# Question about the derivation of an equation in full replica symmetry breaking solution

Using replica method and saddle point method, the free energy of a magnetic system can be expressed as $$-\beta[f]=\lim_{n\to0}\{\frac{-\beta^2J^2}{4n}\sum_{a\ne b}q_{\alpha\beta}^2-\frac{\beta J_0}{2n}\sum_\alpha m_\alpha^2+\frac{\beta^2J^2}{4}+\frac{1}{n}\log\operatorname{Tr}e^L\}$$ $$L=\beta^2J^2\sum_{\alpha<\beta}q_{\alpha\beta}S^\alpha S^\beta+\beta\sum_\alpha(J_0m_\alpha+h)S^\alpha$$ $J$ is the negative interaction energy, $J_0$ is the mean of $J$, $q_{\alpha\beta}$ is $[<S_i^\alpha S_i^\beta>]$, $S$ is the magnetic spin, $m_\alpha$ is $[<S^\alpha>]$, $h$ is the external magnetic field.

What I don't understand is that if we assume $J_0, h, m_\alpha$ all equal to 0, then expand the above equation to fourth order, we can get the following equation: $$\beta[f]=\lim_{n\to0}\frac{1}{n}\{\frac{1}{4}(\frac{T^2}{T_f^2}-1)\operatorname{Tr}Q^2-\frac{1}{6}\operatorname{Tr}Q^3-\frac{1}{8}\operatorname{Tr}Q^4+\frac{1}{4}\sum_{\alpha\ne\beta\ne\gamma}Q_{\alpha\beta}^2Q_{\alpha\gamma}^2-\frac{1}{12}\sum_{\alpha\ne\beta}Q_{\alpha\beta}^4\}$$ $T_f$ is the critical temperature, it equals to $J$.

I just can't see how this equation is derived.

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