Holographic boundary conformal field theory with $T\bar{T}$ deformation

A typical example for Type A $T\bar{T}$ BCFT is given by $T\bar{T}$ BCFT₂ in half flat space. According to our proposal, the bulk dual of the vacuum state for $T\bar{T}$ BCFT in half flat space is given by

	$\displaystyle ds^{2}$	$\displaystyle=\frac{l^{2}}{z^{2}}(-dt^{2}+dz^{2}+dx^{2})$		(9)
		$\displaystyle=d\rho^{2}+l^{2}\cosh^{2}{\frac{\rho}{l}}\left(\frac{-dt^{2}+du^{2}}{u^{2}}\right)\ ,\quad\rho\leq\rho_{0}\ ,\quad z\geq z_{c}\ .$

where $z=\frac{u}{\cosh\frac{\rho}{l}},x=-u\tanh\frac{\rho}{l}$. $z=z_{c}$ is the position of Dirichlet cutoff boundary and $\rho=\rho_{0}$ is the position of EOW brane. The bulk geometry Eq. (9) is the combination of Eq. (3) and Eq. (8), see Fig. 2 for an illustration. The holographic dictionary for deformation parameter $\lambda$ in $T\bar{T}$ BCFT is still given by $\lambda=\frac{8G_{N}}{l}z_{c}^{2}$. As shown in Fig. 2, the $T\bar{T}$ deformation clearly shifts the boundary from $x=0$ to $x=-z_{c}\sinh\frac{\rho_{0}}{l}$, which indicates that the boundary is deformed.

A typical example for Type B $T\bar{T}$ BCFT is given by $T\bar{T}$ BCFT₂ in AdS₂. In this case, the warped factor of background AdS₂ metric will fully suppress the bulk $T\bar{T}$ deformation, so the bulk $T\bar{T}$ deformation does not touch the boundary and will leave the BCFT boundary undeformed Jiang:2019tcq ; Brennan:2020dkw ; Deng:2023pjs . According to our proposal, the holographic dual of $T\bar{T}$ BCFT in AdS₂ is given by pure AdS₃ with a Dirichlet AdS₂ cutoff boundary and a Neumann EOW brane, they intersect with each other at asymptotic boundary. The bulk dual of the vacuum state for $T\bar{T}$ BCFT in AdS₂ is given by

$$ds^{2}=d\rho^{2}+l^{2}\cosh^{2}\frac{\rho}{l}\left(\frac{-dt^{2}+du^{2}}{u^{2}}\right)\ ,\quad-\rho_{c}\leq\rho\leq\rho_{0}\ ,$$

(10)

where $\rho=-\rho_{c}$ is Dirichlet cutoff boundary and $\rho=\rho_{0}$ is EOW brane, see Fig. 3 for an illustration.

The holographic dictionary is derived as follows.

The extrinsic curvature on Dirichlet boundary is

$$K_{ab}=\frac{\tanh{\frac{\rho_{c}}{l}}}{l}h_{ab}=\frac{\tanh{\frac{\rho_{c}}{l}}}{l}\begin{bmatrix}-\frac{l^{2}\cosh^{2}\frac{\rho_{c}}{l}}{u^{2}}&0\\ 0&\frac{l^{2}\cosh^{2}\frac{\rho_{c}}{l}}{u^{2}}\par\end{bmatrix}\ ,\qquad$$

(11)

where $h_{ab}$ is the induced metric on $\rho=-\rho_{c}$ slice. Then the Brown-York tensor is computed as

$$T_{ab}=\frac{1}{8\pi G_{N}}\left(K_{ab}-Kh_{ab}+\frac{1}{l}h_{ab}\right)$$

(12)

and the result is

$$T_{ab}=-\frac{1}{8\pi G_{N}}\frac{1-\tanh{\frac{\rho_{c}}{l}}}{l}\begin{bmatrix}-\frac{l^{2}\cosh^{2}\frac{\rho_{c}}{l}}{u^{2}}&0\\ 0&\frac{l^{2}\cosh^{2}\frac{\rho_{c}}{l}}{u^{2}}\par\end{bmatrix}\ .$$

(13)

