I am trying to read a paper (https://arxiv.org/abs/2206.07725) that models electric and magnetic charges on a lattice model. But I am new to lattice theories, so I might be misunderstanding a lot of things here. I do have a basic understanding of QFT. My confusion stemmed while reading this paper, but I think my question is general enough. So it might be answerable without reference to this paper. But let me describe it for the sake of completeness. In this model, the authors have defined charge current $n_\mu$ on the links ( which is coupled to the gauge field $a_\mu$): $$ S = \frac{1}{4g^2} \sum_P F^2_{\mu\nu} + \frac{iN\theta}{16 \pi^2} \sum_{r,R} F_{\mu\nu}(r)K(r-R)\tilde{F}_{\mu\nu}(R) - iN \sum_l n_\mu a_\mu \hspace{2cm} (3.2)$$ ( Here l are links, and P are plaquettes ). So basically, we are defining the electric charge current $n_\mu$ running over the links. To compute the partition function Z, we are summing over all possible configurations of $n_\mu$.

Note that in this action, there is NO kinetic term for the charged matter carrying electric current. Yet we are summing over all $n_\mu$ configurations (which is akin to doing path integral over the charge carrying field). In the third last line on page 10, authors say:

"This lattice gauge theory contains both $\textbf{dynamical}$ charges and monopoles."

From this, what I understand is that the matter field ( that leads to the current $n_\mu$ ) is not a background field. It is dynamic and hence fluctuating.

But I am still confused. So my confusion is:

  1. Is it okay to path integrate a field without having its kinetic term?
  2. Why is there no kinetic term?
  3. What would it represent physically? I mean, do we have (charged) matter field in the theory or not.

My best guess is that such a theory will indeed represent "dynamic" electric charges i.e. they can be pair produced. But such "electron/positron" particles will NOT propagate since there is no kinetic term for this field.

I might have said many incorrect things. Please feel free to educate me on obvious things as well.


2 Answers 2


I will answer your questions in their field-theoretic form, in a general setting, and then I will explain in what guise they appear, in the referenced paper [1]. $\newcommand{\d}{\mathrm{d}}$


  • It is okay to integrate a field without kinetic terms, these usually represent Lagrange multiplier or discrete gauge fields. The case in [1] is a Lagrange multiplier forcing a gauge field to be discrete.
  • The fields in question in [1] have no kinetic terms because they are not allowed by symmetry, since they are conserved currents.
  • Finally, the charged matter has been integrated out. The conserved current term is the most relevant term in the EFT of the matter, and integrating over it completes integrating out the matter.

Is it okay to integrate a field without kinetic terms?

There are various occasions in which you need to path integrate over a field that has no kinetic terms.

  1. A very common occasion is a Lagrange multiplier. Say you have a field theory, with dynamical fields $\phi$, and you want to enforce the constraint $\mathcal{C}(\phi)$. In the path integral you would write something like $$ Z = \int \mathrm{D}\phi\ \delta[\mathcal{C}(\phi)] \exp(-S[\phi])$$ and then you could represent the delta function as $$\delta[\mathcal{C}(\phi)] = \int \mathrm{D}\lambda \exp\!\left(\mathrm{i}\int \lambda\,\mathcal{C}(\phi)\right).$$ Now $\lambda$ is a new field, cooked up specifically to enforce the constraint $\mathcal{C}(\phi)$. So, here you have it, your theory now involves a path integral with a field with no kinetic terms $$ Z = \int \mathrm{D}\phi\,\mathrm{D}\lambda \exp\!\left(-S[\phi]+\mathrm{i}\int\lambda\;\mathcal{C}(\phi)\right). $$

  2. Another common instance is discrete gauge fields. Namely, if you have a gauge theory with a discrete gauge group, $G$, the gauge fields, $a$, have no curvature (see e.g. this phys.SE answer of mine) and in fact, they take values in the cohomology, $\mathrm{H}^1(M;G)$. So the path integral of a typical gauge theory with a discrete gauge field would be of the form $$Z = \sum_{a\in\mathrm{H}^1(M;G)}\int\mathrm{D}\phi\ \exp\!\left(-S[\phi]+\int a\; w\right),$$ where $w$ is some topological invariant build out of the data that define the theory.

  3. To make this more fun, combine 1. and 2. and come into contact with the paper you're struggling with, if the discrete group is $G=\mathbb{Z}_N$, you can write a $\mathbb{Z}_N$ gauge field (in the spirit of 2.) by considering a good old $\mathrm{U}(1)$ gauge field, $a$ and then forcing it to be a $\mathbb{Z}_N$ gauge field by enforcing the constraints $$ \d{a} = 0 \qquad\text{&}\qquad \mathrm{exp}\!\left(\mathbb{i} N \oint a\right)=1.\tag{1}$$ Usually, this is achieved by coupling the $\mathrm{U}(1)$ gauge-field to a BF term $$ \sim N \int b\ \d{a} \sim N \int \d{b}\ a,\tag{2}$$ as explained, e.g. in this phys.SE question and answer. However, you see that in [1], the same thing can be achieved by coupling $a$ to a conserved current $J$, (e.g. in eq. (2.1) of [1]) such that $\mathrm{d}\star J = 0$. Writing $K=\star J$ the coupling looks like $$\sim N \int K\ a. \tag{3}$$ Locally you can represent $K$ as $\d{\kappa}$ and you see how (3) takes the form of (2), thus enforcing the constraints (1). You can morally view $K$ as a Lagrange multiplier, in the spirit of 1. enforcing the constraint, thus the path integral for this $\mathbb{Z}_N$ theory would look like $$ Z = \int \mathrm{D}a\; \mathrm{D}K\ \exp\!\left(-S[a] + \mathrm{i}\; N \int K\ a\right),$$ where $S[a]$ is a typical action of a $\mathrm{U}(1)$ gauge field, e.g. as is the case in [1] Maxwell with a theta term.

