††institutetext: Center for Joint Quantum Studies and Department of Physics,
School of Science, Tianjin University, Yaguan Road 135, Jinnan District,
300350 Tianjin, P. R. China

# Holographic quantum distances and replica trick

Zi-Qing Xiao and Run-Qiu Yang<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

This paper gives concrete examples to exhibit how to use the replica trick to
calculate the quantum (quasi-)distances holographically. First, we consider
the fidelity and relative entropy between thermal states that are dual to the
Schwarzschild-AdS black holes. Then we generalize our method into the RN-AdS
black holes by adding a U(1) gauge field. We also investigate the fidelity
between states excited by scalar operator in probe limit. In this case, it is
surprising that the fidelity in standard quantization will suffer from new UV
divergence though the usual holographic renormalization has been applied. We
call for deep understanding for such divergence in the future. We also
discover a holographic method to check whether the density matrices of two
holographic states are commutative.

##  1 Introduction

In recent years, quantum information theoretic considerations play an
important role in the study of AdS/CFT or more general gauge/gravity duality
Maldacena:1997re ; Gubser:1998bc ; Witten:1998qj . For instance, Ryu-
Takayanagi (RT) formula Ryu:2006bv ; Hubeny:2007xt  quantifies the
entanglement entropy of the boundary conformal field theories (CFTs), which is
given by the area of a codimension-2 minimal surface in the dual bulk
spacetime. Under some reasonable assumptions, RT formula can be obtained from
the more general Rényi entropy Fursaev:2006ih ; Headrick:2010zt ;
Lewkowycz:2013nqa ; Nakaguchi:2016zqi

$S_{n}\equiv\frac{1}{1-n}\ln\frac{\text{tr}\rho^{n}}{(\text{tr}\rho)^{n}}$

by applying the _replica trick_. Here the index $n$ labels the number of the
replicated boundary CFTs and $\rho$ is the density matrix of the entangling
region. The $n$ is originally assumed to be a positive integer value. In
Lewkowycz:2013nqa , Lewkowycz and Maldacena argued that, for integer index
$n>1$, the CFTs on a branched cover111See Ref. Callan:1994py ; Holzhey:1994we
for how to construct $\mathcal{M}_{n}$ by taking $n$ copies of
$\mathcal{M}_{1}$. The key points here are making suitable identifications for
quantum fields between these copies and obeying right boundary conditions. The
latter, in other words, requires us to regulate the path integral of
$\mathcal{M}_{n}$ by imposing the right field configurations at the
boundaries. $\mathcal{M}_{n}$ over the entangling region still have
corresponding interior gravity duals $\mathcal{B}_{n}$ like the original
Euclidean spacetime $\mathcal{B}_{1}$ does. In the gravity side, there exists
a solution $B_{n}$ that satisfies the Einstein equation along with boundary
condition $\partial\mathcal{B}_{n}=\mathcal{M}_{n}$. Based on the holographic
duality, one should expect that the partition functions of two sides’ theories
are equated each other

$Z[\mathcal{M}_{n}]=Z[\mathcal{B}_{n}]\,,$ (1)

where $Z[\mathcal{M}_{n}]=\text{tr}\rho^{n}$. Note that we are working in the
large $N$ limit such that the bulk geometry can be taken as a solution of
Einstein gravity. Under the saddle point approximation, the partition function
of bulk geometry can be written as the exponential function of corresponding
Einstein-Hilbert action

$Z[\mathcal{M}_{n}]=Z[\mathcal{B}_{n}]\approx\exp(-I_{\text{bulk}}[\mathcal{B}_{n}])\,.$
(2)

Following Rényi entropy’s definition, the direct calculation

$S_{n}=\frac{1}{1-n}\ln\frac{\text{tr}\rho^{n}}{(\text{tr}\rho)^{n}}=\frac{1}{n-1}\left(I_{\text{bulk}}[\mathcal{B}_{n}]-nI_{\text{bulk}}[\mathcal{B}_{1}]\right)$

indicates that Rényi entropy is proportional to the gravitational action of
corresponding bulk geometry. If it is possible to analytically continue the
Rényi entropy $S_{n}$ to non-integer222The Carlson’s theorem shows that such
analytical continuation is unique if it exists. $n$, the von Neumann entropy
that is usually referred as the entanglement entropy in quantum information
theory can be recovered from $S_{n}$ at the $n\rightarrow 1$ limit.

The entanglement entropy concerns about the entanglement property of only one
quantum state but not involves the relationship between two states. There are
other quantities in quantum information nielsen_chuang_2010 , which measures
how close two quantum states are. In other words, these quantities label some
kind of “distance” between quantum states. For example, given quantum states
$\rho$ and $\sigma$, the _trace distance_ is defined as

$D[\rho,\sigma]\equiv\frac{1}{2}\text{tr}\left|\rho-\sigma\right|\,,$ (3)

where $\text{tr}|A|\equiv\sum_{i}\lambda_{i}$ and $\lambda_{i}$ is the
eigenvalue of $\sqrt{A^{\dagger}A}$. Introducing a positive number $n$,
$D[\rho,\sigma]$ can be generalized into $D_{n}[\rho,\sigma]$

$D_{n}[\rho,\sigma]\equiv\frac{1}{2^{1/n}}\sqrt[n]{\text{tr}\left|\rho-\sigma\right|^{n}}\,.$
(4)

For $n=1$, the above formula is just the trace distance. There is other
special choice that is widely applied. By choosing $n=2$, _Hilbert-Schmidt
distance_ can be obtained, which is most convenient $D_{n}$ for calculations.
The second families of quantum distance are based on the _fidelity_

$\text{Fi}[\rho,\sigma]\equiv\text{tr}\sqrt{\rho^{\frac{1}{2}}\sigma\rho^{\frac{1}{2}}}\,.$
(5)

The fidelity is not a distance but it can be used to define other measures of
distance: the _Fubini-Study distance_ $D_{F}[\rho,\sigma]$ and the _Bures
distance_ $D_{B}[\rho,\sigma]$

$D_{F}[\rho,\sigma]\equiv\arccos\text{Fi}[\rho,\sigma],~{}~{}~{}D_{B}[\rho,\sigma]\equiv\sqrt{1-\text{Fi}[\rho,\sigma]^{2}}\,.$
(6)

Due to the well-defined calculations in finite dimensional Hilbert spaces and
the preservation under unitary transformations, trace distance and fidelity
are widely used within the quantum information community. However, their
analytical calculations are extremely difficult in quantum field theory. Ref.
Lashkari:2014yva  makes the first step towards this issue. By virtue of the
certain correlation functions for the vacuum or the thermal state, author of
Ref. Lashkari:2014yva  develop a replica trick to calculate the fidelity for
$(1+1)$ dimensional CFTs. Using similar method, Refs. Zhang:2019wqo ;
Zhang:2019itb ; Zhang:2019kwu  also developed replica method to compute the
trace distance for a class of special states for single short interval in 1+1
dimensional CFTs. Recently, Ref. Kudler-Flam:2022zgm  proposes how to use
replica trick to compute Rényi mutual information in 1+1 CFTs. Though in
higher dimensions the replica trick is still effective, the correlation
functions of interacting theories are highly nontrival to compute. Similar to
holographic descriptions of entanglement, one would wish there exist a
holographic duality to calculate trace distance and fidelity. Actually Ref.
Miyaji:2015woj  argues that the fidelity between a state and its infinitesimal
perturbation state is approximately given by a volume of maximal time slice in
AdS spacetime. However, when the difference between two states is not
infinitesimal, the gravity dual of fidelity given by Ref. Miyaji:2015woj  is
not reliable. In addition, the holographic proof on the proposal of Ref.
Miyaji:2015woj  is absent. See Colin-Ellerin:2020mva ; Dong:2021clv ; Colin-
Ellerin:2021jev ; Dong:2021oad ; Penington:2022dhr  for recent progress
related to replicas in quantum information under the holographic setup.

Although it seems quite difficult to solve this problem completely, this paper
uses some concrete examples to exhibit how to calculate quantum distances in
holography. As a basic perspective in holographic picture, Maldacena argues
that the thermal states in the boundary CFT correspond to Schwarzschild-AdS
black holes. Firstly, similar to the deduction of Rényi entropy by employing
the replica trick, we analytically calculate the fidelity and relative entropy
between thermal states in holography. Next, we want to check our method in the
presence of bulk matters. Here we consider two concrete examples. In the first
one we consider that there is a conserved charge in the boundary theory so the
bulk theory is the Einstein-Maxwell theory. In the second example, an
Einstein-scalar theory is constructed here to gain the insight into the
fidelity between states excited by scalar operator. In probe limit, we derive
the analytic expression of fidelity in this case. Surprisingly, we find that
the fidelity of probe limit is well defined only in the alternative
quantization. The fidelity in standard quantization will suffer from new UV
divergency though the usual holographic renormalization has been applied. To
the best of our knowledge, such kind of divergency has not been discussed well
in the study of holography, and so we call for deep understanding in the
future.

In a physical system described by quantum mechanics, quantum states are
generally the superposition of several eigenstates under a observable
Hermitian operator. It is more troublesome but feasible to check the
commutativity of two quantum mechanics states. As we claimed before, quantum
states in infinite dimensional Hilbert spaces are more tricky. Without more
assumptions, it is difficult to directly write the density matrix for a
general quantum state in quantum field theory. So it is nontrivial question to
determine the commutativity of two quantum states in quantum field theory. At
the end of this paper, we propose a method to answer this question partially
if two quantum states are holographic. We will see whether two holographic
states are commutative or not is related the contribution of corresponding
bulk theory’s action.

The organization of this paper is as follows. In section 2, we consider the
fidelity and relative entropy between thermal states with different
temperatures. In section 3, under the same procedure, we generalize the above
method into the grand canonical ensemble, where we consider the fidelity and
relative entropy of the states which has different chemical potentials. In
section 4, we construct an Einstein-scalar theory in asymptotically AdS
spacetime which is dual to states excited by scalar operator, and derive the
semi-analytic formula of fidelity in the probe limit. Particularly, in the
standard quantization, we show that the fidelity will suffer from a kind of
new UV divergence even though we have added usual counterterms into the
action. We will also show that such kind of new divergence will not appear in
alternative quantization. In the section 5, we will discuss how to check the
commutativity of two density matrices holographically.

##  2 Holographic quantum distance between thermal states

In this section, we show how to holographically get the quantum distances
between thermal states. In addition, we will explain the main idea about how
to use replica trick to holographically compute the expression such as
tr$(\rho_{1}\rho_{2}\cdots\rho_{n})$ for a series of given holographic states
$\\{\rho_{1},\rho_{2},\cdots,\rho_{n}\\}$.

