Quantum circuit model of black hole evaporation

We describe the Hawking radiation using qubits in our model. A qubit has two distinct internal states $\renewcommand{\ket}[1]{|#1\rangle} \ket{0}$ and $\renewcommand{\ket}[1]{|#1\rangle} \ket{1}$. In order to formulate the black hole evaporation, we separate the Hilbert space of the total system to that of the black hole, the Hawking radiation and the Hawking radiation that had been radiated previously. Hence, prepare the Hilbert space of black hole ($\mathsf{BH}$), just radiation ($\mathsf{JR}$), early radiation ($\mathsf{ER}$) (figure 1)

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{H}_{\rm tot} = \mathcal{H}_\textsf{BH}\otimes \mathcal{H}_\textsf{JR}\otimes\mathcal{H}_\textsf{ER}. \nonumber \end{align} \tag{ 2 }$$

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Here, we regard the state $\renewcommand{\ket}[1]{|#1\rangle} \ket{0}$ of $\mathsf{JR}$ and $\mathsf{ER}$ as the 'vacuum state' with no particles.

We treat particles in $\mathsf{BH}$, that are paired with the emitted Hawking particles and fall into the black hole, constitute the black hole's degrees of freedom. This is because particles falling into the black hole can be regarded as a part of the black hole owing to the black hole complementary postulate 4.

Particles come out from the black hole in the evaporation process. For example, a particle in $\mathsf{JR}$ will belong to $\mathsf{ER}$ at the next time step. This process can be realized using the SWAP operation, which exchanges inputted two qubits and is represented by the unitary quantum gate shown in figure 2.

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In the evaporation process, the black hole shrinks by emitting particles and finally disappears. Furthermore, the information of the black hole is transferred to radiated particles. Qubits in $\mathsf{BH}$ gradually change to $\renewcommand{\ket}[1]{|#1\rangle} \ket{0}$ state during the evaporation process. When the evaporation finishes and the black hole disappears completely, all the qubits in $\mathsf{BH}$ become $\renewcommand{\ket}[1]{|#1\rangle} \ket{0}$ state. On the other hand, at the initial stage of evaporation, most of the radiated particles are in $\renewcommand{\ket}[1]{|#1\rangle} \ket{0}$ state (no particles). Thus, radiated particles state will becomes very complicated because information of the black hole is carried away by the radiated particles. We expect that evaporation can be modeled by swapping the information (qubit) of the black hole with a part of radiation's degrees of freedom. In our model, the SWAP are performed twice between $\mathsf{BH}$ and $\mathsf{JR}$, and between $\mathsf{JR}$ and $\mathsf{ER}$.

As a demonstration, we look at the case of the 5 qubit total system and show how the SWAP procedure transfer the information of $\mathsf{BH}$ to radiations. The total system is

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi}=\ket{\textsf{BH}}\otimes \ket{\textsf{JR}}\otimes \ket{\textsf{ER}}, \nonumber \end{align} \tag{ 3 }$$

with $|\textsf{BH}|=2^2$ (2 qubit), $|\textsf{JR}|=2^1$ (1 qubit), $|\textsf{ER}|=2^2$ (2 qubit). The initial state of each subsystem is assumed to be

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle {\ket{\textsf{BH}}=\frac{\ket{0}_{\textsf{BH}_1}\ket{1}_{\textsf{BH}_2}+\ket{1}_{\textsf{BH}_1}\ket{0}_{\textsf{BH}_2}}{\sqrt{2}}},\quad \ket{\textsf{JR}}=\ket{0}_\textsf{JR},\quad \ket{\textsf{ER}}=\ket{0}_{\textsf{ER}_1}\ket{0}_{\textsf{ER}_2}, \nonumber \end{align} \tag{ 4 }$$

and the initial total state $\renewcommand{\ket}[1]{|#1\rangle} \ket{\psi_0}$ is

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi_0}={\frac{1}{\sqrt{2}}}\Bigl[\ket{10}+\ket{01}\Bigr]_{\textsf{BH}_{12}}\ket{0}_\textsf{JR}\ket{00}_{\textsf{ER}_{12}}. \nonumber \end{align} \tag{ 5 }$$

Although the Page curve is the average of entanglement entropy, we focus here on a specific state and compute the non-averaged entanglement entropy. The quantum circuit we introduce is shown in figure 3.

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Let us follow the time steps realized by this circuit.

$\mathsf{At~time~step~1:}$

• 1.  
  Apply the SWAP between the first qubit of $\mathsf{BH}$ and $\mathsf{JR}$,

  $$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi_0}\longmapsto\ket{\psi_{0'}}={\frac{1}{\sqrt{2}}}\ket{0}_{\textsf{BH}_1}\Bigl[\ket{01}+\ket{10}\Bigr]_{\textsf{BH}_2,\textsf{JR}}\ket{00}_{\textsf{ER}_{12}}. \nonumber \end{align} \tag{ 6 }$$
• 2.  
  Apply the SWAP between $\mathsf{JR}$ and the first qubit of $\mathsf{ER}$,

  $$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi_{0'}}\longmapsto\ket{\psi_{1}}={\frac{1}{\sqrt{2}}}\ket{0}_{\textsf{BH}_1}\ket{0}_\textsf{JR}\Bigl[\ket{01}+\ket{10}\Bigr]_{\textsf{BH}_2,\textsf{ER}_1}\ket{0}_{\textsf{ER}_2}. \nonumber \end{align} \tag{ 7 }$$