By using (13), one can rewrite $T_{a}^{a}$ in terms of $T\bar{T}$ operator and the result is

$$T_{a}^{a}=\frac{1}{8\pi G_{N}l}\frac{1}{\cosh^{2}{\frac{\rho_{c}}{l}}}-32\pi G_{N}lT\bar{T}\ .$$

(14)

One can directly identify (14) with the $T\bar{T}$ deformed CFT trace flow equation, living in the curved spacetime

$$T_{a}^{a}=-\frac{c}{24\pi}\mathcal{R}\left[h\right]-4\pi\lambda T\bar{T}\ ,$$

(15)

where the first term comes from trace anomaly and $\mathcal{R}\left[h\right]=-\frac{2}{l^{2}\cosh^{2}{\frac{\rho_{c}}{l}}}$ is the Ricci scalar computed by using $h_{ab}$. Thus one can determine the holographic dictionary to be

$$c=\frac{3l}{2G_{N}}\ ,\quad\lambda=8G_{N}l\ .$$

(16)

To see the cutoff dependence of deformation parameter, we can observe the trace flow equation under background metric $\gamma^{ab}=\cosh^{2}{\frac{\rho_{c}}{l}}h^{ab}$ and it is

$$T_{a}^{a}=\frac{1}{8\pi G_{N}l}-\frac{32\pi G_{N}l}{\cosh^{2}{\frac{\rho_{c}}{l}}}T\bar{T}\ .$$

(17)

Then we compare it with the trace flow equation of CFT living in curved spacetime with metric $\gamma^{ab}$ and we get

$$\lambda=\frac{8G_{N}l}{\cosh^{2}{\frac{\rho_{c}}{l}}}\ .$$

(18)

Now we compute the boundary entropy of a Type A holographic $T\bar{T}$ BCFT by considering $T\bar{T}$ BCFT in a disk. The boundary entropy is given by the amplitude Affleck:1991tk

$$S_{\text{bdy}}^{\mathrm{disk}}=\log g=\log\langle 0|B\rangle\ ,$$

(19)

where $|B\rangle$ is a general boundary state ⁷⁷7Since the $T\bar{T}$ deformation continuously deforms the spectrum and states of a given CFT, we consider this definition of boundary entropy to be equally well-defined for $T\bar{T}$ BCFT. . In holographic set up we can compute boundary entropy from bulk on-shell action Takayanagi:2011zk . According to our proposal, the dual geometry of the $T\bar{T}$ BCFT in a disk is given by

$$ds_{\mathrm{bulk}}^{2}=\frac{l^{2}}{z^{2}}\left(dz^{2}+dr^{2}+r^{2}d\theta^{2}\right)\ ,\quad z\geq z_{c}\$$

(20)

with a spherical EOW brane

$$r^{2}+\left(z-r_{d}\sinh\frac{\rho_{0}}{l}\right)^{2}=r_{d}^{2}\cosh^{2}\frac{\rho_{0}}{l}\ ,$$

(21)

where $r_{d}$ is the disk radius of the BCFT on the asymptotic boundary. The brane induced metric is

$$ds_{\mathrm{brane}}^{2}=\frac{l^{2}}{z^{2}(r_{c}^{2}-z^{2}+2r_{d}z\sinh\frac{\rho_{0}}{l})}\left[r_{d}^{2}\cosh^{2}\frac{\rho_{0}}{l}dz^{2}+\left(r_{d}^{2}-z^{2}+2r_{d}z\sinh\frac{\rho_{0}}{l}\right)^{2}d\theta^{2}\right]\ ,$$

(22)

and the extrinsic curvature of the brane is

	$\displaystyle K$	$\displaystyle=-\frac{z(r)^{2}\left(\left(2r+z(r)z^{\prime}(r)\right)\left(1+z^{\prime}(r)^{2}\right)+rz(r)z^{\prime\prime}(r)\right)}{lrz(r)^{2}(1+z^{\prime}(r)^{2})^{\frac{3}{2}}}$		(23)
		$\displaystyle=\frac{2}{l}\tanh\frac{\rho_{0}}{l}$
		$\displaystyle=2T\ .$

The on-shell action of the bulk bounded by a finite tension brane is ⁸⁸8Note that when we calculate entanglement entropy by bulk on-shell action, we should not add a counter term to cancel the Weyl anomaly, which is different to the calculation of renormalized on-shell action Caputa:2020lpa ; Li:2020zjb ; Apolo:2023vnm . The reason is that entanglement entropy in quantum field theory is always divergent, and the Weyl anomaly will contribute to the entanglement entropy Tian:2023fgf .

