Study of a gauge theory with a pure $Z(N)$ lattice in 4D
This is a solution attempt at the last problem from the 2025 Rudolf Ortvay competition proposed by Dávid Pesznyák. I want to develop my skills in this particular subject, so I paraphrase the problem here as a discussion. Firstly, there is many interest in describing our world with gauge theories for technical reasons such as renormalization of observables, but consider first that such theories involve many symmetries that are natural to rationalize. Question the fact that electrons all have the same charge or that systems with conservation of energy are far simpler to describe. Those can be viewed as symmetries (assumptions) that in many cases describe our world very well up to a given generalization.
A common way to achieve a gauge theory is with a path integral formulation, which can be viewed through the lens of the lattice field theory method, making sense of infinite-dimensional integrals with a small spacetime lattice of length $\Lambda$. The gauge field $U_{\mu}\left(x\right)$ is the integration variable on those path integrals, so that we can informally think of these integrals as a sum of a quantity over all symmetrical configurations. When introducing lattice field theory, these fields live on the edges/links of the lattice in a discrete way.
We will be considering QED whose gauge group is the Lie group $U\left(1\right)$, the group of pure phases $U_{\mu}\left(x\right)=e^{i\varphi}\in U\left(1\right)$ ($\varphi\in\left[0,2\pi\right)$). For the lattice theory we discretize $U\left(1\right)$ by replacing it with $Z\left(N\right)$, where $U_{\mu}\left(x\right)=e^{2\pi in/N}\in Z\left(N\right)$ are its elements with $n=0,\ldots,N-1$. This is due to the approximation when $N\to\infty$, then $Z\left(N\right)\to U\left(1\right)$. The Euclidean action in this new group is $Z\left(N\right)$ is $$ S\left[\left\{ U\right\} \right]\equiv-\beta\sum_{x\in\Lambda,\:\mu<\nu}\Re\left\{ \text{Tr}\left[P_{\mu\nu}\left(x\right)\right]\right\}. $$
To see why, consider the Yang-Mills action $$ S\sim\frac{1}{4g^{2}}\int d^{D}x\text{Tr}\left\{ F_{\mu\nu}F^{\mu\nu}\right\}. $$ The gauge fields $A_{\mu}$ are generators of the group $U\left(1\right)$, consider the parallel transport $\psi\left(x+\epsilon^{\mu}\right)\approx\left(1+i\epsilon n^{\mu}A_{\mu}\left(x\right)\right)\psi\left(x\right)=U\left(x\right)\psi\left(x\right)$ (notice that in a lattice the smallest step is $\epsilon=\Lambda$). Since we are talking about a Lie group, then $$ U_{\mu}\left(x\right)=e^{i\Lambda A_{\mu}\left(x\right)}. $$ Notice now the term $\bar{\psi}\left(x+\epsilon^{\mu}\right)\psi\left(x+\epsilon^{\mu}\right)$ must be invariant under the transformation $g$ (a kinetic term), then since $\psi\left(x+\epsilon^{\mu}\right)=U_{\mu}\left(x\right)\psi\left(x\right)$, $\psi\left(x\right)\to g\left(x\right)\psi\left(x\right)$ and $\bar{\psi}\left(x+\epsilon^{\mu}\right)\to\bar{\psi}\left(x+\epsilon^{\mu}\right)\bar{g}^{-1}\left(x+\epsilon^{\mu}\right)$ we must have $$ \bar{\psi}\left(x+\epsilon^{\mu}\right)U_{\mu}\left(x\right)\psi\left(x\right) \to\bar{\psi}\left(x+\epsilon^{\mu}\right)\bar{g}^{-1}\left(x+\epsilon^{\mu}\right)U_{\mu}^{\prime}\left(x\right)g\left(x\right)\psi\left(x\right) $$ $$ =\bar{\psi}\left(x+\epsilon^{\mu}\right)U_{\mu}\left(x\right)\psi\left(x\right), $$ This allows us to construct another invariant quantity which is called plaquette, corresponding to the smallest closed loop in the lattice, $$ P_{\mu\nu}\left(x\right)=U_{\mu}\left(x\right)U_{\nu}\left(x+\epsilon^{\mu}\right)U_{\mu}\left(x+\epsilon^{\nu}\right)^{*}U_{\nu}\left(x\right)^{*}, $$ where $\left(^{*}\right)$ is the complex conjugation and the number of plaquettes for one site is given by the following argument: in order to construct a plaquette we need to pick two different directions ($\mu$ and $\nu$) and connect the smallest loop between them, and since in 4D we have 4 directions and we choose 2, the number of ways we can do this corresponds to the number of plaquettes in 4D, ${4 \choose 2}=6$, so for all plaquettes, $6M^{4}$. This plaquette term can then be expanded using $A_{\nu}\left(x+\epsilon^{\mu}\right)\approx A_{\nu}\left(x\right)+\epsilon\partial_{\mu}A_{\nu}$ and the Baker-Campbell-Hausdorff formula $$ P_{\mu\nu}\left(x\right) =e^{i\Lambda A_{\mu}\left(x\right)}e^{i\Lambda A_{\nu}\left(x+\epsilon^{\mu}\right)}e^{-i\Lambda A_{\mu}\left(x+\epsilon^{\nu}\right)}e^{-i\Lambda A_{\nu}\left(x\right)} $$ $$ =\text{lots of math} $$ $$ \approx e^{i\Lambda^{2}F_{\mu\nu}\left(x\right)}. $$ By substituting this on $S[{U}]$ we have a term proportional to the Yang-Mills Lagrangian up to some factors and constants in the limit $\Lambda\ll1$.