$\mathsf{At~time~step~2:}$

• 1.  
  Apply the SWAP between the second qubit of $\mathsf{BH}$ and $\mathsf{JR}$,

  $$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi_1}\longmapsto\ket{\psi_{1'}}={\frac{1}{\sqrt{2}}}\ket{00}_{\textsf{BH}_{12}}\Bigl[\ket{01}+\ket{10}\Bigr]_{\textsf{JR},\textsf{ER}_1}\ket{0}_{\textsf{ER}_2}. \nonumber \end{align} \tag{ 8 }$$
• 2.  
  Apply the SWAP between $\mathsf{JR}$ and the second qubit of $\mathsf{ER}$,

  $$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi_{1'}}\longmapsto\ket{\psi_{2}}={\frac{1}{\sqrt{2}}}\ket{00}_{\textsf{BH}_{12}}\ket{0}_\textsf{JR}\Bigl[\ket{10} +\ket{01}\Bigr]_{\textsf{ER}_{12}}. \nonumber \end{align} \tag{ 9 }$$

To check this process reflects the property of evaporating models, we compute the von Neumann entropy of radiations $\mathsf{R}:=\mathsf{JR}\, \cup\, \mathsf{ER}$, $\newcommand{\Tr}{{\rm Tr}} S_\textsf{R} = - \Tr \rho_\textsf{R} \log_2 \rho_\textsf{R}$ at each step of the SWAP (figure 4).

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The typical shape of the Page curve is obtained and the information of $\mathsf{BH}$ is transferred to the radiation by the SWAP operation. The effective size of $\mathsf{BH}$'s Hilbert space at step n is defined by the amount of information

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle |\textsf{BH}_n^{{{\rm eff}}}| := |\textsf{BH}|*2^{-(n-1)}\quad (n\geqslant 1),\qquad |\textsf{BH}_0^{{{\rm eff}}}| := |\textsf{BH}| \nonumber \end{align} \tag{ 10 }$$

where $|\textsf{BH}|$ is the dimension of $\mathsf{BH}$'s Hilbert space determined by the number of qubits. Particles are radiated from $\mathsf{BH}$ by the first SWAP at each step. Thus, this quantum circuit does not have the black hole horizon and the entanglement structure of this circuit is determined only by the $\mathsf{BH}$ initial state. Qubits in $\mathsf{BH}$ are entangled with each other at the initial step. Then, a part of entangled qubit is moved to become $\mathsf{JR}$ and the entanglement between $\mathsf{BH}$ and $\mathsf{R}$ increase. However, the entanglement of the Hawking radiation is originated from the entangled particle pairs created in the vicinity of the black hole horizon and this SWAP circuit is the model of evaporation process without the horizon.

The quantum state of the Hawking radiation is expressed as $\renewcommand{\ket}[1]{|#1\rangle} \prod\nolimits_{i}\sum\nolimits_{n_i=0}^\infty {\rm e}^{-n_i\beta \omega_i/2} \ket{n_i}_\textsf{BH}\otimes\ket{n_i}_\textsf{JR}$ where $\beta=8\pi M$ is the inverse temperature of the black hole and $\omega$ is the frequency of the radiation [1]. The state $\renewcommand{\ket}[1]{|#1\rangle} \ket{n_i}_\textsf{BH}\otimes\ket{n_i}_\textsf{JR}$ represents pairs of the Hawking particles. For $M\omega\gg 0.1$, the state is almost same as the vacuum state without particle excitation and for $M\omega < 0.1$, $\renewcommand{\ket}[1]{|#1\rangle} \ket{1}_\textsf{BH}\otimes\ket{1}_\textsf{JR}$ particle state mainly contributes to the total state of the Hawking radiation. Therefore, the essence of the entanglement structure can be revealed even if two or more particle number states are ignored. Thus, we assume the following entangled state represents the Hawking radiation

$$\begin{align} \newcommand{\bi}{\boldsymbol} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \frac{1}{\sqrt{1+\exp(-8\pi M\omega)}}\biggl[\ket{0}_{\textsf{BH}}\ket{0}_\textsf{JR} +\exp(-4\pi M\omega)\ket{1}_{\textsf{BH}}\ket{1}_\textsf{JR}\biggr]. \label{eq:haw} \nonumber \end{align} \tag{ 11 }$$

From now on, we focus on a specific ω mode of the Hawking radiation. We consider a quantum gate mimicking this state (figure 5). The gate consists of two well-known quantum gates and outputs the entangled state.

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The gate of the right part in the circuit is an unitary gate acting on two qubit called the CNOT gate and works as $\renewcommand{\ket}[1]{|#1\rangle} \ket{x} \ket{y} \to \ket{x} \ket{x\oplus y}$. The gate U in the left part is an unitary gate acting on a single qubit. In terms of the matrix representation, it is expressed as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle U := \left[\begin{array}{@{}cc@{}} \cos{\gamma} & \sin{\gamma} \nonumber \\ \sin{\gamma} & - \cos{\gamma} \label{rotation} \end{array}\right],\quad \tan\gamma=\exp(-4\pi M\omega), \nonumber \end{align} \tag{ 12 }$$

where γ is the squeezing parameter. For $\gamma = \pi/4$, U corresponds to the Hadamard gate H, and the CNOT-U gate will output the maximally entangled EPR state. The gate in figure 5 can output an entangled state and the amount of entanglement depends on the parameter $\gamma$.