Our starting point is that thermal states in $\text{CFT}_{d}$ are dual to
$\text{Schwarzschild-AdS}_{d+1}$ black hole Maldacena:2001kr . Let us start
from the properties of the thermal state. Thermal state can be used to
describe the quantum system which is in contact with a heat bath at
temperature $T=1/\beta$. In the canonical ensemble at temperature $T$, the
density matrix of thermal state reads

$\rho=\frac{e^{-\beta H}}{Z(\beta)}\,.$ (7)

Here $H$ is the Hamiltonian of this system and $Z(\beta)=\text{tr}e^{-\beta
H}$ is the partition function. To illustrate how use replica trick to compute
the quantum distance here, we will consider two thermal systems, of which the
density matrices are

$\rho_{1}=\frac{e^{-\beta_{1}H}}{Z(\beta_{1})}\,,~{}\rho_{2}=\frac{e^{-\beta_{2}H}}{Z(\beta_{2})}\,.$
(8)

In following we will first introduce how to compute the trace of product
$\rho_{1}\rho_{2}$ holographically,

$\rho_{1}\rho_{2}=\frac{e^{-(\beta_{1}+\beta_{2})H}}{Z(\beta_{1})Z(\beta_{2})}\,.$
(9)

This is product will play the essential role in our computations of quantum
distances later on.

Note that for general two different density matrix $\rho_{1,2}$, we can not
guarantee their product $\rho_{1}\rho_{2}$ is hermitian. Then, after
normalized, the $\rho_{1}\rho_{2}/\mathrm{tr}(\rho_{1}\rho_{2})$ is not a
density matrix unless satisfying the condition $[\rho_{1},\rho_{2}]=0$.
However, for thermal state, $\rho_{1,2}$ commute with each other since they
own the same Hamiltonian $H$. The main argument here is that
$\rho_{1}\rho_{2}$ can be regarded as an un-normalized density matrix of a new
thermal state with the inverse temperature
$\tilde{\beta}=\beta_{1}+\beta_{2}$, as depicted in Fig. 1. Note that
$\rho_{1}\rho_{2}$ is not properly normalized.

$\rho_{1}$$\rho_{2}$$\times$$=$$2\beta$$\beta$$3\beta$$\rho_{1}\rho_{2}$$\frac{2\pi}{3}$
Figure 1: The schematic diagram about the replica trick between thermal
states. Here we consider $\rho_{1}\propto e^{-2\beta H}$ and $\rho_{2}\propto
e^{-\beta H}$. Then, the inverse temperature of replicate thermal state is
$3\beta$.

From the bulk perspective, the gravity dual of $\rho_{1}\rho_{2}$ would be a
Schwarzschild-AdS black hole with new inverse temperature $\tilde{\beta}$. See
appendix A for the convention we use here. According to the relation (110)
between temperature $T$ and horizon radius $r_{h}$, this new black hole’s
horizon radius $r_{h}$ can be in principle solved. For convenience, this paper
will focus on planar black hole (black brane). For planar Schwarzschild-AdS
black hole denoted by $k=0$,

$r_{h}=\frac{4\pi L^{2}}{d\beta}\,,\beta=\frac{1}{T}\,.$ (10)

Then we can get the new horizon radius $\tilde{r}_{h}$ through $r_{1,2}$, that
satisfies $r_{1,2}=4\pi L^{2}/d\beta_{1,2}$

$\tilde{r}_{h}=\frac{4\pi L^{2}}{d\tilde{\beta}}=\frac{4\pi
L^{2}}{d(\beta_{1}+\beta_{2})}=\frac{r_{1}r_{2}}{r_{1}+r_{2}}\,.$ (11)

Based on the above calculation, we conclude that the gravity dual of
$\rho_{1}\rho_{2}$ is Schwarzschild-AdS black hole with smaller horizon radius
$\tilde{r}_{h}$. The product between thermal states gives us more “smaller”
black hole as depicted in Fig. 2.

$\beta_{1}$$\beta_{2}$$r_{0}/2$

$\mathcal{M}_{n}$$\mathcal{B}_{n}$$\tilde{r}_{h}~{}$

Figure 2: The gravity dual of the replicate system by multiplying $n$ thermal
states. For general replicate thermal state $\rho_{1}\rho_{2}\cdots\rho_{n}$,
its gravity dual is a Schwarzschild-AdS black hole with horizon located at
$\tilde{r}_{h}=1/(r_{1}^{-1}+r_{2}^{-1}+\cdots+r_{n}^{-1})$.

After a long journey, we are ready to show how to calculate
$\text{tr}(\rho_{1}\rho_{2})$ in Einstein-Hilbert gravity. From the basic
holographic dictionary in Ref. Maldacena:2001kr , we have the relation

$Z_{\text{CFT}}[\beta]=Z_{\text{bulk}}[\beta]\,,$ (12)

that is a most simplest nontrival example of relation (1). Recalling the
relation (2), here we have

$Z[\beta]=\text{tr}e^{-\beta H}=e^{-I[\beta]}\,.$ (13)

Ref. Emparan:1999pm  shows that, the full Euclidean gravitational action
$I[\beta]$ in $d+1$ spacetime dimensions should have three contributions

$I[\beta]=I_{\text{bulk}}+I_{\text{surf}}+I_{\text{c.t}}\,.$ (14)

Here

$I_{\text{bulk}}+I_{\text{surf}}=-\frac{1}{16\pi
G}\int_{\mathcal{B}}d^{d+1}x\sqrt{g}\left[\mathcal{R}+\frac{d(d-1)}{L^{2}}\right]-\frac{1}{8\pi
G}\int_{\mathcal{\partial B}}d^{d}x\sqrt{h}K\,.$ (15)

The first term is the Einstein-Hilbert action with cosmological constant
$-d(d-1)/2L^{2}$. The second term is called Gibbons-Hawking boundary term that
guarantees the well-defined variational principle. The counter term
$I_{\text{c.t}}$ is added to regulate the infinity from the classical action.
See Ref. deHaro:2000vlm ; Skenderis:2002wp ; Balasubramanian:1999re ;
Bianchi:2001kw  for details. Here we consider Schwarzschild-AdS planar black
hole with inverse temperature $\beta$. The regulated action $I[\beta]$ is
given by

$\displaystyle I[\beta]$ $\displaystyle=-\frac{\beta\Omega_{k=0,d-1}}{16\pi
GL^{2}}r_{h}^{d}=-\frac{4^{d-2}\pi^{d-1}L^{2d-2}\Omega_{k=0,d-1}}{d^{d}G}\frac{1}{\beta^{d-1}}=-\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4Gd}$
(16)

Under such holographic description and the saddle point approximation, the
trace of $\rho_{1}\rho_{2}$ can be expressed in geometric quantities

$\displaystyle\text{tr}(\rho_{1}\rho_{2})=\frac{Z(\beta_{1}+\beta_{2})}{Z(\beta_{1})Z(\beta_{2})}$
$\displaystyle=\exp\left(I[\beta_{1}]+I[\beta_{2}]-I[\beta_{1}+\beta_{2}]\right)$
(17)
$\displaystyle=\exp\left[{\frac{\Omega_{k=0,d-1}(\tilde{r}_{h}^{d-1}-r_{1}^{d-1}-r_{2}^{d-1})}{4Gd}}\right]$

Above calculation can be generalized into more general case such as
tr$(\rho_{1}\rho_{2}\rho_{3}\cdots)$.

$\displaystyle\text{tr}(\rho_{1}\rho_{2}\rho_{3}\cdots)$
$\displaystyle=\frac{Z(\beta_{1}+\beta_{2}+\beta_{3}+\cdots)}{Z(\beta_{1})Z(\beta_{2})Z(\beta_{3})\cdots}=\exp\left(I[\beta_{1}]+I[\beta_{2}]+I[\beta_{3}]\cdots-I[\beta_{1}+\beta_{2}+\beta_{3}\cdots]\right)$
(18)
$\displaystyle=\exp\left[{\frac{\Omega_{k=0,d-1}(\tilde{r}_{h}^{d-1}-r_{1}^{d-1}-r_{2}^{d-1}-r_{3}^{d-1}-\cdots)}{4Gd}}\right]$

Here $\tilde{r}_{h}=1/(r_{1}^{-1}+r_{2}^{-1}+r_{3}^{-1}+\cdots)$. This gives
us a formula to compute the trace of $\rho_{1}\dots\rho_{n}$ from holography.

To check our result, one can calculate the Rényi entropy for thermal state
$\rho$ by taking $n$ copies. According to the definition of Rényi entropy, we
can obtain that

$\displaystyle
S_{n}=\frac{1}{1-n}\ln\frac{\text{tr}\rho^{n}}{(\text{tr}\rho)^{n}}$
$\displaystyle=\frac{1}{n-1}\left(I_{\text{bulk}}[\mathcal{B}_{n}]-nI_{\text{bulk}}[\mathcal{B}_{1}]\right)$
(19)
$\displaystyle=\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4G}\frac{n^{d}-1}{d(n^{d}-n^{d-1})}\,.$

Following the spirit of the replica trick, then we analytically continue $n$
into unit one

$S=\lim\limits_{n\rightarrow
1}S_{n}=\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4G}\,.$ (20)

We can see that the Bekenstein-Hawking entropy formula can be precisely
recovered. This is an check of self-consistence on our method. Taken the
lessons we learned above, we will consider various quantum distances between
thermal states in the remaining content. More specifically, we will employ the
method of constructing Rényi entropy, i.e. _replica trick_ , to calculate
these quantum distances.

###  2.1 Fidelity

Given two thermal states $\rho$ and $\sigma$ in the same Hilbert space, we now
is ready to calculate the fidelity
$\text{Fi}(\rho,\sigma)=$tr$\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}$ in
holography. Since the dimension of Hilbert spacetime is infinity in the field
theory, the fidelity of two different states will almost vanish and it will be
more useful to consider the logarithmic fidelity $\ln\text{Fi}(\rho,\sigma)$.
Since the square root is involved into the trace, we cannot directly compute
the fidelity by using integer-order replica shown in above example. To
overcome this issue, we now define a generalized $n$-order log-fidelity for
positive integer $n$

$\ln\text{Fi}_{n}(\rho,\sigma)=\ln\text{tr}\left(\sqrt{\sigma}\rho\sqrt{\sigma}\right)^{n}=\ln\text{tr}(\rho\sigma)^{n}\,.$
(21)

Then, we can analytically continue $n$ into real number. If
$\ln\text{Fi}_{n}(\rho,\sigma)$ is analytic for $n>0$, the fidelity can be
recovered

$\ln\text{Fi}(\rho,\sigma)=\lim_{n\rightarrow
1/2}\ln\text{Fi}_{n}(\rho,\sigma)\,.$ (22)

Generalizing the procedure for $\text{tr}\rho_{1}\rho_{2}$ into
$\text{tr}(\rho_{1}\rho_{2})^{n}$

$\displaystyle\text{tr}(\rho_{1}\rho_{2})^{n}=\frac{Z(n\beta_{1}+n\beta_{2})}{\left[Z(\beta_{1})Z(\beta_{2})\right]^{n}}$
$\displaystyle=\exp\left(nI[\beta_{1}]+nI[\beta_{2}]-I[n\beta_{1}+n\beta_{2}]\right)$
(23)
$\displaystyle=\exp\left[{\frac{\Omega_{k=0,d-1}(\tilde{r}_{h}^{d-1}-nr_{1}^{d-1}-nr_{2}^{d-1})}{4Gd}}\right]$

here $\tilde{r}_{h}(n)=1/[n(r_{1}^{-1}+r_{2}^{-1})]$, that is an inverse
proportional function of copies number $n$. As just claimed before, we then
analytically continue $n$ into $\frac{1}{2}$ to obtain the fidelity

$\ln\text{Fi}(\rho,\sigma)=\lim_{n\rightarrow
1/2}\ln\text{Fi}_{n}(\rho,\sigma)={\frac{\Omega_{k=0,d-1}\left[2\tilde{r}_{h}^{d-1}-r_{1}^{d-1}-r_{2}^{d-1}\right]}{8Gd}}\,.$
(24)

here $\tilde{r}_{h}$ take the value $\tilde{r}_{h}=2r_{1}r_{2}/(r_{1}+r_{2})$.
As a consistent check, we can confirm that $F(\rho,\sigma)=1$ if
$\beta_{1}=\beta_{2}$.

