radius of three arcs are $R_{l},\ R,\ R_{r}$. In the following, we will start
with $R_{r}$ and $\sigma_{r}$, and finally express $\sigma_{l}$ with $R_{r}$
and $\sigma_{r}$. Then we can express the geodesic length only with
$\sigma_{l}$ and $\sigma_{r}$.

We will calculate the geodesic in several steps. (i) Assuming $|{\rm
OD}|=\alpha$, and under the constraint $R_{r}>\frac{\sigma_{r}}{2}$, we have

$\begin{split}|{\rm
DF}|^{2}=[(\sigma_{r}-R_{r})\cos{(\psi_{1}+\psi_{2})}-\alpha\sin{\psi_{2}}]^{2}&+[(\sigma_{r}-R_{r})\sin{(\psi_{1}+\psi_{2})}+\alpha\cos{\psi_{2}}]^{2}=R_{r}^{2},\\\
\alpha=(R_{r}-\sigma_{r})\sin{\psi_{1}}+&\sqrt{R_{r}^{2}-(R_{r}-\sigma_{r})^{2}\cos^{2}{\psi_{1}}}.\end{split}$
(154a) (ii) Solving the equation of the line ${\rm FD}$, we have the point
${\rm G}=(x_{0},0)$ with
$x_{0}=\alpha\sin{\psi_{2}}+\frac{(\sigma_{r}-R_{r})\cos{(\psi_{1}+\psi_{2})}-\alpha\sin{\psi_{2}}}{-(\sigma_{r}-R_{r})\sin{(\psi_{1}+\psi_{2})}-\alpha\cos{\psi_{2}}}(-\alpha\cos{\psi_{2}}).$
(154b) (iii) We define $|{\rm DG}|=\beta$ and $(\cdot)_{x,y}$ is
$x,y$-coordinate of the point $(\cdot)$, then $\beta=R_{r}\frac{{\rm
D}_{y}-{\rm G}_{y}}{{\rm D}_{y}-{\rm
F}_{y}}=\frac{\alpha\cos{\psi_{2}}}{\alpha\cos{\psi_{2}-[-(\sigma_{r}-R_{r})\sin{(\psi_{1}+\psi_{2})}]}}R_{r}.$
(154c) (iv) With $|{\rm OC}|=\gamma$, then $|{\rm CG}|=|{\rm DG}|$ with the
constraint $\beta>x_{0}$ gives
$\gamma=\sqrt{\beta^{2}-x_{0}^{2}\cos^{2}{\psi_{2}}}-x_{0}\sin{\psi_{2}}.$
(154d) (v) With ${\rm E}$ located at $(x_{1},\
y_{1}=\tan{(\psi_{1}+\psi_{2})}x_{1})$, it is also on the line ${\rm CG}$,
which requires
$x_{1}=\frac{\gamma\cos{\psi_{2}}}{\gamma\cos{\psi_{2}}+(\gamma\sin{\psi_{2}}+x_{0})\tan{(\psi_{1}+\psi_{2})}}x_{0}.$
(154e) (vi) Finally we have $|{\rm CE}|=|{\rm AE}|$, which gives
$\sigma_{l}=-\frac{x_{1}}{\cos{(\psi_{1}+\psi_{2})}}+\sqrt{\gamma^{2}+\frac{x_{1}}{\cos{(\psi_{1}+\psi_{2})}}\left(\frac{x_{1}}{\cos{(\psi_{1}+\psi_{2})}}-2\gamma\sin{\psi_{1}}\right)},$
(154f) where we use the constraint $\sigma_{l}>0$.

For a fixed $\sigma_{r}$, and any given $R_{r}$, we can express all variables
above, including $\sigma_{l}$. Then the length of the geodesic can be
considered as below.

For ${\rm AC}$, ${\rm A}=(-\sigma_{l},\ \varepsilon)$ and ${\rm
C}=(\gamma\sin{\psi_{1}},\ \gamma\cos{\psi_{1}})$, so the length is $d_{\rm
AC}=L_{1}\cosh^{-1}{\left(\frac{\sigma_{l}^{2}+\gamma^{2}+2\sigma_{l}\gamma\sin{\psi_{1}}}{2\varepsilon\gamma\cos{\psi_{1}}}\right)}\approx
L_{1}\log{\left(\frac{\sigma_{l}^{2}+\gamma^{2}+2\sigma_{l}\gamma\sin{\psi_{1}}}{\varepsilon\gamma\cos{\psi_{1}}}\right)}.$
(155a) Similarly, for ${\rm CD}$, ${\rm C}=(-\gamma\sin{\psi_{2}},\
\gamma\cos{\psi_{2}})$ and ${\rm D}=(\alpha\sin{\psi_{2}},\
\alpha\cos{\psi_{2}})$, so the length is $d_{\rm
CD}=L_{2}\cosh^{-1}{\left(\frac{(\alpha+\gamma)^{2}-2\alpha\gamma\cos^{2}{\psi_{2}}}{2\alpha\gamma\cos^{2}{\psi_{2}}}\right)}.$
(155b) For ${\rm DB}$, ${\rm D}=(-\alpha\sin{\psi_{1}},\
\alpha\cos{\psi_{1}})$ and ${\rm B}=(\sigma_{r},\ \varepsilon)$, so the length
is $d_{\rm
DB}=L_{1}\cosh^{-1}{\left(\frac{\sigma_{r}^{2}+\alpha^{2}+2\sigma_{r}\alpha\sin{\psi_{1}}}{2\varepsilon\alpha\cos{\psi_{1}}}\right)}\approx
L_{1}\log{\left(\frac{\sigma_{r}^{2}+\alpha^{2}+2\sigma_{r}\alpha\sin{\psi_{1}}}{\varepsilon\alpha\cos{\psi_{1}}}\right)}.$
(155c)

Therefore, the total length of the geodesic gives the entanglement entropy

$\begin{split}&S_{R_{r}(\sigma_{l}),\sigma_{r}}=\frac{d_{\rm AC}+d_{\rm
CD}+d_{\rm DB}}{4G_{(3)}}\\\
&=\frac{c_{1}}{6}\log{\left(\frac{\sigma_{l}^{2}+\gamma^{2}+2\sigma_{l}\gamma\sin{\psi_{1}}}{\varepsilon\gamma\cos{\psi_{1}}}\frac{\sigma_{r}^{2}+\alpha^{2}+2\sigma_{r}\alpha\sin{\psi_{1}}}{\varepsilon\alpha\cos{\psi_{1}}}\right)}+\frac{c_{2}}{6}\cosh^{-1}{\left(\frac{(\alpha+\gamma)^{2}-2\alpha\gamma\cos^{2}{\psi_{2}}}{2\alpha\gamma\cos^{2}{\psi_{2}}}\right)}.\end{split}$
(156)

From the result above, it seems that the prefactor of $\log$-term is
$\frac{c_{1}}{3}$, and the contribution of region 2 is only a constant. Later
we will show that it is not correct, and for different parameters there are
also two cases.

####  A.2.2 Numerical results of the general case

The equations above cannot be solved analytically with general variables
$\sigma_{l}$ and $\sigma_{r}$, and only numerical calculation is available.
Here we show some numerical results.

We first consider the marginal case $L_{1}>L_{2}$ and $\psi_{1}<\psi_{2}$.
With (154) and (155), we consider $L_{1}=2,\ L_{2}=1,\ T=1,\ \varepsilon=1$
and $\sigma_{r}=10$. Numerically, we can tune $R_{r}\in[5,20]$, and calculate
the corresponding $\sigma_{l}$ and the total geodesic length. In the
calculation, we will find that there is a range $R_{r}\in(R_{r}^{\rm
min},R_{r}^{\rm max})$ for the solution to exist, and $R_{r}^{\rm
min}>\frac{\sigma_{r}}{2}$.

Figure 24:  (a) Numerical and analytical results of the length $d$ of the
geodesic with respect to $\sigma_{l}$ for the marginal case. The parameters
are $L_{1}=2,\ L_{2}=1,\ T=1,\ \varepsilon=1$, $\sigma_{r}=10$ and
$R_{r}\in(R_{r}^{\rm min},R_{r}^{\rm max})$. (b) Numerical and analytical
results of the length $d$ of the geodesic with respect to $\sigma_{l}$ for the
irrelevant case. The parameters are $L_{1}=1,\ L_{2}=2,\ T=1,\ \varepsilon=1$,
$\sigma_{r}=10$ and $R_{r}\in(R_{r}^{\rm min},\infty)$.

Figure 25:  AdS dual of CFT with interface defect for infinite system and its
geodesic. (a) General CFT/CFT geodesic for $\psi_{2}>\psi_{1}$. (b) Geodesic
for large $\sigma_{l}$ limit case with $\psi_{2}>\psi_{1}$.

For $R_{r}=R_{r}^{\rm min}$, we have $\sigma_{l}=0$. The geodesic length is
shown in figure 24 (a). For $L_{1}<L_{2}$ and $\psi_{1}>\psi_{2}$, the
geodesic length is shown in figure 24 (b). With (154) and (155), we consider
$L_{1}=1,\ L_{2}=2,\ T=1,\ \varepsilon=1$ and $\sigma_{r}=10$. In the
numerical calculation, there is no $R_{r}^{\rm max}$ and $R_{r}^{\rm
min}=\frac{\sigma_{r}}{2}$.

It is unusual that $R_{r}^{\rm min}>\frac{\sigma_{r}}{2}$, which means in
figure 23 (a) although there is a finite length of $|{\rm DO}|$, the length of
$|{\rm AO}|=0$. In the following, we will first prove the existence of
$R_{r}^{\rm min}>\frac{\sigma_{r}}{2}$ for $L_{1}>L_{2}$ case. Actually, there
are two possibilities. One is $|{\rm CE}|=|{\rm OE}|\neq 0$, and the other is
$|{\rm OC}|=0$. We can prove that only the second one is possible. If $|{\rm
CE}|=|{\rm OE}|\neq 0$, then $\angle{\rm ECO}=\angle{\rm EOC}$, so $\angle{\rm
CGO}=(\frac{\pi}{2}-\psi_{2})-(\frac{\pi}{2}-\psi_{1})=\psi_{1}-\psi_{2}<0$,
which means there will be no crossing point ${\rm G}$ on the positive x-axis
and no solution for ${\rm F}$. For the second case we have $|{\rm DG}|=|{\rm
OG}|$, $\angle{\rm DGO}=\pi-2(\frac{\pi}{2}-\psi_{2})=2\psi_{2}$ and
$\angle{\rm GOF}=\psi_{1}+\psi_{2}<\angle{\rm DGO}$, so there always exists
the crossing point ${\rm F}$, which means the existence of $R_{r}^{\rm
min}>\frac{\sigma_{r}}{2}$.

While, for $L_{1}<L_{2}$ and $\psi_{1}>\psi_{2}$, we have $\angle{\rm
GOF}=\psi_{1}+\psi_{2}>\angle{\rm DGO}$, so there is no solution for crossing
point and no $R_{r}^{\rm min}>\frac{\sigma_{r}}{2}$. It can also be shown as
follows with figure 23 (b). With one fixed end point at the defect, we can
prove that ${\rm DG}\mathrel{/\mskip-2.5mu/}{\rm CFT_{right}}$ for another
side, which means for finite $\sigma_{r}$ we cannot have $\sigma_{l}=0$. (i)
Assuming $|{\rm CA}|=\alpha$, with $\angle{\rm
CGA}=(\frac{\pi}{2}-\psi_{2})-\angle{\rm
GCA}=(\frac{\pi}{2}-\psi_{2})-\angle{\rm EAC}=\psi_{1}-\psi_{2}$, we have the
equation of line ${\rm CG}$ that
$\frac{y-\alpha\cos{\psi_{2}}}{x+\alpha\sin{\psi_{2}}}=-\tan{(\psi_{1}-\psi_{2})}$.
(ii) Setting $y=0$ we have $x_{\rm
G}=x_{0}=\frac{\alpha\cos{\psi_{2}}}{\tan{\psi_{1}-\psi_{2}}}-\alpha\sin{\psi_{2}}$.
(iii) Defining $|{\rm AD}|=\beta$, with $|{\rm CG}|=|{\rm DG}|$ we have the
equation $\beta^{2}-2\beta x_{0}\sin{\psi_{2}}=\alpha^{2}+2\alpha
x_{0}\sin{\psi_{2}}$ and the solution $\alpha=\beta-2x_{0}\sin{\psi_{2}}$.
(iv) Then we can calculate the slope of the line ${\rm DG}$, which gives

$\tan{\angle{\rm
DGA}}=\frac{\beta\cos{\psi_{2}}}{x_{0}-\beta\sin{\psi_{2}}}=\tan{(\psi_{1}+\psi_{2})}.$
(157)

Because the slope of right CFT is $\tan{(\psi_{1}+\psi_{2})}$, it means ${\rm
CFT_{left}}\mathrel{/\mskip-2.5mu/}{\rm DG}$.

To summarize, we have shown that for $L_{1}>L_{2}$, $R_{r}^{\rm
min}>\frac{\sigma_{r}}{2}$, while for $L_{1}<L_{2}$, $R_{r}^{\rm
min}=\frac{\sigma_{r}}{2}$. Besides, we can calculate $R_{r}^{\rm min}$
analytically for $L_{1}>L_{2}$. For $R_{r}=R_{r}^{\rm min}$, we have $|{\rm
DG}|=|{\rm OG}|$. In figure 23 (a), for $\triangle{\rm DFO}$, we have $|{\rm
DF}|=R_{r}$ and $|{\rm OF}|=\sigma_{r}-R_{r}$. With $|{\rm DG}|=|{\rm OG}|$,
we have $\angle{\rm ODF}=\angle{\rm DOG}=\frac{\pi}{2}-\psi_{2}$, and
$\angle{\rm DOF}=\frac{\pi}{2}+\psi_{1}$. Then the law of sines gives

$\frac{R_{r}}{\sin{\left(\frac{\pi}{2}+\psi_{1}\right)}}=\frac{\sigma_{r}-R_{r}}{\sin{\left(\frac{\pi}{2}-\psi_{2}\right)}},$
(158)

which means $R_{r}^{\rm
min}=\frac{\sigma_{r}\cos{\psi_{1}}}{\cos{\psi_{1}}+\cos{\psi_{2}}}>\frac{\sigma_{r}}{2}$.
If we plug the parameter above in it, we will get $R_{r}^{\rm min}\approx
6.6667$, which is consistent with the numerical results.