The strength of the entanglement between the created Hawking particle pairs is determined by the mass of the black hole at the considering time step and we must determine the relation between the parameter $\gamma$ and the black hole mass. For this purpose, we consider the correspondence between the dimension of the $\mathsf{BH}$ Hilbert space ($\mathsf{BH}$ qubit number) and the black hole mass M. From the formula of the Bekenstein–Hawking entropy (1)

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{\rm BH}(M)=4\pi \left(\frac{M}{m_{\rm pl}}\right)^2 = \log_2 |\textsf{BH}^{\rm eff}|, \nonumber \end{align} \tag{ 13 }$$

we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle M_n = \frac{m_{\rm pl}}{2}\sqrt{\frac{\log_2{|\textsf{BH}_n^{\rm eff}|}}{\pi}}. \nonumber \end{align} \tag{ 14 }$$

Note that $| \textsf{BH}_n^{\rm eff} |$ is just the effective size of $\mathsf{BH}$ Hilbert space and not the actual number of the $\mathsf{BH}$ states. Using these relations, in the natural unit, we obtain

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \gamma_n = \mathrm{Arctan}\left[\exp\left(-2\omega \sqrt{\pi \log_2 |\textsf{BH}_n^{\rm eff}|}\right)\right]. \label{gamma} \nonumber \end{align} \tag{ 15 }$$

With this squeezing parameter $\gamma_n$, the CNOT-U gate in figure 5 mimics the the Hawking radiation (11).

We combine the SWAP gate and the CNOT-U gate to construct a model of the evaporation process of the Hawking radiation. We have already confirmed that the initial state of $\mathsf{BH}$ is transferred to $\mathsf{JR}$ and $\mathsf{ER}$ by the SWAP circuit and the Page curve is obtained. In the following analysis, we focus on the property of the entanglement caused by the Hawking radiation to clarify how the dynamics of the evaporation is connected with the quantum information process and the firewall argument. Hence, the initial state is assumed to be non-entangled state

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{{{\rm init}}} = \ket{000 \cdots 000}_\textsf{BH} \ket{0}_\textsf{JR} \ket{000 \cdots 000}_\textsf{ER}. \nonumber \end{align} \tag{ 16 }$$

The quantum circuit of our model is presented in figure 6.

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In this model, the initial mass of the black hole corresponds to qubit number of $\mathsf{BH}$. The qubit number can not make so large due to increase of the simulation time. We denote the black hole mass at a certain step n as $M_{n} \omega$. Because $M_n\propto \sqrt{\log_2 |\textsf{BH}_n|}$, the black hole mass at arbitrary step n is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle M_n = M_{{{\rm init}}}\sqrt{\frac{\log_2|\textsf{BH}_n|}{\log_2|\textsf{BH}_{{{\rm init}}}|}} = M_{{{\rm init}}} \sqrt{1-\frac{n-1}{N_\textsf{BH}}}, \nonumber \end{align} \tag{ 17 }$$

where $M_{\rm init}$ is the initial mass of the black hole and $N_\textsf{BH}$ is the qubit number of $\mathsf{BH}$ used for analysis. The only parameters contained in this model are the value of $M_{n} \omega$ and n. In our analysis, we choose n as the step at the Page time.

As in section 2.2, we check the amount of information transferred from the $\mathsf{BH}$ to the emitted particles and investigate whether the black hole in our model evaporates. For this purpose, we evaluate the entanglement entropy between $\mathsf{BH}$ and $\mathsf{R}=\mathsf{JR}\, \cup\, \mathsf{ER}$. The evolution of this quantity provides the Page curve, which is an indicator of the information transfer from $\mathsf{BH}$ to radiation. The result is shown for $M_4\, \omega = 0.1$ and 0.01 in figure 7. Here, M₄ denotes the mass of the black hole at step 4.

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For $M_4 \omega=0.01$ (small $M_4 \omega$), the Page curve has a symmetric shape with respect to the middle point of the whole time steps. On the other hand, for $M_4 \omega=0.1$ (large $M_4\omega$), the entanglement entropy between $\mathsf{BH}$ and $\mathsf{R}$ is smaller compared to the small $M_4\omega$ case during whole period of the evaporation process. For large $M_4\omega$, the black hole radiates Hawking particles with weak entanglement (see equation (11)). Thus the evaporation process is not random enough as Page assumed. However, our model results in Page curves for both values of $M_4\omega$ because the degree of freedom of $\mathsf{BH}$ decreases as the time step proceeds due to the action of the SWAP gate. The entanglement entropy becomes maximum at step 4 and this time step is the Page time of our model.

