# RG of Ising model whose Hamiltonian is represented with Kronecker delta

Let $$H$$ be hamiltonian, $$i$$ the index of a spin, and $$S_i = \pm 1$$ the $$i$$-th spin's value. When 1D Ising model's hamiltonian is represented as

$$H = - J \sum _i S_i S_{i + 1}\ \ \ (J > 0),$$

it is known (e.g. see Renormalization Group: 1d Ising Model) that RG gives two fixed points $$K \equiv \beta J = 0, \infty$$ and only $$K = 0$$ is stable.

Question: What if the hamiltonian is represented as

$$H = - J \sum _i \delta _{S_i, S_{i + 1}}\ \ \ (J > 0),$$

where $$\delta _{i, j}$$ is Kronecker delta, and $$S_i \in \{1, 2\}$$.

My calculation gives

$$K_{n + 1} = K_n - \ln 2,$$

which means there are two fixed points $$K = \pm \infty$$ and only $$K = - \infty$$ is stable. At least apparently, this result is different from the above one. And, even though we start from $$J > 0$$ (ferromagnetic), repeated renormalization leads $$J \to - \infty$$ (anti-ferromagnetic).

I have no idea which is true:

1. My calculation is incorrect.

2. My calculation is correct and the difference comes simply from the difference of looks of hamiltonians.

Could anyone please give an idea?

• I just found my careful mistake. The correct version of calculation (not technical nor difficult) gave the physically same result as that led from the first hamiltonian (without Kronecker delta). – ynn Jul 27 at 13:49