# Path integral for fermion on circle

I am trying to understand talk by Edward Witten Nonsupersymmetric D-Branes and the Kitaev Fermion Chain. More concretely, I wanna to understand this slide:

If I try to calculate such path integral I do following:

$$Z(S^1) = \int_{NS} D\psi\; e^{i\frac{1}{2}\int_0^T dt \psi \frac{d}{dt} \psi}$$

1) I need find eigenfunctions and eigenvalues for $$i\frac{d}{dt}$$ with antiperiodic boundary conditions: $$i\frac{d}{dt} \psi_n = \lambda_n \psi_n, \;\;\;\;\;\;\;\;\;\; \psi_n(T) = -\psi_n(0)$$ $$\psi_n(t) = e^{i\frac{2\pi(n+1/2)}{T}t}, \;\;\;\; \lambda_n= - \frac{2\pi(n+1/2)}{T},\;\;\;n\in \mathbb{Z}$$

Note that fermions is complex. What I need to do to calculate such integral for Majorana fermion?

2) I choose $$T=2\pi$$. So I need calculate: $$\prod_{n\in \mathbb{Z}} (n+1/2) = \prod_{n>0} (n+1/2) \prod_{n≥0} (-n+1/2) = 2\prod_{n≥0} (n+1/2) \times \frac{1}{2}\prod_{n≥0}(-1) (n+1/2) = (-1)^{1+\sum_{1}^{+\infty}1} \;e^{2\sum^{+\infty}_{n=0}\ln(n+1/2)}$$

We regularise using $$\zeta$$-function (using this):

$$-(-1)^{\zeta(0)} e^{-2\zeta^\prime(0,1/2)} = 2i$$

Where I have mistake? How to obtain $$\sqrt{2}$$?

3) I think that I missed $$\prod_{n\in\mathbb{Z}} (-1) = (-1)^{\sum_{n\in \mathbb{Z}}1} = ? (-1)^{\sum_{n>0}1} = (-1)^{\zeta(0)} = -i$$

And so I obtain $$Z(S^1)= \det^{1/2}_{AP}(i\frac{d}{dt}) = \sqrt{2}$$

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.
• Thank you! Why did you take square root from determinant? – Nikita Jan 3 '20 at 23:59
• Well you're fermions seem to be real (there is no conjugation in your kinetic term). If they're complex get rid of the root, but then you would integrate over $\psi$ and $\psi^\star$. – lux Jan 4 '20 at 0:10
• I found some mistakes in my calculations. But I obtain $2i$. Could you give some comment about it? – Nikita Jan 4 '20 at 0:20
• You would have to tell me where you disagree with my calculations – lux Jan 4 '20 at 1:31

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.

--

$$^{\dagger}$$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.

• What's wrong with following: $\prod_{n\in Z} = a^{\sum_{n>0}1 }= a^{-1/2}$ ? – Nikita Jan 4 '20 at 14:55
• Where did the $n\leq 0$ part go? – Qmechanic Jan 4 '20 at 15:00
• I use that powersets is the same.. – Nikita Jan 4 '20 at 15:02
• The above heuristic calculus doesn't work well under cardinality equivalence. – Qmechanic Jan 4 '20 at 15:14