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.

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  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.


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