###  2.2 Relative entropy

Relative entropy is a measure of the distinguishability of two states. It will
not suffer from the ultraviolet divergence that carries no information about
the state. Given two density matrices, the relative entropy is defined as

$S(\rho||\sigma)=\text{tr}(\rho\ln\rho)-\text{tr}(\rho\ln\sigma)\,.$ (25)

Here $\rho$ and $\sigma$ are both properly normalized. The first term on the
right hand of equal sign is just the entanglement entropy of $\rho$. In order
to compute the second term in holography, we can imitate the construction from
Rényi entropy to entanglement entropy

$\text{tr}(\rho\ln\sigma)=\lim_{n\rightarrow
1}\frac{1}{n-1}\ln\text{tr}(\rho\sigma^{n-1})\,.$ (26)

Still considering two thermal states $\rho_{1}$ and $\rho_{2}$,
$\text{tr}(\rho_{1}\rho_{2}^{n-1})$ is given by

$\displaystyle\text{tr}(\rho_{1}\rho_{2}^{n-1})=\frac{Z(\beta_{1}+(n-1)\beta_{2})}{Z(\beta_{1})Z(\beta_{2})^{n-1}}$
$\displaystyle=\exp\left(I[\beta_{1}]+(n-1)I[\beta_{2}]-I[\beta_{1}+(n-1)\beta_{2}]\right)$
(27)
$\displaystyle=\exp\left[{\frac{\Omega_{k=0,d-1}(\tilde{r}_{h}^{d-1}-r_{1}^{d-1}-(n-1)r_{2}^{d-1})}{4Gd}}\right]$

here $\tilde{r}_{h}(n)=4\pi L^{2}/d\left[\beta_{1}+(n-1)\beta_{2}\right]$. The
second term in Eq. (25) can be recovered by analytically continue $n$ into $1$

$\text{tr}(\rho_{1}\ln\rho_{2})=\lim_{n\rightarrow
1}\frac{1}{n-1}\ln\text{tr}(\rho_{1}\rho_{2}^{n-1})=-\frac{\Omega_{k=0,d-1}r_{2}^{d-1}}{4Gd}-\frac{\Omega_{k=0,d-1}r_{1}^{d-1}}{4Gd}\frac{\beta_{2}(d-1)}{\beta_{1}}\,.$
(28)

Combining the above result with entanglement entropy of $\rho_{1}$, the
relative entropy between thermal states $\rho_{1}$ and $\rho_{2}$ is given by

$S(\rho_{1}||\rho_{2})=\text{tr}(\rho_{1}\ln\rho_{1})-\text{tr}(\rho_{1}\ln\rho_{2})=-\frac{\Omega_{k=0,d-1}r_{1}^{d-1}}{4G}+\frac{\Omega_{k=0,d-1}r_{2}^{d-1}}{4Gd}+\frac{\Omega_{k=0,d-1}r_{1}^{d-1}}{4Gd}\frac{\beta_{2}(d-1)}{\beta_{1}}\,.$
(29)

It should be noted that the relative entropy vanishes when $\rho_{1}$ equates
to $\rho_{2}$, which is a expected result no matter what method is used here.

##  3 Add some matter: Maxwell field as the first step

In this section, we want to generalize our replica trick into the charged
black hole. For boundary field theory, we consider the grand canonical
ensemble instead of the canonical ensemble. Given the conserved charge $Q$, we
can do the similar construction for the matrix density of the states

$\rho=\frac{e^{-\beta(H-\mu Q)}}{Z(\beta,\mu)}\,.$ (30)

Here $\mu$ is the chemical potential. The gravity dual of such state (30) is
charged black hole in AdS spacetime. Since charge $Q$ is conserved, it should
satisfy $[Q,H]=0$. By virtue of such commutativity, we will find that the
density matrices of different temperatures and charges are also commutative to
each others. Thus we can generalize our above replica trick from neutral
system into the charged one. Considering states $\rho_{1},\rho_{2}$

$\rho_{1}=\frac{e^{-\beta_{1}(H-\mu_{1}Q)}}{Z(\beta_{1},\mu_{1})}\,,\rho_{2}=\frac{e^{-\beta_{2}(H-\mu_{2}Q)}}{Z(\beta_{2},\mu_{2})}$
(31)

Following the spirit of the previous section, we reconsider $\rho_{1}\rho_{2}$
as a new state

$\begin{split}\rho_{1}\rho_{2}=\frac{e^{-\tilde{\beta}(H-\tilde{\mu}Q)}}{Z(\beta_{1},\mu_{1})Z(\beta_{2},\mu_{2})}\\\
\tilde{\beta}=\beta_{1}+\beta_{2},~{}\tilde{\mu}=\frac{\mu_{1}\beta_{1}+\mu_{2}\beta_{2}}{\beta_{1}+\beta_{2}}\,.\end{split}$
(32)

We can see that the replicate state $\rho_{1}\rho_{2}$ have the same
construction with the original two states. In holography, this claim means the
replicate state $\rho_{1}\rho_{2}$ prepared by (31) is still dual to AdS-RN
black hole with new parameters ${\tilde{\beta},\tilde{\mu}}$ defined by (32).

In gravity side, considering the Einstein-Maxwell theory, the total action
(14) need to add the contribution from the Maxwell field

$I_{\text{Maxwell}}=-\frac{1}{16\pi
G}\int_{\mathcal{B}}d^{d+1}x\sqrt{g}F_{\mu\nu}F^{\mu\nu}\,.$ (33)

If we just consider the RN black hole, the form of the metric coincide with
the neutral one (104) but with

$f(r)=\frac{r^{2}}{L^{2}}+k-\frac{f_{0}}{r^{d-2}}+\frac{2(d-2)\mu^{2}r_{h}^{2(d-2)}}{(d-1)r^{2(d-2)}}\,.$
(34)

Here $\mu$ is the chemical potential of the black hole and $r_{h}$ is the
largest root of $f(r)=0$. For RN black hole, the Maxwell field strength tensor
has a form

$F_{\mu\nu}=-\frac{Q}{r^{d-1}}(d\tau)_{\mu}\wedge(dr)_{\nu},~{}Q=(d-2)\mu
r_{h}^{d-2}\,.$ (35)

Here $Q$ is the total charge of RN black hole. Notice that
$F_{\mu\nu}F^{\mu\nu}=2Q^{2}/r^{2(d-1)}$, the direct calculation for
$I_{\text{Maxwell}}$ reads

$\displaystyle I_{\text{Maxwell}}$ $\displaystyle=-\frac{1}{16\pi
G}\int_{\mathcal{B}}d^{d+1}x\sqrt{g}F_{\mu\nu}F^{\mu\nu}$ (36)
$\displaystyle=-\frac{\beta\Omega_{k,d-1}Q^{2}r_{h}^{2-d}}{8\pi G(d-2)}$
$\displaystyle=-\frac{(d-2)\beta\Omega_{k,d-1}\mu^{2}r_{h}^{d-2}}{8\pi G}$

$\left\\{r_{h},\beta,\mu\right\\}$ is not independent of each other. For
$k=0$, we have a relation among $\left\\{r_{h},\beta,\mu\right\\}$:

$r_{h}=\frac{\sqrt{2(d-1)L^{2}\left[\beta^{2}d(d-2)^{2}\mu^{2}+2\pi^{2}(d-1)L^{2}\right]}+2\pi(d-1)L^{2}}{d(d-1)\beta}$
(37)

According to the replica trick (32), we should consider the gravity dual of
replicate state $\rho_{1}\rho_{2}$ as a RN-AdS black hole with thermodynamic
parameters $\left\\{\tilde{\beta},\tilde{\mu}\right\\}$. By employing the
holographic dictionary

$Z[\beta,\mu]=\text{tr}e^{-\beta(H-\mu Q)}=e^{-I[\beta,\mu]}\,,$

we can use the on shell action to express $\text{tr}(\rho_{1}\rho_{2})$

$\text{tr}(\rho_{1}\rho_{2})=\frac{Z(\tilde{\beta},\tilde{\mu})}{Z(\beta_{1},\mu_{1})Z(\beta_{2},\mu_{2})}=\exp\left(I[\beta_{1},\mu_{1}]+I[\beta_{2},\mu_{2}]-I[\tilde{\beta},\tilde{\mu}]\right)\,.$
(38)

For instance, if we set
$\left\\{\beta_{1}=\beta_{2}=\beta\,,\mu_{1}=\mu_{2}=\mu\right\\}$, the
parameters of replicate RN black hole are

$\tilde{\beta}=\beta_{1}+\beta_{2}=2\beta,~{}\tilde{\mu}=\frac{\mu_{1}\beta_{1}+\mu_{2}\beta_{2}}{\beta_{1}+\beta_{2}}=\mu\,.$
(39)

In general the action $Z(\beta,\mu)$ will be a complicated nonlinear function
of $\beta$ and $\mu$. For simplicity of calculation, we can ignore the back
reaction to bulk geometry from the Maxwell field if $\mu\ll 1$. In this sense,

$r_{h}\approx\frac{4\pi L^{2}}{d\beta}\,.$ (40)

Then we obtain that

$\displaystyle I[\beta,\mu]$
$\displaystyle=-\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4Gd}-\frac{(d-2)\Omega_{k=0,d-1}\mu^{2}r_{h}^{d-3}L^{2}}{2Gd}+\mathcal{O}(\mu^{4})$
(41) $\displaystyle I[2\beta,\mu]$
$\displaystyle=-\frac{\Omega_{k=0,d-1}\tilde{r}_{h}^{d-1}}{4Gd}-\frac{(d-2)\Omega_{k=0,d-1}\mu^{2}\tilde{r}_{h}^{d-3}L^{2}}{2Gd}+\mathcal{O}(\mu^{4})$

Using the fact that $\tilde{r}_{h}$ is the half of $r_{h}$ and neglect the
terms smaller than $\mathcal{O}(\mu^{2})$, it is straight to calculate
$\text{tr}(\rho_{1}\rho_{2})$

$\displaystyle\text{tr}(\rho_{1}\rho_{2})$
$\displaystyle=\exp\left(2I[\beta,\mu]-I[2\beta,\mu]\right)$ (42)
$\displaystyle=\exp\left[{\frac{\Omega_{k=0,d-1}r_{h}^{d-1}(1/2^{d-1}-2)}{4Gd}}+\frac{(d-2)\Omega_{k=0,d-1}\mu^{2}L^{2}r_{h}^{d-3}(1/2^{d-3}-2)}{2Gd}\right]$

Finally, we generalize our result into $n$ copies and calculate the Rényi
entropy