Now understanding that $P_{\mu\nu}\left(x\right)\in Z\left(N\right)\approx U\left(1\right)$, we are able to represent it with $P_{\mu\nu}\left(x\right)=e^{2\pi in_{\mu\nu}/N}$ where $n_{\mu\nu}$ it the number obtained from expanding the plaquette in terms of operators from $Z\left(N\right)$, that is, $n_{\mu\nu}=n_{\mu}\left(x\right)+n_{\nu}\left(x+\epsilon^{\mu}\right)-n_{\mu}\left(x+\epsilon^{\nu}\right)-n_{\nu}\left(x\right)$. In a $Z\left(N\right)$ gauge theory each link variable isn’t an $N\times N$ matrix at all, but simply $U_{\mu}\left(x\right)=e^{2\pi in_{\mu}\left(x\right)/N}$, $n_{\mu}\left(x\right)\in\left\{ 0,\ldots,N-1\right\} $, which you can think of as a 1x1 "matrix". So that we can write $$ S\left[\left\{ U\right\} \right]=-\beta\sum_{x\in\Lambda,\:\mu<\nu}\Re\left\{ e^{2\pi in_{\mu\nu}/N}\right\} =-\beta\sum_{x\in\Lambda,\:\mu<\nu}\cos\left(\frac{2\pi in_{\mu\nu}}{N}\right). $$ We can use a computer to calculate $n_\mu(x)$ and $n_{\mu\nu}(x)$ in a finite lattice represented by, for example, NumPy tensors. I implemented a code, as requested by the problem statement, that measures every 10th iteration of the average plaquette, defined as $$ \langle P\rangle\equiv1-\frac{1}{6M^{4}}\sum_{x\in\Lambda,\:\mu<\nu}\Re\left\{ \text{Tr}\left[P_{\mu\nu}\left(x\right)\right]\right\} =1-\frac{1}{6M^{4}}\sum_{x\in\Lambda,\:\mu<\nu}\cos\left(\frac{2\pi in_{\mu\nu}}{N}\right). $$ Lets take a minute to talk about the Metropolis algorithm which is used to simulate a finite lattice. Here is a great video explaining it:
The main message we want to extract from this video is that the Metropolis-Hasting algorithm is going to reproduce other distribution based on outputs of a known distribution. In this case the problem statement assumes this distribution to be proportional to $\exp\big(S[\{U\}]\big)$. A good guess if you look at Feynman path integrals, which have this factor. Other concepts we can obtain from the video is detailed balance, a condition in which the probability of starting at state $X$ ($\pi\left(x\right)$) and going to a state $Y$ ($\pi\left(x\right)g\left(y|x\right)$, $\rightarrow$) is the same as the probability of starting at $Y$ ($\pi\left(y\right)$) and going to $X$ ($\pi\left(y\right)g\left(x|y\right)$, $\leftarrow$). The random processes of the Metropolis algorithm must respect the following restriction $$ \alpha\left(y|x\right)=\min\left(1,\frac{\pi\left(y\right)g\left(x|y\right)}{\pi\left(x\right)g\left(y|x\right)}\right), $$ where $\alpha\left(y|x\right)$ is a probability function for “accepting a change" to $Y$ when at state $X$ (notice it is in $\left[0,1\right]$). This function reflects how much “imbalance” there is in the system, for example $\pi\left(y\right)g\left(x|y\right)<\pi\left(x\right)g\left(y|x\right)$ gives values for $\alpha$ closer to 1 as this inequality becomes an equality. And if the system is imbalanced, you simply accept the change in hopes it will tip the scale a bit. The trick for the Metropolis algorithm is that when this scale is even, it can then mimic the known distribution we fed, and that is exactly what we will do to prepare, or thermalize, our simulation. There are important computational tricks I am glancing over, and one of them is the format of our acceptance function. It should be something like $$ \alpha\left(y|x\right)=\min\left(1,e^{\Delta S}\right), $$ where $\Delta S$ is the difference between the action before the proposed update and the new action. Computationally this is very interesting because we don't need to worry about normalization factors of the distribution (a feature which the Metropolis algorithm is known for) and $\Delta S$ is easy to compute as if you look at the action $S\left[\left\{ U\right\} \right]$ in terms of the angles, only few of these cosines is going to change, and the others that remain are going to cancel out in the difference. A system with $M=4$ sites per dimension has its continuous generator group of translations replaced by a discretized group $Z(N)$, i.e, our lattice, where $N$ is the number of discrete parts. In the simulation we made $N\in[3,10]$. The coupling constant $\beta$ is varied for each measurement of the average plaquettes, and each measurement is made with 10 updates of the ensemble simulation. As a result, we plot a graph $\langle P \rangle$ vs $\beta$ for various values of $N$:
Again, this is me trying to develop my skills (ᴗ_ ᴗ。) If you spotted a mistake please don't be shy and e-mail me :)
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