# High efficient multipartite entanglement purification using
hyperentanglement

Lan Zhou,1 Pei-Shun Yan,2 Wei Zhong,2 Yu-Bo<EMAIL_ADDRESS>1
School of Science, Nanjing University of Posts and Telecommunications,
Nanjing, 210003, China
2Institute of Quantum Information and Technology, Nanjing University of Posts
and Telecommunications, Nanjing, 210003, China

###### Abstract

Multipartite entanglement plays an important role in controlled quantum
teleportation, quantum secret sharing, quantum metrology and some other
important quantum information branches. However, the maximally multipartite
entangled state will degrade into the mixed state because of the noise. We
present an efficient multipartite entanglement purification protocol (EPP)
which can distill the high quality entangled states from low quality entangled
states for $N$-photon systems in a Greenberger-Horne-Zeilinger (GHZ) state in
only linear optics. After performing the protocol, the spatial-mode
entanglement is used to purify the polarization entanglement and one pair of
high quality polarization entangled state will be obtained. This EPP has
several advantages. Firstly, with the same purification success probability,
this EPP only requires one pair of multipartite GHZ state, while existing EPPs
usually require two pairs of multipartite GHZ state. Secondly, if consider the
practical transmission and detector efficiency, this EPP may be extremely
useful for the ratio of purification efficiency is increased rapidly with both
the number of photons and the transmission distance. Thirdly, this protocol
requires linear optics and does not add additional measurement operations, so
that it is feasible for experiment. All these advantages will make this
protocol have potential application for future quantum information processing.

###### pacs:

03.67.Lx

##  I Introduction

Entanglement plays an important role in quantum communication and computation.
Quantum teleportation teleportation , quantum key distributionQKD , dense
codingdensecoding , quantum secure direct communicationQSDC1 ; QSDC2 ; QSDC3 ,
distributed quantum computing computation , distributed secure quantum machine
learning DSQML , and other important branches all require the parties to share
the entanglement. Besides the bipartite entanglement, multipartite
entanglement also plays an important role in controlled quantum teleportation
cteleportation1 ; cteleportation2 , quantum secret sharingQSS1 ; QSS2 ; QSS3 ,
quantum state sharingQSTS1 ; QSTS2 ; QSTS3 , quantum metrologymetrology1 ;
metrology2 , and so on. Recently, the multipartite entangled state named
Greenberger-Horne-Zeilinger (GHZ) states was been used in some important
quantum communication experiment, such as long-distance measurement-device-
independent multiparty quantum communication chenzb , equitable multiparty
quantum communication without a trusted third party jeong1  and quantum
teleportation of shared quantum secret jeong2 . The GHZ state also have been
realized with superconducting system superconduct1 ; superconduct2 , trapped
ions ion , and photonic system photon . The detection of the multipartite
entanglement structure has also been reported detection .

In a practical application, the quantum system should inevitably interact with
its environment, and the environment noise will degrade the entanglement. In
general, the decoherence will make the maximally entangled state become a
mixed state. The degraded entanglement will decrease the efficiency of the
quantum communication and it also will make the quantum communication become
insecure repeater . Entanglement purification is a powerful tool to distill
the high quality entangled states from the low quality entangled states
purification1 ; purification2 ; purification3 ; purification4 ; purification5
; addpurification1 ; addpurification3 ; experiment2 ; purification6 ;
purification7 ; purification8 ; purification9 ; purification11 ;
purification12 ; purification13 ; purification14 ; purification15 ;
purification16 ; purification17 ; purification18 ; purification19 ;
purification20 ; purification21 ; purification22 ; purification23 ;
purification24 ; purification25 ; purification26 ; addpurification2 ;
purification27 ; addpurification4 ; shengprl . Entanglement purification has
been widely discussed since Bennett et al. proposed the first entanglement
purification protocol (EPP) purification1 . EPPs for bipartite system were
proposed in photonic system purification3 ; purification4 ; purification5 ;
purification7 ; purification8 ; purification9 , electionsaddpurification1 ,
quantum dotspurification11 ; purification12 , atoms experiment2 ;
purification13  and so on. For example, in 2001, Pan et al. proposed the
polarization EPP with only feasible linear optics elements purification3 . In
2008, the EPP with spontaneous parametric down conversion (SPDC) source based
on cross-Kerr nonlinearity was proposed purification7 . In this EPP, the
purified high quality entangled state can be remained for further application
and the remained entangled states can also be repeated to perform the
purification to obtain the higher entangled states. The experiments of
entanglement purification in optical system were also reported purification4 .
In 2017, Chen et al. realized the nested entanglement purification for quantum
repeaters. In this experiment, the entanglement purification and entanglement
swapping can be realized simultaneously purification17 . On the other hand,
the double-pair noise components from the SPDC source can be eliminated
automatically. This work was extended to the multi-copy cases addpurification2
. The optimal entanglement purification was also investigated purification26 .
Recently, the first high efficient and long-distance entanglement purification
using hyperentanglement was demonstrated shengprl . The hyperentanglement was
first distributed to 11 km and the spatial entanglement was used to
purification polarization entanglement. The authors also demonstrated its
powerful application in entanglement-based QKD.

