Eq. (15) — $S(\Lambda(|\phi\rangle\langle\phi|))=S(\Omega)$, $\Omega_{kl}=\langle\phi,X_l^\dagger X_k\phi\rangle$

Initializing...

Eq. (15) — $S(\Lambda(|\phi\rangle\langle\phi|))=S(\Omega)$, $\Omega_{kl}=\langle\phi,X_l^\dagger X_k\phi\rangle$ · Prove2Me