$S_{n}=\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4Gd}\frac{n^{d}-1}{n^{d}-n^{d-1}}+\frac{(d-2)\Omega_{k=0,d-1}\mu^{2}L^{2}r_{h}^{d-3}}{2Gd}\frac{n^{d-2}-1}{n^{d-2}-n^{d-3}}\,.$
(43)

By taking the limit $n\rightarrow 1$, the Von Neumann entropy can be recovered

$S=\lim\limits_{n\rightarrow
1}S_{n}=\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4G}+\frac{(d-2)^{2}\Omega_{k=0,d-1}\mu^{2}L^{2}r_{h}^{d-3}}{2Gd}\,.$
(44)

One can also get the above result according to Bekenstein-Hawking formula.
Since we consider the chemical potential $\mu$ as a infinitesimal, the horizon
(37) of RN-AdS black hole denoted by $r_{\text{RN}}$ can be separated into two
parts

$r_{\text{RN}}=r_{h}+\frac{2(d-2)^{2}L^{2}\mu^{2}}{d(d-1)r_{h}}\,.$ (45)

Here $r_{h}$ is the largest root of $f(r)=0~{}\text{for}~{}\mu=0$. Following
the Bekenstein-Hawking formula, the entropy is described by the area of
horizon. Up to the order $\mathcal{O}(\mu^{2})$ we obtain

$\displaystyle S$
$\displaystyle=\frac{\Omega_{k=0,d-1}r_{\text{RN}}^{d-1}}{4G}$ (46)
$\displaystyle=\frac{\Omega_{k=0,d-1}r_{h}^{d-1}}{4G}+\frac{(d-2)^{2}\Omega_{k=0,d-1}\mu^{2}L^{2}r_{h}^{d-3}}{2Gd}\,.$

We see that two formulas (44) and (46) are same, which is what we expect and
shows the consistence of our methods.

##  4 Fidelity between states excited by scalar operator

In this section, we consider how to compute the quantum distance between
states that are excited by scalar operator. In holography, the scalar operator
$\mathcal{O}_{\phi}$ in boundary CFT is dual to a bulk scalar field $\phi$.
Considering a $(d+1)$-dimensional Einstein-scalar theory, such theory’s action
is

$I=-\frac{1}{16\pi
G}\int_{\mathcal{B}}d^{d+1}x\sqrt{g_{E}}\left[\mathcal{R}+\frac{d(d-1)}{L^{2}}-\frac{1}{2}\nabla_{\mu}\phi\nabla^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}\right]\,.$
(47)

The parameter $m$ is the mass of the scalar field. Different from the
Minkowski spacetime, $m^{2}$ is negative in AdS spacetime. Despite this,
$m^{2}$ can not be too negative but above the Breitenlohner-Freedman bound
$m_{\text{BF}}^{2}\equiv-d^{2}/4L^{2}$. For dual boundary CFT, the
corresponding conformal dimension $\Delta_{\mathcal{O}}$ for scalar operator
is $(d\pm\sqrt{d^{2}+4m^{2}L^{2}})/2$. The “$+$” corresponding to standard
quantization and “$-$” stands for alternative quantization. We can get the
equations of motion following the action (47)

$\displaystyle\mathcal{R}_{\mu\nu}-\frac{1}{2}\mathcal{R}g_{\mu\nu}-\frac{d(d-1)}{2L^{2}}g_{\mu\nu}$
$\displaystyle=-\frac{1}{4}\left(\nabla_{\rho}\phi\nabla^{\rho}\phi+m^{2}\phi^{2}\right)$
(48) $\displaystyle\nabla_{\mu}\nabla^{\mu}\phi-m^{2}\phi$
$\displaystyle=0\,.$

In most general cases, we should solve this coupled gravitational system since
the backreaction of the scalar field can not be neglected. In principle, given
the proper boundary conditions, these equations can be solved analytically or
numerically. We consider the thermal states in the canonical ensemble.
However, we can neglect the backreaction of the scalar field in the bulk, if
the change of the total energy $\Delta E$ induced from the scalar operator
$\mathcal{O}_{\phi}$ is much smaller than the mass of black hole. This is the
so-called _probe limit_. In the probe limit, the gravity and scalar field will
decouple. That means in contact with a heat bath at $\beta$, the bulk geometry
remains to be Schwarzschild-AdS black hole. Then, we just analyze the behavior
of scalar field in this fixed black hole background. What’s more, we only need
to explore the additional contributions from the scalar field to holographic
quantum distance. In this paper, we focus on the fidelity between states
excited by scalar operator since it is easy to obtain. The main reason why it
is tricky to calculate other quantum distances will be explained soon.

###  4.1 Set up

In this section, we adopt the Poincaré coordinate
$\left\\{z,\tau,x^{i}\right\\}$. Under this gauge, the $n$-copies Euclidean
Schwarzschild-AdS black hole’s metric reads

$ds^{2}=\frac{1}{z^{2}}\left[f(z)d\tau^{2}+\frac{dz^{2}}{f(z)}+\sum_{i=1}^{d-1}dx_{i}^{2}\right],\quad
f(z)=\frac{1}{L^{2}}-(2nz)^{d}f_{0}\,.$ (49)

Here the factor $2n$ is induced from the definition of fidelity where
$\text{tr}(\rho\sigma)^{n}$ is involved. Based on our previous discussion, for
two states $\rho$ and $\sigma$ with same inverse temperature $\beta$, the dual
black hole to compute $\text{tr}(\rho\sigma)^{n}$ will have inverse
temperature $2n\beta$. Then, we write the equation of motion for scalar field
in this background:

$\frac{1}{\sqrt{g_{E}}}\frac{\partial}{\partial
x^{\mu}}\left(\sqrt{g_{E}}g^{\mu\nu}\frac{\partial}{\partial
x^{\nu}}\right)\phi-m^{2}\phi=0,\quad
g_{E}=\frac{\Omega_{0,d-1}}{z^{2(d+1)}}\,.$ (50)

More specifically,

$\frac{z^{2}\ddot{\phi}}{f}+z^{2}f\phi^{\prime\prime}+\left[z^{2}f^{\prime}-(d-1)zf\right]\phi^{\prime}-m^{2}\phi=0\,,$
(51)

where a prime/dot denotes the derivative with respect to $z$ or $\tau$,
respectively. In general case, the Eq. (51) does not have analytical solutions
so we have to solve this equation numerically. Here we assume the boundary can
keep the asymptotic AdS behavior such that

$\phi(\tau,z)\rightarrow\tilde{\phi}_{s}(\tau)z^{d-\Delta}+\tilde{\phi}_{v}(\tau)z^{\Delta}+\dots\,,~{}~{}~{}\Delta=\frac{d+\sqrt{d^{2}+4m^{2}L^{2}}}{2}\,.$
(52)

Other requirement is that $\phi$ should be continuous near the horizon
$z_{h}$. Here $\tilde{\phi}_{s}(\tau)$ and $\tilde{\phi}_{v}(\tau)$ are
independent of $z$. In holography, both $\tilde{\phi}_{s}(\tau)$ and
$\tilde{\phi}_{v}(\tau)$ can be considered as a source if the mass parameter
satisfies $m_{\text{BF}}^{2}<m^{2}<m_{\text{BF}}^{2}+1/L^{2}$. It needs to
note that $\tilde{\phi}_{s}$ and $\tilde{\phi}_{v}$ will be both nonzero if
the scalar field is not zero in the bulk. To keep the asymptotically AdS
structure and make the probe limit self-consistent we have to require
$\phi^{2}\rightarrow 0$ as $z\rightarrow 0$. This requires that we have to set
$d>\Delta$ and so we have to set:

$m_{\text{BF}}^{2}<m^{2}<0\,.$ (53)

The selection of source corresponds different dual field theories, see Ref.
Klebanov:1999tb . In gravity side, this selection determines how to set the
boundary condition for Eq. (51). For instance, if considering
$\tilde{\phi}_{s}(\tau)$ as source, we must fix $\tilde{\phi}_{s}(\tau)$ at
$z\rightarrow 0$ correspondingly. This is called _standard quantization
scheme_ for the scalar field. While in this paper, we employ the _alternative
quantization scheme_ that considers $\tilde{\phi}_{v}(\tau)$ as the source.
The reason why we choose “non-standard” quantization scheme will be explicitly
explained. It should be noted that the scalar field obeys second order
equation of motion (51), but we only give one boundary condition such as
fixing $\tilde{\phi}_{v}(\tau)$ in alternative quantization scheme. However,
we must discard one of the independent solutions since this solution blows up
at the horizon $z_{h}$. Hence, the regularity of Euclidean spacetime makes the
bulk scalar field uniquely specified.

The next step is to solve Eq. (51) since the final result for action (47) is
dependent of $\tilde{\phi}_{v}(\tau)$ and $\tilde{\phi}_{s}(\tau)$. Then we
can assume the general solution of $\phi(\tau,z)$ by the method of separation
of variables

$\phi(\tau,z)=\sum_{j=0}^{\infty}R_{j}(z)T_{j}(\tau)\,.$ (54)

Substituting this into Eq. (51), we can get the asymptotic behavior of general
solution for scalar field

$\displaystyle\lim_{z\rightarrow 0}\phi(\tau,z)$
$\displaystyle=\lim_{z\rightarrow 0}\sum_{j=0}^{\infty}T_{j}(\tau)R_{j}(z)$
(55) $\displaystyle T_{0}(\tau)$ $\displaystyle=A_{0}\,,T_{j}(\tau)=A_{j}\cos
w_{j}\tau+B_{j}\sin w_{j}\tau\,,$ $\displaystyle\lim_{z\rightarrow 0}R_{j}(z)$
$\displaystyle=\phi_{s}^{(j)}z^{d-\Delta}(1+\cdots)+\phi_{v}^{(j)}z^{\Delta}(1+\cdots)\,.$

and the radial equation

$\begin{split}R_{j}^{\prime\prime}+\left(\frac{f^{\prime}}{f}-\frac{d-1}{z}\right)R_{j}^{\prime}-\left(\frac{m^{2}}{z^{2}f}+\frac{w_{j}^{2}}{f^{2}}\right)R_{j}=0\\\
f(z)=\frac{1}{L^{2}}-(2nz)^{d}f_{0}\,,~{}~{}j=0,1,2,\dots{}\end{split}$ (56)

Here $\phi_{s,v}^{(j)}$ is denoted by the asymptotic behavior of $R_{j}(z)$.
The frequency $w_{j}$ take some discrete values given by

$w_{j}=\frac{2\pi j}{\tilde{\tau}}\,,\quad T(\tau)=T(\tau+\tilde{\tau})$ (57)

where $\tilde{\tau}$ is the period in boundary condition (52). According to
Fourier transformation, coefficients $A_{0},A_{n},B_{n}$ are determined by

$\displaystyle A_{0}\phi_{v}^{(0)}$
$\displaystyle=\frac{1}{\tilde{\tau}}\int_{0}^{\tilde{\tau}}\tilde{\phi}_{v}(\tau)\mathrm{d}\tau$
(58) $\displaystyle A_{j}\phi_{v}^{(j)}$
$\displaystyle=\frac{2}{\tilde{\tau}}\int_{0}^{\tilde{\tau}}\tilde{\phi}_{v}(\tau)\cos
w_{j}\tau\mathrm{d}\tau$ $\displaystyle B_{j}\phi_{v}^{(j)}$
$\displaystyle=\frac{2}{\tilde{\tau}}\int_{0}^{\tilde{\tau}}\tilde{\phi}_{v}(\tau)\sin
w_{j}\tau\mathrm{d}\tau$