	$\displaystyle I_{E}=$	$\displaystyle I_{\mathrm{bulk}}+I_{\mathrm{brane}}+I_{\mathrm{ct}}$		(24)
	$\displaystyle=$	$\displaystyle-\frac{1}{16\pi G}\int dx^{3}\sqrt{g}\left(R+\frac{2}{l^{2}}\right)-\frac{1}{8\pi G}\int d^{2}x\sqrt{h}(K-T)-\frac{1}{8\pi G}\int d^{2}x\sqrt{h}(K-\frac{1}{l})$
	$\displaystyle=$	$\displaystyle-\frac{1}{16\pi G}\int_{z_{c}}^{r_{d}(\cosh\frac{\rho_{0}}{l}+\sinh\frac{\rho_{0}}{l})}dz\int_{0}^{\sqrt{r_{d}^{2}\cosh^{2}\frac{\rho_{0}}{l}-(z-r_{d}\sinh\frac{\rho_{0}}{l})^{2}}}dr\frac{-4}{l^{2}}\frac{l^{3}}{z^{3}}2\pi r$
		$\displaystyle-\frac{1}{8\pi G}\int_{z_{c}}^{r_{d}(\cosh\frac{\rho_{0}}{l}+\sinh\frac{\rho_{0}}{l})}dz\frac{1}{l}\tanh\frac{\rho_{0}}{l}2\pi\frac{l^{2}}{z^{2}}r_{d}\cosh\frac{\rho_{0}}{l}$
		$\displaystyle-\frac{1}{8\pi G}\int_{0}^{\sqrt{r_{d}^{2}\cosh^{2}\frac{\rho_{0}}{l}-(z_{c}-r_{d}\sinh\frac{\rho_{0}}{l})^{2}}}dr\frac{1}{l}2\pi r\frac{l^{2}}{z_{c}^{2}}$
	$\displaystyle=$	$\displaystyle-\frac{l}{4G}\left(\frac{\rho_{0}}{l}+\log\frac{r_{d}}{z_{c}}\right)\ .$

As in Ref. Takayanagi:2011zk , we assume that the boundary entropy of the boundary state dual to a zero tension brane is zero, meaning that we need to subtract the on-shell action within a reference zero tension brane to calculate the holographic boundary entropy. For $T\bar{T}$ BCFT in a disk, the boundary of the disk will shift in the AdS bulk along the $T\bar{T}$ deformation. Then we have two choices for the reference zero tension brane. The first choice is the zero tension brane of the BCFT, as shown in Fig. 4. This choice implies that we treat the field theory degrees of freedom on the cutoff disk, located between the BCFT zero tension and finite tension brane, as boundary degrees of freedom. Here, the boundary is defined as the intersection of the cutoff disk and the zero tension brane.

The second choice is a new zero tension brane which intersects with the $T\bar{T}$ BCFT boundary, as shown in Fig. 5. This choice implies we regard the field theory degrees of freedom on the cutoff disk, located between the BCFT zero tension and finite tension brane, as bulk degrees of freedom. Now we calculate the boundary entropy separately for each of these two choices.

For the first choice, the on-shell action within the BCFT zero tension brane is

$$I_{E}|_{\rho_{0}\rightarrow{0}}=-\frac{l}{4G}\log\frac{r_{d}}{z_{c}}\ ,$$

(25)

Then the boundary entropy is given by disk partition function as

$$S_{\mathrm{bdy}}^{\mathrm{disk}}=-(I_{E}-I_{E}|_{\rho_{0}\rightarrow{0}})=\frac{\rho_{0}}{4G}\ .$$

(26)

For a BCFT, the boundary entropy can also be extracted from entanglement entropy as

$$S_{\mathrm{bdy}}=S^{\mathrm{BCFT}}([0,L])-\frac{1}{2}S^{\mathrm{CFT}}([-L,L])\ ,$$

