Fermionic Harmonic Oscillator Partition Function

Question

I am reading Nakahara Geometry, Topology, and Physics. In the section on fermionic harmonic oscillator, after some math, the partition function is given by $$\begin{aligned} Z(\beta) &=\mathrm{e}^{\beta \omega / 2} \lim _{N \rightarrow \infty} \prod_{k=-N / 4}^{N / 4}\left[\mathrm{i}(1-\varepsilon \omega) \frac{\pi(2 n-1)}{\beta}+\omega\right] \\ &=\mathrm{e}^{\beta \omega / 2} \mathrm{e}^{-\beta \omega / 2} \prod_{k=1}^{\infty}\left[\left(\frac{2 \pi(n-1 / 2)}{\beta}\right)^{2}+\omega^{2}\right] \\ &=\prod_{k=1}^{\infty}\left[\frac{\pi(2 k-1)}{\beta}\right]^{2} \prod_{n=1}^{\infty}\left[1+\left(\frac{\beta \omega}{\pi(2 n-1)}\right)^{2}\right], \end{aligned}$$ where $\varepsilon = \beta/N$. I don't understand how it goes from the first line to the second line.

References:

1. M. Nakahara, Geometry, Topology and Physics, 2003; section 1.5.10 p. 69.

Answer 1

Hints:

1. First of all, there is a typo in Nakahara: The integer $n$ should be $k$ in the first 2 lines (but not in the 3rd line).

2. Secondly, pull the factor $(1-\varepsilon \omega)$ outside the square bracket. It becomes $(1-\varepsilon \omega)^{N / 2+1}\to e^{-\beta\omega/2}$ for $N\to\infty$, which is the second factor in the second line. Here we have used that $\varepsilon =\beta/N$, and a well-known representation of the exponential function.

3. In the modified square bracket product, use the identity $(a+ib)(a-ib)=a^2+b^2$.

References:

1. M. Nakahara, Geometry, Topology and Physics, 2003; section 1.5.10 p. 69.