Exact chiral Majorana field with central charge one half

For $r=n+\frac{1}{2}>0$, $\epsilon ({\xi }_r{\xi }_{-r}\mathrm{\Omega })=1$, while every other mode pairing has zero vacuum coefficient. Therefore, in $\mathrm{\mid }z\mathrm{\mid }>\mathrm{\mid }w\mathrm{\mid }$, $\epsilon (\xi (z)\xi (w)\mathrm{\Omega })={\sum }_{n=0}^{\mathrm{\infty }}{z}^{-n-1}{w}^n=\frac{1}{z-w}\mathrm{.}$ The phase $\psi =i\xi$ gives the required negative contraction. Differentiation yields $⟨\psi (z)\psi (w){⟩}_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}}=-s^{-1},⟨\psi (z)\mathrm{\partial }\psi (w){⟩}_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}}=-s^{-2},⟨\mathrm{\partial }\psi (z)\psi (w){⟩}_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}}=s^{-2},⟨\mathrm{\partial }\psi (z)\mathrm{\partial }\psi (w){⟩}_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}}=2s^{-3},s=z-w\mathrm{.}$

In $T(z)\psi (w)$, contracting the first odd factor across the second gives a minus sign, while the second factor is already adjacent and gives a plus sign. Thus the two single contractions of $\frac{1}{2}:\psi (z)\mathrm{\partial }\psi (z):\psi (w)$ are $\frac{1}{2}(-⟨\psi (z)\psi (w)⟩\mathrm{\partial }\psi (z)+⟨\mathrm{\partial }\psi (z)\psi (w)⟩\psi (z))=\frac{1}{2}(\frac{\mathrm{\partial }\psi (z)}{s}+\frac{\psi (z)}{s^2})\mathrm{.}$ Expanding $\psi (z)=\psi (w)+s\mathrm{\partial }\psi (w)+O(s^2)$ and $\mathrm{\partial }\psi (z)=\mathrm{\partial }\psi (w)+O(s)$ gives $T(z)\psi (w)\sim \frac{\psi (w)}{2s^2}+\frac{\mathrm{\partial }\psi (w)}{s}\mathrm{.}$ Thus the formal stress OPE has the primary-field singular form with coefficient $h=\frac{1}{2}$. This identifies the Virasoro weight in the declared stress-OPE convention; this algebraic calculation does not establish the finite passive-coordinate covariance defined by def-cft-2d-primary-field.

For $T(z)T(w)$, write $A=\psi (z)$, $B=\mathrm{\partial }\psi (z)$, $C=\psi (w)$, and $D=\mathrm{\partial }\psi (w)$. Move the annihilation modes to the right using [A2]. The possible single pairings AC, AD, BC, BD cross respectively 1, 2, 0, 1 odd factors, so their signs are $-,+,+,-$; the two double pairings AC·BD and AD·BC have odd and even crossing parity, so their signs are $-,+$. These exhaust the pairings of the four odd factors, with the uncontracted fully normal-ordered term regular at $z=w$. The double pairings therefore equal $-(-\frac{1}{s})\left(\frac{2}{s^3}\right)+(-\frac{1}{s^2})\left(\frac{1}{s^2}\right)=\frac{1}{s^4}\mathrm{.}$ Multiplication by the two stress prefactors $1\mathrm{/}2$ gives the central term $1\mathrm{/}(4s^4)$. The single contractions are $\frac{1}{4}(\frac{:\mathrm{\partial }\psi (z)\mathrm{\partial }\psi (w):}{s}-\frac{:\mathrm{\partial }\psi (z)\psi (w):}{s^2}+\frac{:\psi (z)\mathrm{\partial }\psi (w):}{s^2}-\frac{2:\psi (z)\psi (w):}{s^3})\mathrm{.}$ Using fermionic antisymmetry and the Taylor expansions $:\mathrm{\partial }\psi (z)\psi (w):=-2T(w)-2s\mathrm{\partial }T(w)+O(s^2),:\psi (z)\mathrm{\partial }\psi (w):=2T(w)+O(s^2),:\psi (z)\psi (w):=-2sT(w)-s^2\mathrm{\partial }T(w)+O(s^3),$ the pole terms combine to $2T(w)\mathrm{/}{s}^2+\mathrm{\partial }T(w)\mathrm{/}s$; the first single-contraction term has no pole because $:\mathrm{\partial }\psi (w)\mathrm{\partial }\psi (w):=0$. Therefore $T(z)T(w)\sim \frac{1\mathrm{/}4}{(z-w{)}^4}+\frac{2T(w)}{(z-w{)}^2}+\frac{\mathrm{\partial }T(w)}{z-w}\mathrm{.}$ Comparison with [F2] yields $c\mathrm{/}2=1\mathrm{/}4$, hence $c=\frac{1}{2}$.

For fixed integer mode index $m$, the coefficient of $z^{-m-2}$ is $L_m=\frac{1}{2}{\sum }_{r+s=m}(-s-\frac{1}{2}):{\psi }_r{\psi }_s:$. Terms with both indices negative have only finitely many solutions; terms with both positive also have only finitely many solutions. In a mixed-sign term, the positive-index annihilation operator can act nontrivially on a fixed exterior basis vector only when its label belongs to that vector's finite occupied set, leaving finitely many terms. Hence each stress mode is defined on all of $\mathcal{F}$. The finite-particle mode sums establish the formal OPE and its two boundary coefficients $h=\frac{1}{2}$, $c=\frac{1}{2}$ on the specified domain. No conclusion about a full nonchiral theory, a spin-structure sum, analytic convergence, or a physical path integral follows.