda_L \phi^4

has two distinct minimal for $m_L^2 < 0$ whose position depends on $m_L^2$ and $\lambda_L$. It should also not be too surprising that there are qualitative differences between the case where $V(\phi)$ has a single minimum at $V(0)=0$ and the case where it has two separate minima. Though classically the distinction is just whether $m_L^2$ is positive or negative, the fact that the dividing line is not at zero will only find an explanation later on. If we want to numerically find the precise point of the phase transition then it would be nice to have a better observable than $|\bar\phi|$, for example one that has a simple peak at the point of the transition. Such an observable is the susceptibility, which you might have seen in the Ising model as magnetic susceptibility. The only difference is that $\phi\in \pm 1$ in the Ising model. There, the average $\bar\phi$ is the total magnetization. The susceptibility is the change of the magnetization if a constant external field is applied. In the $\phi^4$ theory it is not really justified to call $\bar\phi$ a magnetization, but we can still talk about the change in an external field. Technically, this means we add a source term $-E \mapsto -E + J\bar\phi$ to the exponent of the partition function.

Definition:With $Z(J) = \sum_\mu \exp(-E_\mu + J\bar\phi)$, the susceptibility $\chi$ is the quantity \begin{equation} \chi = \frac{\partial\langle \bar\phi\rangle}{\partial J} \Big|_{J=0}. \end{equation}

We expect that it is only close to the phase transition that a small change in an external field will affect a large change in $\langle\bar\phi\rangle$. However, the definition is not very useful to compute $\chi$ from a lattice simulation. Fortunately, we can rewrite it as \begin{equation} \begin{split} \chi =&\; \frac{\partial}{\partial J} \frac{\sum_\mu \bar\phi e^{-E_\mu +J\bar\phi}} {\sum_\mu e^{-E_\mu +J\bar\phi}} = \frac{\partial}{\partial J} \frac{1}{Z(J)} \frac{\partial}{\partial J} Z(J) \big|_{J=0} \\ =&\; \left[ \frac{1}{Z(J)} \frac{\partial^2 Z(J)}{\partial J^2} – \left( \frac{1}{Z(J)} \frac{\partial Z(J)}{\partial J} Z(J) \right)^2 \right]_{J=0} \\ =&\; \langle \bar\phi^2 \rangle – \big( \langle\bar\phi\rangle \big)^2, \end{split} \end{equation} which is just the variance of the $\bar\phi$ observable. And the variance is of course easily approximated using the sample variance. The result is plotted in the figure below. It shows the susceptibility for fixed $\lambda_L=1.0$ and varying $-2\leq m_L^2 \leq 0$ and three different lattice sizes. See also the notes about the lattice simulation code if you want to try it out yourself!