I'm trying to understand the discussion in this book on the fermionization of the 2D Ising model. The transfer matrix for this model becomes $T = \theta\tilde{\theta}$ where: $$\theta = e^{\beta \sum_{x}\sigma_{x}^{(1)}\sigma_{x+1}^{(1)}} \quad \mbox{and} \quad \tilde{\theta} = e^{\tilde{\beta}\sum_{x}\sigma_{x}^{(3)}}$$ where $\sigma^{(i)}_{x}$ are Pauli matrices at each $x$: $$\sigma^{(1)} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma^{(2)} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \quad \mbox{and} \quad \sigma^{(3)} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ Then, the following change of variables is made: $$\Gamma_{-\frac{1}{2},0} := \sigma_{0}^{(1)} \quad \Gamma_{x-\frac{1}{2},x} := \bigg{(}\prod_{0}^{x-1}\sigma_{x'}^{(3)}\bigg{)}\sigma_{x}^{(1)} \quad (x\ge 1)$$ and: $$\Gamma_{0,\frac{1}{2}} := \sigma_{0}^{(2)} \quad \mbox{and} \quad \Gamma_{x,x+\frac{1}{2}} :=\bigg{(}\prod_{0}^{x-1}\sigma_{x'}^{(3)}\bigg{)}\sigma_{x}^{(2)} \quad (x\ge 1)$$ After some manipulations, we get: $$\theta(\beta) = e^{-i\beta \sum_{x}\Gamma_{x,x+\frac{1}{2}}\Gamma_{x+\frac{1}{2},x+1}} \quad \mbox{and} \quad \tilde{\theta}(\beta) = e^{-i\tilde{\beta}\sum_{x}\Gamma_{x-\frac{1}{2},x}\Gamma_{x,x+\frac{1}{2}}}$$ Finally, one more change of variables $\Gamma_{x-\frac{1}{2},x} = a_{x}+a^{\dagger}_{x}$ and $\Gamma_{x,x+\frac{1}{2}} = i(a_{x}-a^{\dagger}_{x})$ leads to: \begin{eqnarray}\tilde{\theta}(\beta) = e^{\tilde{\beta}\sum_{x}(a^{\dagger}_{x}a_{x}-a_{x}a^{\dagger}_{x})} = \prod_{x}(e^{-\tilde{\beta}}+2\sin\tilde{\beta}a^{\dagger}_{x}a_{x})\tag{1}\label{1}\end{eqnarray}
Then, the author states:
Accordingly, using the same symbol, the corresponding integral kernel is: \begin{eqnarray}\tilde{\theta}(\tilde{\beta},\tilde{\xi},\xi) = \prod_{x}(e^{-\tilde{\beta}}+2\sin\tilde{\beta}\tilde{\xi}_{x}\xi_{x})e^{\tilde{\xi}_{x}\xi_{x}}\tag{2}\label{2}\end{eqnarray}
Question: What is done to pass from (\ref{1}) to (\ref{2})? I don't follow the reasoning.
