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(S1)=∫NSDψei12∫T0dtψddtψ
1) I need find eigenfunctions and eigenvalues for iddt with antiperiodic boundary conditions: iddtψn=λnψn,ψn(T)=−ψn(0) ψn(t)=ei2π(n+1/2)Tt,λn=−2π(n+1/2)T,n∈Z
Note that fermions is complex. What I need to do to calculate such integral for Majorana fermion?
2) I choose T=2π. So I need calculate: ∏n∈Z(n+1/2)=∏n>0(n+1/2)∏n≥0(−n+1/2)=2∏n≥0(n+1/2)×12∏n≥0(−1)(n+1/2)=(−1)1+∑+∞11e2∑+∞n=0ln(n+1/2)
We regularise using ζ-function (using this):
−(−1)ζ(0)e−2ζ′(0,1/2)=2i
Where I have mistake? How to obtain √2?
3) I think that I missed ∏n∈Z(−1)=(−1)∑n∈Z1=?(−1)∑n>01=(−1)ζ(0)=−i
And so I obtain Z(S1)=det
Answer
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}.
Comments:
- 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].
- 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.
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