Now we discuss the critical value $R_{r}^{\rm max}>\sigma_{r}$ for
$L_{1}>L_{2}$. The key is if the crossing point ${\rm G}$ exists in figure 25
(a), which corresponds to ${\rm DF}\mathrel{/\mskip-2.5mu/}{x-{\rm axis}}$,
then $\angle{\rm DFB}=\psi_{1}+\psi_{2}$, and $\angle{\rm FDB}=\angle{\rm
FBD}=\frac{\pi-(\psi_{1}+\psi_{2})}{2}$. Because we require $\angle{\rm
FBD}<\angle{\rm
FOD}=\psi_{2}+(\frac{\pi}{2}-(\psi_{1}+\psi_{2}))=\frac{\pi}{2}-\psi_{1}$, it
means $\psi_{1}<\psi_{2}$. This result is consistent with the conclusion that,
for $\psi_{1}<\psi_{2}$, there exists $R_{r}^{\rm max}$. While for
$\psi_{1}>\psi_{2}$, we can show that there does not exist $R_{r}^{\rm max}$
in later numerical results. Similarly, we can also calculate $R_{r}^{\rm max}$
analytically. With figure 25 (b), consider $\triangle{\rm FDO}$ and ${\rm
DF}\mathrel{/\mskip-2.5mu/}{x-{\rm axis}}$, where $|{\rm FD}|=|{\rm
DB}|=R_{r}$ and $|{\rm FO}|=R_{r}-\sigma_{r}$. Because $\angle{\rm
FDO}=\frac{\pi}{2}-\psi_{2}$, $\angle{\rm FOD}=\frac{\pi}{2}-\psi_{1}$, with
the law of sines we have

$\frac{R_{r}}{\sin{\left(\frac{\pi}{2}-\psi_{1}\right)}}=\frac{R_{r}-\sigma_{r}}{\sin{\left(\frac{\pi}{2}-\psi_{2}\right)}},$
(159)

which gives $R_{r}^{\rm
max}=\frac{\sigma_{r}\cos{\psi_{1}}}{\cos{\psi_{1}}-\cos{\psi_{2}}}>\sigma_{r}$.
If we plug the parameters above in it, we will get $R_{r}^{\rm max}=20$ and it
is consistent with numerical results.

There are several remarks for both cases. (i) Although there is a finite range
for $R_{r}$, for any $\sigma_{l}$ there always exists a solution, which means
$\sigma_{l}$ ranges from zero to infinity. (ii) We plot the geodesic length in
a log-linear plot in figure 24. For $L_{1}>L_{2}$ the prefactor of log term
for large $R_{r}$ is $3$, while for $L_{1}<L_{2}$ the prefactor is $2$. It can
be understood as $3=L_{1}+L_{2}$ and $2=\frac{3}{2}L_{1}+\frac{1}{2}L_{1}$.

####  A.2.3 Analytical approximation of the general case

In the following, we consider the limit case $\sigma_{l}\gg\sigma_{r}$ for
different phases, which are shown in figure 26.

Figure 26:  AdS dual of CFT with interface defect for infinite system and its
geodesic. Here ${\rm A}$ is the crossing point of left CFT and geodesic, and
${\rm E}$ is the crossing point of left CFT and line ${\rm CG}$. For the
discussion above, we consider the CFT boundary of $AdS_{2}$ is on a single
line $x=0$.

For $\psi_{1}<\psi_{2}$ in figure 26 (a), the limit case is ${\rm
DF}\mathrel{/\mskip-2.5mu/}{\rm OG}$. We first solve the geodesic. Considering
$\triangle{\rm FDB}$, (similar to figure 22 (a)), we have $\angle{\rm
DFB}=\psi_{1}+\psi_{2},\angle{\rm FDO}=\frac{\pi}{2}-\psi_{2}$ and $\angle{\rm
FOD}=\frac{\pi}{2}-\psi_{1}$. With the law of sines and ${\rm DF}=R_{r}$, we
have

$\frac{R_{r}}{\sin{(\frac{\pi}{2}-\psi_{1})}}=\frac{R_{r}-\sigma_{r}}{\sin{(\frac{\pi}{2}-\psi_{2})}}=\frac{|\rm
OD|}{\sin{(\psi_{1}+\psi_{2})}}.$ (160a) The solutions are
$R_{r}=\frac{\cos{\psi_{1}}\sigma_{r}}{(\cos{\psi_{1}}-\cos{\psi_{2}})}$ and
$|{\rm
OD}|=\frac{\sin{(\psi_{1}+\psi_{2})}\sigma_{r}}{(\cos{\psi_{1}}-\cos{\psi_{2}})}=A_{2}\sigma_{r}$.
It means for $\sigma_{l}\rightarrow\infty$, $|{\rm
OD}|\sim\mathcal{O}(\sigma_{r})={\rm const.}$ with a fixed $\sigma_{r}$, and
$\angle{\rm GDH}=\angle{\rm GHD}=\frac{\pi}{2}$. So $|{\rm OH}|\approx|{\rm
OD}|\cos{(\frac{\pi}{2}-\psi_{2})}\sim\mathcal{O}(\sigma_{r})={\rm const.}$.
Therefore, for $\triangle{\rm EBC}$, we have $\angle{\rm GCH}=\angle{\rm
GHC}\approx\angle{\rm GOC}=\frac{\pi}{2}-\psi_{2}\approx\angle{\rm GCO}$ and
$\angle{\rm COE}=\frac{\pi}{2}+\psi_{1}$. Now with the law of sines (similar
to figure 22 (b)), we have
$\frac{R_{l}}{\sin{(\frac{\pi}{2}+\psi_{1})}}=\frac{\sigma_{l}-R_{l}}{\sin{(\frac{\pi}{2}-\psi_{2})}}=\frac{|\rm
OC|}{\sin{(\psi_{2}-\psi_{1})}}.$ (160b)

The solutions are
$R_{l}=\frac{\cos{\psi_{1}}\sigma_{l}}{(\cos{\psi_{1}}+\cos{\psi_{2}})}$ and
$|{\rm
OC}|=\frac{\sin{(\psi_{2}-\psi_{1})}\sigma_{l}}{(\cos{\psi_{1}}+\cos{\psi_{2}})}=A_{1}\sigma_{l}$.

Now we can consider the length of the geodesic. For ${\rm AC}$ with ${\rm
A}=(-\sigma_{l},\varepsilon)$ and ${\rm C}=(|{\rm OC}|\sin{\psi_{1}},|{\rm
OC}|\cos{\psi_{1}})$, we have

$\begin{split}d_{\rm AC}=&L_{1}\cosh^{-1}{\frac{(|{\rm
OC}|\sin{\psi_{1}}+\sigma_{l})^{2}+\varepsilon^{2}+(|{\rm
OC}|\cos{\psi_{1}})^{2}}{2\varepsilon|{\rm OC}|\cos{\psi_{1}}}}\\\
\approx&L_{1}\cosh^{-1}{\frac{\sigma_{l}}{2\varepsilon}\frac{A_{1}^{2}+1+2A^{2}\sin{\psi_{1}}}{A_{1}\cos{\psi_{1}}}}\approx
L_{1}\log{\frac{\sigma_{l}}{\varepsilon}\frac{A_{1}^{2}+1+2A_{1}^{2}\sin{\psi_{1}}}{A_{1}\cos{\psi_{1}}}}.\end{split}$
(161a) For ${\rm CD}$ with ${\rm C}=(-|{\rm OC}|\sin{\psi_{2}},|{\rm
OC}|\cos{\psi_{2}})$ and ${\rm D}=(|{\rm OD}|\sin{\psi_{2}},|{\rm
OD}|\cos{\psi_{2}})$, we have $\begin{split}d_{\rm
CD}=&L_{2}\cosh^{-1}{\frac{(|{\rm OC}|+|{\rm
OD}|)^{2}\sin^{2}{\psi_{2}}+(|{\rm OC}|\cos{\psi_{2}})^{2}+(|{\rm
OD}|\cos{\psi_{2}})^{2}}{2|{\rm OC}||{\rm OD}|\cos^{2}{\psi_{2}}}}\\\
=&L_{2}\cosh^{-1}{\frac{A_{1}^{2}\sigma_{l}^{2}+A_{2}^{2}\sigma_{r}^{2}+2A_{1}A_{2}\sigma_{l}\sigma_{r}\sin^{2}{\psi_{2}}}{2A_{1}A_{2}\sigma_{l}\sigma_{r}\cos^{2}{\psi_{2}}}}.\end{split}$
(161b) Under the limit $\sigma_{l}\gg\sigma_{r}$, we have $d_{\rm CD}\approx
L_{2}\cosh^{-1}{\frac{A_{1}^{2}\sigma_{l}^{2}}{2A_{1}A_{2}\sigma_{l}\sigma_{r}\cos^{2}{\psi_{2}}}}\approx
L_{2}\log{\frac{\sigma_{l}}{\sigma_{r}}\frac{1}{A_{2}\cos^{2}{\psi_{2}}}}.$
(161c) For ${\rm DB}$ with ${\rm D}=(-|{\rm OD}|\sin{\psi_{1}},|{\rm
OD}|\cos{\psi_{1}})$ and ${\rm b}=(\sigma_{r},\varepsilon)$, we have
$\begin{split}d_{\rm DB}=&L_{1}\cosh^{-1}{\frac{(|{\rm
OD}|\sin{\psi_{1}}+\sigma_{r})^{2}+\varepsilon^{2}+(|{\rm
OD}|\cos{\psi_{1}})^{2}}{2\varepsilon|{\rm OD}|\cos{\psi_{1}}}}\\\
\approx&L_{1}\cosh^{-1}{\frac{\sigma_{r}}{2\varepsilon}\frac{A_{2}^{2}+1+2A_{2}^{2}\sin{\psi_{1}}}{A_{2}\cos{\psi_{1}}}}\approx
L_{1}\log{\frac{\sigma_{r}}{\varepsilon}\frac{A_{2}^{2}+1+2A_{2}^{2}\sin{\psi_{1}}}{A_{2}\cos{\psi_{1}}}}.\end{split}$
(161d)

Then the total length of the geodesic is

$d\approx
L_{1}\log{\frac{\sigma_{l}}{\varepsilon}\frac{A_{1}^{2}+1+2A_{1}^{2}\sin{\psi_{1}}}{A_{1}\cos{\psi_{1}}}}+L_{2}\log{\frac{\sigma_{l}}{\sigma_{r}}\frac{1}{A_{2}\cos^{2}{\psi_{2}}}}+L_{1}\log{\frac{\sigma_{r}}{\varepsilon}\frac{A_{2}^{2}+1+2A_{2}^{2}\sin{\psi_{1}}}{A_{2}\cos{\psi_{1}}}},$
(162)

where the prefactor of $\log{\sigma_{l}}$ is $L_{1}+L_{2}$. Actually, we can
compare this analytical result (162) with the numerical results (156), and
they are consistent as shown in figure 24 (a).

Now we consider another case $\psi_{1}>\psi_{2}$, which is more complicated
and shown in figure 26 (b). For the limit case $\sigma_{l}\gg\sigma_{r}$, we
expect that ${\rm CG}\mathrel{/\mskip-2.5mu/}{\rm OE}$, which means $|{\rm
OC}|\ll\sigma_{l}$. While ${\rm DG}$ will always cross ${\rm OG}$ at a finite
angle $\angle{\rm DGO}$, which means $|{\rm OD}|\gg\sigma_{r}$. Besides, with
numerical results, we may expect that $\frac{|\rm OC|}{|\rm
OD|}\sim\mathcal{O}(1)$ for $\sigma_{l}\rightarrow\infty$. The key difference
is that here we cannot directly get $|{\rm OC}|$ with $\sigma_{l}$ and $|{\rm
OD}|$ with $\sigma_{r}$. Instead, we expect $\sigma_{r}\ll
R_{r},\alpha,x_{0},\beta,\gamma\ll x_{1},\sigma_{l}$, which can be shown
below.

Here we first define $|{\rm OD}|=x$. (i) With $|{\rm OD}|\gg\sigma_{r}$, we
have $\angle{\rm FDB}=\angle{\rm DBF}\approx\angle{\rm
DOF}=\frac{\pi}{2}-\psi_{1}$, $\angle{\rm DFO}=2\psi_{1}$ and $\angle{\rm
DGO}=\psi_{1}-\psi_{2}$. For $\triangle{\rm OGD}$ we apply the law of sines,
then we have

$\frac{|{\rm OG}|}{\sin{(\frac{\pi}{2}-\psi_{1})}}=\frac{|{\rm
DG}|}{\sin{(\frac{\pi}{2}+\psi_{2})}}=\frac{|{\rm
DO}|}{\sin{(\psi_{1}-\psi_{2})}}.$ (163)

The solutions are $|{\rm
OG}|=\frac{\cos{\psi_{1}}x}{\sin{(\psi_{1}-\psi_{2})}}$ and $|{\rm
DG}|=\frac{\cos{\psi_{2}}x}{\sin{(\psi_{1}-\psi_{2})}}$. (ii) With ${\rm
CG}\mathrel{/\mskip-2.5mu/}{\rm OE}$, we have $\angle{\rm CGO}=\angle{\rm
GOA}=\psi_{1}+\psi_{2}$, and $\angle{\rm GCO}=\frac{\pi}{2}-\psi_{1}$. Using
the law of sines for $\triangle{\rm OCG}$ we have

$\frac{|{\rm OG}|}{\sin{(\frac{\pi}{2}-\psi_{1})}}=\frac{|{\rm
OC}|}{\sin{(\psi_{1}+\psi_{2})}}=\frac{|{\rm
CG}|}{\sin{(\frac{\pi}{2}-\psi_{2})}}.$ (164)

The solutions are $|{\rm
OC}|=\frac{\sin{(\psi_{1}+\psi_{2})}x}{\sin{(\psi_{1}-\psi_{2})}}$ and $|{\rm
CG}|=\frac{\cos{\psi_{2}}x}{\sin{(\psi_{1}-\psi_{2})}}$. (iii) Finally, with
$\angle{\rm DFO}=2\psi_{1}$ and $\angle{\rm DOF}=\frac{\pi}{2}-\psi_{1}$, we
have $\frac{x}{\sin{2\psi_{1}}}=\frac{|{\rm
DF}|}{\sin{(\frac{\pi}{2}-\psi_{1})}}$. So $|{\rm
DF}|=\frac{\cos{\psi_{1}}x}{\sin{2\psi_{1}}}$. To be consistent with the
notations above, we have $|{\rm OD}|=\alpha,|{\rm OG}|=x_{0},|{\rm
OC}|=\gamma,|{\rm CG}|=\beta,|{\rm FD}|=R_{r}$. Therefore, $|{\rm
OC}|=\frac{\sin{(\psi_{1}+\psi_{2})}\sin{2\psi_{1}}}{\sin{(\psi_{1}-\psi_{2})}\cos{\psi_{1}}}R_{r}$
and $|{\rm OD}|=\frac{\sin{2\psi_{1}}}{\cos{\psi_{1}}}R_{r}$.