Why is there no kinetic term?

In the case of [1], there is no kinetic term, simply because it is not allowed by symmetry/it would vanish. $J$ is a conserved current, or equivalently $K$ satisfies $\mathrm{d}K=0$, so any kinetic terms would be zero.

Physical interpretation

You should see the action (2.1) or equivalently (3.2) of [1] as representing an effective action of the gauged charge $N$ matter. To unpack what I mean by that, forget all these actions and let's try to build them from scratch. Take some fields with a $\mathrm{U}(1)$ symmetry, so they come with a conserved current $J$, or equivalently $K=\star J$ such that $\d K=0$. As before you should imagine $K$ as $\d{\kappa}$ in the absence of topological obstructions.

Now, you want to write an effective theory of this conserved quantity, so you start writing the most relevant terms. Had you had no massless quantities the most relevant thing you could write would be $$ S_\text{eff} \sim \int K^2 = \int K\wedge \star K,$$ as it contains the least number of derivatives. But in the case of [1], you have coupled your theory to a $\mathrm{U}(1)$ gauge field. So there is something more relevant you can write. That is $$ \sim \int K\ a.$$ That's exactly what they wrote, that is the $J^\mu a_\mu$ term in (2.1) or the $n_\mu a_\mu$ term in (3.2). Therefore, the full effective action at this scale is (keeping in mind the charge, $N$ of the matter fields) $$S_\text{eff}[a,K] = S[a] + \mathrm{i}\; N\int K\ a.$$ Last step: why stop here? Since you have massless degrees of freedom, nothing stops you from integrating out the matter fields completely. That's what they are doing by integrating over $J$/summing over $n$. The final effective action, the response theory, is what remains after that $$ S_{\text{eff}}[a] = -\log\int\mathrm{D}K \exp(-S_\text{eff}[a,K]).$$

The fact that the matter is condensing, is important to guarantee that $\kappa$ in $\int K\; a = \int \d\kappa\; a$ fluctuates wildly, thus allowing the Lagrange multiplier interpretation of the first section. This is explained in more detail in appendix A, section A.2 of [1].

The lattice version

The lattice version of the story is just trying to imitate the field-theoretic story I explained above. So if you discretise each of my equations you should end up with (3.1) and (3.2) of [1].

References: [1] Theory of oblique topological insulators, arxiv:2206.07725

  • 1
    $\begingroup$ Your answer suggests that the form of the effective action is constrained by symmetry and RG principles. This seems to be an opposing view to my answer where I claim that the matter field being condensed (due to the particular values of its coupling constants) is crucial for that effective action. If it were not condensed, it would be in a different phase and that contradicts your point that this effective action is forced by symmetry/RG. A simple example is given by the Abelian-Higgs model (see fig. 2/3 of journals.aps.org/prd/abstract/10.1103/PhysRevD.19.3682). $\endgroup$ Nov 7, 2022 at 18:38
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    $\begingroup$ @NandagopalManoj Thanks, that is a good comment. But actually, our answers do not disagree. The form of the effective action is indeed imposed by symmetry/RG, but the condensation is crucial to ensure that $\kappa$ (where $\mathrm{d}\kappa=K$) fluctuates a lot, thus Higgsing $\mathrm{U}(1)$ to $\mathbb{Z}_N$. See also appendix A.2 in [1]. Condensed/uncondensed phases are distinguished by their unbroken one-form symmetries. This is the same as a confinement/deconfinement phase transition as per the paper you linked. $\endgroup$ Nov 7, 2022 at 19:23

The crux of the answers to your questions is that the matter field is in the (Bose) condensed phase (as mentioned in the beginning of section 3.1). You are right that one should include "hopping" and "chemical potential" terms for the charges, which look like (in the Hamiltonian picture, ignoring the gauge fields for clarity) \begin{equation} H_{\text{matter}} = -t \sum_{\langle ij\rangle}\cos(\theta_i - \theta_j) - \mu \sum_i n_i^2 . \end{equation} Where $i,j$ are vertices where your charges live. The $\theta \in U(1)$ is the conjugate variable to the integer charge density, satisfying $[\theta_i,(n_j)_{\mu=0}] = i\delta_{ij}$. The condensed phase occurs when $\mu \to 0$, we set all $\theta_i = \theta$ to minimize energy, so the $\cos()$ term turns into a constant. This state corresponds to large fluctuations in the $n$ (charge density) variables, where each charge configuration is given equal weight in the partition function. Therefore, in your effective Lagrangian, we do not add a hopping / chemical potential term which effectively realizes an equal weight superposition of charge configurations.

Related comment, in the Lagrangian you mentioned, one must sum over $n_\mu$ configurations that satisfy the continuity equation $\Delta_\mu n_\mu = 0$, which correspond to closed worldlines of the charged particles.

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    $\begingroup$ They actually do mention it in the paper. Equation (3.1) contains a $\delta(\Delta_\mu n_\mu)$ term enforcing the continuity equation. $\endgroup$ Nov 7, 2022 at 13:50
  • $\begingroup$ Thanks, I did not notice that. Edited that now. $\endgroup$ Nov 7, 2022 at 18:32

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