To examine the evaporation rate of our model, we calculate the expectation value of the emitted particle number. In our model, the Hawking radiation is expressed by the state (11) and this expression results in the expectatin value of the particle number

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \langle n \rangle = \frac{1}{{\rm e}^{8\pi M\omega}+1}. \label{eq:expect-a} \nonumber \end{align} \tag{ 18 }$$

Although this distribution is equivalent to that of the Hawking radiation with the fermion field [1], we are treating the bosonic particles and the fermionic distribution is result of our assumption for the state of the Hawking radiation. For the state $\renewcommand{\ket}[1]{|#1\rangle} \ket{\psi} = \sum\nolimits_{ijk} C_{ijk} \ket{i}_\textsf{BH}\ket{j}_\textsf{JR}\ket{k}_\textsf{ER}$, the expectation value of the particle number of $\mathsf{JR}$ is expressed as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \langle n_\textsf{JR} \rangle = \sum_{ik} |C_{i1k}|^2. \label{eq:expect-b} \nonumber \end{align} \tag{ 19 }$$

We checked these formulas for $M_4 \omega =1, ~0.1, ~0.01, ~0.001$ (figure 8). As the black hole evaporates, the mass decreases and the time step advances from right to left. In our model, $\langle n_\textsf{JR} \rangle$ well coincides with the formula (18) except at the last step with $M_4\omega\geqslant 0.1$. For $M_4\omega\geqslant 0.1$, $\langle n_\textsf{JR} \rangle<0.3$ and this means the emitted pair of Hawking particles is weakly entangled. On the other hand, for $M_4\omega\leqslant0.01$, $\langle n_\textsf{JR} \rangle\approx 0.5$ and this means the Hawking pair is nearly maximally entangled.

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At the last step, the both formula coincides for $M_4 \omega\leqslant 0.01$ but differs much for $M_4 \omega\geqslant 0.1$. This disagreement is caused by the different treatment of the Hawking radiation at the last step in our circuit model. In the evaporation process, $M\rightarrow 0$ is realized at the last step of evaporation. On the other hand, in our model, there is no CNOT-U gate at the last step (see figure 6), so sufficient radiation will not be created at the last step. This is the reason why large disagreement appeared at the last step for $M_4\omega\geqslant 0.1$. However, if the number of qubits is increased and make the interval of the time step sufficiently small, this difference is expected to be reduced.

Let us analyze the entanglement measure in our model. As we have already introduced the entanglement entropy as the measure of entanglement in section 2.2, we introduce here other two measures needed for our analysis. The mutual information of the bipartite system $\rho_{\rm AB}$ is defined as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I({{\rm A}}\!:\!{{\rm B}})= S_{\rm A} + S_{\rm B} - S_{{{\rm A}}\cup{{\rm B}}}. \label{mutual-info} \nonumber \end{align} \tag{ 20 }$$

This quantity evaluates how the system is close to the product state. If the mutual information is zero, the state is the product state without classical and quantum correlations. The negativity [13, 14] is defined by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{N}({{\rm A}}\!:\!{{\rm B}})= \frac{1}{2}\left(\sum_i|\lambda_i^T|-1\right),\quad \sum_i\lambda_i^T=1,\label{def:negativity} \nonumber \end{align} \tag{ 21 }$$

where $\lambda_i^T$ is the eigenvalue of the partially transposed state $\rho^{T_{\rm A}}_{\rm AB}$. The partial transposition $T_{\rm A}$ for the state $\renewcommand{\bra}[1]{\langle#1|} \renewcommand{\ket}[1]{|#1\rangle} \rho = \sum C_{ab:a'b'}\ket{a}\bra{a'}\otimes \ket{b}\bra{b'}$ is defined by $\renewcommand{\bra}[1]{\langle#1|} \renewcommand{\ket}[1]{|#1\rangle} \rho^{T_{\rm A}}:= \sum C_{ab:a'b'}\ket{a'}\bra{a}\otimes \ket{b}\bra{b'}$. For a separable bipartite state which has only classical correlations, $\lambda_i^T>0$ owing to Peres's partial transpose criterion [15] and we have $\mathcal{N}=0$. Thus $\mathcal{N}>0$ means the state is non-separable (entangled). The negativity is an entanglement monotone function, and can be used to quantify the entanglement between A and B even if the combined system A$\cup$B is not pure. $\mathcal{N}>0$ is the necesary and sufficient non-separable condition for $2\otimes2$ and $2\otimes 3$ bipartite quantum systems.

We calculate these entanglement measures between $\mathsf{BH}$ and $\mathsf{JR}$, $\mathsf{JR}$ and $\mathsf{ER}$, $\mathsf{ER}$ and $\mathsf{BH}$ for $M_4\, \omega = 0.1, ~0.01$. The result is shown in figure 9. For $M_4 \omega=0.01$, the negativity between $\mathsf{BH}$ and $\mathsf{JR}$ becomes zero at step 5 after the Page time and $\mathsf{BH}$ and $\mathsf{JR}$ are separable. In other words, for the Hawking radiation with sufficiently low frequencies, a structure similar to the firewall appears between $\mathsf{BH}$ and $\mathsf{JR}$. However, contrary to the original proposed firewall by AMPS, the classical correlations remains in our firewall-like structure and we do not expect high energy phenomena associated to it. On the other hand, for $M_4 \omega=0.1$, the values of mutual information and negativity between $\mathsf{BH}$ and $\mathsf{JR}$ remain nonzero, and $\mathsf{BH}$ and $\mathsf{JR}$ are entangled until the end of the evaporation. For sufficiently small $M_4 \omega$, nearly maximally entangled pair of the Hawking particles is created by the CNOT-U gate and the qubit strongly entangled with other qubit goes through the CNOT-U gate again after the Page time. However, due to the limited amount of entanglement between qubits, which is known as the monogamy property of the entanglement, it is impossible to add more entanglement and no entangled pairs can be created when passing through the second CNOT-U gate after the Page time. If this reasoning for small $M_4 \omega$ is correct, the emergence of the firewall-like structure can be explained.