###  4.2 Fidelity in the probe limit

In following we will first focus on alternative quantization, i.e. treating
the coefficient $\phi_{v}$ as the source term. As previously defined, given
two density matrices $\rho$ and $\sigma$ that are both excited by scalar
operator $\mathcal{O_{\phi}}$ with different values of source
$\tilde{\phi}_{v}$, a generalized $n$-order log-fidelity for positive integer
$n$ is

$\ln\text{Fi}_{n}(\rho,\sigma)=\ln\text{tr}\left(\sqrt{\sigma}\rho\sqrt{\sigma}\right)^{n}=\ln\text{tr}(\rho\sigma)^{n}\,.$
(59)

Then, we can analytically continue $n$ into $1/2$ to recover the fidelity

$\ln\text{Fi}(\rho,\sigma)=\lim_{n\rightarrow
1/2}\ln\text{Fi}_{n}(\rho,\sigma)\,.$ (60)

Next we construct the corresponding distribution $\tilde{\phi}_{v}(\tau)$ for
$\text{tr}(\rho\sigma)^{n}$ in the bulk, see Fig. (3) for illustration of
$\text{tr}(\rho\sigma)^{2}$. In the bulk gravity side, the boundary condition
for scalar field $\phi$ will be dependent of euclidean time

$\tilde{\phi}_{v}(\tau)=\left\\{\begin{aligned} \rho_{v}\quad
0<\tau\leq\beta\\\ \sigma_{v}\quad\beta<\tau\leq 2\beta\\\ \rho_{v}\quad
2\beta<\tau\leq 3\beta\\\ \dots{\\\ }\rho_{v}\quad 2(n-2)\beta<\tau\leq
2(n-1)\beta\\\ \sigma_{v}\quad 2(n-1)\beta<\tau\leq 2n\beta\\\
\end{aligned}\right\\}\,.$ (61)

Here $\rho_{v}$ and $\sigma_{v}$ are constants. We already set
$\beta_{1}=\beta_{2}=\beta$. To simplify the discussion, in this paper we just
set $\sigma_{v}=0$.

$\rho_{v}$$\sigma_{v}$$\rho_{v}$$\sigma_{v}$ Figure 3: The schematic diagram
about the corresponding bulk geometry with the boundary field configuration
$\rho\sigma\rho\sigma$. Here the red line labels the distribution of state
$\rho$ and blue dashed line labels the state $\sigma$. Here subscript $v$
follows from treating the coefficient $\phi_{v}$ as the source term under
alternative quantization. As setting $\beta_{1}=\beta_{2}=\beta$, we can see
that the period $\tilde{\tau}=2\beta$.

In principle, we need to know the value of bulk scalar field in every
spacetime point. Since the total action (47) is a spacetime volume integral.
However, according to Gauss theorem and using the equation of motion, the on-
shell action for scalar field can be expressed by the boundary term

$\displaystyle I_{m}$ $\displaystyle=\frac{1}{16\pi
G}\int_{\mathcal{B}_{n}}\mathrm{d}^{d+1}x\sqrt{g}\left[\frac{1}{2}\nabla_{\mu}\left(\phi\nabla^{\mu}\phi\right)-\frac{1}{2}\phi\nabla_{\mu}\nabla^{\mu}\phi+\frac{1}{2}m^{2}\phi^{2}\right]$
(62) $\displaystyle=\frac{1}{16\pi
G}\int_{\mathcal{B}_{n}}\mathrm{d}^{d+1}x\sqrt{g}\left[\frac{1}{2}\nabla_{\mu}\left(\phi\nabla^{\mu}\phi\right)\right]$
$\displaystyle=\frac{1}{16\pi
G}\int_{\partial\mathcal{B}_{n}}\mathrm{d}^{d}x\sqrt{h}\left(\frac{1}{2}n_{\mu}\phi\nabla^{\mu}\phi\right)\,.$

Here $n^{\mu}$ is the outward point normal vector of AdS boundary
$\partial\mathcal{B}_{n}$. In Poincaré coordinate, the only nonzero component
is

$n^{r}=-z\sqrt{f(z)}\,,$ (63)

so this boundary integral can reduce to

$I_{m}=-\frac{\Omega_{0,d-1}}{32\pi
G}\int\mathrm{d}\tau\left(\frac{f\phi\partial_{z}\phi}{z^{d-1}}\right)\,.$
(64)

Note that the above action is not regulated. In principle, according to
spacetime dimension $d$ and conformal dimension $\Delta$, we will
correspondingly select different forms of counter term deHaro:2000vlm ;
Skenderis:2002wp ; Balasubramanian:1999re ; Bianchi:2001kw . In this paper, we
will focus on 4-dimensional Einstein-scalar theory while fixing $m^{2}=-2$.
For $\left\\{d=3,\Delta=2\right\\}$, the corresponding counter term is

$I_{\text{c.t}}=-\frac{1}{16\pi G}\int
d^{3}x\sqrt{h}\left[\phi(n^{\mu}\partial_{\mu}\phi)+\frac{1}{2}\phi^{2}\right]\,.$
(65)

Therefore, the regulated scalar field action $\tilde{I}_{\rho\sigma}$ can be
easily calculated

$\displaystyle\tilde{I}_{\rho\sigma}$ $\displaystyle=I_{m}+I_{\text{c.t}}$
(66) $\displaystyle=\frac{1}{32\pi
G}\int\mathrm{d}^{3}x\left(\phi_{s}^{(0)}A_{0}+\sum_{i=1}^{\infty}\phi_{s}^{(i)}B_{i}\sin
w_{i}\tau\right)\left(\phi_{v}^{(0)}A_{0}+\sum_{j=1}^{\infty}\phi_{v}^{(j)}B_{j}\sin
w_{j}\tau\right)$ $\displaystyle=\frac{2n\beta\Omega_{0,2}}{32\pi
G}\left(\phi_{s}^{(0)}\phi_{v}^{(0)}A_{0}^{2}+\frac{1}{2}\sum_{j=1}^{\infty}\phi_{s}^{(j)}\phi_{v}^{(j)}B_{j}^{2}\right)$
$\displaystyle=\frac{n\beta\Omega_{0,2}}{16\pi
G}\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}+\sum_{j=1}^{\infty}\frac{(1-\cos
j\pi)^{2}}{j^{2}\pi^{2}}\frac{\rho_{v}^{2}\phi_{s}^{(j)}}{2\phi_{v}^{(j)}}\right]$
$\displaystyle\equiv\frac{n\beta\Omega_{0,2}}{16\pi
G}\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}+\Xi[\rho_{v},n]\right]\,.$

Here

$\Xi[\rho_{v},n]:=\sum_{j=1}^{\infty}\frac{(1-\cos
j\pi)^{2}}{j^{2}\pi^{2}}\frac{\rho_{v}^{2}\phi_{s}^{(j)}}{2\phi_{v}^{(j)}}\,.$
(67)

Before we go further, it needs to emphasize that divergency of “naked” action
arises only when we approaches to UV boundary $z\rightarrow 0$. The
counterterm then is applied to cancel such divergency. We will see later that,
since the boundary is inhomogeneous, there is an other different divergency of
on-shell action appears in standard quantization even if we do not take the
limit $z=0$. This kind of new divergency cannot be cancelled according to the
known holographic renormalization schema deHaro:2000vlm ; Skenderis:2002wp ;
Balasubramanian:1999re ; Bianchi:2001kw .

We now return to Eq. (66). Notice that $\Xi$ is dependent on the copies number
$n$, since the radial equation is related to $n$, i.e.
$f(z)=1/L^{2}-(2nz)^{d}f_{0}$. This property is very different from the usual
cases that $\tilde{\phi}_{v}$ is independent of the Euclidean time $\tau$. If
$\Xi[\rho_{v},\frac{1}{2}]$ is convergent, then we would analytically continue
$n$ into $1/2$. In other words, we need to solve the radial equation (56) with
$f(z)=1/L^{2}-z^{d}f_{0}$. Finally, the log-fidelity reads

$\displaystyle\ln\text{Fi}(\rho,\sigma)$ $\displaystyle=\lim_{n\rightarrow
1/2}\ln\frac{\text{tr}(\rho\sigma)^{n}}{(\text{tr}\rho)^{n}(\text{tr}\sigma)^{n}}$
(68) $\displaystyle=\lim_{n\rightarrow
1/2}\ln\exp[-\tilde{I}_{\rho\sigma}+n\tilde{I}_{\rho}]$
$\displaystyle=\lim_{n\rightarrow 1/2}\frac{n\beta\Omega_{0,2}}{16\pi
G}\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}-\Xi[\rho_{v},n]\right]$
$\displaystyle=\frac{\beta\Omega_{0,2}}{32\pi
G}\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}-\Xi[\rho_{v},\frac{1}{2}]\right]\,.$

Here $\tilde{I}_{\rho}$ is the regulated action for $\text{tr}\rho$

$\displaystyle\tilde{I}_{\rho}$ $\displaystyle=\frac{1}{32\pi
G}\int\mathrm{d}^{3}x\left[\phi_{s}^{(0)}\phi_{v}^{(0)}A_{0}^{2}\right]$ (69)
$\displaystyle=\frac{1}{32\pi
G}\int\mathrm{d}^{3}x\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{\phi_{v}^{(0)}}\right]$
$\displaystyle=\frac{\beta\Omega_{0,2}}{32\pi
G}\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{\phi_{v}^{(0)}}\,.$

Note again that Eq. (56) can be solved for arbitrary real number $n$. Thus, we
can numerically compute the value of $\Xi[\rho_{v},n]$ in the Eq. (67) and
then obtain the fidelity numerically.