(27)

where $S^{\mathrm{BCFT}}([0,L])$ is the vacuum entanglement entropy of interval $[0,L]$ for a BCFT in half flat space and $S^{\mathrm{CFT}}([-L,L])$ is the vacuum entanglement entropy of the CFT in the full flat space. According to AdS/BCFT, we find that this implies the holographic boundary entropy is simply given by the minimal surface $\gamma_{I}$ between the zero-tension brane and the finite-tension brane, i.e.

$$S^{\mathrm{RT}}_{\mathrm{bdy}}=\frac{\mathrm{Area}(\gamma_{I})}{4G}\ .$$

(28)

In our case of $T\bar{T}$ BCFT in a disk, the minimal surface $\gamma_{I}$ between the finite-tension brane and the new zero-tension brane is determined by rotational symmetry, as shown in Fig. 4. The boundary entropy computed by $\gamma_{I}$ is

$$S_{\mathrm{bdy}}^{\mathrm{RT}}=\frac{1}{4G}\int_{r_{d}}^{r_{d}(\cosh\frac{\rho_{0}}{l}+\sinh\frac{\rho_{0}}{l})}dz\frac{l}{z}=\frac{\rho_{0}}{4G}\ .$$

(29)

We find that this result agrees with the one obtained from the disk partition function. Notice that in this case, the boundary entropy is the same as BCFT boundary entropy without $T\bar{T}$ deformation. This indicates in Type A $T\bar{T}$ BCFT, if we regard the degree of freedom in the extended interval between BCFT zero tension brane and finite tension brane as boundary degrees of freedom, then the total boundary entropy will be conserved along $T\bar{T}$ deformation.

To quantify the effect of $T\bar{T}$ deformation on the boundary, we take the second choice. In this case, we need to solve the new zero tension brane which intersects with the $T\bar{T}$ BCFT boundary, as shown in Fig. 5.

The on-shell action of the bulk bounded by the new zero tension brane is

$$I_{E}|_{\rho_{0}\rightarrow{0},r_{d}\rightarrow{\sqrt{r_{d}(r_{d}+2z_{c}\sinh\frac{\rho_{0}}{l})}}}=-\frac{l}{8G}\log\frac{r_{d}(r_{d}+2z_{c}\sinh\frac{\rho_{0}}{l})}{z_{c}^{2}}\ .$$

(30)

Then the boundary entropy is obtained as

	$\displaystyle S_{\mathrm{bdy}}^{\mathrm{disk}}$	$\displaystyle=-(I_{E}-I_{E}|_{\rho_{0}\rightarrow{0},r_{d}\rightarrow{\sqrt{r_{d}(r_{d}+2z_{c}\sinh\frac{\rho_{0}}{l})}}})$		(31)
		$\displaystyle=\frac{\rho_{0}}{4G}-\frac{l}{8G}\log(1+\frac{2z_{c}}{r_{d}}\sinh\frac{\rho_{0}}{l})\ .$

In this case, we observe that the boundary entropy is no longer constant; it depends on the $T\bar{T}$ deformation parameter, or equivalently, on $z_{c}$. This result clearly indicates that the boundary of the BCFT is deformed, and we can use boundary entropy to quantify the amount of boundary deformation. The boundary entropy can also be computed by the minimal surface $\gamma_{I}$ between finite tension brane and the new zero tension brane, as shown in Fig. 5. The result is

$$S_{\mathrm{bdy}}^{\mathrm{RT}}=\frac{1}{4G}\int_{\sqrt{r_{d}(r_{d}+2z_{c}\sinh\frac{\rho_{0}}{l})}}^{r_{d}(\cosh\frac{\rho_{0}}{l}+\sinh\frac{\rho_{0}}{l})}dz\frac{l}{z}=\frac{\rho_{0}}{4G}-\frac{l}{8G}\log(1+\frac{2z_{c}}{r_{d}}\sinh\frac{\rho_{0}}{l})\ .$$

(32)

We can see it agrees with the result obtained by disk partition function.

