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?

  • $\begingroup$ 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). $\endgroup$ – ynn Jul 27 at 13:49

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