For multipartite system, Murao et al. described the first EPP with controlled-
not (CNOT) gate multipurification1 . In 2003, Dur et al. described the EPP for
Graph state multipurification2 . In 2007, this protocol was extended to high-
dimension multipartite system with the generalized CNOT gate
multipurification3 . In 2008, the multipartite EPP for polarization entangled
states with cross-Kerr nonlinearity was proposed multipurification4 . In 2011,
Deng proposed the multipartite EPP using entanglement link from subspace
multipurification5 . In his protocol, the discussed items in conventional EPPs
still have entanglement in a subspace and they can be reused with entanglement
link. There are another kind of EPP for multipartite entanglement system,
named deterministic EPP multipurification6 ; multipurification7 . In these
EPPs, they exploit the hyperentanglement to perform the purification. Such
EPPs are based on the condition that the spatial mode entanglement is robust
and it does not suffer from the noise. Therefore, the spatial mode
entanglement or the frequency entanglement can be completely transformed to
the polarization entanglement. In conventional EPPs, they all require two
copies of low quality entangled states. After performing the CNOT or similar
operations, one pair of high quality entangled state is remained, if the
purification is successful. On the other hand, if the purification is a
failure, both pairs should be discarded.

In this paper, we will describe an efficient EPP for multipartite polarization
entangled systems in a GHZ state, inspired the idea of Ref.shengprl  .
Different from existing EPPs for multipartite system, this protocol only
requires one pair of hyperentangled state. By performing the CNOT gate between
two degrees of freedom, the spatial entanglement is consumed. Therefore, if
the protocol is successful, one can obtain a high quality polarization
entangled state. This EPP is based on linear optics and it is also feasible
for current experiment condition.

This protocol is organized as follows. In Sec.II, we describe this EPP for
bit-flip error. In Sec.III, we describe this protocol for phase-flip error. In
Sec.IV, we extend this EPP to a general case for arbitrary N-photon GHZ state.
In Sec.V, we present a discussion. Finally, in Sec. VI, we will provide a
conclusion.

##  II Multipartite entanglement purification for bit-flip error

In this section, we describe this EPP with a simple example. The three-photon
GHZ states can be written as follows.

$\displaystyle|\Phi_{0}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}\pm|V\rangle_{A}|V\rangle_{B}|V\rangle_{C}),$
$\displaystyle|\Phi_{1}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}\pm|V\rangle_{A}|V\rangle_{B}|H\rangle_{C}),$
$\displaystyle|\Phi_{2}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|H\rangle_{A}|V\rangle_{B}|H\rangle_{C}\pm|V\rangle_{A}|H\rangle_{B}|V\rangle_{C}),$
$\displaystyle|\Phi_{3}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|V\rangle_{A}|H\rangle_{B}|H\rangle_{C}\pm|H\rangle_{A}|V\rangle_{B}|V\rangle_{C}).$
(1)

Here $|H\rangle$ denotes the horizonal polarization and $|V\rangle$ denotes
the vertical polarization of the photon, respectively.

Figure 1: Schematic drawing showing the principle of entanglement purification
for bit-flip error. HWP45 is the half-wave plate setting as 45∘. The PBS is
the polarization beam splitter which can transmit the $|H\rangle$ polarization
and reflect the $|V\rangle$ polarization photon. The polarizing beam
displacers (BD) can couple $|H\rangle$ and $|V\rangle$ polarization components
from different spatial modes.

The spatial mode GHZ states can be written as follows.

$\displaystyle|\phi_{0}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}\pm|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}),$
$\displaystyle|\phi_{1}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}\pm|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}),$
$\displaystyle|\phi_{2}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}\pm|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}),$
$\displaystyle|\phi_{3}^{\pm}\rangle_{ABC}=\frac{1}{\sqrt{2}}(|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}\pm|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}).$
(2)

Here a1, a2, b1, b2, c1 and c2 are the spatial modes as shown in Fig. 1. The
PBS is the polarization beam splitter which can transmit the $|H\rangle$
polarization and reflect the $|V\rangle$ polarization photon. The polarizing
beam displacers (BD) can couple $|H\rangle$ and $|V\rangle$ polarization
components from different spatial modes. HWP45 is the half-wave plate setting
as 45∘. It can convert $|H\rangle$ polarization to $|V\rangle$ and $|V\rangle$
to $|H\rangle$ , respectively. The entanglement source $S$ emits a pair of
hyperentangled GHZ state of the form

$\displaystyle|\Psi\rangle_{ABC}=|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}.$
(3)

The hyperentangled GHZ state is distributed to Alice, Bob and Charlie,
respectively. During distribution, if a bit-flip error occurs on both the
polarization and spatial modes entangled state, the original state
$|\Psi\rangle_{ABC}$ will become a mixed state as

$\displaystyle\rho_{ABC}=\rho^{P}_{ABC}\otimes\rho^{S}_{ABC}.$ (4)

Here $\rho^{P}_{ABC}$ and $\rho^{S}_{ABC}$ are the mixed state in polarization
and spatial modes. They can be written as