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According to our numerical calculation, the firewall-like behavior becomes remarkable as $M_4\omega$ becomes small. To obtain comprehensive understanding of this structure, we analytically evaluate the state around the Page time. We introduce the matrix representation of the U gate and the CNOT gate,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle U_n= \left[\begin{array}{@{}cc@{}} \cos\gamma_n & \sin\gamma_n \nonumber \\ \sin\gamma_n & -\cos\gamma_n \end{array}\right],\quad \mathrm{CNOT}= \left[\begin{array}{@{}cc@{}} \mathbb{I}_2 & 0 \nonumber \\ 0 & X \end{array}\right], \nonumber \end{align} \tag{ 22 }$$

where $\mathbb{I}_2$ is the $2\times 2$ identity matrix, $\newcommand{\e}{{\rm e}} X=\left[\begin{array}{@{}cc@{}} 0 & 1 \\ 1 & 0\end{array}\right]$ and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \gamma_n=\mathrm{Arctan}\left[\exp\left(-4\pi M_4\omega\sqrt{\frac{7-n}{3}}\right)\right],\quad 1\leqslant n\leqslant 7. \nonumber \end{align} \tag{ 23 }$$

Then the matrix representation of the CNOT-U gate is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{\rm CNOT\mbox{-}U}}=\left[\begin{array}{@{}cc@{}} \mathbb{I}_2 & 0 \nonumber \\ 0 & X\end{array}\right](\mathrm{U}_n\otimes\mathbb{I}_2) =\left[\begin{array}{@{}cccc@{}} \cos\gamma_n & 0 & \sin\gamma_n & 0 \nonumber \\ 0 & \cos\gamma_n & 0 & \sin\gamma_n \nonumber \\ 0 & \sin\gamma_n & 0 & -\cos\gamma_n \nonumber \\ \sin\gamma_n & 0 & -\cos\gamma_n & 0 \end{array}\right]. \nonumber \end{align} \tag{ 24 }$$

We prepare $\renewcommand{\ket}[1]{|#1\rangle} \ket{{{\rm init}}} = \ket{000000}_\textsf{BH}\ket{0}_\textsf{JR}\ket{000000}_\textsf{ER}$ as the initial state and consider the action of the quantum circuit on this state.

Before the Page time: The Page time corresponds to step 4 and the circuit up to just before step 4 is equivalent to the following circuit diagram (figure 10). We rearranged order of qubit and changed the number of labels from those defined in the original circuit in figure 6.

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The state of the total system is

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi^{{{\rm before~step~4}}}} = \ket{{{\rm squ}}}_{\textsf{BH}_1,\textsf{ER}_1}\ket{{{\rm squ}}}_{\textsf{BH}_2,\textsf{ER}_2} \ket{{{\rm squ}}}_{\textsf{BH}_3,\textsf{ER}_3} \ket{0}_{\textsf{BH}_{456},\textsf{JR},\textsf{ER}_{456}}. \nonumber \end{align} \tag{ 25 }$$

At step 4: The circuit just before the Page time to step 4 (the Page time) can be drawn as figure 11. Only $\textsf{BH}_1$, $\textsf{JR}$ and $\textsf{ER}_1$ get involved.

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The state at step 4 is obtained as

$$\begin{align} \newcommand{\bi}{\boldsymbol} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi^4}_{\textsf{BH}_1,\textsf{JR},\textsf{ER}_1} &= {{\rm CNOT}}_{\textsf{BH}_1,\textsf{JR}}\, U_{\textsf{BH}_1}\ket{{{\rm squ}}}_{\textsf{BH}_1,\textsf{ER}_1}\ket{0}_{\textsf{JR}}\nonumber \\ &= {{\rm CNOT}}_{\textsf{BH}_1,\textsf{JR}}\biggl(\cos\gamma_1\Bigl[\cos\gamma_4\ket{0}+\sin\gamma_4\ket{1}\Bigr]_{\textsf{BH}_1}\ket{0}_{\textsf{ER}_1} \nonumber \\ &\quad +\sin\gamma_1\Bigl[\sin\gamma_4\ket{0}+\cos\gamma_4\ket{1}\Bigr]_{\textsf{BH}_1}\ket{1}_{\textsf{ER}_1} \biggr)\ket{0}_\textsf{JR} \nonumber \\ &= \cos\gamma_1\Bigl[\cos\gamma_4\ket{000}+\sin\gamma_4\ket{101}\Bigr]_{\textsf{BH}_1, \textsf{JR}, \textsf{ER}_1} \nonumber \\ & \quad +\sin\gamma_1\Bigl[\sin\gamma_4\ket{010}-\cos\gamma_4\ket{111}\Bigr]_{\textsf{BH}_1, \textsf{JR}, \textsf{ER}_1}. \label{step4} \nonumber \end{align} \tag{ 26 }$$

In the low frequency limit $M_4\omega\rightarrow 0$, squeezing parameters are $\gamma_1, \gamma_4\rightarrow \pi/4$ and