However, the results above are not enough to solve the values of all unknown
variables using $\sigma_{l}$ and $\sigma_{r}$. We need (154). With the
discussion above, $\sigma_{l}\gg R_{r}\gg\sigma_{r}$. By symmetry, we assume
that
$R_{r}\sim\mathcal{O}(\sqrt{\sigma_{l}\sigma_{r}})\sim\mathcal{O}(\sqrt{\sigma_{l}})$
for $\sigma_{l}\rightarrow\infty$ and a fixed $\sigma_{r}$. This assumption
can also be checked numerically. Then to simplify equations (154), we can
first rescale all variables with $\frac{1}{\sigma_{l}}$, which means
$\sigma_{l}^{\prime}=1$, $\sigma_{r}^{\prime}=\frac{\sigma_{r}}{\sigma_{l}}\ll
1$ and
$R_{r}^{\prime}=\frac{R_{r}}{\sigma_{l}}=c\cdot\sqrt{\frac{\sigma_{r}}{\sigma_{l}}}$,
where $c$ is an unknown variable to be determined. Now if we can get $c$, then
all variables can be solved. The constraint for $c$ comes from
$\sigma_{l}^{\prime}=1$. For the limit case, with (154f) we have
$\sigma_{l}^{\prime}\approx-\frac{2x_{1}^{\prime}}{\cos{(\psi_{1}-\psi_{2})}}$
at the zero order. Then with the Taylor expansion of (154), we can finally
express $x_{1}^{\prime}$ to the zero order of $\sigma_{r}^{\prime}$, which
gives

$x_{1}^{\prime}=-c^{2}\cdot\frac{\sin^{2}{\psi_{1}}\sin{[2(\psi_{1}+\psi_{2})]}}{\sin{(\psi_{1}-\psi_{2})}}+\mathcal{O}(\sigma_{r}^{\prime}).$
(165)

So we have $1=\sigma_{l}^{\prime}\approx
c^{2}\cdot\frac{2\sin^{2}{\psi_{1}}\sin{[2(\psi_{1}+\psi_{2})]}}{\sin{(\psi_{1}-\psi_{2})}\cos{(\psi_{1}+\psi_{2})}}$,
which means

$c=\frac{1}{2\sin{\psi_{1}}}\sqrt{\frac{\sin{(\psi_{1}-\psi_{2})}}{\sin{(\psi_{1}+\psi_{2})}}}.$
(166)

Now we have $|{\rm OC}|=\frac{\sin{(\psi_{1}+\psi_{2})}|{\rm
OD}|}{\sin{(\psi_{1}-\psi_{2})}}=\frac{\sin{(\psi_{1}+\psi_{2})}}{\sin{(\psi_{1}-\psi_{2})}}\cdot
2\sin{\psi_{2}}R_{r}=\sqrt{\frac{\sin{(\psi_{1}+\psi_{2})}}{\sin{(\psi_{1}-\psi_{2})}}}\sqrt{\sigma_{l}\sigma_{r}}=b\sqrt{\sigma_{l}\sigma_{r}}$
and $|{\rm OD}|=b^{-1}\sqrt{\sigma_{l}\sigma_{r}}$.

Finally, we can calculate the length of the geodesic, which is similar to
(161). For ${\rm AC}$ we have

$\begin{split}d_{\rm AC}=&L_{1}\cosh^{-1}{\frac{(|{\rm
OC}|\sin{\psi_{1}}+\sigma_{l})^{2}+\varepsilon^{2}+(|{\rm
OC}|\cos{\psi_{1}})^{2}}{2\varepsilon|{\rm OC}|\cos{\psi_{1}}}}\\\
\approx&L_{1}\log{\frac{b^{2}\sigma_{l}\sigma_{r}+\sigma_{l}^{2}+2b\sigma_{l}^{3/2}\sigma_{r}^{1/2}\sin{\psi_{1}}}{b(\sigma_{l}\sigma_{r})^{1/2}\cos{\psi_{1}}\varepsilon}}\approx
L_{1}\left(\log{\left(\frac{b^{-1}\sigma_{r}^{-1/2}\sigma_{l}^{3/2}}{\cos{\psi_{1}}\varepsilon}\right)}+2b\sin{\psi_{1}}\sqrt{\frac{\sigma_{r}}{\sigma_{l}}}\right).\end{split}$
(167a) For ${\rm CD}$ we have $\begin{split}d_{\rm
CD}=&L_{2}\cosh^{-1}{\frac{(|{\rm OC}|+|{\rm
OD}|)^{2}\sin^{2}{\psi_{2}}+(|{\rm OC}|\cos{\psi_{2}})^{2}+(|{\rm
OD}|\cos{\psi_{2}})^{2}}{2|{\rm OC}||{\rm OD}|\cos^{2}{\psi_{2}}}}\\\
=&L_{2}\cosh^{-1}{\frac{(b^{-1}+b)^{2}\sin^{2}{\psi_{2}}+(b^{-2}+b^{2})\cos^{2}{\psi_{2}}}{2\cos^{2}{\psi_{2}}}}.\end{split}$
(167b) For ${\rm DB}$ we have $\begin{split}d_{\rm
DB}=&L_{1}\cosh^{-1}{\frac{(|{\rm
OD}|\sin{\psi_{1}}+\sigma_{r})^{2}+\varepsilon^{2}+(|{\rm
OD}|\cos{\psi_{1}})^{2}}{2\varepsilon|{\rm OD}|\cos{\psi_{1}}}}\\\
\approx&L_{1}\log{\frac{b^{-2}\sigma_{l}\sigma_{r}+\sigma_{r}^{2}+2b^{-1}\sigma_{l}^{1/2}\sigma_{r}^{3/2}\sin{\psi_{1}}}{b^{-1}(\sigma_{l}\sigma_{r})^{1/2}\cos{\psi_{1}}\varepsilon}}\approx
L_{1}\left(\log{\left(\frac{b^{-1}\sigma_{r}^{1/2}\sigma_{l}^{1/2}}{\cos{\psi_{1}}\varepsilon}\right)}+2b\sin{\psi_{1}}\sqrt{\frac{\sigma_{r}}{\sigma_{l}}}\right).\end{split}$
(167c)

Then the total length of the geodesic is

$d\approx
L_{1}\left(\log{\left(\frac{\sigma_{l}^{2}}{\varepsilon^{2}}\frac{b^{-2}}{\cos^{2}{\psi_{1}}}\right)}+4b\sin{\psi_{1}}\sqrt{\frac{\sigma_{r}}{\sigma_{l}}}\right)+L_{2}\cosh^{-1}{\left(\frac{(b^{-1}+b)^{2}-2\cos^{2}{\psi_{2}}}{2\cos^{2}{\psi_{2}}}\right)},$
(168)

where the prefactor of $\log{\sigma_{l}}$ is
$\frac{3}{2}L_{1}+\frac{1}{2}L_{1}$. Actually, we can compare this analytical
result (168) and numerical results (156), and they are consistent in figure 24
(b).

###  A.3 ETW/defect geodesic with an ETW brane and two regions

####  A.3.1 Special case with $\sigma_{l}\rightarrow 0$

In the following, we consider the effect of the ETW brane. More specifically,
the two end points of the geodesic are located at $\sigma_{l}\rightarrow 0$
and the ETW brane, as shown in figure 27.

Figure 27:  AdS dual of CFT with interface defect for half infinite system and
ETW brane. Two end points of the geodesic are located on the defect and the
ETW brane.

It is obvious that there are more than one solution for the geodesic equation
with an endpoint on the ETW, so we need to find a geodesic with the minimal
length. Similar to appendix A.2.1, we start by assuming $|{\rm OA}|=r$. With
$\angle{\rm ACO}=\frac{\pi}{2}-\psi_{2}$ and $\angle{\rm
COB}=\frac{\pi}{2}+\psi_{1}$, we have $\angle{\rm CBO}=\psi_{2}-\psi_{1}$. It
is obvious that for $\psi_{2}<\psi_{1}$ the crossing point of lines ${\rm CA}$
and ${\rm OB}$ will be on the left of the point ${\rm O}$, which means there
is no solution for the geodesic we want.

Therefore, in the following, we just consider $\psi_{2}>\psi_{1}$. (i)
Starting from figure 27 (a), we assume $|{\rm OA}|=r$, and the equation of the
line ${\rm AB}$ is
$y-r\sin{(\psi_{1}+\psi_{2})}=\tan{(\pi+\psi_{1}-\psi_{2})}(x-r\cos{(\psi_{1}+\psi_{2})})$.
Setting $y=0$ we have

$x_{\rm
B}=\beta=\frac{-r\sin{(\psi_{1}+\psi_{2})}}{\tan{(\psi_{1}-\psi_{2})}}+r\cos{(\psi_{1}+\psi_{2})}.$
(169a) (ii) Assuming $|{\rm OC}|=\alpha$, because the point ${\rm C}$ is
located at the line ${\rm AB}$, we can solve and get
$\alpha=\frac{r\sin{(\psi_{1}+\psi_{2})}-r\tan{(\psi_{1}-\psi_{2})}\cos{(\psi_{1}+\psi_{2})}}{\cos{\psi_{1}}+\sin{\psi_{1}}\tan{(\psi_{1}-\psi_{2})}}.$
(169b) (iii) In $\triangle{\rm BCO}$, using the law of sines we have $|{\rm
BC}|=\frac{\beta\cos{\psi_{1}}}{\cos{\psi_{2}}}$. Assuming ${\rm
D}=(\sigma,\delta)$, because $|{\rm BC}|=|{\rm BD}|$, we have
$\delta=\sqrt{\beta^{2}\left(\frac{\cos{\psi_{1}}}{\cos{\psi_{2}}}\right)^{2}-(\sigma-\beta)^{2}}.$
(169c)

Now we can express the geodesic length with the variables above. For ${\rm
OC}$, with ${\rm C}=(\alpha\sin{\psi_{2}},\alpha\cos{\psi_{2}})$ and ${\rm
O}=(0,\varepsilon)$ we have

$d_{\rm
OC}=L_{2}\cosh^{-1}{\frac{(\alpha\sin{\psi_{1}})^{2}+\varepsilon^{2}+(\alpha\cos{\psi_{1}})^{2}}{2\varepsilon\alpha\cos{\psi_{2}}}}\approx
L_{2}\log{\frac{\alpha}{\varepsilon\cos{\psi_{2}}}}.$ (170a) For ${\rm CD}$,
with ${\rm C}=(-\alpha\sin{\psi_{1}},\alpha\cos{\psi_{1}})$ and ${\rm
D}=(\sigma,\delta)$ we have $d_{\rm
OC}=L_{1}\cosh^{-1}{\frac{(-\alpha\sin{\psi_{1}}-\sigma)^{2}+\delta^{2}+(\alpha\cos{\psi_{1}})^{2}}{2\delta\alpha\cos{\psi_{1}}}}.$
(170b)

Therefore, the total geodesic length is

$d=L_{2}\log{\frac{\alpha}{\varepsilon\cos{\psi_{2}}}}+L_{1}\cosh^{-1}{\frac{(-\alpha\sin{\psi_{1}}-\sigma)^{2}+\delta^{2}+(\alpha\cos{\psi_{1}})^{2}}{2\delta\alpha\cos{\psi_{1}}}}.$
(171)

Besides, we also mention that without a defect (region 2), the geodesic will
cross the ETW brane perpendicularly. So, the length is

$d_{0}=L_{1}\cosh^{-1}{\frac{\sigma^{2}+\sigma^{2}+\varepsilon^{2}}{2\sigma\varepsilon}}\approx
L_{1}\log{\frac{2\sigma}{\varepsilon}}.$ (172)

Now what we need to do is to minimize (171) with respect to $r$. Actually,
with numerical check we can find that, after minimizing (171), the
corresponding $r=r_{0}$ will lead $\beta=\sigma$, which means the geodesic
will cross the ETW brane perpendicularly and the point ${\rm B}$ is also
located at the ETW. In the following, we will prove that the geodesic will
always cross the ETW brane perpendicularly, and express the total geodesic
length analytically.

The calculation is tedious. With (169), we can have $\beta=\beta(r),\
\alpha=\alpha(r),\ \delta=\delta(\beta)$. Then we plug these equations into
(171) and calculate $\frac{\partial d}{\partial r}$. For the perpendicular
case we have $\beta=\sigma$, which means

$r=r_{0}=\frac{\sigma}{\cos{(\psi_{1}+\psi_{2})}-\frac{\sin{(\psi_{1}+\psi_{2})}}{\tan{(\psi_{1}-\psi_{2})}}}.$
(173)

With $r=r_{0}$ in $\frac{\partial d}{\partial r}$ and simplifying it with
$\psi_{1}<\psi_{2}$, we have

$\left.\frac{\partial d}{\partial
r}\right|_{r=r_{0}}=-\frac{2\cos{\psi_{2}}\sin{\psi_{2}}(L_{2}\csc{(\psi_{1}-\psi_{2})}-L_{1}\cos{\psi_{2}}\csc{(\psi_{1}-\psi_{2})}\sec{\psi_{1}})}{\sigma}.$
(174)

With the junction condition (16a) we have
$\frac{L_{1}}{\cos{\psi_{1}}}=\frac{L_{2}}{\cos{\psi_{2}}}$, so
$\left.\frac{\partial d}{\partial r}\right|_{r=r_{0}}=0$, which means that
$r=r_{0}$ corresponds to the minimal point of $d(r)$. Therefore, the minimal
geodesic will cross the ETW brane perpendicularly. Actually, we will show
later that this property of minimal geodesics will also be true for a more
complicate case, and give an intuitive proof in appendix A.4.1.

For the minimal geodesic above with $r=r_{0}$, we can plug it into (171) and
calculate the length of the minimal geodesic in figure 27 (b). Firstly, we
consider $\triangle{\rm COB}$ with the law of sines, then we have

$\frac{\sigma}{\sin{(\frac{\pi}{2}-\psi_{2}})}=\frac{|\rm
CB|}{\sin{(\frac{\pi}{2}+\psi_{1}})}=\frac{|\rm
OC|}{\sin{(\psi_{2}-\psi_{1})}},$ (175)

which leads $|{\rm CB}|=|{\rm
BD}|=\frac{\sigma\cos{\psi_{1}}}{\cos{\psi_{2}}}$ and $|{\rm
OC}|=\frac{\sigma\sin{(\psi_{2}-\psi_{1})}}{\cos{\psi_{2}}}$. For $|{\rm OC}|$
we have ${\rm O}=(0,\varepsilon)$ and ${\rm C}=(|{\rm OC}|\sin{\psi_{2}},|{\rm
OC}|\cos{\psi_{2}})$, so

$d_{\rm OC}=L_{2}\cosh^{-1}{\frac{(|{\rm
OC}|\sin{\psi_{2}})^{2}+\varepsilon^{2}+(|{\rm
OC}|\cos{\psi_{2}})^{2}}{2\varepsilon(|{\rm OC}|\cos{\psi_{2}})}}\approx
L_{2}\log{\left(\frac{\sigma}{\varepsilon}\frac{\sin{(\psi_{2}-\psi_{1})}}{\cos^{2}{\psi_{2}}}\right)}.$
(176a) For $|{\rm CD}|$ we have ${\rm C}=(-|{\rm OC}|\sin{\psi_{1}},|{\rm
OC}|\cos{\psi_{1}})$ and ${\rm D}=(\sigma,|{\rm BD}|)$, so
$\begin{split}d_{\rm CD}=&L_{1}\cosh^{-1}{\frac{(-|{\rm
OC}|\sin{\psi_{1}}-\sigma)^{2}+(|{\rm OC}|\cos{\psi_{1}})^{2}+|{\rm
BD}|^{2}}{2|{\rm OC}|\cos{\psi_{1}}|{\rm BD}|}}\\\
=&L_{1}\cosh^{-1}{\left(\frac{\sin^{2}{(\psi_{2}-\psi_{1})}+\cos^{2}{\psi_{2}}+\cos^{2}{\psi_{1}}+2\sin{(\psi_{2}-\psi_{1})}\sin{\psi_{1}}\cos{\psi_{2}}}{2\sin{(\psi_{2}-\psi_{1})}\cos^{2}{\psi_{1}}}\right)}.\end{split}$
(176b)

Therefore, the total length is

$d=L_{2}\log{\left(\frac{\sigma}{\varepsilon}\right)}+d_{\rm sub},$ (177)

with the sub-leading term

$\begin{split}d_{\rm
sub}=&L_{2}\log{\left(\frac{\sin{(\psi_{2}-\psi_{1})}}{\cos^{2}{\psi_{2}}}\right)}\\\
&+L_{1}\cosh^{-1}{\left(\frac{\sin^{2}{(\psi_{2}-\psi_{1})}+\cos^{2}{\psi_{2}}+\cos^{2}{\psi_{1}}+2\sin{(\psi_{2}-\psi_{1})}\sin{\psi_{1}}\cos{\psi_{2}}}{2\sin{(\psi_{2}-\psi_{1})}\cos^{2}{\psi_{1}}}\right)}.\end{split}$
(178)

Finally, we have corresponding entanglement entropy
$S_{\sigma}=\frac{c_{2}}{6}\log{\frac{\sigma}{\varepsilon}}+S_{\rm sub}$ and
$S_{\rm sub}=\frac{d_{\rm sub}}{4G_{(3)}}$.

