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One-dimensional Ising model

The Hamiltonian is defined as $$ H = E(\mathbf{s}) \equiv - J\sum_i s_i s_{i+1} - h\sum_i s_i . $$ 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.

The partition function is defined as $$ Z \equiv \sum_{\mathbf{s}} e^{-\beta E(\mathbf{s})}, $$ which can be rewritten into a matrix form: \begin{eqnarray*} Z & = & \prod_i \left( \sum_{s_i=\pm 1} \right) e^{\sum_i \beta s_i(J s_{i+1} + h)} \\ & = & \prod_i \left( \sum_{s_i=\pm 1} e^{\beta s_i(J s_{i+1} + h)} \right)\\ & = & \operatorname{tr} \left(T^N\right) \end{eqnarray*} where $$ T = \begin{pmatrix} e^{\beta(J-h)} & e^{\beta(-J-h)} \\ e^{\beta(-J+h)} & e^{\beta(J+h)} \end{pmatrix} $$ is the transfer matrix.

The characteristic polynomial is $$ \lambda^2 - \left(e^{\beta (J+h)}+e^{\beta (J-h)}\right) \lambda + e^{2\beta J} - e^{-2\beta J}. $$ Since the discriminant \begin{eqnarray*} & & \left(e^{\beta (J+h)}+e^{\beta (J-h)}\right)^2 - 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) \\ & = & \left(e^{\beta (J+h)}-e^{\beta (J-h)}\right)^2 + 4 e^{-2\beta J} \\ & > & 0 . \end{eqnarray*} The characteristic polynomial has two real roots, that is, the transfer matrix $T$ has two real eigenvalues.

The two eigenvalues are \begin{eqnarray*} \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] \\ & = & e^{\beta J}\left[\cosh(\beta h)\pm\sqrt{\sinh^2(\beta h)+e^{-4\beta J}}\right] \end{eqnarray*} and the partition function is given by $$ Z = \lambda_+^N+\lambda_-^N . $$

The eigenvalues become $$ \lambda_\pm = e^{\beta}\pm e^{-\beta}. $$ The partition function is $$ Z = \left(e^\beta+e^{-\beta}\right)^N+\left(e^\beta-e^{-\beta}\right)^N . $$

Generally, the average energy $U = \langle E\rangle$ can be calculated from the $\beta$-derivative of $-\log Z$: \begin{eqnarray} U & = & \langle E\rangle \\ & = & Z^{-1} \sum_{\mathbf{s}} E e^{-\beta E} \\ & = & - Z^{-1} \frac{\partial}{\partial\beta} Z \\ & = & - \frac{partial}{\partial\beta}\log Z \end{eqnarray}