$\displaystyle\rho^{P}_{ABC}=F_{1}|\Phi_{0}^{+}\rangle_{ABC}\langle\Phi_{0}^{+}|+(1-F_{{}_{1}})|\Phi_{1}^{+}\rangle_{ABC}\langle\Phi_{1}^{+}|,$
(5)

and

$\displaystyle\rho^{S}_{ABC}=F_{2}|\phi_{0}^{+}\rangle_{ABC}\langle\phi_{0}^{+}|+(1-F_{{}_{2}})|\phi_{1}^{+}\rangle_{ABC}\langle\phi_{1}^{+}|.$
(6)

From Eq.(4), the mixed state $\rho_{ABC}$ can be described as follows. With
the probability of $F_{1}\otimes F_{2}$, it is in the state
$|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$. With the
probability of $(1-F_{1})(1-F_{2})$, it is in the state
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{1}^{+}\rangle_{ABC}$. With the
probability of $F_{1}(1-F_{2})$ and $(1-F_{{}_{1}})F_{2}$, they are in the
states $|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{1}^{+}\rangle_{ABC}$ and
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$, respectively.
The first case $|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$
can be described as

$\displaystyle|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$ (7)
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|V\rangle_{C})$
$\displaystyle\otimes$
$\displaystyle\frac{1}{\sqrt{2}}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C})$
$\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{a_{1}}|H\rangle_{b_{1}}|H\rangle_{c_{1}}+|H\rangle_{a_{2}}|H\rangle_{b_{2}}|H\rangle_{c_{2}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{1}}|V\rangle_{b_{1}}|V\rangle_{c_{1}}+|V\rangle_{a_{2}}|V\rangle_{b_{2}}|V\rangle_{c_{2}})$
$\displaystyle\rightarrow$
$\displaystyle\frac{1}{2}(|V\rangle_{a_{3}}|V\rangle_{b_{3}}|V\rangle_{c_{3}}+|H\rangle_{a_{6}}|H\rangle_{b_{6}}|H\rangle_{c_{6}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{5}}|V\rangle_{b_{5}}|V\rangle_{c_{5}}+|H\rangle_{a_{4}}|H\rangle_{b_{4}}|H\rangle_{c_{4}}).$

From Fig. 1, items $|V\rangle_{a_{3}}|V\rangle_{b_{3}}|V\rangle_{c_{3}}$ and
$|H\rangle_{a_{4}}|H\rangle_{b_{4}}|H\rangle_{c_{4}}$ will couple in the BD1,
BD2, and BD3, respectively, and finally become the polarization entangled
state $|\Phi_{0}^{+}\rangle$ in the output modes D1D2D3. On the other hand,
items $|H\rangle_{a_{6}}|H\rangle_{b_{6}}|H\rangle_{c_{6}}$ and
$|V\rangle_{a_{5}}|V\rangle_{b_{5}}|V\rangle_{c_{5}}$ will also couple in the
BD4, BD5 and BD6 and become the polarization entangled state
$|\Phi_{0}^{+}\rangle$ in the output modes D4D5D6.

The second case $|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{1}^{+}\rangle_{ABC}$
can evolve as

$\displaystyle|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{1}^{+}\rangle_{ABC}$ (8)
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|H\rangle_{C})$
$\displaystyle\otimes$
$\displaystyle\frac{1}{\sqrt{2}}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C})$
$\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{a_{1}}|H\rangle_{b_{1}}|V\rangle_{c_{2}}+|H\rangle_{a_{2}}|H\rangle_{b_{2}}|V\rangle_{c_{1}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{1}}|V\rangle_{b_{1}}|H\rangle_{c_{2}}+|V\rangle_{a_{2}}|V\rangle_{b_{2}}|H\rangle_{c_{1}})$
$\displaystyle\rightarrow$
$\displaystyle\frac{1}{2}(|V\rangle_{a_{3}}|V\rangle_{b_{3}}|H\rangle_{c_{4}}+|H\rangle_{a_{6}}|H\rangle_{b_{6}}|V\rangle_{c_{5}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{5}}|V\rangle_{b_{5}}|H\rangle_{c_{6}}+|H\rangle_{a_{4}}|H\rangle_{b_{4}}|V\rangle_{c_{3}}).$

Items $|V\rangle_{a_{3}}|V\rangle_{b_{3}}|H\rangle_{c_{4}}$ and
$|H\rangle_{a_{4}}|H\rangle_{b_{4}}|V\rangle_{c_{3}}$ will become
$|\Phi_{1}^{+}\rangle$ in the output modes D1D2D3. Items
$|H\rangle_{a_{6}}|H\rangle_{b_{6}}|V\rangle_{c_{5}}$ and
$|V\rangle_{a_{5}}|V\rangle_{b_{5}}|H\rangle_{c_{6}}$ will also become
$|\Phi_{1}^{+}\rangle$ in the output modes D4D5D6.