####  A.3.2 General case with $\sigma_{l}\neq 0$

Before we only considered the case with $\sigma_{l}\rightarrow 0$, now we
consider a more general case where $\sigma_{l}\neq 0$. Similar to the method
above, in figure 28 (a) we assume $|{\rm OA}|=r,\ |{\rm OC}|=\alpha$.

Figure 28:  AdS dual of CFT with interface defect for half infinite system and
ETW brane. The two end points are located on the left CFT and the ETW brane.

(i) Because $|{\rm OA}|=|{\rm AC}|$, we can solve
$(\sigma_{l}+r)^{2}=\left[r\cos{(\psi_{1}+\psi_{2})}+\alpha\sin{\psi_{1}}\right]^{2}+\left[r\sin{(\psi_{1}+\psi_{2})}-\alpha\cos{\psi_{1}}\right]^{2}$
and get

$\alpha=r\sin{\psi_{2}}+\sqrt{(r+\sigma_{l})^{2}-r^{2}\cos^{2}{\psi_{2}}}.$
(179a) (ii) Then we have the equation of the line ${\rm AC}$ and $B=(x_{0},0)$
on it, so
$x_{0}=[-r\sin{(\psi_{1}+\psi_{2})}]\frac{-\alpha\sin{\psi_{1}}-r\cos{(\psi_{1}+\psi_{2})}}{\alpha\cos{\psi_{1}}-r\sin{(\psi_{1}+\psi_{2})}}+r\cos{(\psi_{1}+\psi_{2})}.$
(179b) (iii) Assuming ${\rm E}=(\sigma_{r},\beta)$, with $|{\rm BC}|=|{\rm
BE}|$, we have
$(-\alpha\sin{\psi_{1}}-x_{0})^{2}+(\alpha\cos{\psi_{1}})^{2}=(\sigma_{r}-x_{0})^{2}+\beta^{2}$,
which leads to
$\beta=\sqrt{2\alpha\sin{\psi_{1}}x_{0}+\alpha^{2}-\sigma_{r}^{2}+2\sigma_{r}x_{0}}.$
(179c)

Now we can calculate the length of the geodesic. For ${\rm DC}$, with ${\rm
D}=(-\sigma_{l},\varepsilon)$ and ${\rm
C}=(\alpha\sin{\psi_{2}},\alpha\cos{\psi_{2}})$, we have

$d_{\rm
DC}=L_{2}\cosh^{-1}{\frac{(\alpha\sin{\psi_{2}}+\sigma_{l})^{2}+\varepsilon^{2}+(\alpha\cos{\psi_{2}})^{2}}{2\varepsilon\alpha\cos{\psi_{2}}}}\approx
L_{2}\log{\frac{\alpha^{2}+2\alpha\sigma_{l}\sin{\psi_{2}}+\sigma_{l}^{2}}{\varepsilon\alpha\cos{\psi_{2}}}}.$
(180a) For ${\rm CE}$, with ${\rm
C}=(-\alpha\sin{\psi_{1}},\alpha\cos{\psi_{1}})$ and ${\rm
E}=(\sigma_{r},\beta)$, we have $d_{\rm
CE}=L_{1}\cosh^{-1}{\frac{(\alpha\sin{\psi_{1}}+\sigma_{r})^{2}+(\alpha\cos{\psi_{1}})^{2}+\beta^{2}}{2\alpha\beta\cos{\psi_{1}}}}=L_{1}\cosh^{-1}{\frac{\alpha^{2}+\beta^{2}+2\alpha\sigma_{r}\sin{\psi_{1}}+\sigma_{r}^{2}}{2\alpha\beta\cos{\psi_{1}}}}.$
(180b)

Therefore, the total length is

$d=L_{2}\log{\frac{\alpha^{2}+2\alpha\sigma_{l}\sin{\psi_{2}}+\sigma_{l}^{2}}{\varepsilon\alpha\cos{\psi_{2}}}}+L_{1}\cosh^{-1}{\frac{\alpha^{2}+\beta^{2}+2\alpha\sigma_{r}\sin{\psi_{1}}+\sigma_{r}^{2}}{2\alpha\beta\cos{\psi_{1}}}}.$
(181)

Similar to the method above, we should minimize (181) with respect to $r$.
Numerically, it is easy to check that the minimal geodesic crosses the ETW
brane perpendicularly. But analytical proof is difficult.

In the following, we will take the assumption above and calculate the length
analytically. With $x_{0}=\sigma_{r}$, expressing $\alpha$ with $r$ by (179b)
and plugging it into (179a), we will get an effective cubic equation for $r$.
What we need to do is solving this equation which has an analytical solution.
The effective cubic equation is $ar^{3}+br^{2}+cr+d=0$ with factors

$\begin{split}a=&-2\sigma_{r}\cos{\psi_{1}}\cos{\psi_{2}}+2\sigma_{l}\cos^{2}{\psi_{2}}+2\sigma_{r}\cos^{2}{\psi_{2}}\cos{(\psi_{1}+\psi_{2})},\\\
b=&\
\sigma_{r}^{2}\cos^{2}{\psi_{1}}-4\sigma_{l}\sigma_{r}\cos{\psi_{1}}\cos{\psi_{2}}+\sigma_{l}^{2}\cos^{2}{\psi_{2}}-\sigma_{r}^{2}\cos^{2}{\psi_{2}},\\\
c=&\
2\sigma_{l}\sigma_{r}^{2}\cos^{2}{\psi_{1}}-2\sigma_{l}^{2}\sigma_{r}\cos{\psi_{1}}\cos{\psi_{2}},\\\
d=&\ \sigma_{l}^{2}\sigma_{r}^{2}\cos^{2}{\psi_{1}}.\end{split}$ (182)

This equation has analytical solutions, and we will not show them here. After
solving this equation, we also need to choose one true solution from three
solutions. One criterion is that the corresponding $\alpha,\ \beta$ and
$x_{0}$ are positive. Finally, plugging the solution into (179) and (181), we
will get the final minimal length.

Here we give some remarks about the results above. (i) In the discussion we
don’t consider the relation between $L_{1}$ and $L_{2}$, which plays an
important role before when we take $\sigma_{l}\rightarrow 0$. With reference
Anous_2022 , for any given $\sigma_{l}$ and $\sigma_{r}$, we can have a
nontrivial solution. Then for our case, we can consider there is an ETW brane
across the center of the arc in the right region 1. Therefore, the solution
always exists in this case. (ii) However, the prefactor of $\log$-term may be
different for different relation between $L_{1}$ and $L_{2}$, and it may also
be different at different limits. For example, $\sigma_{r}\rightarrow\infty$
and $\sigma_{l}\rightarrow\infty$ may have different prefactors of
$\log$-term.

###  A.4 ETW/defect geodesic with an ETW brane and three regions

Similar to the case above, now we consider the most complicated case with
three regions and an ETW brane, which is shown in figure 29.

Figure 29:  AdS dual of CFT with interface defect for half infinite system and
ETW brane. The geodesic crosses two interface branes and ends on the ETW
brane.

Now we construct the equations for solving this problem. (i) Assume $|{\rm
OB}|=\alpha$ and $|{\rm OE}|=\beta$, $|{\rm AE}|=|{\rm BE}|$ leads to
$(\beta+\sigma_{l})^{2}=[-\alpha\sin{\psi_{2}}-\beta\cos{(\psi_{1}+\psi_{2})}]^{2}+[\alpha\cos{\psi_{2}}-\beta\sin{(\psi_{1}+\psi_{2})}]^{2}$.
The solution of equation above is

$\beta=\frac{\alpha^{2}-\sigma_{l}^{2}}{2(\sigma_{l}+\alpha\sin{\psi_{1}})}.$
(183a) (ii) Then with the equation of the line ${\rm BE}$ and point ${\rm
F}=(x_{0},0)$ on it, we can get
$x_{0}=(-\alpha\cos{\psi_{2}})\frac{\beta\cos{(\psi_{1}+\psi_{2})}+\alpha\sin{\psi_{2}}}{\beta\sin{(\psi_{1}+\psi_{2})}-\alpha\cos{\psi_{2}}}-\alpha\sin{\psi_{2}}.$
(183b) (iii) With $|{\rm OC}|=\gamma$, $|{\rm CF}|=|{\rm BF}|$ leads to
$(\gamma\sin{\psi_{2}}-x_{0})^{2}+(\gamma\cos{\psi_{2}})^{2}=(-\alpha\sin{\psi_{2}}-x_{0})^{2}+(\alpha\cos{\psi_{2}})^{2}$,
which has solution $\gamma=\alpha+2x_{0}\sin{\psi_{2}}.$ (183c) (iv)Then
assuming $|{\rm OG}|=\delta$, because ${\rm
G}=(\delta\cos{(\psi_{1}+\psi_{2})},-\delta\sin{(\psi_{1}+\psi_{2})})$ on the
line ${\rm CF}$, we have the equation
$\frac{-\delta\sin{(\psi_{1}+\psi_{2})}}{\delta\cos{(\psi_{1}+\psi_{2})}-x_{0}}=\frac{\gamma\cos{\psi_{2}}}{\gamma\sin{\psi_{2}}-x_{0}}$.
The solution is $\delta=\frac{\gamma
x_{0}\cos{\psi_{2}}}{\gamma\cos{\psi_{1}}-x_{0}\sin{(\psi_{1}+\psi_{2})}}.$
(183d) (v) Finally assume $|{\rm DH}|=\xi$, $|{\rm CG}|=|{\rm DG}|$ leads
$\xi^{2}+(\sigma_{r}-\delta)^{2}=[\gamma\sin{\psi_{2}}-\delta\cos{(\psi_{1}+\psi_{2})}]^{2}+[\gamma\cos{\psi_{2}}_{\delta}\cos{(\psi_{1}+\psi_{2})}]^{2}$.
The solution is
$\xi=\sqrt{\gamma^{2}+(2\delta-\sigma_{r})\sigma_{r}+2\delta\gamma\sin{\psi_{1}}}.$
(183e)

Now we can calculate the length of the geodesic. For ${\rm AB}$, with ${\rm
A}=(-\sigma_{l},\varepsilon)$ and ${\rm
B}=(\alpha\sin{\psi_{1}},\alpha\cos{\psi_{1}})$ we have

$d_{\rm
AB}=L_{1}\cosh^{-1}{\left(\frac{(-\sigma_{l}-\alpha\sin{\psi_{1}})^{2}+\varepsilon^{2}+(\alpha\cos{\psi_{1}})^{2}}{2\varepsilon\alpha\cos{\psi_{1}}}\right)}\approx
L_{1}\log{\frac{(\sigma_{l}+\alpha\sin{\psi_{1}})^{2}+(\alpha\cos{\psi_{1}})^{2}}{\varepsilon\alpha\cos{\psi_{1}}}}.$
(184a) For ${\rm BC}$, with ${\rm
B}=(-\alpha\sin{\psi_{2}},\alpha\cos{\psi_{2}})$ and ${\rm
C}=(\gamma\sin{\psi_{2}},\gamma\cos{\psi_{2}})$ we have $d_{\rm
BC}=L_{2}\cosh^{-1}{\left(\frac{(\gamma+\alpha)^{2}\sin^{2}{\psi_{2}}+(\alpha\cos{\psi_{2}})^{2}+(\gamma\cos{\psi_{2}})^{2}}{2\gamma\alpha\cos^{2}{\psi_{2}}}\right)}=L_{2}\cosh^{-1}{\left(\frac{(\gamma+\alpha)^{2}-2\alpha\gamma\cos^{2}{\psi_{2}}}{2\gamma\alpha\cos^{2}{\psi_{2}}}\right)}.$
(184b) For ${\rm CD}$, with ${\rm
C}=(-\gamma\sin{\psi_{1}},\gamma\cos{\psi_{1}})$ and ${\rm
C}=(\sigma_{r},\xi)$ we have $d_{\rm
CD}=L_{1}\cosh^{-1}{\left(\frac{(-\gamma\sin{\psi_{1}}-\sigma_{r})^{2}+(\gamma\cos{\psi_{1}})^{2}+\xi^{2}}{2\gamma\xi\cos{\psi_{1}}}\right)}.$
(184c)

Therefore, the total length of the geodesic is

$\begin{split}d=&L_{1}\log{\frac{(\sigma_{l}+\alpha\sin{\psi_{1}})^{2}+(\alpha\cos{\psi_{1}})^{2}}{\varepsilon\alpha\cos{\psi_{1}}}}+L_{1}\cosh^{-1}{\left(\frac{(-\gamma\sin{\psi_{1}}-\sigma_{r})^{2}+(\gamma\cos{\psi_{1}})^{2}+\xi^{2}}{2\gamma\xi\cos{\psi_{1}}}\right)}\\\
&+L_{2}\cosh^{-1}{\left(\frac{(\gamma+\alpha)^{2}-2\alpha\gamma\cos^{2}{\psi_{2}}}{2\gamma\alpha\cos^{2}{\psi_{2}}}\right)}.\end{split}$
(185)

Now what we need to do is minimizing the length of geodesic (185) with respect
to $\alpha$. Similar to the results above, we can check numerically that the
minimal geodesic will cross the ETW brane perpendicularly. Therefore, in the
following we will assume $\delta=\sigma_{r}$. Plugging this condition into
(183), finally we will get a 5th order equation for $\alpha$, which can only
be solved numerically. After getting the solution, we can plug it into (185)
and get the total length of the minimal geodesic.

####  A.4.1 Proof of perpendicular crossing of a geodesic

Before, we show that for some simple cases we can prove analytically and
numerically that, with ETW brane, the minimal geodesic will end on the ETW
brane perpendicularly. And for more complicate cases, we may only check this
property numerically. Here we give an intuitive proof without any calculation
for this property.