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi^4}_{\textsf{BH}_1, \textsf{JR}, \textsf{ER}_1}\longrightarrow H_\textsf{JR}\frac{1}{\sqrt{2}}\Bigl[\ket{000}+\ket{111}\Bigr]_{\textsf{BH}_1, \textsf{JR}, \textsf{ER}_1}, \nonumber \end{align} \tag{ 27 }$$

where $\newcommand{\e}{{\rm e}} H_\textsf{JR}:=\left[\begin{array}{@{}cc@{}} 1 & 1 \\ 1 & -1 \end{array}\right]$ acts on the qubit of $\mathsf{JR}$. The appeared state $\renewcommand{\ket}[1]{|#1\rangle} (\ket{000}+\ket{111})/\sqrt{2}$ is the GHZ state, which is the maximally entangled triqubit state. If one qubit in the GHZ state is traced out, the remaining subsystem with two qubits becomes separable. Since an unitary operator acting only on a single qubit does not affect the entanglement structure, $\mathsf{BH}$ and $\mathsf{JR}$ become separable at step 4 in the low frequency limit if $\mathsf{ER}$ is traced out. As we will see, this leads to the firewall-like structure at step 5.

At before step 5: The circuit diagram is figure 12. Qubits pass through the SWAP gate from step 4 to step 5.

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At step 4, one qubit in $\mathsf{BH}$ and two qubits in $\mathsf{JR}$ and $\mathsf{ER}$ are entangled. At just before step 5, one qubit in $\mathsf{BH}$ is $\renewcommand{\ket}[1]{|#1\rangle} \ket{0}$ state, and one qubit in $\mathsf{JR}$ and two qubit in $\mathsf{ER}$ are entangled.

At step 5: The quantum circuit up to step 5 is shown in figure 13.

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The state at step 5 is obtained as

$$\begin{align} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi^5}_{\textsf{BH}_{12},\textsf{JR},\textsf{ER}_{124}} = \ket{0}_{\textsf{BH}_1}\ket{\psi^5}_{\textsf{BH}_2,\textsf{JR},\textsf{ER}_{124}}, \nonumber \end{align} \tag{ 28 }$$

where

$$\begin{align} \newcommand{\bi}{\boldsymbol} \newcommand{\e}{{\rm e}} \renewcommand{\ket}[1]{|#1\rangle} \displaystyle \ket{\psi^5}_{\textsf{BH}_2,\textsf{JR},\textsf{ER}_{124}} &=\mathrm{CNOT}_{\textsf{BH}_2, \textsf{JR}}\,U_{\textsf{BH}_2}\ket{{{\rm squ}}}_{\textsf{BH}_2, \textsf{ER}_2}\ket{\psi^4}_{\textsf{JR}, \textsf{ER}_{14}} \nonumber \\ &=\mathrm{CNOT}_{\textsf{BH}_2, \textsf{JR}}\biggl[\cos\gamma_2\left(\cos\gamma_5\ket{00}+\sin\gamma_5\ket{10}\right) \nonumber \\ &\quad \qquad \qquad \qquad +\sin\gamma_2\left(\sin\gamma_5\ket{01}-\cos\gamma_5\ket{11}\right)\biggr]_{\textsf{BH}_2, \textsf{ER}_2}\ket{\psi^4}_{\textsf{JR}, \textsf{ER}_{14}} \nonumber \\ &=\cos\gamma_1\cos\gamma_2\cos\gamma_4\cos\gamma_5\biggl(\ket{0}_{\textsf{BH}_2}\Bigl[\ket{0}+\tan\gamma_2\tan\gamma_5\ket{1}\Bigr]_{\textsf{ER}_2} \nonumber \\ &\qquad \times\Bigl[\ket{000}+\tan\gamma_4\ket{101}+\tan\gamma_1\tan\gamma_4\ket{010}-\tan\gamma_1\ket{111}\Bigr]_{\textsf{JR}, \textsf{ER}_{14}} \nonumber \\ &\qquad +\ket{1}_{\textsf{BH}_2}\Bigl[\tan\gamma_5\ket{0}-\tan\gamma_5\ket{1}\Bigr]_{\textsf{ER}_2} \nonumber \\ &\qquad \times \Bigl[\ket{100}+\tan\gamma_4\ket{001}+\tan\gamma_1\gamma_4\ket{110}-\tan\gamma_1\ket{011}\Bigr]_{\textsf{JR}, \textsf{ER}_{14}} \biggr). \nonumber \end{align} \tag{ 29 }$$

As we have obtain the state at step 4 and step 5, it is possible to investigate the entanglement structure by evaluating the mutual information and the negativity between $\mathsf{BH}$ and $\mathsf{JR}$.

4.2.1. At step 4.

The reduced density matrix $\rho^4_{\textsf{BH}_1\cup\textsf{JR}}$ at step 4 is

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\Tr}{{\rm Tr}} \renewcommand{\ket}[1]{|#1\rangle} \renewcommand{\bra}[1]{\langle#1|} \displaystyle \rho^4_{\textsf{BH}_1\cup\textsf{JR}}&=\Tr_{\textsf{ER}}\left[\ket{\psi^4}\bra{\psi^4}_{\textsf{BH}_1, \textsf{JR}, \textsf{ER}_1}\right] \nonumber \\ &=\left(\cos^2\gamma_1\cos^2\gamma_4+\sin^2\gamma_1\sin^2\gamma_4\right) \ket{00}\bra{00}+\frac{1}{2}\cos 2\gamma_1\sin 2\gamma_4\ket{00}\bra{11} \nonumber \\ &\quad +\frac{1}{2}\cos 2\gamma_1\sin 2\gamma_4\ket{11}\bra{00}+\left(\cos^2\gamma_1\sin^2\gamma_4+\sin^2\gamma_1\cos^2\gamma_4\right)\ket{11}\bra{11} \nonumber \\ &= \frac{1}{2}\left[\begin{array}{@{}cccc@{}} 1+\cos2\gamma_1\cos2\gamma_4 & 0 & 0 & \cos 2\gamma_1\sin 2\gamma_4 \nonumber \\ 0 & 0 & 0 & 0 \nonumber \\ 0 & 0 & 0 & 0 \nonumber \\ \cos 2\gamma_1\sin 2\gamma_4 & 0 & 0 & 1-\cos 2\gamma_1\cos 2\gamma_4 \end{array}\right]. \nonumber \end{align} \tag{ 30 }$$

