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--- technical:ising1d [2022/10/29 09:27] – [Simple case 1: $J = 1$, $h = 0$] chunchung
+++ technical:ising1d [2022/10/29 15:10] (current) – [Simple case 1: $J = 1$, $h = 0$] chunchung
@@ Line 1: / Line 1: @@
 ======One-dimensional Ising model======
 The Hamiltonian is defined as
-$$ H = E(\mathbf{s}) \equiv  - J\sum_i s_i s_{i+1} - h\sum_i s_i . $$
+\begin{equation} H = E(\mathbf{s}) \equiv  - J\sum_i s_i s_{i+1} - h\sum_i s_i . \end{equation}
 where the state vector is $\mathbf{s}=(s_0,s_1,\ldots,s_{N-1})$ with $s_i = \pm 1$ for $i = 0\ldots N-1$ belonging to the least residue system modulo $N$, i.e., $\mathbb{Z}_N$, and corresponding to the periodic boundary condition.
 ====Partition function====
 The partition function is defined as
-$$ Z \equiv \sum_{\mathbf{s}} e^{-\beta E(\mathbf{s})}, $$
+\begin{equation} Z \equiv \sum_{\mathbf{s}} e^{-\beta E(\mathbf{s})}, \end{equation}
 which can be rewritten into a matrix form:
-\begin{eqnarray*}
+\begin{align}
-Z & = &  \prod_i \left( \sum_{s_i=\pm 1} \right) e^{\sum_i \beta s_i(J s_{i+1} + h)} \\
+Z & =  \prod_i \left( \sum_{s_i=\pm 1} \right) e^{\sum_i \beta s_i(J s_{i+1} + h)} \notag \\
-  & = & \prod_i \left( \sum_{s_i=\pm 1} e^{\beta s_i(J s_{i+1} + h)} \right)\\
+  & =  \prod_i \left( \sum_{s_i=\pm 1} e^{\beta s_i(J s_{i+1} + h)} \right) \notag \\
-  & = & \operatorname{tr} \left(T^N\right)
+  & =  \operatorname{tr} \left(T^N\right)
-\end{eqnarray*}
+\end{align}
 where
-$$ T = \begin{pmatrix}
+\begin{equation}
+T = \begin{pmatrix}
  e^{\beta(J-h)} & e^{\beta(-J-h)} \\
  e^{\beta(-J+h)} & e^{\beta(J+h)}
-\end{pmatrix} $$
+\end{pmatrix}
+\end{equation}
 is the transfer matrix.
 ====Characteristic polynomial====
 The characteristic polynomial is
-$$ \lambda^2 - \left(e^{\beta (J+h)}+e^{\beta (J-h)}\right) \lambda + e^{2\beta J} - e^{-2\beta J}. $$
+\begin{equation} \lambda^2 - \left(e^{\beta (J+h)}+e^{\beta (J-h)}\right) \lambda + e^{2\beta J} - e^{-2\beta J}. \end{equation}
-Since the discriminant \begin{eqnarray*}
+Since the discriminant \begin{align}
-&  & \left(e^{\beta (J+h)}+e^{\beta (J-h)}\right)^2 - 4 \left(e^{2\beta J} - e^{-2\beta J}\right) \\
+\mathrm{Disc}_\lambda & = \left(e^{\beta (J+h)}+e^{\beta (J-h)}\right)^2 - 4 \left(e^{2\beta J} - e^{-2\beta J}\right) \notag \\
-& = & e^{2\beta(J+h)} + e^{2\beta(J-h)} + 2 e^{2\beta J} - 4 \left(e^{2\beta J} - e^{-2\beta J}\right) \\
+& = e^{2\beta(J+h)} + e^{2\beta(J-h)} + 2 e^{2\beta J} - 4 \left(e^{2\beta J} - e^{-2\beta J}\right) \notag \\
-& = & \left(e^{\beta (J+h)}-e^{\beta (J-h)}\right)^2 + 4 e^{-2\beta J} \\
+& = \left(e^{\beta (J+h)}-e^{\beta (J-h)}\right)^2 + 4 e^{-2\beta J} \notag \\
-& > & 0 .
+& >  0,
-\end{eqnarray*}
+\end{align}
-The characteristic polynomial has two real roots, that is, the transfer matrix $T$ has two real eigenvalues.
+the characteristic polynomial has two real roots, that is, the transfer matrix $T$ has two real eigenvalues.
 ====Eigenvalues====
 The two eigenvalues are
-\begin{eqnarray*}
+\begin{align}
-\lambda_\pm & = & \frac{1}{2} \left[ \left( e^{\beta(J+h)}+e^{\beta(J−h)} \right) \pm \sqrt{ \left(e^{\beta (J+h)}-e^{\beta (J-h)}\right)^2 + 4 e^{-2\beta J} } \right] \\
+\lambda_\pm &= \frac{1}{2} \left[ \left( e^{\beta(J+h)}+e^{\beta(J−h)} \right) \pm \sqrt{ \left(e^{\beta (J+h)}-e^{\beta (J-h)}\right)^2 + 4 e^{-2\beta J} } \right] \notag \\
-& = & e^{\beta J}\left[\cosh(\beta h)\pm\sqrt{\sinh^2(\beta h)+e^{-4\beta J}}\right]
+ &= e^{\beta J}\left[\cosh(\beta h)\pm\sqrt{\sinh^2(\beta h)+e^{-4\beta J}}\right]
-\end{eqnarray*}
+\end{align}
 and the partition function is given by
-$$ Z = \lambda_+^N+\lambda_-^N . $$
+\begin{equation} Z = \lambda_+^N+\lambda_-^N . \end{equation}
 ====Simple case 1: $J = 1$, $h = 0$====
 The eigenvalues become
-$$ \lambda_\pm = e^{\beta}\pm e^{-\beta}. $$
+\begin{equation} \lambda_\pm = e^{\beta}\pm e^{-\beta}. \end{equation}
 The partition function is
-$$ Z = \left(e^\beta+e^{-\beta}\right)^N+\left(e^\beta-e^{-\beta}\right)^N $$
+\begin{equation} Z = \left(e^\beta+e^{-\beta}\right)^N+\left(e^\beta-e^{-\beta}\right)^N . \end{equation}
-average energy is given
+Generally, the average energy $U = \langle E\rangle$ can be calculated from the $\beta$-derivative of $-\log Z$:
+\begin{align}
+U &= \langle E\rangle \notag \\
+&= Z^{-1} \sum_{\mathbf{s}} E e^{-\beta E} \notag \\
+&= - Z^{-1} \frac{\partial}{\partial\beta} Z \notag \\
+&= - \frac{\partial}{\partial\beta}\log Z \\
+&= \frac{N\left[\left(e^\beta+e^{-\beta}\right)^{N-1}\left(e^\beta-e^{-\beta}\right)+\left(e^\beta+e^{-\beta}\right)\left(e^\beta-e^{-\beta}\right)^{N-1}\right]}{\left(e^\beta+e^{-\beta}\right)^N+\left(e^\beta-e^{-\beta}\right)^N} \notag \\
+&= N\frac{r+r^{N-1}}{1+r^N}
+\end{align}
+where $r = \left(e^\beta-e^{-\beta}\right)/\left(e^\beta+e^{-\beta}\right)<1$ is the ratio between the two eigenvalues.