Now we compute the energy spectrum for a $T\bar{T}$ BCFT in a finite interval. Let us start from the bulk dual of a high temperature CFT state, which is the BTZ black hole

$$ds^{2}=l^{2}\left(\frac{f(z)}{z^{2}}d\tau^{2}+\frac{dz^{2}}{f(z)z^{2}}+\frac{dx^{2}}{z^{2}}\right)\ ,$$

(33)

where $f(z)=1-\left(\frac{z}{z_{H}}\right)^{2}$, $\tau\sim\tau+2\pi z_{H}$, the temperature of the black hole is $T_{\mathrm{CFT}}=\frac{1}{2\pi z_{H}}$. We begin by first considering the addition of a conformal boundary to the CFT at $(x,z)=(0,0)$ and solving for the EOW brane. By assuming constant tension and solving the Neumann boundary condition, the trajectory of the EOW brane is determined to be Takayanagi:2011zk

$$x=-z_{H}\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)\ .$$

(34)

Consequently, after the Type A deformation, the field theory exists on the cutoff boundary $z=z_{c}$, and the boundary of the field theory shifts from $(x,z)=(0,0)$ to $(x,z)=(-z_{H}\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right),z_{c})$. The corresponding bulk picture is illustrated in F.G.6.

To compute the energy spectrum of Type A $T\bar{T}$ BCFT in finite interval, we consider the spatial coordinate $x$ to be periodic $x\sim x+2\pi l$ and set $x=l\theta$ where $-\pi\leqslant\theta\leqslant\pi$. We add two boundaries to the original CFT. One sits at $x=0$, the other locates at $x=l\pi$, as shown in Fig. 7. We aim to match the energy spectrum for $T\bar{T}$ BCFT bulk.

We first focus on deriving the energy spectrum from the gravity side. To start, we recall the gravity calculation of energy spectrum without the boundary. The energy can be straightforwardly obtained by first calculating the energy density from the Brown-York tensor of the cutoff boundary

$$e=u^{i}u^{j}T_{ij}=\frac{u^{i}u^{j}}{8\pi G}\left(K_{ij}-Kh_{ij}+h_{ij}\right)\ ,$$

(35)

where $u^{i}$ is the unit vector that is normal to the constant $\tau$ slice of the cutoff boundary

$$u^{\tau}=\frac{z_{c}}{l\sqrt{f(z_{c})}}\ ,\quad u^{\theta}=0\ .$$

(36)

Then we get

$$\begin{split}e=\frac{1}{8\pi Gl}\left(1-\sqrt{f(z_{c})}\right)=\frac{1}{8\pi Gl}\left(1-\sqrt{1-\frac{z_{c}^{2}}{z_{H}^{2}}}\right)=\frac{1}{8\pi Gl}\left(1-\sqrt{1-\frac{8G_{N}Mz_{c}^{2}}{l^{2}}}\right)\ ,\end{split}$$

(37)

where we have used $z_{H}^{2}=\frac{l^{2}}{8G_{N}M}$. The energy can be obtained by subsequently integrating over the spatial region. The result is

$$E=\int_{0}^{2\pi}d\theta\sqrt{g_{\theta\theta}}e=\frac{l}{4G_{N}z_{c}}\left(1-\sqrt{1-\frac{8G_{N}Mz_{c}^{2}}{l^{2}}}\right)\ .$$

(38)

By multiplying the energy with the proper length $L=\int_{0}^{2\pi}d\theta\sqrt{g_{\theta\theta}}=\frac{2\pi l^{2}}{z_{c}}$, one can further get the dimensionless quantity

$$\mathcal{E}=EL=\frac{\pi l^{3}}{2G_{N}z_{c}^{2}}\left(1-\sqrt{1-\frac{8G_{N}Mz_{c}^{2}}{l^{2}}}\right)\ .$$

(39)

Next we consider the energy spectrum after the EOW brane is introduced. We note that the energy density remains unchanged, as the boundary does not affect the Brown-York tensor; only the spatial region changes due to the introduction of boundaries. Thus we can get the energy with boundaries as

$$\begin{split}E&=\int^{\pi+z_{H}/l\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)}_{-z_{H}\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)}d\theta\frac{l^{2}}{z_{c}}e\\ &=\int^{\pi+z_{H}/l\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)}_{-z_{H}\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)}\frac{l}{8\pi G_{N}z_{c}}\left[1-\sqrt{1-\frac{8G_{N}Mz_{c}^{2}}{l^{2}}}\right]d\theta\\ &=\frac{l}{8\pi G_{N}z_{c}}\left[\pi+2z_{H}/l\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)\right]\cdot\left[1-\sqrt{1-\frac{8G_{N}Mz_{c}^{2}}{l^{2}}}\right]\ .\end{split}$$