To obtain the mutual information $I(\textsf{BH}_1\!:\!\textsf{JR})$, we prepare the reduce states

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\Tr}{{\rm Tr}} \displaystyle \rho^4_{\textsf{BH}_1}&=\Tr_{\textsf{JR}}\left[\rho^4_{\textsf{BH}_1\cup\textsf{JR}}\right]=\frac{1}{2}\left[\begin{array}{@{}cc@{}}1+\cos 2\gamma_1 \cos 2\gamma_4 & 0 \nonumber \\ 0 & 1-\cos 2\gamma_1 \cos 2\gamma_4 \end{array}\right],\nonumber \\ \rho^4_{\textsf{JR}}&=\Tr_{\textsf{BH}_1}\left[\rho^4_{\textsf{BH}_1\cup\textsf{JR}}\right]=\frac{1}{2}\left[\begin{array}{@{}cc@{}}1+\cos 2\gamma_1 \cos 2\gamma_4 & 0 \nonumber \\ 0 & 1-\cos 2\gamma_1 \cos 2\gamma_4 \end{array}\right]. \nonumber \end{align} \tag{ 31 }$$

Eigenvalues of these states are

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda_{\textsf{BH}_1\cup\textsf{JR}}=\frac{1}{2}\left(1\pm\cos 2\gamma_1\right),\quad \lambda_{\textsf{BH}_1}=\lambda_{\textsf{JR}}=\frac{1}{2}(1\pm\cos 2\gamma_1\cos 2\gamma_4), \nonumber \end{align} \tag{ 32 }$$

and the mutual information is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I(\textsf{BH}_1\!:\!\textsf{JR})= -\sum_{i=\textsf{BH}_1, \textsf{JR}}\lambda_i\log_2\lambda_i +\sum_{i=\textsf{BH}_1\cup\textsf{JR}}\lambda_i\log_2\lambda_i. \nonumber \end{align} \tag{ 33 }$$

The red line in the left panel of figure 14 shows the mutual information as a function of $M_4\omega$. It behaves as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I\sim \left\{\begin{array}{@{}ll@{}} 1+16\pi^2(M_4\omega)^2 &\quad {{\rm for}} \quad M_4\omega\ll 0.1 \nonumber \\ 16\sqrt{2}\pi (M_4\omega){\rm e}^{-8\pi M_4\omega} &\quad {{\rm for}} \quad M_4\omega\gg 0.1. \end{array}\right. \nonumber \end{align} \tag{ 34 }$$

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The negativity is obtained using eigenvalues of the partial transposed state $\rho^{4T_\textsf{JR}}_{\textsf{BH}\cup\textsf{JR}}$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{1}{2}\left(1\pm\cos 2\gamma_1\cos 2\gamma_4\right),\quad \pm\frac{1}{2}\cos 2\gamma_1\sin 2\gamma_4, \nonumber \end{align} \tag{ 35 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{N}(\textsf{BH}_1\!:\!\textsf{JR})=\frac{1}{2}\cos 2\gamma_1\sin 2\gamma_4. \nonumber \end{align} \tag{ 36 }$$

The red line in the right panel of figure 14 shows the negativity as a function of $M_4\omega$. The negativity at step 4 has a peak at $M_4\omega\approx 0.1$ and behaves as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{N}\sim \left\{\begin{array}{@{}ll@{}} 2\sqrt{2}\pi M_4\omega &\quad {{\rm for}} \quad M_4\omega\ll 0.1 \nonumber \\ {\rm e}^{-4\pi M_4\omega} &\quad {{\rm for}} \quad M_4\omega\gg 0.1. \end{array}\right. \nonumber \end{align} \tag{ 37 }$$

As we have mentioned, $\renewcommand{\ket}[1]{|#1\rangle} \ket{\psi^4}_{\textsf{BH}_1, \textsf{JR}, \textsf{ER}_1}$ becomes the GHZ state for $M_4\omega\rightarrow 0$. Corresponding to this, the state $\rho^4_{\textsf{BH}_1\cup\textsf{JR}}$ becomes separable in this limit. The von Neuman entropies for this state are $S(\textsf{BH}_1)=S(\textsf{JR})=S(\textsf{BH}_1\!\cup\!\textsf{JR})=1$ (bit). Thus the mutual information has the value 1 (bit), which means $\textsf{BH}_1$ and $\mathsf{JR}$ have the perfect classical correlation.

