Renormalization Group

Consider the ising model.

$\mathcal{H}=-J\sum _{i=1}^{N-1}{s}_i{s}_{i+1}-\tilde{B}\sum _{i=1}^N{s}_i-\tilde{C}$.

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

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, $Z(N,K,B)=\sum _{\{s_i\}}\exp (\sum _{i}K{s}_i{s}_{i+1}+B/2(s_i+s_{i+1}))$. 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 $Z=\sum _{\{s_i\}}\exp (\sum _i{K}^{\prime }{s}_i{s}_{i+2}+\frac{B}{2}(s_i+s_{i+2})+2c)$. Then, per term, $\exp ((k+B/2)(s_i+s_{i+1})+B)+\exp ((-k+\frac{B}{2})(s_i+s_{i+1})-B)=\exp (k^{\prime }{s}_i{s}_{i+1}+\frac{B^{\prime }}{2}(s_i+s_{i+2})+2c)$. This must hold for $s_i=\pm 1,s_{i+2}=\pm 2$. This gives our first equation [1], for $s_i=s_{i+2}=1$, $\exp (2K+2B)+\exp (-2K)=\exp (K^{\prime }+B^{\prime }+2c)$. This gives our second equation [2], for $s_i=s_{i+2}=-1$, $\exp (-2K)+\exp (2K-2B)=\exp (K^{\prime }-B^{\prime }+2c)$. This gives our third [3], for $s_i=-s_{i+2}=-1$, $\exp (B)+\exp (B)=\exp (-K^{\prime }+2c)$. Solving for $K^{\prime }=K^{\prime }(K,B)$, $B^{\prime }=B^{\prime }(K,B)$, and $c=c(K,B)$. Dividing [1] by [2]: $4\exp (2B^{\prime })=\frac{\exp (2K)\exp (2B)+\exp (-2K)}{\exp (2K)\exp (-2B)+\exp (-2K)}=\frac{\exp (B)(\exp (2K+B)+\exp (-2K+B))}{\exp (-B)(\exp (2K-B)+\exp (-(2K-B))}=\exp (2B)\frac{\cosh (2K+B)}{\cosh (2K-B)}$. The third equation gives, $\exp (2C)=\exp (K^{\prime })2\cosh B$. From the first equation, $\exp (B)2\cosh (2K+B)=\exp (K^{\prime }+B^{\prime }+2C)=\exp (2K^{\prime }+B^{\prime })2\cosh B$. From the second equation, $\exp (-B)2\cosh (2K-B)=\exp (K^{\prime }+B^{\prime }+2C)=\exp (2K^{\prime }-B^{\prime })2\cosh B$. So, $\exp (4K^{\prime })=\frac{\cosh (2K+B)\cosh (2K-B)}{{\cosh }^{2}B}$ and $\exp (8C)=\exp (4K^{\prime })2^4{\cosh }^4(B)=16\cosh (2K+8)\cosh (2K-B){\cosh }^2(B)$. Since we have $B^{\prime }(K,B)$, we have determined $K^{\prime }(K,B)$ and thus $C(K,B)$. Then, $K^{\prime }=\frac{1}{4}\ln \frac{\cosh (2K+B)\cosh (2K-B)}{{\cosh }^{2}B}$. Then, $B^{\prime }=B+\frac{1}{2}\ln \cosh (2KB)$. Then, $C=\frac{1}{8}\ln 16\cosh (2K+B)\cosh (2K-B){\cosh }^2(B)$. For $B=0$, $K^{\prime }=\frac{1}{2}\ln \cosh (2K)$ So, $Z(N,K,B)=\exp (NC(K,B))Z(\frac{N}{2},K^{\prime },B^{\prime })$. 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, $\zeta (K)=\frac{1}{2}\ln f(K)+\frac{1}{2}\zeta (K^{\prime })$. So, $\zeta (K^{\prime })=2\zeta (k)-\ln (2\sqrt{\cosh (2K)})$.