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, \(\zeta(K') = 2\zeta(k) - \ln(2\sqrt{\cosh(2K)})\).