For example, considering the most complicate case shown in figure 29, the
geodesic starts from the point ${\rm A}$ with fixed $\sigma_{l}$. Now we want
to find the minimal geodesic from A to the ETW brane. We can construct another
geometry that, besides the original part, we add another part which is
symmetric with respect to ETW brane. Then we can image that, when we minimize
the geodesic with end points A and its symmetric point, the corresponding
geodesic length will always twice of the original geodesic. If the original
geodesic is not perpendicular to ETW brane, then the new geodesic in doubled
geometry will not be smooth, which means it is not a “geodesic”. Then we can
have a shorter one. Therefore, with ETW brane, the geodesic will always be
perpendicular to the ETW brane.

##  Appendix B Boundary entropy with path integral

In this section, we focus on the boundary entropy induced by the interface
brane. By evaluating the partition function, we can obtain the boundary
entropy without UV information, and the results are universal.

###  B.1 A review of a single region boundary entropy and the conformal
transformation

Before discussing our cases, we first review the results in reference
Fujita_2011 . They consider a single CFT region with a boundary. The
corresponding action is

$I=\frac{1}{16\pi G_{N}}\int_{N}\sqrt{-g}(R-2\Lambda)+\frac{1}{8\pi
G_{N}}\int_{Q}\sqrt{-h}(K-T),$ (186)

where $R$ denotes Ricci scalar, $\Lambda$ is cosmological constant, $K$ is
extrinsic curvature, and $T$ is the tension of the interface brane. Generally,
we have $\Lambda=-\frac{d(d-1)}{2L^{2}}$ where $L$ is the ${\rm AdS}_{d+1}$
radius (here we take $d=2$ and $\Lambda=-\frac{1}{L^{2}}$). Besides, we have
$R=\frac{2d+2}{d-1}\Lambda$, so for $d=2$ we have
$R=6\Lambda=-\frac{6}{L^{2}}$. While, for the integral on $Q$, we need to
solve the equation of motion and fix the location of brane.

In the action we only have one region, so we only consider the second junction
condition, which requires

$K_{ab}=(K-T)h_{ab}.$ (187)

Taking trace gives $K=\frac{d}{d-1}T$, it means $K=2T$ for $d=2$. Besides, we
still have

$K_{ab}=\frac{1}{L}\tanh{\left(\frac{\rho}{L}\right)}g_{ab},$ (188)

which gives $K=g^{ba}K_{ab}=\frac{d}{L}\tanh{\left(\frac{\rho}{L}\right)}$.
For $d=2$, we have $LT=\tanh{\left(\frac{\rho}{L}\right)}$. Now we can
simplify the action to (with Euclidean time $\tau$)

$I=\frac{-1}{16\pi
G_{N}}\left(-\frac{4}{L^{2}}\int_{N}\sqrt{g}+K\int_{Q}\sqrt{h}\right).$ (189)

We keep $K$ in action instead of $T$ as because we can consider $K$ as a
variable that belongs to one region. Then, we calculate $\int_{N}\sqrt{g}$ and
$\int_{Q}\sqrt{h}$, where $h$ is the induced metric on the brane.

Before calculating the volume, we introduce a UV cutoff to make it finite. To
realize it, we apply a special conformal transformation (SCT) for the CFT. The
general form of transformation is Berenstein_1999

$x_{\mu}^{\prime}=\frac{x_{\mu}+c_{\mu}x^{2}}{1+2c\cdot x+c^{2}x^{2}},$ (190)

where we define $c_{\mu}=(c_{x},c_{\tau},c_{z}=0)$, $c\cdot
x=\sum_{\mu}c_{\mu}x_{\mu}$, $c^{2}=\sum_{\mu}c_{\mu}^{2}$ and
$x^{2}=\sum_{\mu}x_{\mu}^{2}$. In the equation (3.58) of reference Anous_2022
, the corresponding transformation is $c_{\mu}=(0,c_{\tau}=c,0)$. While, in
the equation (3.5) of reference Fujita_2011 , they consider a general
transformation to $(x,\tau,z)$

$\begin{split}x^{\prime}=\frac{x+c_{x}(x^{2}+\tau^{2}+z^{2})}{1+2(c_{x}x+c_{\tau}\tau)+(c_{x}^{2}+c_{\tau}^{2})(x^{2}+\tau^{2}+z^{2})},\\\
\tau^{\prime}=\frac{\tau+c_{\tau}(x^{2}+\tau^{2}+z^{2})}{1+2(c_{x}x+c_{\tau}\tau)+(c_{x}^{2}+c_{\tau}^{2})(x^{2}+\tau^{2}+z^{2})},\\\
z^{\prime}=\frac{z}{1+2(c_{x}x+c_{\tau}\tau)+(c_{x}^{2}+c_{\tau}^{2})(x^{2}+\tau^{2}+z^{2})}.\end{split}$
(191)

Here we briefly discuss the SCT in $d=2$ CFT. Defining $x_{\mu}=(x,\tau)$,
$b=b_{x}+{\rm i}b_{\tau}$ and $z=x+{\rm i}\tau$, the transformation
$x_{\mu}=\frac{x_{\mu}+b_{\mu}x^{2}}{1+2b\cdot x+b^{2}x^{2}}$ can be expressed
as $z^{\prime}=\frac{z+bz\bar{z}}{1+(b\bar{z}+\bar{b}z)+b\bar{b}z\bar{z}}$. On
the other hand, the transformation above can be written as
$z^{\prime}=\frac{z}{1+\bar{b}z}=\frac{1}{z^{-1}+\bar{b}}$. It is because

$z^{\prime}=\frac{z\bar{z}}{\bar{z}+\bar{b}z\bar{z}}=\frac{z\bar{z}(z+bz\bar{z})}{(\bar{z}+\bar{b}z\bar{z})(z+bz\bar{z})}=\frac{z\bar{z}(z+bz\bar{z})}{z\bar{z}+\bar{b}z^{2}\bar{z}+bz\bar{z}^{2}+b\bar{b}z\bar{z}z\bar{z}}=\frac{z+bz\bar{z}}{1+(b\bar{z}+\bar{b}z)+b\bar{b}z\bar{z}}.$
(192)

Then it is straightforward to check that the lines $x=x_{0}$ and
$\tau=\tau_{0}$ will be mapped to circles except for some special lines that
depend on $b$.

Now we consider the SCT in the AdS space. To simplify the problem, we consider
that $c_{\mu}=(c_{0},0,0)$, and the location of brane is
$x=z\sinh{\delta}+x_{0},\delta=\frac{\rho}{L}$. Then we can directly check
that under (191), the brane is mapped to

$(x^{\prime}-d_{1})^{2}+\tau^{\prime 2}+(z^{\prime}-d_{2})^{2}-d_{3}=0,$ (193)

where $d_{1}=\frac{1+2c_{0}x_{0}}{2c_{0}(1+c_{0}x_{0})}$,
$d_{2}=-\frac{\sinh{\delta}}{2c_{0}(1+c_{0}x_{0})}$ and
$d_{3}=\frac{\cosh^{2}{\delta}}{4c_{0}^{2}(1+c_{0}x_{0})^{2}}$. It is a sphere
with radius $R=\sqrt{d_{3}}=\frac{\cosh{\delta}}{|2c_{0}(1+c_{0}x_{0})|}$ and
center $(d_{1},0,d_{2})$. For $z^{\prime}=0$ and $\tau^{\prime}=0$, we have
$x^{\prime}=\frac{1}{c_{0}},\frac{x_{0}}{1+c_{0}x_{0}}$, which means
$(\frac{1}{c_{0}},0,0)$ is a fixed point for any $x_{0}$. Besides, after
translation of $x^{\prime}$, we can set $d_{1}=0$. And for $x_{0}=0$ we have
$r_{D}=\frac{1}{2|c_{0}|}$ in Fujita_2011 . Therefore, with a proper
$c_{\mu}$, we can transform the region $x<x_{0}$ to a disk.

After the SCT, the CFT region is on a disk with radius $r_{D}$ in figure 30.
While the disk radius is not important for boundary entropy.

Figure 30:  Location of the brane after conformal transformation.

And the location of the center is $z=r_{D}\sinh{\frac{\rho^{*}}{L}}$. It is
because conformal transformation won’t change the angle. The angle between the
brane and the asymptotic CFT boundary is $\frac{\pi}{2}-\psi$, and
$\tan{\psi}=\sinh{\frac{\rho^{*}}{L}}$. This is also consistent with the
result before, where the location of center is $z=\sqrt{d_{3}}$ and
corresponding $r_{D}=\frac{1}{|2c_{0}(1+c_{0}x_{0})|}$. 202020In the following
subsection, to simplify the problem, we may add an ETW brane artificially,
which only acts as a UV-cutoff. However, from later results we will find that
the value of radius $r_{D}$ will not change the boundary entropy, which means
we can consider the region bounded by an ETW brane at
$x_{0}\rightarrow+\infty$. Then we can apply the corresponding conformal
transformation with $c_{0}=-\frac{1}{x_{0}+\delta}$ and $\delta\gtrsim 0$. It
means the region on the left of $x=-\frac{1}{c_{0}}=x_{0}+\delta$ will be
mapped to a disk on the right of $x^{\prime}=\frac{1}{c_{0}}=-(x_{0}+\delta)$,
and the corresponding radius is $r_{D}=\frac{(x_{0}+\delta)^{2}}{2\delta}$.
Assuming $\delta=1$ and $x_{0}\gg 1$, we have
$r_{D}\approx\frac{x_{0}^{2}}{2}$. The corresponding brane after SCT will
cross x-axis on $x=\frac{1}{c_{0}}=-(x_{0}+\delta)\approx-x_{0}$ and
$x=\frac{1}{c_{0}}+2r_{D}\approx x_{0}^{2}-x_{0}\approx x_{0}^{2}$, which
means the disk region will cover the whole space for $x\rightarrow+\infty$.

Now we discuss the integral $\int_{N}\sqrt{g}$ and $\int_{Q}\sqrt{h}$. With
metric (15), we have $\sqrt{g}=\left(\frac{L}{z}\right)^{3}$. Defining
$\bar{z}=r_{D}(\sinh{\frac{\rho^{*}}{L}}+\cosh{\frac{\rho^{*}}{L}})=r_{D}e^{\rho^{*}/L}$,
we have

$\begin{split}\int_{N}\sqrt{g}=&L^{3}\int_{0}^{\bar{z}}{\rm d}z\
\frac{1}{z^{3}}\int{\rm d}^{2}S=L^{3}\int_{0}^{\bar{z}}{\rm d}z\
\frac{1}{z^{3}}\pi\left((r_{D}\cosh{\frac{\rho^{*}}{L}})^{2}-(z-r_{D}\sinh{\frac{\rho^{*}}{L}})^{2}\right)\\\
=&\pi L^{3}\int_{0}^{\bar{z}}{\rm d}z\
\frac{1}{z^{3}}\left(r_{D}^{2}-z^{2}+2zr_{D}\sinh{\frac{\rho^{*}}{L}}\right)=\left.\pi
L^{3}\left(-\frac{r_{D}^{2}}{2}z^{-2}-2r_{D}\sinh{\frac{\rho^{*}}{L}}z^{-1}-\log{z}\right)\right|_{\varepsilon}^{\bar{z}}\\\
=&\pi
L^{3}\left(\frac{r_{D}^{2}}{2\varepsilon^{2}}-\frac{1}{2}e^{-2\rho^{*}/L}+2r_{D}\sinh{\frac{\rho^{*}}{L}}(\frac{1}{\varepsilon}-\frac{1}{r_{D}}e^{-\rho^{*}/L})+\log{\frac{\varepsilon}{r_{D}}e^{-\rho^{*}/L}}\right).\end{split}$
(194)

As for $\int_{Q}\sqrt{h}$, because Q satisfies

$x^{2}+\tau^{2}+(z-r_{D}\sinh{\frac{\rho^{*}}{L}})^{2}=r_{D}^{2}\cosh^{2}{\frac{\rho^{*}}{L}},$
(195)

the induced metric is

$\begin{split}{\rm d}s^{2}=&\frac{L^{2}}{z^{2}}\left({\rm d}z^{2}+{\rm
d}\tau^{2}+\frac{1}{x^{2}}\left[(z-r_{D}\sinh{\frac{\rho^{*}}{L}}){\rm
d}z+\tau{\rm d}\tau\right]^{2}\right)\\\
=&\frac{L^{2}}{z^{2}}\left(\left(1+\frac{(z-r_{D}\sinh{\frac{\rho^{*}}{L}})^{2}}{x^{2}}\right){\rm
d}z^{2}+\left(1+\frac{\tau^{2}}{x^{2}}\right){\rm
d}\tau^{2}+\frac{2\tau}{x^{2}}\left(z-r_{D}\sinh{\frac{\rho^{*}}{L}}\right){\rm
d}z{\rm d}\tau\right).\end{split}$ (196)

Therefore, we have volume form

$\begin{split}\sqrt{h}=&\frac{L^{2}}{z^{2}}\sqrt{\left(1+\frac{(z-r_{D}\sinh{\frac{\rho^{*}}{L}})^{2}}{x^{2}}\right)\left(1+\frac{\tau^{2}}{x^{2}}\right)-\left(\frac{\tau}{x^{2}}\left(z-r_{D}\sinh{\frac{\rho^{*}}{L}}\right)\right)^{2}}\\\
=&\frac{L^{2}}{z^{2}}\sqrt{\frac{r_{D}^{2}+z_{0}^{2}}{r_{D}^{2}-\tau^{2}-z^{2}+2zz_{0}}}.\end{split}$
(197)

To simplify the notation, in the following we will denote
$z_{0}=r_{D}\sinh{\frac{\rho^{*}}{L}}$, then the integral region is $Q:\
\tau\in[0,\sqrt{z_{0}^{2}+r_{D}^{2}-(z-z_{0})^{2}}],\
z\in[0,\sqrt{z_{0}^{2}+r_{D}^{2}}+z_{0}]$. Therefore, the integral is

$\begin{split}\int_{Q}\sqrt{h}=&4\int_{0}^{\sqrt{z_{0}^{2}+r_{D}^{2}}+z_{0}}{\rm
d}z\int_{0}^{\sqrt{z_{0}^{2}+r_{D}^{2}-(z-z_{0})^{2}}}{\rm
d}\tau\frac{L^{2}}{z^{2}}\sqrt{\frac{r_{D}^{2}+z_{0}^{2}}{r_{D}^{2}-\tau^{2}-z^{2}+2zz_{0}}}\\\
=&4\int_{0}^{\sqrt{z_{0}^{2}+r_{D}^{2}}+z_{0}}{\rm d}z\ \frac{\pi
L^{2}}{2z^{2}}\sqrt{r_{D}^{2}+z_{0}^{2}}=2\pi
L^{2}\sqrt{r_{D}^{2}+z_{0}^{2}}\left(\frac{1}{\varepsilon}+\frac{z_{0}-\sqrt{r_{D}^{2}+z_{0}^{2}}}{r_{D}^{2}}\right)\\\
=&2\pi
L^{2}r_{D}\cosh{\frac{\rho^{*}}{L}}\left(\frac{1}{\varepsilon}+\frac{-e^{-\rho^{*}/L}}{r_{D}}\right).\end{split}$
(198)

