# Path integral for fermion on circle

I agree with your eigenvalues but I'm not sure that your calculation of the Determinant via the Zeta function came out right.

I would split into $$n$$ positive and negative, as $$\zeta_{\lambda}(s) := \left(\frac{2\pi}{T}\right)^{-s}\sum_{n = -\infty}^{\infty} \left(n + \frac{1}{2}\right)^{-s} = \left(\frac{2\pi}{T}\right)^{-s} \left[ \zeta\left(s, \frac{1}{2}\right) + \sum_{n = 0}^{-\infty} \left(n + \frac{1}{2}\right)^{-s} - \frac{1}{2^{-s}}\right]$$ Sending $$n \rightarrow -n$$ in the middle term gets you $$\zeta_{\lambda}(s) = \left(\frac{2\pi}{T}\right)^{-s} \left[ \zeta\left(s, \frac{1}{2}\right) +(-1)^{-s}\zeta\left(s, -\frac{1}{2}\right) - \frac{1}{2^{-s}}\right].$$

From here we will use $$\textrm{Det}_{AP}\left\{i \frac{d}{dt}\right\} = \textrm{e}^{-\zeta'_{\lambda}(0)}$$. We find $$\zeta_{\lambda}'(s) = - \ln\left( \frac{2\pi}{T}\right) \left(\frac{2\pi}{T}\right)^{-s}\left[ \zeta\left(s, \frac{1}{2}\right) +(-1)^{-s}\zeta\left(s, -\frac{1}{2}\right) - \frac{1}{2^{-s}}\right] + \left(\frac{2\pi}{T}\right)^{-s}\left[ \zeta'\left(s, \frac{1}{2}\right) +i\pi (-1)^{-s}\zeta\left(s, -\frac{1}{2}\right) + (-1)^{-s}\zeta'\left(s, -\frac{1}{2}\right) - \ln(2)2^{s}\right]$$ so that $$\zeta_{\lambda}'(0) = -\ln\left(\frac{2\pi}{T}\right) \left[\zeta\left(0, \frac{1}{2}\right) + \zeta\left(0, -\frac{1}{2}\right) - 1 \right] + \left[\zeta'\left(0, \frac{1}{2}\right) + i\pi \zeta\left(0, -\frac{1}{2}\right) + \zeta'\left(0, -\frac{1}{2}\right) - \ln2\right].$$

Finally we need $$\zeta(0, 1/2) = 0$$, $$~\zeta(0, -1/2) = 1$$, $$~\zeta'(0, 1/2) = -1/2 \ln2$$ and $$\zeta'(0, -1/2) = 1/2\ln2 - i\pi$$. The first term vanishes along with the $$T$$ dependence whilst the second one evaluates to $$\zeta_{\lambda}'(0) = -\frac{1}{2} \ln2 + i\pi + 1/2 \ln2 - i\pi - \ln2 = -\ln2,$$ so that $$\textrm{e}^{-\zeta_{\lambda}'(0) } = 2$$. The normalisation of the path integral will be $$\textrm{Det}_{AP}\left\{i \frac{d}{dt}\right\} ^\frac{1}{2} = \sqrt{2}$$.

1. Firstly the $$T$$ dependence goes due to the invariance of the action under $$t \rightarrow \mu t$$ which allows, for example, the rescaling $$t = Tu$$ with $$u \in [0,1]$$.
2. The value $$2$$ counts the degrees of freedom of a real fermion, and is better calculated using coherent states to form the trace of the operator $$\textrm{e}^{-i\hat{H}}$$ that is being calculated.

Well, for what it's worth, OP's heuristic$$^{\dagger}$$ zeta function evaluation works in principle:

\begin{align}{\rm Det}(i\frac{d}{dt}) ~=~&\prod_{n\in\mathbb{Z}}\lambda_n ~=~\prod_{n\in\mathbb{Z}}(-(n+1/2))\cr ~=~&\left[ \prod_{n\in\mathbb{Z}}(-1/2)\right]\left[ \prod_{n\in\mathbb{Z}}(2n+1)\right]\cr~\stackrel{(2)}{=}~& \prod_{n\in\mathbb{Z}}(2n+1)~=~ \frac{\prod_{n\in\mathbb{Z}\backslash\{0\}} n}{\prod_{n\in\mathbb{Z}\backslash\{0\}}2n}\cr~=~& \frac{1}{\prod_{n\in\mathbb{Z}\backslash\{0\}}2}~=~\frac{2}{\prod_{n\in\mathbb{Z}}2}~\stackrel{(2)}{=}~2. \end{align} \tag{1} In eq. (1) we divided into even and odd integers and used the formula \begin{align} \prod_{n\in\mathbb{Z}}a~=~&\left[ \prod_{n\in\mathbb{Z}_{<0}}a\right]a\left[ \prod_{n\in\mathbb{Z}_{>0}}a\right] ~=~a\left[ \prod_{n\in\mathbb{N}}a\right]^2\cr~=~&a^{1+2\zeta(0)}~=~a^0~=~ 1, \qquad a~\in~\mathbb{C}\backslash\{0\}. \end{align} \tag{2} Finally, the Gaussian Grassmann integral yields the Pfaffian $${\rm Pf}(i\frac{d}{dt})~=~\sqrt{{\rm Det}(i\frac{d}{dt})}~\stackrel{(1)}{=}~\sqrt{2},\tag{3}$$ i.e. the square root.

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$$^{\dagger}$$Nota bene: In order to correctly mimic the corresponding rigorous calculation, one should refrain from performing any frivolous Hilbert hotel move on the index set of the infinite product.