###  4.3 Numerical result

After the previous derivation for fidelity, we then show the corresponding
numerical process and result. The only remaining unsolved equation is the
radial equation

$\begin{split}R_{j}^{\prime\prime}+\left(\frac{f^{\prime}}{f}-\frac{d-1}{z}\right)R_{j}^{\prime}-\left(\frac{m^{2}}{z^{2}f}+\frac{w_{j}^{2}}{f^{2}}\right)R_{j}=0\\\
f(z)=\frac{1}{L^{2}}-(2nz)^{d}f_{0}\,,~{}~{}j=0,1,2,\dots{}\end{split}$ (70)

We need to get the value of $\phi_{s}^{(j)}/\phi_{v}^{(j)}$ in
$\Xi[\rho_{v},n]$ for every frequency $w_{j}$. For $j=0$, the frequency is
vanished, i.e. $w_{0}=0$. The general solution for $R_{0}$ is just the static
case

$\displaystyle R_{0}=$
$\displaystyle\phi_{s}^{(0)}z^{d-\Delta}{{}_{2}F_{1}}\left(1-\frac{\Delta}{d}\,,1-\frac{\Delta}{d}\,;2-\frac{2\Delta}{d}\,;f_{0}L^{2}n^{d}z^{d}\right)$
(71) $\displaystyle+$
$\displaystyle\phi_{v}^{(0)}z^{\Delta}{{}_{2}F_{1}}\left(\frac{\Delta}{d}\,,\frac{\Delta}{d}\,;\frac{2\Delta}{d}\,;f_{0}L^{2}n^{d}z^{d}\right)\,.$

Continuity condition requires that $R_{0}$ should be finite at horizon $z_{h}$
which satisfies

$\frac{1}{L^{2}}-f_{0}n^{d}z_{h}^{d}=0\,.$ (72)

By virtue of

${{}_{2}F_{1}}(a\,,b\,;c\,;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\,,$
(73)

$R_{0}(z_{h})$ can be expressed as

$R_{0}(z_{h})=\phi_{s}^{(0)}z_{h}^{d-\Delta}\frac{\Gamma(2-\frac{2\Delta}{d})\Gamma(0)}{\Gamma(1-\frac{\Delta}{d})\Gamma(1-\frac{\Delta}{d})}+\phi_{v}^{(0)}z_{h}^{\Delta}\frac{\Gamma(\frac{2\Delta}{d})\Gamma(0)}{\Gamma(\frac{\Delta}{d})\Gamma(\frac{\Delta}{d})}\,.$
(74)

Notice that $R_{0}(z_{h})$ is singular since $\Gamma(0)$ is infinite. We need
to eliminate this singularity. Extract $\Gamma(0)$ outside the square bracket

$R_{0}(z_{h})=\Gamma(0)\left[\phi_{s}^{(0)}z_{h}^{d-\Delta}\frac{\Gamma(2-\frac{2\Delta}{d})}{\Gamma(1-\frac{\Delta}{d})\Gamma(1-\frac{\Delta}{d})}+\phi_{v}^{(0)}z_{h}^{\Delta}\frac{\Gamma(\frac{2\Delta}{d})}{\Gamma(\frac{\Delta}{d})\Gamma(\frac{\Delta}{d})}\right]\,.$
(75)

Therefore, the square bracket must vanish and the ratio
$\phi_{s}^{(0)}/\phi_{v}^{(0)}$ is

$\frac{\phi_{s}^{(0)}}{\phi_{v}^{(0)}}=-\frac{\Gamma(\frac{2\Delta}{d})\Gamma^{2}(1-\frac{\Delta}{d})}{\Gamma(2-\frac{2\Delta}{d})\Gamma^{2}(\frac{\Delta}{d})}z_{h}^{2\Delta-d}=\frac{\pi
d}{2\Delta-d}\frac{\sin{\frac{2\pi\Delta}{d}}\Gamma^{2}(\frac{2\Delta}{d})}{\sin^{2}{\frac{\pi\Delta}{d}}\Gamma^{4}(\frac{\Delta}{d})}z_{h}^{2\Delta-d}\,.$
(76)

For $j=1,2,3,\dots$, Eq. (70) does not have analytical solutions so we have to
solve the radial equation numerically. Notice that $\Xi[\rho_{v},n]$ is the
sum of an infinite series. Although we cannot analytically get the explicit
expression of $\phi_{s}^{(j)}/\phi_{v}^{(j)}$, the convergence or not of
$\Xi[\rho_{v},n]$ can be obtained from the comparison with the scalar field in
thermal AdS spacetime given $f_{0}=0$ in metric (49). Considering the scalar
field with the same boundary condition (61) in thermal AdS spacetime, the
corresponding radio $\phi_{s}^{(j)}/\phi_{v}^{(j)}$ can be analytically
solved:

$\left.\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}\right|_{\text{thermal
AdS}}=\frac{\Gamma(\Delta-\frac{d}{2})}{(w_{j}/2)^{2\Delta-d}\Gamma(\frac{d}{2}-\Delta)}=-\frac{\sin[\pi(\Delta-\frac{d}{2})](\Delta-\frac{d}{2})\Gamma(\Delta-\frac{d}{2})^{2}}{\pi(w_{j}/2)^{2\Delta-d}}\,.$
(77)

Notice that in thermal AdS spacetime $\Xi[\rho_{v},\frac{1}{2}]_{\text{thermal
AdS}}$ is convergent since the general term is proportional to $1/j^{3}$. To
determine whether $\Xi[\rho_{v},\frac{1}{2}]$ is convergent, we give a
concrete numerical result to check the change of $\delta_{j}$ with $j$

$\delta_{j}\equiv\frac{\phi^{(j)}_{s}}{\phi^{(j)}_{v}}-\left.\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}\right|_{\text{thermal
AdS}}$ (78)

Figure 4: The curve-fitting result of $\delta_{j}$. Here we use a logarithmic
function $b+\ln j^{a}$ to fit their difference.

Here we fix the following parameters

$m^{2}=-2,~{}d=3,~{}L^{2}=1,~{}f_{0}=1\,.$

From above parameters, we can obtain $\beta=4\pi/3$ and $\Delta=2$. We solve
the equation (70) numerically and then read the radio
$\phi^{(j)}_{s}/\phi^{(j)}_{v}$ from the asymptotic behavior of the radial
equation.

According to the fitting result, i.e. Fig. (4), we conclude that the
difference between $\frac{\phi^{(j)}_{s}}{\phi^{(j)}_{v}}$ and
$\left.\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}\right|_{\text{thermal AdS}}$ is
$j^{a}$ order. Since $a\approx-3.81678$, the difference between two series is
convergent. It should be noted that the numerical result of $a$ is very close
to $-4$. In practice, we can confirm that

$\frac{\phi^{(j)}_{s}}{\phi^{(j)}_{v}}-\left.\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}\right|_{\text{thermal
AdS}}=\mathcal{O}(1/w_{j}^{2\Delta})$ (79)

with a analytical approximation method. See appendix B for more details about
the solution for radial equation.

Then, according to the comparison test, $\Xi[\rho_{v},\frac{1}{2}]$ must be
convergent under the alternative quantization scheme since in this scheme
$\Xi[\rho_{s},\frac{1}{2}]_{\text{thermal AdS}}$ is convergent. The final
numerical result reads

$\displaystyle\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}$
$\displaystyle\approx-1.29053\rho_{v}^{2}$ (80)
$\displaystyle\Xi[\rho_{v},\frac{1}{2}]$
$\displaystyle\approx-0.0176151\rho_{v}^{2}$

###  4.4 Comment on standard quantization

We have shown that the on-shell action of replica spacetime is convergent in
alternative quantization. However, under the standard quantization the
situation is very different. In this case, we still have Eqs. (66)-(69), but
the $\rho_{v}$ is replaced by $\rho_{s}$ and $\Xi$ should be replaced by

$\Xi[\rho_{s},n]:=\sum_{j=1}^{\infty}\frac{(1-\cos
j\pi)^{2}}{j^{2}\pi^{2}}\frac{\rho_{s}^{2}\phi_{v}^{(j)}}{2\phi_{s}^{(j)}}\,.$
(81)

We will find that $\Xi[\rho_{s},\frac{1}{2}]_{\text{thermal AdS}}$ is
divergent. Specifically, $\Xi[\rho_{s},\frac{1}{2}]_{\text{thermal AdS}}$ is a
_harmonic series_ under the standard quantization since the general term in
$\Xi[\rho_{s},\frac{1}{2}]_{\text{thermal AdS}}$ becomes

$\Xi\sim\sum\frac{\phi^{(j)}_{v}}{j^{2}\phi^{(j)}_{s}}\sim\frac{1}{j}\,.$ (82)

The high frequency scalar modes with $w_{j}\gg 1$ will be not suppressed
enough and so will play more and more important roles. When it comes to the
computation of action, this effect will cause the infinity of
$\Xi[\rho_{v},n]$ in the standard quantization scheme. While under alternative
quantization scheme, these UV contributions from high frequency modes are
suppressed enough. It should be noted that this behavior is quite common when
considering inhomogeneous boundary condition for bulk matter. It needs to
emphasize that such as divergency appearing in standard quantization cannot be
removed by using the usual so called “holographic renormalization”, since one
can realize that this divergency will appear even we move the AdS boundary
into finite $z$. We note similar thing was also noticed by Ref. Yang:2022ibq ,
which finds that generalized susceptibly in terms of momentum integration will
be divergent in standard quantization.

One may seek to regulate this infinity in standard quantization scheme. For
instance, given a UV cutoff $k$, we seem to be able to extract divergence from
the harmonic series

$\sum_{j=1}^{k}\frac{1}{j}=\ln k+\gamma+\mathcal{O}(1/k)\,~{}~{}k\gg 1$ (83)

here $\gamma$ is Euler-Mascheroni constant. So the divergence here is
logarithmic divergence. When the mode $j$ is high enough, the probe limit and
even including the classical gravity approximation will be broken, so we
should take the full backreaction and even full UV quantum gravity theory into
account. If we believe the theory is UV complete, then “unknown” high energy
physics should just cancel such a logarithmic divergency. Then we will throw
away $\ln k$, which will be interpreted as the contribution from the
deformation of spacetime due to these high frequency modes. However, this
procedure is only effective for the set of parameters we have selected here.
When we fine-tune these parameters, such as $m^{2}$, the above procedure is
broken and we need to regulate such infinity again in this new set of
parameters. From the Eq. (125) in appendix B one can see that,

$\frac{\phi_{v}^{(j)}}{\phi_{s}^{(j)}}\sim w_{j}^{2\Delta-d}\,.$ (84)

for large mode $j$. Thus, in standard quantization, we can find from Eq. (81)
that

$\Xi\sim\sum j^{2\Delta-d}\,.$ (85)

Since $2\Delta-d=\sqrt{d^{2}+4m^{2}L^{2}}>0$, we see that $\Xi$ will be
divergent. It is still an open question for us about how to regularized and
renormalize such a divergency systematically. We leave the general
regularization under the standard quantization scheme for an important future
problem.

##  5 Commutativity of $\rho$ and $\sigma$

In this section, we will gives a holographic method to check if the density
matrices of two holographic states are commutative to each others. In quantum
field theory, we can’t usually write the density matrix of quantum states
precisely. Assume that the quantum state $\rho$ is excited by some operator
$\mathcal{O_{\rho}}(x)$ and other quantum state $\sigma$ is excited by
operator $\mathcal{O_{\sigma}}(y)$. Here ${x,y}$ label the points in the
background spacetime. It should be noted that even employing the same
operator, the commutator $[\mathcal{O}(x),\mathcal{O}(y)]$ is not necessarily
vanish for different spacetime points. However, as mentioned before, there
exist a gravity dual for the trace of density matrix’s product. Based on the
holographic description, is it possible to determine whether two operators are
commutative? In this section, we propose a feasible strategy to answer this
question.