Finally we get the integral of action (189)

$\begin{split}I=&\frac{-1}{16\pi G_{N}}\left[-\frac{4}{L^{2}}\left(\pi
L^{3}\left(\frac{r_{D}^{2}}{2\varepsilon^{2}}-\frac{1}{2}e^{-2\rho^{*}/L}+2r_{D}\sinh{\frac{\rho^{*}}{L}}(\frac{1}{\varepsilon}-\frac{1}{r_{D}}e^{-\rho^{*}/L})+\log{(\frac{\varepsilon}{r_{D}}e^{-\rho^{*}/L})}\right)\right)\right.\\\
&\left.+K\left(2\pi
L^{2}r_{D}\cosh{\frac{\rho^{*}}{L}}\left(\frac{1}{\varepsilon}+\frac{-e^{-\rho^{*}/L}}{r_{D}}\right)\right)\right]\\\
=&\frac{4\pi L}{16\pi
G_{N}}\left[\left(\frac{r_{D}^{2}}{2\varepsilon^{2}}-\frac{1}{2}e^{-2\rho^{*}/L}+2r_{D}\sinh{\frac{\rho^{*}}{L}}(\frac{1}{\varepsilon}-\frac{1}{r_{D}}e^{-\rho^{*}/L})+\log{(\frac{\varepsilon}{r_{D}}e^{-\rho^{*}/L})}\right)\right.\\\
&\left.+\frac{KL}{2}r_{D}\cosh{\frac{\rho^{*}}{L}}\left(-\frac{1}{\varepsilon}+\frac{e^{-\rho^{*}/L}}{r_{D}}\right)\right]\\\
=&\frac{L}{4G_{N}}\left[\frac{r_{D}^{2}}{2\varepsilon^{2}}-\frac{1}{2}e^{-2\rho^{*}/L}+2r_{D}\sinh{\frac{\rho^{*}}{L}}(\frac{1}{\varepsilon}-\frac{1}{r_{D}}e^{-\rho^{*}/L})+\log{(\frac{\varepsilon}{r_{D}}e^{-\rho^{*}/L})}+r_{D}\sinh{\frac{\rho^{*}}{L}}\left(-\frac{1}{\varepsilon}+\frac{e^{-\rho^{*}/L}}{r_{D}}\right)\right]\\\
=&\frac{L}{4G_{N}}\left(\frac{r_{D}^{2}}{2\varepsilon^{2}}+\log{\frac{\varepsilon}{r_{D}}}-\frac{\rho^{*}}{L}-\frac{1}{2}+\frac{r_{D}}{\varepsilon}\sinh{\frac{\rho^{*}}{L}}\right),\end{split}$
(199)

where we use $LT=\tanh{\left(\frac{\rho}{L}\right)}$ in the third equation.
The final result is exactly the equation (3.7) in reference Fujita_2011 . The
corresponding boundary entropy is

$S_{\rm bdy}=-\left[I(\rho^{*})-I(0)\right]=\frac{\rho^{*}}{4G_{N}}.$ (200)

With $LT=\tanh{\left(\frac{\rho}{L}\right)}$, we have $S_{\rm
bdy}=\frac{L\tanh^{-1}{LT}}{4G_{N}}=\frac{c\tanh^{-1}{(LT)}}{6}$. Later we
will find it is universal and valid for other cases. Besides, we can also
express boundary entropy as $S_{\rm
bdy}=\frac{c\rho^{*}}{6L}=\frac{c}{6}\tanh^{-1}{(\sin{\psi})}$, where $\psi$
is defined below (15). With (47b), we find that $S_{\rm
bdy}=\frac{c}{6}\log{\left(\tan{(\frac{\psi}{2}+\frac{\pi}{4})}\right)}$,
which is consistent with (49), where the prefactor $1/3$ is because there are
two branes and three regions for that case.

Figure 31:  (a) AdS bulk dual of single CFT region before and after conformal
transformation. (b) Single CFT region on the boundary before and after
conformal transformation.

Let’s consider two branes. Under the same SCT, on the boundary CFT two branes
will be mapped to two circles which are tangent at the fixed point, which is
shown in figure 31. From the figure 31 (a), we can define the angle $\rho$
which corresponds to the inner angle in the blue region. It means we define
the direction of the ETW brane to be outwards to the bulk region. Then the
corresponding boundary entropy is $S_{\rm
bdy}=\frac{\rho^{*}_{1}+\rho^{*}_{2}}{4G_{N}}$. We can consider the integral
as two parts. One part is the larger disk and the other is the smaller disk,
and the corresponding actions have opposite signs. For each action, we can
calculate the boundary entropy with the method above. It is worth mentioning
that if we set the larger ETW brane perpendicularly with angle $\rho^{*}=0$,
then it will not contribute to the boundary entropy.

###  B.2 Boundary entropy for more than one region

Figure 32:  (a) AdS bulk dual of two CFT regions before and after conformal
transformation. (b) Two CFT regions on the boundary before and after conformal
transformation. Here we show ETW with blue line, and for limit case we can
take ETW to infinity.

In the following, we consider two regions with different AdS radii that are
separated by an interface brane, as shown in figure 32. We expect that the
boundary entropy of this case is also $S_{\rm
bdy}=\frac{\rho^{*}_{1}+\rho^{*}_{2}}{4G_{N}}$, where $\rho^{*}_{1},\
\rho^{*}_{2}$ are the corresponding angles of the brane in region 1 and 2.
Actually, we can obtain a result consistent with (49).

We first consider figure 32 (a). If we connect two half planes of CFT, then
obviously two bulk AdS regions will have an overlap. The proper way to
understand it is in figure 32 (a), where we use some magic glue to connect two
parts. Then we must discuss two bulk regions individually with a proper
connection condition and a local coordinate transformation. For example, we
should fix $z$ and $\tau$ but translate $x$ when we go through the brane from
one region to another. After special conformal transformation, we have the
geometry shown in figure 32. To be concrete, we discuss the action (9). With
connection condition (11b), we have $\Delta K=2T$, and the action can be
written as

$\begin{split}S_{{\rm EH}}=&-\frac{1}{16\pi
G_{(3)}}\left[\int_{\mathcal{M}_{1}}{\rm d}^{3}x\
\sqrt{g_{1}}\left(R_{1}+\frac{2}{L^{2}_{1}}\right)+\int_{\mathcal{M}_{2}}{\rm
d}^{3}x\
\sqrt{g_{2}}\left(R_{2}+\frac{2}{L^{2}_{2}}\right)+\int_{\mathcal{S}}{\rm
d}^{2}y\ \sqrt{h}(K_{1}-K_{2})\right]\\\ =&-\frac{1}{16\pi
G_{(3)}}\left[\int_{\mathcal{M}_{1}}{\rm d}^{3}x\
\sqrt{g_{1}}\left(R_{1}+\frac{2}{L^{2}_{1}}\right)+\int_{\mathcal{S}}{\rm
d}^{2}y\ \sqrt{h}K_{1}+\int_{\mathcal{M}_{2}}{\rm d}^{3}x\
\sqrt{g_{2}}\left(R_{2}+\frac{2}{L^{2}_{2}}\right)\right.\\\
&\left.-\int_{\mathcal{S}}{\rm d}^{2}y\ \sqrt{h}K_{2}\right]=-\frac{1}{16\pi
G_{(3)}}\sum_{\alpha={1,2}}\left[\int_{\mathcal{M}_{\alpha}}{\rm d}^{3}x\
\sqrt{g_{\alpha}}\left(R_{\alpha}+\frac{2}{L^{2}_{\alpha}}\right)+\int_{\mathcal{S}_{\alpha}}{\rm
d}^{2}y\ \sqrt{h}K_{\alpha}\right]=\sum_{\alpha={1,2}}S_{\alpha}.\end{split}$
(201)

In reference Anous_2022  the left region is $1$, the right region is $2$ and
the direction of brane and the extrinsic curvature $K_{1,2}$ are from $1$ to
$2$. But here we introduce the surface $\mathcal{S}_{1,2}$, where
$\mathcal{S}_{1}=\mathcal{S}=-\mathcal{S}_{2}$. Therefore, directions of
$\mathcal{S}_{1,2}$ are defined from inside to outside. With this, we have
$S_{{\rm EH}}=\sum_{\alpha={1,2}}S_{\alpha}$, and $S_{\alpha}$ is the same as
(189). So from the form of the action we can also show that the boundary
entropy can be expressed as the sum of contributions from the two regions.
Besides, here we only consider the brane between two regions, but not the
larger ETW brane, because we set the corresponding angle $\rho=0$.

##  Appendix C Different phases and phase transition for measurements

In this appendix, we give (i) a detailed derivation of the geodesic solutions
in AdS and black hole spacetime, and (ii) more details on the geodesic length
calculation in the bubble phase.

###  C.1 Convention of metric and corresponding geodesics

With the metric above, we can calculate the geodesic equation and its length
with the following method. The geodesic equation is

$\frac{{\rm d}^{2}x^{\mu}}{{\rm
d}\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{{\rm d}x^{\alpha}}{{\rm
d}\lambda}\frac{{\rm d}x^{\beta}}{{\rm d}\lambda}=f(\lambda)\frac{{\rm
d}x^{\mu}}{{\rm d}\lambda},$ (202)

where $f(\lambda)$ is a function to be determined. 212121Don’t confuse with
the function $f(r)$ in the metric (56). With the metric (56) and the
coordinate $(t=t_{0},r=r(x),x=x)$, plugging Christoffel coefficient into
(202), we have

$\ddot{x}+\frac{2}{r}\dot{r}\dot{x}=f(x)\dot{x},\\\ $ (203a)
$\ddot{r}+\frac{r}{L^{2}(\mu-1)-r^{2}}\dot{r}^{2}+r(\mu-1-\frac{r^{2}}{L^{2}})\dot{x}^{2}=f(x)\dot{r}.$
(203b)

Therefore, with a fixed $f(x)$, we only need to solve the differential
equation

$\ddot{r}+\frac{r^{2}-2[L^{2}(\mu-1)-r^{2}]}{r[L^{2}(\mu-1)-r^{2}]}\dot{r}^{2}+r(\mu-1-\frac{r^{2}}{L^{2}})=0.$
(204)

The general solution for this differential equation has two parameters, which
we denote as $c_{1},c_{2}$. For two situations $\mu>1$ and $\mu<1$, the
general solution is

$r=\frac{1}{\sqrt{c_{1}-(\frac{1}{L^{2}(\mu-1)}-c_{1})\sinh^{2}{[\sqrt{\mu-1}(x+c_{2})]}}},\qquad\mu>1,$
(205a)
$r=\frac{1}{\sqrt{c_{1}-(\frac{1}{L^{2}(1-\mu)}+c_{1})\sin^{2}{[\sqrt{1-\mu}(x+c_{2})]}}},\qquad\mu<1.$
(205b)

Here we only take positive branch because $r>0$. For $\mu>1$, we have the
constraint $0<c_{1}<\frac{1}{L^{2}(\mu-1)}$, while for $\mu<1$, we have the
constraint $c_{1}>0$. It is obvious that for the limit $\mu\rightarrow 1$, the
geodesic equation will be a circle with redefinition $\tilde{z}=1/r$. However,
there is no unique solution for differential equation (204). With parameter
shift $x\rightarrow x+\frac{\pi}{\sqrt{\mu-1}}$, (205b) will change $\sin$ to
$\cos$, which is equivalent to the shift $c_{2}\rightarrow
c_{2}+\frac{\pi}{2\sqrt{\mu-1}}$. While in (205a), another solution is given
by parameter shift $x\rightarrow x+\frac{{\rm i}\pi}{2\sqrt{\mu-1}}$, which is
equivalent to changing $\sinh$ to ${\rm i}\cosh$. But we require $c_{2}$ is
real, which means these two geodesics are not equivalent. Therefore, the
second solution for $\mu>1$ is

$r^{-2}+(\frac{1}{L^{2}(\mu-1)}-c_{1}^{\prime})\sinh^{2}{[\sqrt{\mu-1}(x+c_{2})]}=\frac{1}{L^{2}(\mu-1)},$
(206)

where we redefine $c_{1}^{\prime}=\frac{2}{L^{2}(\mu-1)}-c_{1}$.

Now we can calculate the geodesic length,

$d=\int{\rm d}l\ \sqrt{g}=\int{\rm
d}s=\int\sqrt{\frac{r^{\prime}(x)}{f}+r^{2}}\ {\rm d}x,$ (207)

where $f$ is the function defined in metric (56), and $r^{\prime}(x)$ is the
derivative of the geodesic. Then the integral can be solved and the final
result is

$d=L\tanh^{-1}{\left[\frac{\tanh{(\sqrt{\mu-1}(x+c_{2}))}}{L\sqrt{c_{1}(\mu-1)}}\right]},\qquad\mu>1,$
(208a)
$d=L\tanh^{-1}{\left[\frac{\tan{(\sqrt{1-\mu}(x+c_{2}))}}{L\sqrt{c_{1}(1-\mu)}}\right]},\qquad\mu<1.$
(208b)

Actually, in (205) and (208) we can notice that the solutions of two cases are
related. For example, starting from (208a), if we now consider $\mu<1$, then
$\sqrt{\mu-1}={\rm i}\sqrt{1-\mu}$ and ${\rm i}$ in $\tanh$ will change it to
$-{\rm i}\tan$.

Assuming two end points located at
$(r=\frac{1}{\varepsilon},x=\pm\frac{1}{2}x_{0})$, for $\mu<1$, we have
parameter
$c_{1}=[\varepsilon^{2}+\frac{1}{L^{2}(1-\mu)}\sin^{2}{(\sqrt{1-\mu}\frac{x_{0}}{2})}]/\cos^{2}{(\sqrt{1-\mu}\frac{x_{0}}{2})}$
and $c_{2}=0$. Then the total geodesic length is

$\begin{split}\Delta
d=&2L\tanh^{-1}{\left[\frac{\tan{(\sqrt{1-\mu}\frac{x_{0}}{2})}}{L\sqrt{c_{1}(1-\mu)}}\right]}\approx
2L\tanh^{-1}{\left[1-\frac{L^{2}(1-\mu)}{2\sin^{2}{(\sqrt{1-\mu}\frac{x_{0}}{2})}}\varepsilon^{2}+\mathcal{O}(\varepsilon^{3})\right]}\\\
\approx&2L\cosh^{-1}{\left(\frac{\sin{(\sqrt{1-\mu}\frac{x_{0}}{2})}}{L\sqrt{1-\mu}\varepsilon}\right)}\approx
2L\log{\left(\frac{2\sin{(\sqrt{1-\mu}\frac{x_{0}}{2})}}{L\sqrt{1-\mu}\varepsilon}\right)},\end{split}$
(209)

Therefore, for $x_{0}\ll\frac{1}{\sqrt{1-\mu}}$, we retain $\Delta
d=2L\log{\frac{x_{0}/L}{\varepsilon}}$.