On the other hand, the cases
$|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{1}^{+}\rangle_{ABC}$ and
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$ cannot make the
three photons in the output modes D1D2D3 or D4D5D6. For example,
$|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{1}^{+}\rangle_{ABC}$ will lead the
three photons become $|\Phi_{1}^{+}\rangle$ in output modes D1D2D6 or D4D5D3.
Case $|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$ will become
$|\Phi_{0}^{+}\rangle$ in the output modes D1D2D6 or D4D5D3. In this way, by
selecting the output modes D1D2D3 or D4D5D6 each having a photon, they can
ultimately obtain a new mixed state

$\displaystyle\rho^{\prime}_{ABC}=F^{\prime}|\Phi_{0}^{+}\rangle_{ABC}\langle\Phi_{0}^{+}|+(1-F^{\prime})|\Phi_{1}^{+}\rangle_{ABC}\langle\Phi_{1}^{+}|.$
(9)

Here $F^{\prime}$ is

$\displaystyle F^{\prime}=\frac{F_{1}F_{2}}{F_{1}F_{2}+(1-F_{1})(1-F_{2})}.$
(10)

Obviously, if $F_{1}>\frac{1}{2}$ and $F_{2}>\frac{1}{2}$, we can obtain
$F^{\prime}>F_{1}$ and $F^{\prime}>F_{2}$. In this way, we complete the
purification of bit-flip error and the success probability is
$F_{1}F_{2}+(1-F_{1})(1-F_{2})$.

##  III Multipartite entanglement purification for phase-flip error

In this section, we will describe the purification of phase-flip error. If a
phase-flip error occurs in polarization part and spatial mode part, the mixed
state can be written as

$\displaystyle\varrho_{ABC}=\varrho^{P}_{ABC}\otimes\varrho^{S}_{ABC}.$ (11)

Here

$\displaystyle\varrho^{P}_{ABC}=F_{3}|\Phi_{0}^{+}\rangle_{ABC}\langle\Phi_{0}^{+}|+(1-F_{3})|\Phi_{0}^{-}\rangle_{ABC}\langle\Phi_{0}^{-}|,$
(12)

and

$\displaystyle\varrho^{S}_{ABC}=F_{4}|\phi_{0}^{+}\rangle_{ABC}\langle\phi_{0}^{+}|+(1-F_{4})|\phi_{0}^{-}\rangle_{ABC}\langle\phi_{0}^{-}|.$
(13)

Figure 2: Schematic drawing showing the principle of entanglement purification
for phase-flip error. The HWP22.5 can transform the $|H\rangle$ polarization
to $\frac{1}{\sqrt{2}}(|H\rangle+|V\rangle)$ and $|V\rangle$ polarization to
$\frac{1}{\sqrt{2}}(|H\rangle-|V\rangle)$. The BS is the 50:50 beam splitter.
$|a_{1}\rangle\rightarrow\frac{1}{\sqrt{2}}(|a_{1}\rangle+|a_{2}\rangle)$ and
$|a_{2}\rangle\rightarrow\frac{1}{\sqrt{2}}(|a_{1}\rangle-|a_{2}\rangle)$.
Therefore, the HWP22.5 and BS both act as the role of Hadamard operation for
polarization and spatial mode qubits, respectively.

The principle of phase-flip error is shown in Fig.2. Before purification, they
should transform the phase-flip error to bit-flip error using the setup
$T_{i},i=1,2,3$. Here HWP22.5 can perform the Hadamard operation and make
$|H\rangle\rightarrow\frac{1}{\sqrt{2}}(|H\rangle+|V\rangle)$ and
$|V\rangle\rightarrow\frac{1}{\sqrt{2}}(|H\rangle-|V\rangle)$. The beam
splitter (BS) can also act as the role of Hadamard operation for spatial mode
qubit. It can make
$|a_{1}\rangle\rightarrow\frac{1}{\sqrt{2}}(|a_{1}\rangle+|a_{2}\rangle)$ and
$|a_{2}\rangle\rightarrow\frac{1}{\sqrt{2}}(|a_{1}\rangle-|a_{2}\rangle)$.
After performing the Hadamard operation, the GHZ states in polarization and
spatial mode as shown in Eq. (1) and (2) can be rewritten as

