# $1+1D$ $U(1)$ gauge theory is a quantum mechanical system

1. The E&M Hamiltonian Lagrangian density without matter in 1+1D is$$^1$$ $${\cal L}_H~=~-E_1 \dot{A}_1 ~-~{\cal H}_0 ~-~ A_0 \underbrace{\partial_1 E_1}_{\text{ Gauss law}}, \tag{i}$$ where the Hamiltonian density is $${\cal H}_0~=~\frac{1}{2} E_1^2 -\theta E_1. \tag{ii}$$

2. Next if we use the Coulomb gauge $$\partial_1 A_1=0$$ together with Gauss law $$\partial_1 E_1=0$$, we see that the canonical field pair $$(A_1,E_1)$$ does not depend on space $$x^1$$, only on time $$t$$.

3. After eliminating the Lagrange multiplier field $$A_0$$, the field theory model (i) effectively becomes a point mechanical model with a single canonical pair $$(A_1(t),E_1(t))$$. This fact answers OP's title question.

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$$^1$$ Eq. (i) is before gauge-fixing & after possibly ignoring spatial boundary terms. We use a Minkowski metric with positive spatial component and negative temporal component. The momentum variable to $$A_1$$ is minus $$E_1$$.

Maybe there is a more vivid and rigorous argument in the continuum theory, but I am not aware of them, so I 'll present an argument from lattice gauge theory.

The discrete version of gauge theory, which in continuum limit gives ordinary Maxwell (Yang-Mills action) is so called Wilson action: $$S_G = \sum_{\Box} \beta \ (1 - \frac{1}{N_c}\text{Tr } U_{\Box}) \qquad U_{\Box} = U_\mu (x) U_\nu (x + \mu) U_\mu^{\dagger} (x + \nu) U_\nu^{\dagger} (x)$$ Where $$U_\mu (x)$$ is the link poiniting in the $$\mu$$ direction from the site, located at $$x$$. The gauge transformations act on the links in following way: $$U_\mu^{'} (n) = \Omega (n) U_\mu (n) \Omega^{\dagger} (n + \mu)$$ By an appropriate gauge transformation, any configuration can be put to so dubbed temporal gauge : $$U_0 (n) = \mathbb{1}$$

Looking at the Wilson action, one sees, that the links on different timeslices decouple from each other, and we have independent 1-dimensional spin chains, which can be interpreted as (0 + 1) dimensional field theory in Euclidean space.