I have been learning about lattice gauge theories, in particular about the Ising gauge theory on the 2D square lattice. The Hamiltonian for a system with no matter fields is given by (for eg. from this book, Section 9.6, Field Theories of Condensed Matter Physics ) $$ \mathcal{H} = - g \sum_{\vec{x},j} \sigma^x_j(\vec{x}) - \frac{1}{g} \sum_{\boxed{}} \prod_{\vec{x},j \in \boxed{}} \sigma^z_j(\vec{x}) $$ where $\sigma^x_j, \sigma^z_j$ are the Pauli spin operators acting along a link labelled in the direction of the basis vector $\vec{e_j}$ from the site $\vec{x}$ and the $\boxed{}$ refers to a plaquette of the lattice.

I am able to understand that when $g \to 0$ or $\infty$, what the ground state will be. For example, in the $g \to \infty$ limit, all the spins in the x-direction will align themselves in the ground state. However, I am not able to figure out how I should proceed, (i.e. to find the ground state configuration) when one of the terms in the Hamiltonian is not exactly zero, but something close to it (say, at $ O(g^2)$).


One uses simple perturbation theory, but the parts of the Hamiltonian chosen as the ground state and the perturbation are different for $g\rightarrow 0$ and for $1/g\rightarrow 0$.

In the context of Ising model and similar spin systems these approximations are referred to as high-temperature expansion and low-temperature expansion, one can check, e.g., this review by Wu.

  • $\begingroup$ So this would mean that I treat the g term as the perturbing Hamiltonian in the limit $g \to 0$ and vice versa. However, even in that case, the unperturbed Hamiltonian is still a function of g, which would imply that my unperturbed ground state energy is still a function of g itself. I wasn't sure about this being valid, for I thought that $\mathcal{H}_0$ should not depend on the parameter that I am expanding in (kind of how couplings are expanded in QFTs). Can you kindly clarify? $\endgroup$ – curioussoul1234 Nov 28 '20 at 19:27
  • $\begingroup$ What really matters is the ratio of energies: one term is much bigger than the other. The Hamiltonian was explicitly written in this way, with no other coefficients, to underscore the two approximations. You could change notation: $\lambda=1/g$ in one term and $g\rightarrow g^2\lsmbda$ in the other, and expand in the orders of $g^2$. $\endgroup$ – Vadim Nov 28 '20 at 20:57

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