$\displaystyle|\Psi_{0}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|H\rangle_{A}|V\rangle_{B}|V\rangle_{C}$
$\displaystyle+$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|H\rangle_{C}),$
$\displaystyle|\Psi_{0}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|H\rangle_{A}|V\rangle_{B}|H\rangle_{C}$
$\displaystyle+$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|V\rangle_{C}),$
$\displaystyle|\Psi_{1}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|H\rangle_{A}|V\rangle_{B}|V\rangle_{C}$
$\displaystyle-$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|V\rangle_{C}-|V\rangle_{A}|V\rangle_{B}|H\rangle_{C}),$
$\displaystyle|\Psi_{1}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|H\rangle_{A}|V\rangle_{B}|H\rangle_{C}$
$\displaystyle-$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|H\rangle_{C}-|V\rangle_{A}|V\rangle_{B}|V\rangle_{C}),$
$\displaystyle|\Psi_{2}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}-|H\rangle_{A}|V\rangle_{B}|V\rangle_{C}$
$\displaystyle+$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|V\rangle_{C}-|V\rangle_{A}|V\rangle_{B}|H\rangle_{C}),$
$\displaystyle|\Psi_{2}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}-|H\rangle_{A}|V\rangle_{B}|H\rangle_{C}$
$\displaystyle+$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|H\rangle_{C}-|V\rangle_{A}|V\rangle_{B}|V\rangle_{C}),$
$\displaystyle|\Psi_{3}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}-|H\rangle_{A}|V\rangle_{B}|V\rangle_{C}$
$\displaystyle-$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|H\rangle_{C}),$
$\displaystyle|\Psi_{3}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}-|H\rangle_{A}|V\rangle_{B}|H\rangle_{C}$
$\displaystyle-$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|V\rangle_{C}),$
$\displaystyle|\psi_{0}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}$
$\displaystyle+$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}),$
$\displaystyle|\psi_{0}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}$
$\displaystyle+$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}),$
$\displaystyle|\psi_{1}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}$
$\displaystyle-$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}-|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}),$
$\displaystyle|\psi_{1}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}$
$\displaystyle-$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}-|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}),$
$\displaystyle|\psi_{2}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}-|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}$
$\displaystyle+$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}-|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}),$
$\displaystyle|\psi_{2}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}-|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}$
$\displaystyle+$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}-|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}),$
$\displaystyle|\psi_{3}^{+}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}-|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}$
$\displaystyle-$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}),$
$\displaystyle|\psi_{3}^{-}\rangle_{ABC}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}-|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}$
$\displaystyle-$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}).$

After performing the Hadamard operations, the mixed state in Eq.(11) can be
rewritten as

$\displaystyle\sigma_{ABC}=\sigma^{P}_{ABC}\otimes\sigma^{S}_{ABC}.$ (16)

Here

$\displaystyle\sigma^{P}_{ABC}=F_{3}|\Psi_{0}^{+}\rangle_{ABC}\langle\Psi_{0}^{+}|+(1-F_{3})|\Psi_{0}^{-}\rangle_{ABC}\langle\Psi_{0}^{-}|,$
(17)

and

$\displaystyle\sigma^{S}_{ABC}=F_{4}|\psi_{0}^{+}\rangle_{ABC}\langle\psi_{0}^{+}|+(1-F_{4})|\psi_{0}^{-}\rangle_{ABC}\langle\psi_{0}^{-}|.$
(18)

Therefore, $\sigma_{ABC}$ can be described as follows. With the probability of
$F_{3}F_{4}$, it is in the state
$|\Psi_{0}^{+}\rangle_{ABC}\otimes|\psi_{0}^{+}\rangle_{ABC}$. With the
probability of $F_{3}(1-F_{4})$, it is in the state
$|\Psi_{0}^{+}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$. With the
probability of $(1-F_{3})F_{4}$, it is in the state
$|\Psi_{0}^{-}\rangle_{ABC}\otimes|\psi_{0}^{+}\rangle_{ABC}$. With the
probability of $(1-F_{3})(1-F_{4})$, it is in the state
$|\Psi_{0}^{-}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$. We first
discuss the case
$|\Psi_{0}^{+}\rangle_{ABC}\otimes|\psi_{0}^{+}\rangle_{ABC}$. It will evolve
as

