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A summary of the spatial CSI situation as of now: if $\lvert\Psi(\boldsymbol r_1,\boldsymbol r_2)\rvert$ is positive semi-definite, then there is no CS violation anywhere. PSD of the two-photon wavefunction (kernel) can be defined by the requirement that every finite sampled matrix $[\lvert\Psi(\boldsymbol r_i,\boldsymbol r_j)\rvert]_{i,j=1}^n$, for all $n\in\mathbb N$ and all choices of points, be a PSD matrix. In that case, Mercer's theorem guarantees that such a kernel is PSD, which, we recall, is equivalent to all eigenvalues being non-negative, or to $\lvert\Psi\rvert=V^\dagger V$ for some $V$. The case $n=2$ is precisely the CSI, since
$$M_\mathrm{abs}=\begin{pmatrix}\lvert\Psi(\boldsymbol r_1,\boldsymbol r_1)\rvert&\lvert\Psi(\boldsymbol r_1,\boldsymbol r_2)\rvert\\ \lvert\Psi(\boldsymbol r_1,\boldsymbol r_2)\rvert&\lvert\Psi(\boldsymbol r_2,\boldsymbol r_2)\rvert\end{pmatrix},\qquad\det M_\mathrm{abs}=\lvert\Psi_{11}\Psi_{22}\rvert-\lvert\Psi_{12}\rvert^2,$$
and here, because we sample the modulus $|\Psi|$ rather than the wavefunction $\Psi$ (which is what CSI looks at), the two statements are equivalent: the CSI is violated at $(\boldsymbol r_1,\boldsymbol r_2)$ iff $\det M_\mathrm{abs}<0$. Note that PSD demands all minors, whereas CS constrains only those with $n=2$, so a beam may satisfy CS everywhere without $\lvert\Psi\rvert$ being PSD. There might be CSI extension to $n>2$.
12:10 (CET).