Considering two density matrix $\rho$ and $\sigma$ in the same Hilbert space,
the commutativity of $\rho$ and $\sigma$ indicates

$\rho\sigma=\sigma\rho\,.$ (86)

One may think we can use their trace to judge whether they are commutative,
but the property of trace $\text{tr}(AB)=\text{tr}(BA)$ rejects this idea.
Instead, considering a auxiliary matrix $A$

$A=\rho\sigma-\sigma\rho\,,$ (87)

if $A=0$ then we can get

$\text{tr}(\rho\sigma\rho\sigma)=\text{tr}(\rho^{2}\sigma^{2})\,.$ (88)

This result is so straightforward since we can change the position of two
density matrix arbitrarily. Conversely given the condition (88),
$[\rho,\sigma]=0$ is expected to be proved. Firstly, it is not hard to find
that

$\text{tr}(\rho\sigma\rho\sigma)=\text{tr}(\rho^{2}\sigma^{2})~{}\Longrightarrow~{}\text{tr}(A^{2})=0\,.$
(89)

Since $A=-A^{\dagger}$, $A$ can be diagonalized with pure imaginary eigenvalue

$A=\sum_{n}ia_{n}|n\rangle\langle n|,~{}a_{n}\in\mathbb{R}$ (90)

$\left\\{|n\rangle\right\\}$ is a complete orthonormal set.
$\text{tr}(A^{2})=0$ indicates that $a_{n}=0$ for all $n$, by virtue of anti-
hermiticity we actually obtain that

$\text{tr}(\rho\sigma\rho\sigma)=\text{tr}(\rho^{2}\sigma^{2})~{}\Longleftrightarrow~{}[\rho,\sigma]=0\,.$
(91)

From the perspective of holography, there are two bulk geometries
corresponding to boundary condition $\rho\sigma\rho\sigma$ or
$\rho^{2}\sigma^{2}$, as depicted in Fig. 5. In order to determine whether
states $\rho$ and $\sigma$ are commutative, we should calculate these bulk
geometries’ partition function respectively. Recall that under the saddle
point approximation, the partition function of bulk geometry can be written as
the exponential function of corresponding action. So we only need to check
whether their corresponding geometric action are equal, i.e.

$S_{\rho\sigma\rho\sigma}\overset{\text{?}}{=}S_{\rho^{2}\sigma^{2}}\,.$ (92)

Furthermore, in probe limit we only need to compare the sector of scalar
field,

$\tilde{I}_{\rho\sigma\rho\sigma}\overset{\text{?}}{=}\tilde{I}_{\rho^{2}\sigma^{2}}\,.$
(93)

Without losing generality, we continue to use the scalar field model in
previous section. In order to compute the fidelity between $\rho$ and
$\sigma$, we already calculate the action of bulk gravity that corresponds to
the boundary field $(\rho\sigma)^{n}$. For a detailed discussion, we will
present such result in the previous section again

$\displaystyle\tilde{I}_{\rho\sigma\rho\sigma}$
$\displaystyle=\frac{\beta\Omega_{0,2}}{8\pi
G}\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}+\sum_{j=1}^{\infty}\frac{(1-\cos
j\pi)^{2}}{j^{2}\pi^{2}}\frac{\rho_{v}^{2}\phi_{s}^{(j)}}{2\phi_{v}^{(j)}}\right]\,.$
(94)

Here, we need to set $n=2$ to recover $\tilde{I}_{\rho\sigma\rho\sigma}$ from
the expression (66). Then we need to calculate the action
$\tilde{I}_{\rho^{2}\sigma^{2}}$ for boundary field configuration
$\rho^{2}\sigma^{2}$. Following the same process in previous section, we
calculate the action $\tilde{I}_{\rho^{2}\sigma^{2}}$ but with boundary
condition that is different from (61) in $n=2$

$\tilde{\phi}_{v}(\tau)=\left\\{\begin{aligned} \rho_{v}\quad 0<\tau\leq
2\beta\\\ \sigma_{v}\quad 2\beta<\tau\leq 4\beta\\\ \end{aligned}\right\\}\,.$
(95)

$\rho$$\sigma$$\rho$$\sigma$ (a) $\rho\sigma\rho\sigma$

$\rho$$\rho$$\sigma$$\sigma$ (b) $\rho^{2}\sigma^{2}$

Figure 5: The schematic diagram about two boundary conditions relate to the
commutativity of $\rho$ and $\sigma$. Here the red line labels the
distribution of state $\rho$ and blue dashed line labels the state $\sigma$.
For example, figure 5(a) shows schematically that the bulk geometry with the
boundary field configuration $\rho\sigma\rho\sigma$.

It is straightforward that period $\tilde{\tau}_{\rho^{2}\sigma^{2}}$ is two
times than $\tilde{\tau}_{\rho\sigma\rho\sigma}$. And it can be shown that
$\tilde{I}_{\rho^{2}\sigma^{2}}$ has the same form as
$\tilde{I}_{\rho\sigma\rho\sigma}$

$\displaystyle\tilde{I}_{\rho^{2}\sigma^{2}}$
$\displaystyle=\frac{\beta\Omega_{0,2}}{8\pi
G}\left[\frac{\rho_{v}^{2}\phi_{s}^{(0)}}{4\phi_{v}^{(0)}}+\sum_{j=1}^{\infty}\frac{(1-\cos
j\pi)^{2}}{j^{2}\pi^{2}}\frac{\rho_{v}^{2}\phi_{s}^{(j)}}{2\phi_{v}^{(j)}}\right]\,.$
(96)

However, it should be noted that $\phi_{s}^{(j)}/\phi_{v}^{(j)}$ has different
values in formula (94) and (96). More specifically, we know that
$w^{(j)}_{\rho\sigma\rho\sigma}=2\pi j/\tilde{\tau}_{\rho^{2}\sigma^{2}}$ and

$\tilde{\tau}_{\rho^{2}\sigma^{2}}=2\tilde{\tau}_{\rho\sigma\rho\sigma}=4\beta\,.$
(97)

Then we can obtain that
$w^{(j)}_{\rho\sigma\rho\sigma}=2w^{(j)}_{\rho^{2}\sigma^{2}}$. In previous
section, we already get the value of $\phi_{s}^{(j)}/\phi_{v}^{(j)}$ for
$\tilde{I}_{\rho\sigma\rho\sigma}$ and $j$ can only take the odd term, i.e.
$j=1,3,5,7,\cdots$. Since
$w^{(j)}_{\rho\sigma\rho\sigma}=2w^{(j)}_{\rho^{2}\sigma^{2}}$, based on
radial equation (70) for $\tilde{I}_{\rho\sigma\rho\sigma}$, we only take
$j=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2},\cdots$ to get the exact
value of $\tilde{I}_{\rho^{2}\sigma^{2}}$. Why can we do this because
$\phi_{s}^{(j)}/\phi_{v}^{(j)}$ in radial equation (70) changes continuously
with the continuous value of parameter $j$. Considering the thermal AdS
background, $\tilde{I}_{\rho\sigma\rho\sigma}$ is related to the sum of series
(67)

$\Xi_{\rho\sigma\rho\sigma}:=\sum_{j=1,3,5,\cdots}^{\infty}\frac{\phi_{s}^{(j)}}{2\phi_{v}^{(j)}}=-\frac{\sin[\pi(\Delta-\frac{d}{2})](\Delta-\frac{d}{2})\Gamma(\Delta-\frac{d}{2})^{2}}{2\pi(w_{j}/2)^{2\Delta-d}}\,.$
(98)

According to the above discussion, $\Xi$ for $\tilde{I}_{\rho^{2}\sigma^{2}}$
can be easily seen that

$\Xi_{\rho^{2}\sigma^{2}}=\frac{\Xi_{\rho\sigma\rho\sigma}}{2^{2\Delta-d}}\,.$
(99)

So we conclude that
$\tilde{I}_{\rho\sigma\rho\sigma}\neq\tilde{I}_{\rho^{2}\sigma^{2}}$ and
$\rho,\sigma$ are not commutative to each other in our case333It is noted that
$\Xi_{\rho^{2}\sigma^{2}}=\Xi_{\rho\sigma\rho\sigma}$ if $\Delta=d/2$.
However, as we claimed before, the mass parameter should satisfy
$m_{\text{BF}}^{2}<m^{2}<m_{\text{BF}}^{2}+1/L^{2}$ if both
$\tilde{\phi}_{s}(\tau)$ and $\tilde{\phi}_{v}(\tau)$ can be considered as the
source..

Another example is considering the Maxwell field $F_{\mu\nu}$ probing the
thermal states which are corresponding to AdS-Schwarzschild black hole. In the
probe limit, Maxwell field $F_{\mu\nu}$ is decoupled with gravity sector
$g_{\mu\nu}$. Therefore, the only different contribution of
$\text{tr}(\rho\sigma\rho\sigma)$ and $\text{tr}(\rho^{2}\sigma^{2})$ is the
contribution from the Maxwell filed:

$I_{\text{Maxwell}}=-\frac{1}{16\pi
G}\int_{\mathcal{B}}d^{d+1}x\sqrt{g}F_{\mu\nu}F^{\mu\nu}\,.$ (100)

As we have calculated in section 3, the field strength is determined by

$F_{\mu\nu}=-\frac{Q(\tau)}{r^{d-1}}(d\tau)_{\mu}\wedge(dr)_{\nu},~{}Q(\tau)=(d-2)\mu(\tau)r_{h}^{d-2}\,.$
(101)

Here $r_{h}$ is determined by $\pi L^{2}/d\beta$. In this case, the chemical
potential is time dependent. For $\text{tr}(\rho\sigma\rho\sigma)$, the
boundary condition of $\mu(\tau)$ is

$\mu(\tau)=\left\\{\begin{aligned} \mu_{\rho}\quad\quad 0<\tau\leq\beta\\\
\mu_{\sigma}~{}\quad\beta<\tau\leq 2\beta\\\ \mu_{\rho}\quad 2\beta<\tau\leq
3\beta\\\ \mu_{\sigma}\quad 3\beta<\tau\leq 4\beta\\\
\end{aligned}\right\\}\,.$ (102)

Here $\mu_{\rho}$ and $\mu_{\sigma}$ are constants. Although $\mu(\tau)$ is
time dependent, the field strength has the same nonvanishing component
$F_{rt}$. The direct calculation for $I_{\text{Maxwell}}$ for both
$\text{tr}(\rho\sigma\rho\sigma)$ and $\text{tr}(\rho^{2}\sigma^{2})$ reads

$\displaystyle I_{\text{Maxwell}}$ $\displaystyle=-\frac{1}{16\pi
G}\int_{\mathcal{B}}d^{d+1}x\sqrt{g}F_{\mu\nu}F^{\mu\nu}$ (103)
$\displaystyle=-\frac{2(d-2)\beta\Omega_{k,d-1}(\mu_{\rho}^{2}+\mu_{\sigma}^{2})r_{h}^{d-2}}{8\pi
G}\,.$

So we check the consistency of conclusion from section 3 in the probe limit,
which shows any two charged thermal state in the same Hilbert space are
commutative.

As shown in the previous two examples, this gives us a universal444At least in
the probe limit. holographic approach to check if the density matrices of two
holographic states are commutative to each other. As we investigated in this
section, the commutativity of $\rho$ and $\sigma$ are holographically
determined by the difference between two bulk geometries dual to
$\rho\sigma\rho\sigma$ or $\rho^{2}\sigma^{2}$.

