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The passage around this formula

is a real number in units of energy for every fine state $\rho$ : how much the coarse description, evaluated through $H_c$ pulled back along $\Lambda$ , disagrees with the fine description’s own $H_f$ , for that specific state. Because expectation values determine a Hermitian operator uniquely — two Hermitian operators with the same trace against every density matrix are the same operator — the strongest statement follows at once: $\Delta$ $E_\Lambda(\rho)$ = 0 for every fine state $\rho$ if and only if $D_\Lambda$ = 0 as an operator identity, that is, if and only if $\Lambda^\dagger(H_c)$ = $H_f$ exactly. Energy is preserved for every possible input under that one algebraic condition, never as a statistical…

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