(40)

The bulk picture tells us that the boundaries only truncate the spatial circle while preserving the energy density. This suggests we can quantify their impact by recognizing that the effect is simply to rescale the spatial length. The rate of rescaling is

$$\begin{split}f(\lambda)\equiv\frac{1}{2\pi}\left[\pi+\frac{2z_{H}}{l}\operatorname{arcsinh}\left(\frac{lTz_{c}}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)\right]=\frac{1}{2}+\frac{z_{H}}{\pi l}\operatorname{arcsinh}\left(\frac{lT\sqrt{\frac{\lambda l}{8G_{N}}}}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)\ .\end{split}$$

(41)

Finally we can multiply the energy with the spatial length to get the dimensionless quantity

$$\begin{split}\mathcal{E}&=E\cdot\int^{\pi+z_{H}/l\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)}_{-z_{H}\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)}d\theta\frac{l^{2}}{z_{c}}\\ &=\frac{l^{3}}{8\pi G_{N}z_{c}^{2}}\left[\pi+2z_{H}/l\cdot\operatorname{arcsinh}\left(\frac{lTz}{z_{H}\sqrt{1-l^{2}T^{2}}}\right)\right]^{2}\cdot\left[1-\sqrt{1-\frac{8G_{N}Mz_{c}^{2}}{l^{2}}}\right]\ .\end{split}$$

(42)

Now let us compute the energy spectrum from the field theory side. Without boundaries, the spatial length of field theory is $L=2\pi l$. After adding boundaries, the spatial length of deformed theory is $L_{b}=2\pi lf(\lambda)=Lf(\lambda)$, where $f(\lambda)$ is the rescaling rate. The stress energy tensor of $T\bar{T}$ BCFT in finite interval have zero momentum, i.e. $T_{\tau x}=T_{x\tau}=0$ Cardy:2018sdv ; Brizio:2024doe . Then we can separately compute the expectation value of $T\bar{T}$ operator as

$$\begin{split}\langle n|T\bar{T}|n\rangle&=\frac{1}{8}\langle n|T|n\rangle\langle n|\bar{T}|n\rangle-\frac{1}{8}\langle n|\Theta|n\rangle\langle n|\Theta|n\rangle\\ &=-\frac{1}{4}\left(\left\langle n\left|T_{\tau\tau}\right|n\right\rangle\left\langle n\left|T_{xx}\right|n\right\rangle\right)\ ,\end{split}$$

(43)

and expectation value of components of stress tensor is

$$\begin{split}\left\langle n\left|T_{\tau\tau}\right|n\right\rangle&=\frac{E_{n}}{L_{b}}=\frac{E_{n}}{Lf(\lambda)}\ ,\\ \left\langle n\left|T_{xx}\right|n\right\rangle&=\frac{\partial E_{n}}{\partial L_{b}}=\frac{\partial E_{n}}{\partial L}\cdot\frac{1}{f(\lambda)}\ .\end{split}$$

(44)

From the definition of $T\bar{T}$ deformation

$$S_{T\bar{T}}=2\pi\lambda\int d^{2}xT\bar{T}\ ,$$

(45)

we have

$$H_{T\bar{T}}=2\pi\lambda\int^{Lf(\lambda)}_{0}d\theta\ T\bar{T}\ ,$$

(46)

thus

$$\begin{split}\frac{\partial}{\partial\lambda}\langle H_{T\bar{T}}\rangle&=2\pi Lf(\lambda)\langle T\bar{T}\rangle+2\pi L\lambda f^{{}^{\prime}}(\lambda)\langle T\bar{T}\rangle|_{Lf(\lambda)}\\ &=2\pi Lf(\lambda)\langle T\bar{T}\rangle+\frac{f^{{}^{\prime}}(\lambda)}{f(\lambda)}\langle H_{T\bar{T}}\rangle\ ,\end{split}$$