##  6 Summary

To summary, this paper uses the holographic approach to calculate the fidelity
and relative entropy in some concrete examples. By employing the replica tick,
we can analytically calculate these quantum distance between thermal states.
The key point here is that for two thermal states, the product of their
respective density matrices is still a thermal state. In gravity side, this
indicates that the gravity dual of the replicated thermal state is continue to
be Schwarzschild-AdS black hole. We also investigate our approach under the
presence of matter field. By the virtue of $U$(1) invariance for the Maxwell
field, these conclusions for the thermal state can be generalized into the
charged system within the grand canonical ensemble. The similar step can be
also generalized into other cases. Without loss of generality, an Einstein-
scalar theory is constructed here to gain the insight into the process of
calculating the fidelity. For bulk matter, the replicated boundary field will
cause the inhomogeneous boundary condition for the equation of motion. In
general, the physical system we consider here will be time-dependent since the
bulk scalar field has inhomogeneous boundary condition. For simplification, we
ignore the backreaction from the scalar field to the metric. In other words,
the question becomes how to solve the scalar field under the background of
fixed Schwarzschild-AdS black hole. Then, we solve the corresponding scalar
field’s equation of motion in the probe limit. According to the standard
holographic renormalization, we derive the analytic expression of the
fidelity. The contribution from the scalar operator is a sum of an infinite
series. Then we give a numerical result to intuitively feel this contribution
here. It is found that on-shell action is divergent under the standard
quantization scheme while convergent under the alternative quantization
scheme. We observed that this divergency is caused by high frequency modes in
Fourier space. While these UV contributions from high frequency modes are
suppressed under the alternative quantization scheme. We have concluded that
it is a quite common behavior in terms of the bulk matter under inhomogeneous
boundary condition. At the end of this paper, a holographic method is
discovered to check whether the density matrices of two holographic states are
commutative. As examples, we use scalar and Maxwell field to demonstrate this
method in the probe limit. The result is expected that two holographic states
excited by scalar operator are generally not commutative and two charged
thermal states are commutative. However, it should be noted that we do not
really solve the Einstein field equation with full backreaction from time-
dependent chemical potential. To the best of our knowledge, the exact bulk
solution for time-dependent chemical potential is still unclear and worth to
explore.

Although this paper studies the fidelity between the states that are excited
by a scalar operator, it is also interesting to generalize the procedure into
cases of other operators. Further, it is meaningful to generalize our method
to study the quantum distance in the subsystem that is a more practical
physical system. As an important future problem, more general regularization
under the standard quantization scheme is needed to be explore. In the
presence of the matter field, the probe limit is very critical for our
calculation. In principle, someone can numerically calculate the quantum
distance without neglecting the backreaction of the matter field. What is
more, the divergency here though is also a UV divergency, it is not caused by
AdS boundary and so the usual holographic renormalization cannot not be used
to remove such divergency. We leave this problem in the future and call for
deep understanding.

##  Appendix A Euclidean Schwarzschild-AdSd+1 black hole

In $(d+1)$-dimensional AdS spacetime, the metric for neutral black is given by

$ds^{2}=-f(r)dt^{2}+\frac{dr^{2}}{f(r)}+\frac{r^{2}}{L^{2}}d\Sigma^{2}_{k,d-1}$
(104)

with

$f(r)=\frac{r^{2}}{L^{2}}+k-\frac{f_{0}}{r^{d-2}}\,.$ (105)

Where $L$ denote the AdS radius and $(d-1)$-dimensional metric
$d\Sigma^{2}_{k,d-1}$ is defined as

$d\Sigma_{k,d-1}^{2}=\begin{cases}L^{2}d\Omega_{d-1}^{2}&\text{ for }k=+1\\\
d\ell_{d-1}^{2}\equiv\sum_{i=1}^{d-1}dx_{i}^{2}&\text{ for }k=0\\\
L^{2}d\Xi_{d-1}^{2}&\text{ for }k=-1\end{cases}\,.$ (106)

The parameter $k={-1,0,+1}$ corresponds hyperbolic, planar and spherical
horizon geometries respectively. Where
$d\Omega_{d-1}^{2}/d\ell_{d-1}^{2}/d\Xi_{d-1}^{2}$ is the unit metric on
$(d-1)$-dimensional spherical/planar/hyperbolic space. In the following
discussion, $\Omega_{k,d-1}$ will be denoted as the dimensionless volume of
$d\Sigma_{k,d-1}^{2}/L^{2}$. Here $f_{0}$ is interpreted as the mass parameter
since that is related to the mass of black hole Emparan:1999pm

$M=\frac{(d-1)\Omega_{k,d-1}}{16\pi G_{N}}f_{0}\,.$ (107)

One can also get the following relation from the blackening factor
$f(r_{h})=0$

$f_{0}=r_{h}^{d-2}\left(\frac{r_{h}^{2}}{L^{2}}+k\right)\,.$ (108)

The Euclidean Schwarzschild solution is obtained from the ordinary
Schwarzschild metric by operating the Wick rotation $t\rightarrow-i\tau$. In
order to regulate the conical singularity located at $r_{h}$, the imaginary
time $\tau$ should be identified as

$\tau=\tau+\beta$ (109)

here the period $\beta=1/T$ under the natural unit $k_{B}=1$. $T$ is the
temperature of the black hole, which is given by

$T=\frac{1}{4\pi}\left.\frac{\partial f}{\partial
r}\right|_{r=r_{h}}=\frac{1}{4\pi
r_{h}}\left(d\frac{r_{h}^{2}}{L^{2}}+k(d-2)\right)\,.$ (110)

##  Appendix B Details about the solution for radial equation

After variable separation, we obtain the following equation, i.e. radial
equation(70)

$R_{j}^{\prime\prime}+\left(\frac{f^{\prime}}{f}-\frac{d-1}{z}\right)R_{j}^{\prime}-\left(\frac{m^{2}}{z^{2}f}+\frac{w_{j}^{2}}{f^{2}}\right)R_{j}=0\,,f(z)=1-z^{d}\,.$
(111)

In this paper, we assumes horizon locates at $z=z_{h}=1$ and
$w_{j}=\frac{dj}{2n}$. At the horizon we have following asymptotic solution

$R_{j}=a_{1}(1-z)^{w_{j}/d}(1+\cdots)+a_{2}(1-z)^{-w_{j}/d}(1+\cdots)\,.$
(112)

Near the AdS boundary we have following asymptotically behavior:

$R_{j}=\phi_{s}^{(j)}z^{d-\Delta}(1+\cdots)+\phi_{v}^{(j)}z^{\Delta}(1+\cdots)\,.$
(113)

The horizon at the Euclidean black hole will be a smooth tip and scalar field
should be finite here. Thus, we then have following boundary condition

$a_{2}=0\,.$ (114)

In subsection 4.3, we need to determine whether the series $\Xi[\rho_{v},1/2]$
converges in different quantization schemes. According to comparison test, we
obtain that

$\frac{\phi^{(j)}_{s}}{\phi^{(j)}_{v}}-\left.\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}\right|_{\text{thermal
AdS}}=\mathcal{O}(j^{a})\,,a=-3.81678\approx-4\,.$

We note that the numerical result of $a$ is very close to $-4$. In fact, we
can analytically confirm this through making a variable transformation for
radial equation,

$s=\int_{0}^{z}\frac{w_{j}}{f(x)}dx\,,~{}~{}R_{j}=Z_{j}(s)z(s)\,.$ (115)

Then the radial equation is reduced into

$\frac{d^{2}}{ds^{2}}Z(s)-\left[V(s)+1\right]Z(s)=0\,,~{}~{}V(s)=\frac{f(z)\left[m^{2}+(d-1)f(z)-zf^{\prime}(z)\right]}{w_{j}^{2}z^{2}}$
(116)

where $z$ is the function of $s$ defined by Eq. (115) and the boundary
condition at horizon becomes

$Z(s)\rightarrow 0\,,~{}~{}s\rightarrow\infty\,.$ (117)

From the definition (115) we see that $s=w_{j}z+\mathcal{O}(z^{d+1})$,
$f(z)=1-(s/w_{j})^{d}+\mathcal{O}(s/w_{j})^{d+1}$ and

$Z[s]=\phi_{v}^{(j)}w_{j}^{1-\Delta}s^{\Delta-1}(1+\cdots)+\phi_{s}^{(j)}w_{j}^{\Delta+1-d}s^{d-\Delta-1}(1+\cdots)$
(118)

near the AdS boundary.

In principle, Eq. (116) has no analytical solution. However, we can find an
analytically approximated solution, which approaches to the exact solution
when $w_{j}$ becomes large enough. We note that

$V(s)\rightarrow V^{(0)}(s)=\frac{m^{2}+d-1}{s^{2}}$ (119)

when $w_{j}\gg 1$. We then use $V^{(0)}(s)$ to replace the potential $V(s)$
and obtain the analytical solution

$Z_{j}^{(0)}(s)=b_{1}\sqrt{s}I_{\Delta-d/2}(s)+b_{2}\sqrt{s}K_{\Delta-d/2}(s)\,.$
(120)

Here $I_{v}(s)$ and $K_{v}(s)$ are the first and second kinds of modified
Bessel functions. To match the boundary condition (117), we have to set
$b_{1}=0$. For small $s$, we have

$2K_{v}(s)=\Gamma(v)\left(\frac{s}{2}\right)^{-v}(1+\cdots)+\Gamma(-v)\left(\frac{s}{2}\right)^{v}(1+\cdots)$
(121)

if $v$ is not an integer. We then have

$Z^{(0)}_{j}(s)=\frac{b_{2}}{2}\left[\Gamma(\Delta-d/2)2^{\Delta-d/2}s^{2-\Delta}(1+\cdots)+\Gamma(d/2-\Delta)2^{d/2-\Delta}s^{\Delta-1}(1+\cdots)\right]$
(122)

for small $s$. Using the relationship $s=w_{j}z+\mathcal{O}(z^{d+1})$ for
small $z$ and relationship (118), we then obtain

$\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}=-\frac{\sin[\pi(\Delta-\frac{d}{2})](\Delta-\frac{d}{2})\Gamma(\Delta-\frac{d}{2})^{2}}{\pi(w_{j}/2)^{2\Delta-d}}\,.$
(123)

For finite but large $c_{n}$, we can find that
$z=s/w_{j}+\mathcal{O}(1/w_{j}^{d+1})$ and so

$V(s)=V^{(0)}(s)+\frac{(m^{2}+d)s^{d-2}}{w_{j}^{d}}+\cdots\,.$ (124)

This shows that

$\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}=-\frac{\sin[\pi(\Delta-\frac{d}{2})](\Delta-\frac{d}{2})\Gamma(\Delta-\frac{d}{2})^{2}}{\pi(w_{j}/2)^{2\Delta-d}}\left[1+\mathcal{O}(w_{j}^{-d})\right]\,.$
(125)

We note that the solution in thermal AdS, i.e. $f(z)=1$ is equal to the
analytical solution (120) with $V^{(0)}$. In other words, the scalar modes
with $w_{j}\gg 1$ can only probe the UV region which is far from the horizon.
According to this fact and the above formula (125), we can get the difference

$\frac{\phi^{(j)}_{s}}{\phi^{(j)}_{v}}-\left.\frac{\phi_{s}^{(j)}}{\phi_{v}^{(j)}}\right|_{\text{thermal
AdS}}=\mathcal{O}(1/w_{j}^{2\Delta})\,.$ (126)

So we conclude that $a=-3.81678\approx-4$ is not a coincidence, which is
consistent with the previous analysis.

## Acknowledgements

This work is supported by the Natural Science Foundation of China under Grant
No. 12005155.

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