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I want to calculate for two interarcting spins and an applied magnetic field the entropy. The values for the spins are $s_1= \pm 1$ and $s_2 = \pm 1$ and the Hamiltonian is given by \begin{align} H = - \varepsilon s_1 s_2 \, . \end{align} I can use \begin{align} Z= 2 \mathrm{e}^{\beta \varepsilon} \cosh(2 \beta h)\, . \label{eq:Zmg} \end{align} for the system with applied magnetic field. Now I want to calculate the entropy for a strong applied field ($\beta h\gg 1$).

My Idea: The free energy is \begin{align*} F= -k_\mathrm{B}T \ln (Z) = -k_\mathrm{B}T \ln\left(2 \mathrm{e}^{\beta \varepsilon} \cosh(2\beta h) \right) \end{align*} and the entropy \begin{align*} S&= - \frac{\partial F}{\partial T} = k_\mathrm{B}\ln\left(2 \mathrm{e}^{\beta \varepsilon} \cosh(2\beta h) \right) - k_\mathrm{B}T \frac{1}{k_\mathrm{B}T^{2}}\frac{\varepsilon \mathrm{e}^{\beta \varepsilon} \cosh(2 \beta h) + 2 h \mathrm{e}^{\beta \varepsilon} \sinh(2 \beta h) }{\mathrm{e}^{\beta \varepsilon} \cosh(2 \beta h)} \\ &= k_\mathrm{B}\ln\left(2 \mathrm{e}^{\beta \varepsilon} \cosh(2\beta h) \right) -\frac{1}{T} \frac{\varepsilon \mathrm{e}^{\beta \varepsilon} \cosh(2 \beta h) + 2 h \mathrm{e}^{\beta \varepsilon} \sinh(2 \beta h) }{\mathrm{e}^{\beta \varepsilon} \cosh(2 \beta h)} \end{align*} For a strong field the entropy is in my opinion $S=-\infty$, but that doesn't make any sense. Please help me how I get this approximation right.

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If you assume $\beta h \gg 1$, you can simplify $$ \cosh(2\beta h) \rightarrow \frac{1}{2}e^{2\beta h} $$ and find $$ Z = e^{\beta(\varepsilon + 2h)} \\ F = -(\varepsilon + 2h) \\ S = 0 $$ Indeed, the result $S = 0$ can be inferred from the fact that a large magnetic field allows only on possible state - the state with both spins parallel to the magnetic field.

Another way to reasoning goes like this: If $h$ is large, we can assume $\varepsilon \ll h$. Thus, we can ignore the interaction term in the Hamiltonian. Then, the magnetic field $h$ appears only in form of the product $\beta h$. Thus, increasing the magnetic field has the same effect as decreasing the temperature. Thus, $h \rightarrow \infty$ is the same as $T \rightarrow 0$. However, we know that the entropy goes to zero if the temperature goes to zero.

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