(47)

where we have used $\langle H_{T\bar{T}}\rangle=2\pi\lambda Lf(\lambda)\langle T\bar{T}\rangle|_{Lf(\lambda)}=2\pi\lambda Lf(\lambda)\langle T\bar{T}\rangle$ from (46) in the second equality⁹⁹9The $T\bar{T}$ operator in differential equation (47) is the operator before deformation, it does not depend on the deformation parameter $\lambda$ and the position. Thus we have $\langle T\bar{T}\rangle=\langle T\bar{T}\rangle|_{Lf(\lambda)}$.. Then we get

$$\frac{\partial E_{n}}{\partial\lambda}=2\pi Lf(\lambda)\langle T\bar{T}\rangle+\frac{f^{{}^{\prime}}(\lambda)}{f(\lambda)}E_{n}\ .$$

(48)

The final equation becomes

$$2\frac{\partial E_{n}}{\partial\lambda}-\frac{2f^{{}^{\prime}}(\lambda)}{f(\lambda)}E_{n}+\frac{\pi E_{n}}{f(\lambda)}\frac{\partial E_{n}}{\partial L}=0\ .$$

(49)

The solution is

$$E_{n}=\frac{Lf(\lambda)}{\pi\lambda}\left[1-\sqrt{1-\frac{4\pi^{2}\lambda M_{n}}{L^{2}}}\right]\ ,$$

(50)

where $M_{n}=\Delta_{n}+\bar{\Delta}_{n}-\frac{c}{12}$. By multiplying $E_{n}$ with $Lf(\lambda)$, one can therfore get the dimensionless quantity

$$\mathcal{E}_{N}=\frac{L^{2}f(\lambda)^{2}}{\pi\lambda}\left[1-\sqrt{1-\frac{4\pi^{2}\lambda M_{n}}{L^{2}}}\right]\ ,$$

(51)

This result agrees with the gravity outcome under the identification $M_{n}=M\cdot l$ and the holographic dictionary $\lambda=\frac{8G}{l}z_{c}^{2}$.

Now we compute the entanglement entropy (EE) for a $T\bar{T}$ BCFT in AdS₂. We consider the EE of an interval $A$ with $u\in[0,1]$ and $t=0$ in the cutoff boundary, as shown in Fig. 8.

This interval is special as it becomes a radius of the Euclidean AdS₂, making the replica geometry simple. To be specific, the Euclidean AdS₂ is represented as a hyperbolic disk

$$\begin{split}ds_{\mathbb{H}^{2}}^{2}=r^{2}\frac{dt_{E}^{2}+du^{2}}{u^{2}}=r^{2}\left(d\eta^{2}+\sinh^{2}\eta d\phi^{2}\right)\ ,\\ \end{split}$$

(52)

where

$$t_{E}=\frac{\sinh\eta\sin\phi}{\cosh\eta+\cos\phi\sinh\eta}\;\;\;,\;\;\;u=\frac{1}{\cosh\eta+\cos\phi\sinh\eta}\ ,$$

(53)

and $r=l\cosh\frac{\rho_{c}}{l}$ is the AdS radius of the cutoff boundary. This transformation is shown in Fig. 9. The metric of replica space along interval $A$ is

$$ds^{2}_{\mathbb{H}_{n}^{2}}=r^{2}\left(d\eta^{2}+n^{2}\sinh^{2}\eta d\phi^{2}\right)\ .$$

(54)

Using replica trick, the EE of $A$ is obtained as

$$\begin{split}S(A)=\left(1-n\frac{d}{dn}\right)\log Z_{n}\bigg{|}_{n=1}=\left(1-\frac{r}{2}\frac{d}{dr}\right)\log Z\ ,\end{split}$$

(55)

where in the second line we utilize two equations involving variation of the $T\bar{T}$ partition function with respect to $n$

$$\frac{d\log Z_{n}}{dn}\bigg{|}_{n=1}=-\frac{1}{2}\int d^{2}x\sqrt{h}\;\langle T^{a}_{a}\rangle\ ,$$

(56)

and with respect to $r$

$$\frac{d}{dr}\log Z=-\frac{1}{r}\int d^{2}x\sqrt{h}\langle T^{a}_{a}\rangle\ .$$

(57)

Then according to Eq. (55), in order to obtain $S(A)$, we need to evaluate the partition function $\log Z$ for $T\bar{T}$ deformed CFT in $\mathbb{H}^{2}$. This can be accomplished due to the fact that $\mathbb{H}^{2}$ is a maximally symmetric space.