$\displaystyle|\Psi_{0}^{+}\rangle_{ABC}\otimes|\psi_{0}^{+}\rangle_{ABC}$
(19) $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|H\rangle_{A}|V\rangle_{B}|V\rangle_{C}$
$\displaystyle+$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|H\rangle_{C})$
$\displaystyle\otimes$
$\displaystyle\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C}$
$\displaystyle+$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C})$
$\displaystyle=$
$\displaystyle\frac{1}{4}(|H\rangle_{a_{1}}|H\rangle_{b_{1}}|H\rangle_{c_{1}}+|H\rangle_{a_{1}}|H\rangle_{b_{2}}|H\rangle_{c_{2}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{2}}|H\rangle_{b_{1}}|H\rangle_{c_{2}}+|H\rangle_{a_{2}}|H\rangle_{b_{2}}|H\rangle_{c_{1}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{1}}|V\rangle_{b_{1}}|V\rangle_{c_{1}}+|H\rangle_{a_{1}}|V\rangle_{b_{2}}|V\rangle_{c_{2}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{2}}|V\rangle_{b_{1}}|V\rangle_{c_{2}}+|H\rangle_{a_{2}}|V\rangle_{b_{2}}|V\rangle_{c_{1}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{1}}|H\rangle_{b_{1}}|V\rangle_{c_{1}}+|V\rangle_{a_{1}}|H\rangle_{b_{2}}|V\rangle_{c_{2}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{2}}|H\rangle_{b_{1}}|V\rangle_{c_{2}}+|V\rangle_{a_{2}}|H\rangle_{b_{2}}|V\rangle_{c_{1}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{1}}|V\rangle_{b_{1}}|H\rangle_{c_{1}}+|V\rangle_{a_{1}}|V\rangle_{b_{2}}|H\rangle_{c_{2}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{2}}|V\rangle_{b_{1}}|H\rangle_{c_{2}}+|V\rangle_{a_{2}}|V\rangle_{b_{2}}|H\rangle_{c_{1}})$
$\displaystyle\rightarrow$
$\displaystyle\frac{1}{4}(|V\rangle_{a_{3}}|V\rangle_{b_{3}}|V\rangle_{c_{3}}+|V\rangle_{a_{3}}|H\rangle_{b_{6}}|H\rangle_{c_{6}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{6}}|V\rangle_{b_{3}}|H\rangle_{c_{6}}+|H\rangle_{a_{6}}|H\rangle_{b_{6}}|V\rangle_{c_{3}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{3}}|V\rangle_{b_{5}}|V\rangle_{c_{5}}+|V\rangle_{a_{3}}|H\rangle_{b_{4}}|H\rangle_{c_{4}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{6}}|V\rangle_{b_{5}}|H\rangle_{c_{4}}+|H\rangle_{a_{6}}|H\rangle_{b_{4}}|V\rangle_{c_{5}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{5}}|V\rangle_{b_{3}}|V\rangle_{c_{5}}+|V\rangle_{a_{5}}|H\rangle_{b_{6}}|H\rangle_{c_{4}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{4}}|V\rangle_{b_{3}}|H\rangle_{c_{4}}+|H\rangle_{a_{4}}|H\rangle_{b_{6}}|V\rangle_{c_{5}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{5}}|V\rangle_{b_{5}}|V\rangle_{c_{3}}+|V\rangle_{a_{5}}|H\rangle_{b_{4}}|H\rangle_{c_{6}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{4}}|V\rangle_{b_{5}}|H\rangle_{c_{6}}+|H\rangle_{a_{4}}|H\rangle_{b_{4}}|V\rangle_{c_{3}}).$

From Eq. (19), item $|V\rangle_{a_{3}}|V\rangle_{b_{3}}|V\rangle_{c_{3}}$,
$|V\rangle_{a_{3}}|H\rangle_{b_{4}}|H\rangle_{c_{4}}$,
$|H\rangle_{a_{4}}|V\rangle_{b_{3}}|H\rangle_{c_{4}}$ and
$|H\rangle_{a_{4}}|H\rangle_{b_{4}}|V\rangle_{c_{3}}$ will be in the output
modes D1D2D3 and become

$\displaystyle\frac{1}{2}(|V\rangle_{D_{1}}|V\rangle_{D_{2}}|V\rangle_{D_{3}}+|V\rangle_{D_{1}}|H\rangle_{D_{2}}|H\rangle_{D_{3}}$
(20) $\displaystyle+$
$\displaystyle|H\rangle_{D_{1}}|V\rangle_{D_{2}}|H\rangle_{D_{3}}+|H\rangle_{D_{1}}|H\rangle_{D_{2}}|V\rangle_{D_{3}}).$

By performing bit-flip operation on each photon, state in Eq.(20) can be
changed to $|\Psi_{0}^{+}\rangle_{ABC}$. Finally, they can change
$|\Psi_{0}^{+}\rangle_{ABC}$ to $|\Phi_{0}^{+}\rangle_{ABC}$ by adding another
Hadamard operation on each photon. On the other hand, from Eq.(19), by
selecting the output modes D1D5D6, D4D5D3, or D4D2D6, they can obtain the same
state in Eq.(20). In this way, they can also obtain
$|\Phi_{0}^{+}\rangle_{ABC}$.

The case $|\Psi_{0}^{-}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$ can be
evolve as

$\displaystyle|\Psi_{0}^{-}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$
(21) $\displaystyle=$
$\displaystyle\frac{1}{2}(|H\rangle_{A}|H\rangle_{B}|V\rangle_{C}+|H\rangle_{A}|V\rangle_{B}|H\rangle_{C}$
$\displaystyle+$
$\displaystyle|V\rangle_{A}|H\rangle_{B}|H\rangle_{C}+|V\rangle_{A}|V\rangle_{B}|V\rangle_{C})$
$\displaystyle\otimes\frac{1}{2}(|a_{1}\rangle_{A}|b_{1}\rangle_{B}|c_{2}\rangle_{C}+|a_{1}\rangle_{A}|b_{2}\rangle_{B}|c_{1}\rangle_{C}$
$\displaystyle+$
$\displaystyle|a_{2}\rangle_{A}|b_{1}\rangle_{B}|c_{1}\rangle_{C}+|a_{2}\rangle_{A}|b_{2}\rangle_{B}|c_{2}\rangle_{C})$
$\displaystyle\rightarrow\frac{1}{4}(|V\rangle_{a_{3}}|V\rangle_{b_{3}}|H\rangle_{c_{4}}+|V\rangle_{a_{3}}|H\rangle_{b_{6}}|V\rangle_{c_{5}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{6}}|V\rangle_{b_{3}}|V\rangle_{c_{5}}+|H\rangle_{a_{6}}|H\rangle_{b_{6}}|H\rangle_{c_{4}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{3}}|V\rangle_{b_{5}}|H\rangle_{c_{6}}+|V\rangle_{a_{3}}|H\rangle_{b_{4}}|V\rangle_{c_{3}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{6}}|V\rangle_{b_{5}}|V\rangle_{c_{3}}+|H\rangle_{a_{6}}|H\rangle_{b_{6}}|H\rangle_{c_{6}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{5}}|V\rangle_{b_{3}}|H\rangle_{c_{6}}+|V\rangle_{a_{5}}|H\rangle_{b_{6}}|V\rangle_{c_{3}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{4}}|V\rangle_{b_{3}}|V\rangle_{c_{3}}+|H\rangle_{a_{4}}|H\rangle_{b_{6}}|H\rangle_{c_{6}}$
$\displaystyle+$
$\displaystyle|V\rangle_{a_{5}}|V\rangle_{b_{5}}|H\rangle_{c_{4}}+|V\rangle_{a_{5}}|H\rangle_{b_{4}}|H\rangle_{c_{5}}$
$\displaystyle+$
$\displaystyle|H\rangle_{a_{4}}|V\rangle_{b_{5}}|V\rangle_{c_{5}}+|H\rangle_{a_{4}}|H\rangle_{b_{4}}|H\rangle_{c_{4}}).$

