# von Neumann Entropy of Cauchy Slices

In the review "The entropy of Hawking radiation" written by Juan Maldacena and others, which can be found under the following link arXiv:2006.06872, on page 18, within the caption of Figure 6, the authors claim the von Neumann entropy of both the Cauchy slices $$\Sigma$$ and $$\tilde{\Sigma}$$ is identical.

I am not quite able to figure out why this should be the case. I would be thankful if anyone could let me know why it so. I have attached a picture of the figure in question below.

1. Let $$A$$ be a subsystem and $$B$$ its complement (the rest of the system). Suppose that the state of the total system is a pure state $$|\psi\rangle = \sum_n \psi_n|A_n\rangle\otimes|B_n\rangle.$$ Then the reduced density matrices $$\rho_A$$ and $$\rho_B$$ both have the same eigenvalues, namely $$|\psi_n|^2$$, so their von Neumann entropies are equal to each other. If a unitary transformation $$|\psi\rangle\to U|\psi\rangle$$ only affects $$A$$ but not $$B$$, then it doesn't affect the quantities $$|\psi_n|^2$$, so it doesn't change the von Neumann entropy of either $$\rho_A$$ or $$\rho_B$$.
2. Let $$A$$ be the subsystem whose causal diamond is shown in the figure, on the surface $$\Sigma$$. Let $$B$$ be the subsystem on the complement of $$\Sigma$$, shown in the figure as the grey line outside the causal diamond. The states on $$\Sigma$$ and $$\tilde\Sigma$$ are related to each other by a unitary transformation that doesn't affect the state of $$B$$ on the grey line. Such a unitary transformation doesn't change the entropy of $$\rho_A$$, and it doesn't change $$\rho_B$$ at all.