4.2.2. At step 5.

The reduced density matrix $\rho^5_{\textsf{BH}_2\cup\textsf{JR}}$ is

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\Tr}{{\rm Tr}} \renewcommand{\ket}[1]{|#1\rangle} \renewcommand{\bra}[1]{\langle#1|} \displaystyle \rho^5_{\textsf{BH}_2\cup \textsf{JR}}=\Tr_\textsf{ER}\Bigl[\ket{\psi^5}\bra{\psi^5}_{\textsf{BH}_2, \textsf{JR}, \textsf{ER}_{14}}\Bigr]= \left[\begin{array}{@{}cccc@{}} CA & 0 & 0 & EA \nonumber \\ 0 & CB & EB & 0 \nonumber \\ 0 & EB & DB & 0 \nonumber \\ EA & 0 & 0 & DA \end{array}\right], \nonumber \end{align} \tag{ 38 }$$

where

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle &A=\frac{1}{2}(1+\cos 2\gamma_1\cos 2\gamma_4),\quad B=\frac{1}{2}(1-\cos 2\gamma_1\cos 2\gamma_4),\quad C=\frac{1}{2}(1+\cos 2\gamma_2\cos 2\gamma_5), \nonumber \\ &D=\frac{1}{2}(1-\cos 2\gamma_2\cos 2\gamma_5),\quad E=\frac{1}{2}\cos 2\gamma_2\sin 2\gamma_5. \nonumber \end{align*}$$

To obtain the mutual information between $\textsf{BH}_2$ and $\mathsf{JR}$, we calculate eigenvalues of the states $\rho_\textsf{JR}, \rho_{\textsf{BH}_2}$ and $\rho_{\textsf{BH}_2\cup\textsf{JR}}$. Eigenvalues $\lambda_i$ (i  =  $\mathsf{JR}$, $\textsf{BH}_2$, $\textsf{BH}_2\!\cup\!\textsf{JR}$) are

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle &\lambda_\textsf{JR}=\frac{1}{2}(1\pm\cos 2\gamma_1\cos 2\gamma_2\cos 2\gamma_4\cos 2\gamma_5),\quad \lambda_\textsf{BH}=\frac{1}{2}(1\pm\cos 2\gamma_2\cos 2\gamma_5),\nonumber \\ &\lambda_{\textsf{BH}_2\cup\textsf{JR}}=\frac{1}{2}(1\pm\cos 2\gamma_1\cos 2\gamma_4)\cos^2 \gamma_2,~ \frac{1}{2}(1\pm\cos 2\gamma_1\cos 2\gamma_4)\sin^2 \gamma_2. \nonumber \end{align} \tag{ 39 }$$

The mutual information between $\textsf{BH}_2$ and $\mathsf{JR}$ as a function of $M_4\omega$ is shown in figure 14 (the blue line in the left panel). It behaves as $I\approx (40\pi^2/3)(M_4\omega){}^2$ for $M_4\omega\ll0.1$ and $I\rightarrow 0$ for $M_4\omega\rightarrow 0$ limit.

To evaluate the negativity, we obtain the eigenvalues of the partially transposed state $\rho^{T_\textsf{JR}}_{\textsf{BH}_2, \textsf{JR}}$ as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda_i=\frac{(C+D)A\pm\sqrt{(C-D)^2A^2+4E^2B^2}}{2},\quad \frac{(C+D)B\pm\sqrt{(C-D)^2B^2+4E^2A^2}}{2}, \nonumber \end{align} \tag{ 40 }$$

and the negativity is given by $\mathcal{N}=\left(\sum\nolimits_i|\lambda_i|-1\right)/2$. The result is shown in the right panel in figure 14 (the blue line). The negativity becomes exactly zero for $M_4\omega\leqslant 0.041$. Thus, at step 5 (next step to the Page time), our model shows $\textsf{BH}_2$ and $\mathsf{JR}$ become separable for low frequency modes satisfying $M_4\omega < 0.041$ and have only the classical correlation. This separable state corresponds to the firewall. In the $M_4\omega\rightarrow 0$ limit ($\gamma\rightarrow\pi/4$), the reduced density matrix at step 5 is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \rho_{\textsf{BH}_2\cup\textsf{JR}} = \frac{1}{4}(\vert{0}\rangle \langle{0}\vert+\vert{1}\rangle \langle{1}\vert)_\textsf{JR}\otimes (\vert{0}\rangle \langle{0}\vert+\vert{1}\rangle \langle{1}\vert)_{\textsf{BH}_2}=\frac{1}{4}\,\mathbb{I}_{4}. \nonumber \end{align} \tag{ 41 }$$

Thus, $\textsf{BH}_2$ and $\mathsf{JR}$ are in a product state with no classical correlation (random). For $0<M_4\omega\ll 0.041$, the state is separable with weak classical correlations. The schematic diagram representing the quantum states at step 4 and step 5 is shown in figure 15.

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Three qubits in $\mathsf{BH}$, $\mathsf{JR}$ and $\mathsf{ER}$ form the GHZ state at step 4, which is nearly maximally entangled. At step 5, the CNOT-U gate acts on $\mathsf{BH}$ and $\mathsf{JR}$. However, as $\mathsf{BH}$ and $\mathsf{JR}$ are already maximally entangled with other qubits, and no entanglement can be shared between $\mathsf{BH}$ and $\mathsf{JR}$ by the monogamy property of the multipartite entanglement. This is the reason why the entanglement between $\mathsf{BH}$ and $\mathsf{JR}$ is lost and the firewall-like structure arises for $M_4\omega\leqslant 0.041$.