From Eq.(21), items $|V\rangle_{a_{3}}|V\rangle_{b_{3}}|H\rangle_{c_{4}}$,
$|V\rangle_{a_{3}}|H\rangle_{b_{4}}|V\rangle_{c_{3}}$,
$|H\rangle_{a_{4}}|V\rangle_{b_{3}}|V\rangle_{c_{3}}$ and
$|H\rangle_{a_{4}}|H\rangle_{b_{4}}|H\rangle_{c_{4}}$ will be in the output
modes D1D2D3 and become

$\displaystyle\frac{1}{2}(|V\rangle_{D_{1}}|V\rangle_{D_{2}}|H\rangle_{D_{3}}+|V\rangle_{D_{1}}|H\rangle_{D_{2}}|V\rangle_{D_{3}}$
(22) $\displaystyle+$
$\displaystyle|H\rangle_{D_{1}}|V\rangle_{D_{2}}|V\rangle_{D_{3}}+|H\rangle_{D_{1}}|H\rangle_{D_{2}}|H\rangle_{D_{3}}).$

State in Eq.(22) can be changed to $|\Psi_{0}^{-}\rangle_{ABC}$ by adding bit-
flip operation on each photon. By adding another Hadamard operation on each
photon, $|\Psi_{0}^{-}\rangle_{ABC}$ can be converted to
$|\Phi_{0}^{-}\rangle_{ABC}$. On the other hand, from Eq.(21), by selecting
the output modes D1D5D6, D4D5D3, or D4D2D6, they can obtain the same state in
Eq.(22). In this way, they can also obtain $|\Phi_{0}^{-}\rangle_{ABC}$.

The other cases $|\Psi_{0}^{+}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$
and $|\Psi_{0}^{-}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$ will lead
the photons in the output modes D1D2D6, D1D5D3, D4D2D3, or D4D5D6. Therefore,
by selecting the cases that the output modes D1D2D3, D1D5D6, D4D5D3, or D4D2D6
contain one photon, cases
$|\Psi_{0}^{+}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$ and
$|\Psi_{0}^{-}\rangle_{ABC}\otimes|\psi_{0}^{-}\rangle_{ABC}$ can be
eliminated automatically. Finally, with the probability of $F_{3}F_{4}$, they
will obtain $|\Phi_{0}^{+}\rangle_{ABC}$. With the probability of
$(1-F_{3})(1-F_{4})$, they will obtain $|\Phi_{0}^{-}\rangle_{ABC}$. The new
mixed state can be rewritten as

$\displaystyle\sigma^{\prime}_{ABC}=F^{\prime\prime}|\Psi_{0}^{+}\rangle_{ABC}\langle\Psi_{0}^{+}|+(1-F^{\prime\prime})|\Psi_{0}^{-}\rangle_{ABC}\langle\Psi_{0}^{-}|.$

Here $F^{\prime\prime}$ is

$\displaystyle
F^{\prime\prime}=\frac{F_{3}F_{4}}{F_{3}F_{4}+(1-F_{3})(1-F_{4})}.$ (24)

Similar to Eq.(10), $F^{\prime\prime}>F_{3}$ and $F^{\prime\prime}>F_{4}$ if
$F_{3}>\frac{1}{2}$ and $F_{4}>\frac{1}{2}$.

##  IV Arbitrary multipartite entanglement purification

It is easy to extend the EPP to the arbitrary GHZ state. The $m$-photon GHZ
state in polarization can be described as

$\displaystyle|\Phi_{0}^{\pm}\rangle_{m}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|H\rangle_{1}|H\rangle_{2}\cdots|H\rangle_{m}\pm|V\rangle_{1}|V\rangle_{2}\cdots|V\rangle_{m}),$
$\displaystyle|\Phi_{1}^{\pm}\rangle_{m}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|H\rangle_{1}|H\rangle_{2}\cdots|V\rangle_{m}\pm|V\rangle_{1}|V\rangle_{2}\cdots|H\rangle_{m}),$
$\displaystyle\cdots$ , $\displaystyle|\Phi_{2^{m-1}}^{\pm}\rangle_{m}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|V\rangle_{1}|H\rangle_{2}\cdots|H\rangle_{m}\pm|H\rangle_{1}|V\rangle_{2}\cdots|V\rangle_{m}),$

Figure 3: Schematic drawing showing the principle of entanglement purification
for arbitrary GHZ state.