Similarly, for $\mu>1$, we have parameter
$c_{1}=[\varepsilon^{2}+\frac{1}{L^{2}(\mu-1)}\sinh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}]/\cosh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}$
and $c_{2}=0$. Then the total geodesic length is

$\begin{split}\Delta
d=&2L\tanh^{-1}{\left[\frac{\tanh{(\sqrt{\mu-1}\frac{x_{0}}{2})}}{L\sqrt{c_{1}(\mu-1)}}\right]}\approx
2L\tanh^{-1}{\left[1-\frac{L^{2}(\mu-1)}{2\sinh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}}\varepsilon^{2}+\mathcal{O}(\varepsilon^{3})\right]}\\\
\approx&2L\cosh^{-1}{\left(\frac{\sinh{(\sqrt{\mu-1}\frac{x_{0}}{2})}}{L\sqrt{\mu-1}\varepsilon}\right)}\approx
2L\log{\left(\frac{2\sinh{(\sqrt{\mu-1}\frac{x_{0}}{2})}}{L\sqrt{\mu-1}\varepsilon}\right)}.\end{split}$
(210)

Therefore, for $x_{0}\ll\frac{1}{\sqrt{\mu-1}}$, we retain $\Delta
d=2L\log{\frac{x_{0}/L}{\varepsilon}}$. While, for
$x_{0}\gg\frac{1}{\sqrt{\mu-1}}$, we retain $\Delta
d=2L\log{(\frac{1}{\varepsilon L\sqrt{\mu-1}})}+L\sqrt{\mu-1}x_{0}$, which
corresponds to volume-law entanglement.

We focus on the unusual geodesic in (206). Before we argue that for (205a),
$0<c_{1}<\frac{1}{L^{2}(\mu-1)}$, which leads to a circle in some limit.
While, for (206), if $0<c_{1}<\frac{1}{L^{2}(\mu-1)}$, it is similar to
(205a). But we can also have $c_{1}<0$, it is special because corresponding
derivative $\frac{{\rm d}r^{-1}}{{\rm d}x}$ will be larger when $c_{1}$ is
smaller. Besides, there is a fixed point on (206) for
$(x=-c_{2},r^{-1}=\sqrt{\frac{1}{L^{2}(\mu-1)}})$, which is on the horizon of
black hole. It is one reason why we consider it unusual because usual geodesic
will not touch the horizon. Analog to (207) and (208a), we have the length of
geodesic for (206) that

$d=L\tanh^{-1}{\left[\sqrt{2-c_{1}L^{2}(\mu-1)}{\tanh{(\sqrt{\mu-1}(x+c_{2}))}}\right]}.$
(211)

In the following, we compare the length of geodesic for two cases with black
hole metric. If we assume points $(r_{0},\pm\frac{x_{0}}{2})$ on two
geodesics, then $c_{2}=0$, and corresponding parameters $c_{1}^{\rm u}$ for
usual geodesic (205a) and $c_{1}^{\rm n}$ for unusual geodesic (206) are

$c_{1}^{\rm
u}=\frac{r^{-2}+\frac{1}{L^{2}(\mu-1)}\sinh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}}{\cosh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}},\qquad
c_{1}^{\rm
n}=\frac{r^{-2}+\frac{1}{L^{2}(\mu-1)}\left(\sinh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}-1\right)}{\sinh^{2}{(\sqrt{\mu-1}\frac{x_{0}}{2})}}.$
(212)

Plugging it in (208a) and (211), we have

$\Delta d^{\rm
u}=2L\tanh^{-1}{\left[{\left(\frac{L^{2}(\mu-1)r_{0}^{-2}+\sinh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}{\cosh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}\right)}^{-\frac{1}{2}}{\tanh{(\sqrt{\mu-1}(x+c_{2}))}}\right]},$
(213a) $\Delta d^{\rm
n}=2L\tanh^{-1}{\left[{\left(\frac{-L^{2}(\mu-1)r_{0}^{-2}+\cosh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}{\sinh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}\right)}^{\frac{1}{2}}{\tanh{(\sqrt{\mu-1}(x+c_{2}))}}\right]}.$
(213b)

For usual geodesic we have $c_{1}<\frac{1}{L^{2}(\mu-1)}$, which means
$r_{0}^{-2}<c_{1}<\frac{1}{L^{2}(\mu-1)}$. For unusual geodesic we also have
$c_{1}<\frac{1}{L^{2}(\mu-1)}$, which means
$r_{0}^{-2}<\frac{1}{L^{2}(\mu-1)}$. Therefore, we can directly check that,
with $r_{0}^{-2}{L^{2}(\mu-1)}<1$, we have

${\left(\frac{L^{2}(\mu-1)r_{0}^{-2}+\sinh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}{\cosh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}\right)}^{-\frac{1}{2}}<{\left(\frac{-L^{2}(\mu-1)r_{0}^{-2}+\cosh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}{\sinh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}\right)}^{\frac{1}{2}}.$
(214)

Therefore, $\Delta d^{\rm u}<\Delta d^{\rm n}$, which means usual geodesic is
a global minimal point with the shortest length. Besides, take
$r_{0}^{-1}=\varepsilon$, we can simplify (213b) to get

$\begin{split}\Delta
d=&2L\tanh^{-1}{\left[\left(1-\frac{L^{2}(\mu-1)\varepsilon^{2}}{\cosh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}\right)^{\frac{1}{2}}\right]}\approx
2L\tanh^{-1}{\left[1-\frac{L^{2}(\mu-1)}{2\cosh^{2}{[\sqrt{\mu-1}\frac{x_{0}}{2}]}}\varepsilon^{2}+\mathcal{O}(\varepsilon^{3})\right]}\\\
\approx&2L\cosh^{-1}{\left(\frac{\cosh{(\sqrt{\mu-1}\frac{x_{0}}{2})}}{L\sqrt{\mu-1}\varepsilon}\right)}\approx
2L\log{\left(\frac{2\cosh{(\sqrt{\mu-1}\frac{x_{0}}{2})}}{L\sqrt{\mu-1}\varepsilon}\right)}.\end{split}$
(215)

However, different from (210), for $x_{0}\ll\frac{1}{\sqrt{\mu-1}}$, we cannot
retain $\log$-law entanglement. For $x_{0}\gg\frac{1}{\sqrt{\mu-1}}$, we
retain $\Delta d=2L\log{(\frac{1}{\varepsilon
L\sqrt{\mu-1}})}+L\sqrt{\mu-1}x_{0}$, which corresponds to volume law
entanglement.

###  C.2 Details of calculation for the geodesic in the bubble-inside-horizon
phase

With (96) and (97), we mention that the existence of geodesic along horizon
will always lead to a longer geodesic length. Now we compare the two cases
shown in figure 33.

Figure 33:  Two possible geodesics in the region II in the bubble-inside-
horizon phase.

According to (206), assuming $(r,x)=(\varepsilon^{-1},\tilde{x}_{0})$ located
on the geodesic, then from (215) we have

$\Delta
d=L_{2}\tanh^{-1}{\left[\left(1-\frac{L_{2}^{2}(\mu-1)\varepsilon^{2}}{\cosh^{2}{[\sqrt{\mu-1}\tilde{x}_{0}]}}\right)^{\frac{1}{2}}\right]}\approx
L_{2}\log{\left(\frac{2\cosh{(\sqrt{\mu-1}\tilde{x}_{0})}}{L_{2}\sqrt{\mu-1}\varepsilon}\right)}.$
(216)

Now we show that
$L_{2}\log{\left(\frac{2\cosh{(\sqrt{\mu-1}\tilde{x}_{0})}}{L_{2}\sqrt{\mu-1}\varepsilon}\right)}<L_{2}\log{\left(\frac{2\cosh{(\sqrt{\mu-1}\tilde{x}_{1})}}{L_{2}\sqrt{\mu-1}\varepsilon}\right)}+r_{H}(\tilde{x}_{0}-\tilde{x}_{1})$
for $0<\tilde{x}_{1}<\tilde{x}_{0}$. Define function
$f(x)=\sqrt{\mu-1}x-\log{[\cosh{(\sqrt{\mu-1}x)}]}$, then the problem above is
equivalent to $f(\tilde{x}_{1})<f(\tilde{x}_{0})$ for
$\tilde{x}_{1}<\tilde{x}_{0}$. Up to rescaling of $x$ we can define
$\tilde{f}(x)=x-\log{[\cosh{(x)}]}$, and $\tilde{f}^{\prime}(x)=1-\tanh{x}>0$.
Therefore, we only consider the geodesic with three parts as in the main text.

To solve (100) in the main text, we mentioned that $\beta$ must have the same
order of $\delta$ and $|\delta-\beta|\ll\delta$. Here we give a proof. An
intuition argument is that for $|\delta-\beta|\sim\mathcal{O}(\delta)$, the
geodesic corresponds to volume law, while we can find a geodesic with
$\log$-law later. We first construct a geodesic with $\log$-law for
$\varepsilon\ll r_{H}^{-1}\ll\delta\ll 1$. Assume $\delta\approx\beta$ and
$b_{1}^{\prime}=b_{1}$ in (96) and (97), then the total length of part $l_{1}$
and $l_{2}$ is smaller than the length of total unusual geodesic in (215) that

${\rm
Length}[l_{1}+l_{2}]<2L_{2}\log{\left(\frac{2\cosh{(\sqrt{\mu-1}(\delta-\beta))}}{L_{2}\sqrt{\mu-1}\varepsilon}\right)}.$
(217)

Assume $(\delta-\beta)\sim\mathcal{O}(r_{H}^{-1})$, then the right-hand side
of (217) is
$L_{2}\log{\left(2\cosh{(\sqrt{\mu-1}(\delta-\beta))}\right)}\sim\mathcal{O}(1)$
except UV-cutoff and the divergence of $\log{\sqrt{\mu-1}}$. For part $l_{3}$,
the point $(r_{0},x_{0})$ on it with $|\delta-
x_{0}|<2|\delta-\beta|\sim\mathcal{O}(r_{H}^{-1})$, which means
$x_{0}\sim\mathcal{O}(\delta)\gg r_{0}^{-1}$. Therefore, with (209), we have

$\begin{split}{\rm Length}[l_{3}]\approx
L_{1}\log{\left(\frac{2\sin{x_{0}}}{L_{1}r_{0}^{-1}}\right)},\end{split}$
(218)

which corresponds to $\log$-law entanglement for $x_{0}\ll 1$. Therefore, we
have constructed a geodesic with $\log$-law entanglement. Now we prove that
for $|\delta-\beta|\sim\mathcal{O}(\delta)$ the geodesic corresponds to volume
law. We consider the length of part $l_{1}$, which corresponds to equation
(96). Assume $\beta=\eta^{\prime}\delta$ and
$\eta^{\prime}\sim\mathcal{O}(1)<1,(1-\eta^{\prime})\sim\mathcal{O}(1)<1$,
with (215), we have

$\begin{split}{\rm Length}[l_{1}]\approx
L_{2}\log{\left(\frac{2\cosh{(\sqrt{\mu-1}(1-\eta^{\prime})\delta)}}{L_{2}\sqrt{\mu-1}\varepsilon}\right)},\end{split}$
(219)

With limit $\sqrt{\mu-1}\delta\sim\mathcal{O}(\frac{\delta}{r_{H}^{-1}})\gg
1$, the right-hand side of (219) can be simplified as ${\rm r.h.s.}\approx
L_{2}\log{\left(\frac{\exp{(L_{2}^{-1}r_{H}(1-\eta^{\prime})\delta)}}{L_{2}\sqrt{\mu-1}\varepsilon}\right)}=L_{2}\log{\left(\frac{r_{H}^{-1}}{\varepsilon}\right)}+r_{H}(1-\eta^{\prime})\delta$,
which corresponds to volume law entanglement. Therefore, we finish the proof
that $|\delta-\beta|\ll\delta$.

##  Appendix D Tensor network realization of Python’s Lunch

###  D.1 Brief review of Python’s Lunch property

Here we briefly review some basics about complexity and Python’s Lunch. The
first definition is the complexity of a unitary operator $U$, which is defined
as the minimal number of 2-qubit gates $g$ to prepare $U$, e.g.,
$U=g_{n}g_{n-1}...g_{1}$. The corresponding number is
$\mathcal{C}(U)=\mathcal{C}(U^{\dagger})$. Similarly, we can define the
relative complexity between two states $\left|\psi\right>$ and
$\left|\phi\right>$, which is defined as the complexity of the unitary
transformation satisfying $\left|\psi\right>=U\left|\phi\right>$. Besides,
with gauge redundancy there is more than one U, and we must minimize all
possible $\mathcal{C}(U)$. Therefore, we define
$\mathcal{C}(\psi,\phi)=\min{\mathcal{C}(U)}=\mathcal{C}(\phi,\psi)$.

In this article, we always consider the black holes, which can be modeled as a
quantum computer with $N$ qubits and evolve under some all-to-all Hamiltonian
or discrete gates. Here the original theory always has a Hamiltonian, while to
realize some properties we may construct it with some discrete gates. Under
this equivalence, we can consider the complexity of a black hole, or more
generally, a configuration of spacetime geometry. In reference
brown2019pythons , they consider restricted and unrestricted complexity, and
focus on the former. Here we briefly compare these two complexities. When we
consider a two-side black hole, it can be mapped to a maximal entangle state,
where $N$ qubits on the left maximally entangle with $N$ qubits on the right.
Then there may be some unitary evolution for both side. The difference between
restricted and unrestricted complexity is if we require the corresponding $U$
which maps it back to a maximal entangle state can only apply on one side. For
example, we apply $U$ only on right $N$ qubits for restricted case. However,
for the case we are interested in, which is a one-side black hole, this two
definition is the same, and it is the post-selection that will increase the
complexity exponentially.

Now we discuss Python’s Lunch geometry and introduce the conjecture of the
complexity of Python’s Lunch in reference brown2019pythons . A simple example
of Python’s Lunch is shown in figure 34.

Figure 34:  Python’s Lunch geometry with quantum circuit realization. Here we
add some degrees of freedom after unitary evolution on initial system. Then we
apply unitary transformation for the whole system and projective measurement
in some subregion.

