Derivation — C019 (singlet projector in the Pauli basis)

The two-qubit singlet $|\Psi^-\rangle=\frac{1}{\sqrt2}(|!\uparrow\downarrow\rangle -|!\downarrow\uparrow\rangle)$ has projector $$|\Psi^-\rangle\langle\Psi^-|=\frac12\begin{pmatrix}0&0&0&0\\0&1&-1&0\\0&-1&1&0\\ 0&0&0&0\end{pmatrix}$$ in the basis $\{\uparrow\uparrow,\uparrow\downarrow,\downarrow\uparrow,\downarrow\downarrow\}$.

The Bell-state projectors are diagonal in the Pauli-correlator basis. For the singlet: $$\frac14(I\otimes I-X\otimes X-Y\otimes Y-Z\otimes Z).$$ Compute the three correlators on $|\Psi^-\rangle$: $X\otimes X$ and $Y\otimes Y$ both give $-1$ on the singlet, $Z\otimes Z$ gives $-1$ (antiparallel), so $\frac14(1-(-1)-(-1)-(-1))=\frac14\cdot4=1$, the maximal eigenvalue. The three operators $XX,YY,ZZ$ mutually commute and the singlet is their simultaneous $-1$ eigenvector; the four products $\{I,-XX,-YY,-ZZ\}$ sum (with the $1/4$) to the rank-1 projector onto that joint eigenspace. Explicitly the matrix identity $$\tfrac14(I-XX-YY-ZZ)=|\Psi^-\rangle\langle\Psi^-|$$ holds entry-by-entry (verified exactly in sympy_check.py).

Therefore for any state $\rho$, $$F(|\Psi^-\rangle)=\langle\Psi^-|\rho|\Psi^-\rangle =\mathrm{Tr}\big(\rho\,|\Psi^-\rangle\langle\Psi^-|\big) =\tfrac14\big(1-\langle XX\rangle-\langle YY\rangle-\langle ZZ\rangle\big),$$ using $\mathrm{Tr}(\rho I)=1$ and $\langle O\rangle=\mathrm{Tr}(\rho O)$. This is exactly the paper's estimator. Confirmed symbolically.