# Renormalization Group

Consider the ising model.

$$\mathcal{H} = -J\sum_{i=1}^{N-1}s_is_{i+1} - \tilde{B}\sum_{i=1}^Ns_i - \tilde{C}$$.

Let $$K = \frac{J}{kT}$$, $$B = \frac{\tilde{B}}{kT}$$, and $$c = \frac{\tilde{c}}{kT}$$. Then, $$Z = \sum_\alpha \exp(-\beta H = \sum_{s_1=\pm 1}\sum_{s_2=\pm 1}\cdots\sum_{s_N=\pm 1}\exp(-\beta H)$$ $$\beta H = -K\sum s_is_{i+1} - B\sum s_i - c$$.

Performing a sum over the even terms, doubling the spacing, gives the partial trace, we then want to remap them back to the original lattice spacing to understand the original system. Then, $$Z(N,K,B) = \sum_{\{s_i\}}\exp(\sum_i Ks_is_{i+1} + B/2(s_i + s_{i+1}))$$. Keep odd spins $$s_i$$ and even spins $$\sigma_i$$. $$Z = \prod_i\exp(k(s_i\sigma_i+\sigma_i s_{i+2}) + B\sigma_i + \frac{B}{2}(s_i + s_{i+2}))$$. Now we perform the partial trace, $$\sigma_i = \pm 1$$. So, $$Z = \sum_{\{s_i\}} \exp((k+B/2)(s_i + s_{i+2}) + B) + \exp(-(k-B/2)(s_i + s_{i+2}) - B)$$. Then, it must be the case that $$Z = \sum_{\{s_i\}} \exp(\sum_i K's_is_{i+2} + \frac{B}{2}(s_i + s_{i+2}) + 2c)$$. Then, per term, $$\exp((k+B/2)(s_i + s_{i+1}) + B) + \exp((-k+\frac{B}{2})(s_i + s_{i+1}) - B) = \exp(k's_is_{i+1} + \frac{B'}{2}(s_i+s_{i+2}) + 2c)$$. This must hold for $$s_i = \pm 1, s_{i+2} = \pm 2$$. This gives our first equation [1], for $$s_i = s_{i+2} = 1$$, $$\exp(2K+2B) + \exp(-2K) = \exp(K' + B' + 2c)$$. This gives our second equation [2], for $$s_i = s_{i+2} = -1$$, $$\exp(-2K) + \exp(2K-2B) = \exp(K' - B' + 2c)$$. This gives our third [3], for $$s_i = -s_{i+2} = -1$$, $$\exp(B) + \exp(B) = \exp(-K' + 2c)$$. Solving for $$K' = K'(K, B)$$, $$B' = B'(K, B)$$, and $$c = c(K, B)$$. Dividing [1] by [2]: $$4\exp(2B') = \frac{\exp(2K)\exp(2B) + \exp(-2K)}{\exp(2K)\exp(-2B)+\exp(-2K)} = \frac{\exp(B)(\exp(2K+B) + \exp(-2K+B))}{\exp(-B)(\exp(2K-B) + \exp(-(2K-B))} = \exp(2B) \frac{\cosh(2K+B)}{\cosh(2K-B)}$$. The third equation gives, $$\exp(2C) = \exp(K')2\cosh B$$. From the first equation, $$\exp(B)2\cosh(2K+B) = \exp(K' + B' + 2C) = \exp(2K' + B')2\cosh B$$. From the second equation, $$\exp(-B)2\cosh(2K-B) = \exp(K' + B' + 2C) = \exp(2K' - B')2\cosh B$$. So, $$\exp(4K') = \frac{\cosh(2K+B)\cosh(2K-B)}{\cosh^2 B}$$ and $$\exp(8C) = \exp(4K')2^4\cosh^4(B) = 16\cosh(2K+8)\cosh(2K-B)\cosh^2(B)$$. Since we have $$B'(K, B)$$, we have determined $$K'(K, B)$$ and thus $$C(K, B)$$. Then, $$K' = \frac{1}{4}\ln \frac{\cosh(2K+B)\cosh(2K-B)}{\cosh^2B}$$. Then, $$B' = B + \frac{1}{2}\ln \cosh(2KB)$$. Then, $$C = \frac{1}{8}\ln 16\cosh(2K+B)\cosh(2K-B)\cosh^2(B)$$. For $$B = 0$$, $$K' = \frac{1}{2}\ln\cosh(2K)$$ So, $$Z(N,K,B) = \exp(N C(K,B))Z\left(\frac{N}{2}, K', B'\right)$$. Then, $$f(K, B=0) = \exp(NC(K, B)) = 2\sqrt{\cosh(2K}$$.

In the thermodynamic limit, $$F \propto \ln Z = N\zeta(K)$$ but $$\zeta(K)$$ does not depend on $$N$$. So, $$\zeta(K) = \frac{1}{2}\ln f(K) + \frac{1}{2}\zeta(K')$$. So, $$\zeta(K') = 2\zeta(k) - \ln(2\sqrt{\cosh(2K)})$$.

Created: 2024-05-30 Thu 21:19

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