There are two minimal area and one maximal area, and the corresponding numbers
of qubits from left to right are $N$, $(1+\alpha)N$ and $(1+\gamma)N$ with
$\alpha>\gamma>1$. Therefore, using the language of state complexity, we want
to calculate $\mathcal{C}(\left|I\right>\left|0\right>^{\otimes
m_{L}},\left|\psi\right>\left|0\right>^{\otimes m_{R}})$, where $m_{L}=\alpha
N$ and $m_{R}=(\alpha-\gamma)N$. In the reference, authors use Grover
algorithm to construct a decomposition of $U$ with
$\left|\psi\right>\left|0\right>^{\otimes
m_{R}}=U\left|I\right>\left|0\right>^{\otimes m_{L}}$. And the total number of
gates that need to construct $U$ and apply on the system is
$2^{\frac{m_{R}}{2}}\mathcal{C}_{\rm TN}$, where $\mathcal{C}_{\rm TN}$ is the
number of nodes in the tensor network. So the complexity is
$\mathcal{C}(U)=2^{\frac{m_{R}}{2}}\mathcal{C}_{\rm TN}$. Besides, the authors
also argue that for general state $\left|I\right>$, the corresponding
complexity is invariant. They also give a naive argument about the complexity
of unitary transformation $U$, which will lead to a larger number of gates,
but the main conclusion is the same. Starting from the initial state
$\left|I\right>$, we apply the unitary evolution on it and form the left side
of figure 34. Then by adding ancilla to the system, we can apply more unitary
evolution and form the middle part of the diagram. However, for the right-hand
side, we have a smaller system, which means we must apply some measurement and
project out some qubits. Then after applying more unitary evolution, we will
get the final state $\left|\psi\right>$. Totally, we apply $\mathcal{C}_{\rm
TN}$ number of gates in this process, but it needs some measurement which is
non-unitary. To project $m_{R}$ additional qubits on $\left|0\right>^{\otimes
m_{R}}$, we must consider post-selection effect, which means there are
$2^{m_{R}}$ possible outcomes and only one is what we want. Therefore, if we
want to successfully construct a state $\left|\psi\right>$ from
$\left|I\right>$, on average we must repeat this process $2^{m_{R}}$ times.
Finally, we find that the total number of unitary gates we need is
$2^{m_{R}}\mathcal{C}_{\rm TN}$, which is almost the same as Grover algorithm,
and the only difference is the prefactor on exponent.

According to the discussion above, they propose a conjecture about the
complexity of Python’s Lunch. For a Python’s Lunch geometry with min-max-min
areas $\mathcal{A}_{L},\mathcal{A}_{\rm m,ax}$ and $\mathcal{A}_{R}$, and with
$\mathcal{A}_{L}<\mathcal{A}_{R}$, the restricted complexity on the right
system is

$\mathcal{C}_{R}(U)={\rm const.}\times\mathcal{C}_{\rm
TN}\exp{\left[\frac{1}{2}\frac{\mathcal{A}_{\rm
max}-\mathcal{A}_{R}}{4G\hbar}\right]},$ (220)

where $\mathcal{C}_{\rm TN}=\frac{V}{G\hbar l_{\rm AdS}}$ is the volume of
wormhole. If we apply this conjecture to the case in figure 34, then
$\mathcal{A}_{L}\approx N\cdot 4G\hbar$, $\mathcal{A}_{\rm
max}\approx(1+\alpha)N\cdot 4G\hbar$ and
$\mathcal{A}_{R}\approx(1+\gamma)N\cdot 4G\hbar$, and $\mathcal{C}_{R}(U)={\rm
const.}\times\mathcal{C}_{\rm TN}\ e^{[(\alpha-\gamma)N]/2}$. And
$\mathcal{C}_{\rm TN}\geq N\log{N}$. Besides, to be more concrete, two
minimums are well-defined by local minimums, but the maximum is more
complicate. We should first choose one foliation of the geometry which is
known as “sweepout”, and find the global maximum. Then we minimize all
maximums for all foliations, and the minimum of maximum is $\mathcal{A}_{\rm
max}$.

Correspondingly, they also give a covariant version of the conjecture above.
For a covariant Python’s Lunch geometry with min-max-min generalized entropy
$S^{\rm(gen)}_{L}$, $S^{\rm(gen)}_{\rm max}$ and $S^{\rm(gen)}_{R}$, and
$S^{\rm(gen)}_{L}<S^{\rm(gen)}_{R}$, the restricted complexity on the right
system is

$\mathcal{C}_{R}(U)={\rm const.}\times\mathcal{C}_{\rm
TN}\exp{\left[\frac{1}{2}\left(S^{\rm(gen)}_{\rm
max}-S^{\rm(gen)}_{R}\right)\right]},$ (221)

where $\mathcal{C}_{\rm TN}$ is the size of the tensor network.

###  D.2 Python’s Lunch realization

In this section, we try to construct a tensor network to realize Python’s
lunch, which is related to the bubble-inside-horizon phase. We consider a
state dual to such a geometry. The inner AdS space can be considered as the
initial state before measurements, and the black hole geometry can be
considered as the effect of general measurements. Therefore, we can first
construct the inner AdS space with a MERA, and then construct the additional
black hole region with a tensor network which consists of both unitary
evolution and post-selection. We illustrate our construction in figure 35.

Figure 35:  Python’s lunch realization with MERA and additional layers of
tensor network. Here the gray disk corresponds to MERA with the black dots
denoting degrees of freedom on the boundary. The blue dots mean degrees of
freedom after measurements. The orange dots denote ancilla qudits. The green
and purple curves indicate two geodesics.

The inner tensor network of MERA is shown in gray, and the external tensor
network of measurements is shown in blue and orange. We can construct MERA
with unitary circuits and ancilla qubits with the method, e.g., in reference
Haghshenas_2022 , which will be easier to compute the complexity. The key is
how to design the tensor network of quantum quench, and show that for some
parameters it corresponds to the bubble-inside-horizon phase and has
exponentially big complexity, and for other parameters it corresponds to the
bubble-outside-horizon phase with much smaller complexity. A natural
construction is shown in figure 35. There are two layers of tensors for
measurements (from interior to outside, we denote them as the first and second
layers). The bond dimension of each tensor is to be determined in the
following.

We define the bond dimension $D^{(\cdot)}$ and simply use $(\cdot)$ to denote
the bond dimension. Because we hope that, after the weak measurements, the
dimension of the Hilbert space of boundary qubits is invariant, i.e. the
dimensions of the MERA state before and after weak measurements are the same.
For simplicity, we set the bond dimensions of legs in the MERA and of the legs
in the outermost layer of the tensor network to be $1$. Besides, we expect
that there is a min-max-min structure in tensor network for Python’s lunch.
Here the first minimum can be seen as the center of MERA. Then for the circle
with the same center of MERA, the corresponding perimeter is larger for larger
radius, and the maximum may locate near the boundary of MERA. Now if we
increase the radius, then the area will rely on the bond dimension of the legs
between two external layers of tensors, which we define as $\gamma$.
Therefore, $\gamma<1$ will give another minimal area, and the corresponding
geometry can be considered as Python’s lunch. However, if $\gamma=1$, there
will be no maxima and second minimum, which means there doesn’t exist Python’s
lunch. (Here we won’t consider $\gamma>1$ because it may induce additional
post-selection on the second layer.) Now comparing the area of the circle and
the corresponding area in the bubble phase in section C, we can find that the
bubble-inside-horizon phase corresponds to $\gamma<1$, while the bubble-
outside-horizon phase corresponds to $\gamma=1$. And we require tensors in the
weak measurement should also be perfect tensors Pastawski_2015 , which will be
helpful to compute the geodesic and minimal surface. Now we have constructed a
tensor network to realize two phases and Python’s lunch.

Intuitively, from the discussion before, we expect that for $\gamma<1$ the
complexity will be exponentially large, but for $\gamma=1$ it cannot be larger
than a power law. Now we discuss the complexity of figure 35 in details.

Firstly, we discuss the ways of contracting tensors in the tensor network. As
the definition of complexity, we must decompose the tensor network to many
small unitary tensors and apply them to initial qubits one by one. For each
tensor, there are two ways to contract them. One is applying post-selection,
and we choose the maximally entangled state, which can be considered as a
straight line in tensor network and shown in figure 36 (first panel). This
method doesn’t require other condition. Although this method is general, we
may have more post-selection and lead to larger complexity. There is another
way of contracting the tensor by applying it as a unitary transformation.
Actually, we can always divide the legs of tensor into two parts to consider
it as a unitary transformation. For example, when we have a tensor with four
bonds and their dimensions are $(1,\beta,\beta,\gamma)$, we can still consider
it as a unitary tensor by dividing them into two equal parts with bond
dimension $\beta+\frac{1+\gamma}{2}$. However, although a tensor can be seen
as a unitary tensor, we may have additional constraints. Considering one
tensor in the measurement part, because of the rotation symmetry, we require
that two bonds on the tangent direction (angular direction) will be one “in”
and one “out” with the same bond dimension. Besides, tensors in the first
layer will be connected to the boundary of MERA, and the corresponding bond
should also be “in”. Therefore, if we require $\gamma\neq 1$, then the tensor
is not a unitary transformation because the total bond dimension of in and out
are different. It is the reason why we add additional dangling bonds on each
site, and their bond dimension is to be determined to minimize the post-
selection. Because additional bonds will increase the post-selection times, it
is a trade-off for the complexity. To summarize, for the first method we don’t
need additional bonds but post-selection to get maximally entangled state,
while in the second method we need add some additional dangling bonds but may
have less post-selection. So we need to compare them and choose the smaller
one.

Figure 36:  Diagram of equivalence of post-selection on maximally mixed state
and general post-selection on ancilla with dimension $D^{1-\gamma}$.

As shown in figure 36, we first consider the tensors with bond dimension
$(1,\beta,\beta,\gamma)$ in the first layer that connect to the boundary of
MERA. Because of the rotation symmetry, we don’t need to consider two tangent
bonds with dimension $\beta$. If we naively apply post-selection on the bond
with dimension $1$ (to connect to the boundary of MERA), the Hilbert space of
post-selection is $D^{1}\cdot D^{1}=D^{2}$. However, we can use another way
with the property of perfect tensors to reduce the dimension of post-
selection. We can divide the total bond into two equal parts with dimension
$\frac{1+\gamma}{2}<1$. Then we separate the bond with dimension $1$ into two
groups with dimension $\frac{1+\gamma}{2}$ and $\frac{1-\gamma}{2}$, and apply
the tensor as a unitary transformation on the bond with dimension
$\frac{1+\gamma}{2}$. Now what we need to do is applying post-selection on the
left bond with dimension $\frac{1-\gamma}{2}$, which corresponds to the
Hilbert space with dimension $D^{\frac{1-\gamma}{2}}\cdot
D^{\frac{1-\gamma}{2}}=D^{1-\gamma}<D^{2}$. It means that with the first
method we must at least apply post-selection on Hilbert space with dimension
$D^{1-\gamma}$, which is shown in figure 36 (second panel). Now we consider
the second method, where we add a dangling bond with dimension $\alpha$.
Because we want to apply tensor as a unitary transformation on the boundary of
MERA, it requires $\alpha\geq 1-\gamma$. Therefore, with a dangling bond with
dimension $\alpha=1-\gamma$, we must apply post-selection on this bond after
connecting the tensor to MERA boundary as a unitary transformation. It means
the dimension of Hilbert space under post-selection is
$D^{\alpha}=D^{1-\gamma}$, which is shown in figure 36 (fourth panel).
Therefore, we find that for both methods, we must at least apply post-
selection on a Hilbert space with dimension $D^{1-\gamma}$. This equivalence
is shown in figure 36. We can bind two bonds with dimension
$\frac{1-\gamma}{2}+\frac{1-\gamma}{2}$ under post-selection to maximally
entangled state in the first method, and consider them as the additional
dangling bond we add in the second method. This result is also consistent with
the naive argument before, where we consider the times of post-selection as
$(D^{1-\gamma})^{N}$ and $N$ is the number of dangling bonds on one layer.

Besides, the first and last tensors we add for each layer are special, where
the bond with dimension $\beta$ is vital. It is shown in figure 35 that, for
the first layer, we start to construct the network from the tensor located at
A. For this tensor, the bond with dimension $1$ is “in”, and the bonds with
dimension $\beta,\beta$ and $\gamma$ are “out”. It means there are two
additional out bonds. So we may use the dangling bonds to decrease the
dimension of post-selection. For example, if $1+\alpha>\gamma+2\beta$, we can
divide the dangling bond into two parts with dimension
$(-1+\alpha+\gamma+2\beta)/2$ and $(1+\alpha-\gamma-2\beta)/2$ as “in” and
“out” bonds. Then the total dimension of “in” and “out” bond are the same.
Therefore, the first tensor needs post-selection of Hilbert space with
dimension $D^{\frac{(1+\alpha-\gamma-2\beta)}{2}}$. For
$1+\alpha<\gamma+2\beta$, a naive method is directly applying post-selection
on the bond with dimension $1$ to maximally entangled state (without
introducing additional ancilla). For the last tensor located on $B$, we will
find that after connecting it to the whole network, there are two bonds with
dimension $\beta$ left. To connect them, we need to apply post-selection on
them. Therefore, for two layers there are additional post-selections,
nevertheless, the complexity resulted from these post-selections won’t scale
with the total number of qudits in one layer.

The discussion before actually is only valid for the first layer with
$\gamma<1$. For the second layer, the method of connecting tensors is similar.
With rotation symmetry, we focus on the tensor except the first and last one,
and do not need to consider two tangent bonds with dimension $\beta$. Similar
to the first layer, we can set $\alpha^{\prime}=\alpha=1-\gamma$. Then we can
consider two bonds with dimension $\gamma$ and $\alpha^{\prime}$ as “in” bond
and the bond with dimension $1$ as “out” bond. Therefore, we can connect the
tensor by regarding it as a unitary transformation. The difference between the
first and second layers is that, for the first layer, the dangling bonds act
as the Hilbert space under post-selection, while for the second layer, the
dangling bonds act as ancilla qudits. These ancilla qudits are coupled to the
system without the need of post-selection. Besides, the post-selection is also
needed for the first and last tensors, which doesn’t scale with the system
size.

To summarize, for $\gamma<1$, which corresponds to the Python’s lunch
geometry, the complexity of the tensor network is about $\mathcal{C}(U)={(\rm
const.)}\cdot D^{(1-\gamma)N}$, where $N$ is the total system size (the number
of qudits on the boundary), and ${(\rm const.)}$ includes the contribution of
complexity from MERA (and the first and last tensors in two external layers
that do not scale with the system size). While, for $\gamma=1$, which
corresponds to the bubble-outside-horizon phase without Python’s lunch, the
complexity is $\mathcal{C}(U)={(\rm const.)}^{\prime}$ which only comes from
the contribution of MERA (and connecting the first and last tensors in the
angular direction in two external layers).

There are some remarks about the discussion above. (i) Here, we require the
tensors in the tensor network realization of Python’s Lunch to be perfect
tensors. It is because this requirement will be useful when we compute the
geodesic length and regard tensors as unitary transformations. For example,
with Python’s lunch geometry, when we consider the geodesic bounded by two end
points on the boundary, there is a local minimal which is shown in figure 35
with label (b). The length of this geodesic is $d=(\log{D^{\gamma}})\cdot L$
where $L$ is the size of the subsystem, and it corresponds to volume law
entanglement. But the minimal geodesic is labeled in figure 35 by (a) with
length $d=\log{D}\cdot\log{L}+{\rm const.}$ where $\log{L}$ and ${\rm const.}$
terms correspond to the contribution in the AdS region and the boundary region
corresponding to the measurements. And it leads to the area law entanglement.
These geodesics can be gotten with a greedy algorithm for perfect tensor in
reference Pastawski_2015 . For $\gamma=1$, there is no local minimal
geodesics, and the only one has the length $d=\log{D}\cdot\log{L}+{\rm
const.}$. These properties for $\gamma<1$ and $\gamma=1$ are consistent with
the results in section C. (ii) Here we use MERA in reference Haghshenas_2022
figure 2 (b), where we can consider gray circles as ancilla qudits, and each
tensor with two “in” legs with arrows pointing outside and two “out” legs with
arrows pointing inside.

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