Initially, in a maximally symmetric space, the stress tensor is proportional to the metric, i.e. $\langle T_{ab}\rangle=\alpha h_{ab}$. By substituting this into the trace flow equation, we can determine $\alpha$ and obtain

$$\langle T_{ab}\rangle=\frac{1}{\pi\lambda}\left(1-\sqrt{1-\frac{c\lambda}{12r^{2}}}\right)h_{ab}\;\;\;,\;\;\;\langle T_{a}^{a}\rangle=\frac{2}{\pi\lambda}\left(1-\sqrt{1-\frac{c\lambda}{12r^{2}}}\right)\ .$$

(58)

Therefore, the variation of $\log Z$ with respect to $l$ for $\mathbb{H}^{2}$ is

$$\frac{d}{dr}\log Z=\frac{4}{\lambda}\left(r-\sqrt{r^{2}-\frac{c\lambda}{12}}\right)\ .$$

(59)

Note that we have another differential equation for $\log Z$ with respect to the deformation scale $\mu=\frac{1}{\sqrt{\lambda}}$

$$\mu\partial\mu\log Z=-2\lambda\partial_{\lambda}\log Z=-\int d^{2}x\sqrt{h}\langle T^{a}_{a}\rangle=\frac{4r}{\lambda}\left(r-\sqrt{r^{2}-\frac{c\lambda}{12}}\right)\ .$$

(60)

We can obtain $\log Z$ by integrating these two equations and imposing a proper boundary condition. We take the boundary condition to be

$$\begin{split}\log Z|_{\rho_{c}=0}&=-I_{E}|_{\rho_{c}=0}=-\frac{1}{8\pi G_{N}}\int_{\rho=0}d^{2}x\sqrt{\gamma}\frac{1}{l}-I_{\mathrm{bdy}}=\frac{l}{4G_{N}}-I_{\mathrm{bdy}}=\frac{c}{6}-I_{\mathrm{bdy}}\ ,\end{split}$$

(61)

where $I_{\mathrm{bdy}}$ is the on-shell action from $\rho=0$ to $\rho=\rho_{0}$, which is related to the boundary entropy as $I_{\mathrm{bdy}}=-S_{\mathrm{bdy}}=-\frac{\rho_{0}}{4G_{N}}$. Using this boundary condition we obtain $\log Z$

$$\log Z=\frac{c}{6}\log\left[\frac{r}{a}\left(1+\sqrt{1-\frac{c\lambda}{12r^{2}}}\right)\right]-\frac{2r^{2}}{\lambda}\sqrt{1-\frac{c\lambda}{12r^{2}}}+\frac{2r^{2}}{\lambda}+S_{\mathrm{bdy}}\ ,$$

(62)

where $a=\sqrt{\frac{c\lambda}{12}}$ is a finite cutoff of the $T\bar{T}$ deformed theory. We can verify this boundary condition by evaluating the bulk Euclidean on-shell action and the result is

$$\begin{split}I_{E}&=-\frac{1}{16\pi G_{N}}(-2\pi)\int_{0}^{\rho_{c}}d\rho\;l^{2}\cosh^{2}\frac{\rho}{l}(-\frac{4}{l^{2}})\\ &-\frac{1}{8\pi G_{N}}(-2\pi)l^{2}\cosh^{2}\frac{\rho_{c}}{l}(\frac{2}{l}\tanh\frac{\rho_{c}}{l}-\frac{1}{l})+I_{\mathrm{bdy}}\\ &=-\frac{l}{8G_{N}}\left(1+e^{-\frac{2\rho_{c}}{l}}+\frac{2\rho_{c}}{l}\right)\ +I_{\mathrm{bdy}}\ .\end{split}$$

(63)

We find that it exactly equals to $-\log Z$ in Eq. (62) with $a=\sqrt{\frac{c\lambda}{12}}$ and Eq. (16).