On the other hand, the $m$-photon GHZ state in spatial mode can be described
as

$\displaystyle|\phi_{0}^{\pm}\rangle_{m}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|a_{1}\rangle_{1}|b_{1}\rangle_{2}$
$\displaystyle\cdots$
$\displaystyle|m_{1}\rangle_{m}\pm|a_{2}\rangle_{1}|b_{2}\rangle_{2}\cdots|m_{2}\rangle_{m}),$
$\displaystyle|\phi_{1}^{\pm}\rangle_{m}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|a_{1}\rangle_{1}|b_{1}\rangle_{2}$
$\displaystyle\cdots$
$\displaystyle|m_{2}\rangle_{m}\pm|a_{2}\rangle_{1}|b_{2}\rangle_{2}\cdots|m_{1}\rangle_{m}),$
$\displaystyle\cdots$ , $\displaystyle|\phi_{2^{m-1}}^{\pm}\rangle_{m}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|a_{2}\rangle_{1}|b_{1}\rangle_{2}$ (26)
$\displaystyle\cdots$
$\displaystyle|m_{1}\rangle_{m}\pm|a_{1}\rangle_{1}|b_{1}\rangle_{2}\cdots|m_{2}\rangle_{m}).$

As shown in Fig. 3, the entanglement source prepares the $m$-photon
hyperentangled state $|\Psi\rangle_{m}$ of the form

$\displaystyle|\Psi\rangle_{m}=|\Phi_{0}^{+}\rangle_{m}\otimes|\phi_{0}^{+}\rangle_{m}.$
(27)

Such hyperentangled state is distributed to $m$ parties, named Bob1, Bob2,
$\cdots$, and Bobm. After distribution, the initial state becomes a mixed
state as

$\displaystyle\rho_{m}=\rho^{P}_{m}\otimes\rho^{P}_{m}.$ (28)

Here $\rho^{P}_{m}$ can be written as

$\displaystyle\rho^{P}_{m}=F_{1}|\Phi_{0}^{+}\rangle_{m}\langle\Phi_{0}^{+}|+(1-F_{{}_{1}})|\Phi_{1}^{+}\rangle_{m}\langle\Phi_{1}^{+}|.$
(29)

$\rho^{P}_{m}$ can be written as

$\displaystyle\rho^{S}_{m}=F_{2}|\phi_{0}^{+}\rangle_{m}\langle\phi_{0}^{+}|+(1-F_{{}_{2}})|\phi_{1}^{+}\rangle_{m}\langle\phi_{1}^{+}|.$
(30)

The purification is similar as described in above section. By selecting the
output modes D1,D2, $\cdots$, Dm exactly contain one photon, they can
ultimately obtain a high fidelity mixed state in polarization. The fidelity
$F^{\prime}$ is same as it is shown in Eq. (10). On the other hand, if the
phase-flip error occurs, one can also convert it to the bit-flip error, and
perform the purification in a next step. In this way, one can purify the
arbitrary $m$-photon GHZ state.

##  V Discussion

So far, we have completely described this EPP. We first described the EPP for
three-photon GHZ state with a bit-flip error. Then we explained the EPP with a
phase-flip error. In this way, all the errors can be purified. Finally, we
extend this EPP for arbitrary GHZ state and which can be purified in the same
way. In above, we suppose that the bit-flip error occurs on the first qubit.
In a practical transmission, the hyperentanglement in polarization and spatial
mode will suffer from different errors. For example, the bit-flip error occurs
on the first qubit in polarization and which makes the polarization part
become the mixed state in Eq. (5), while the bit-flip error occurs on the
second qubit in spatial mode and makes the spatial part become

$\displaystyle\rho^{\prime
S}_{ABC}=F_{3}|\phi_{0}^{+}\rangle_{ABC}\langle\phi_{0}^{+}|+(1-F_{{}_{3}})|\phi_{2}^{+}\rangle_{ABC}\langle\phi_{2}^{+}|.$
(31)

Therefore, the hyperentangled mixed state can be written as

$\displaystyle\rho^{\prime}_{ABC}=\rho^{P}_{ABC}\otimes\rho^{\prime S}_{ABC}.$
(32)

With the probability of F1F3, it is in the state
$|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$. Such state will
make the three photons in the output modes D1D2D3 or D4D5D6. With the
probability of (1-F${}_{1})$F3, it is in the state
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$. Such state will
make the three photons in the output modes D1D2D6 or D4D5D3. With the
probability of F1(1-F${}_{3})$, it is in the state
$|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{2}^{+}\rangle_{ABC}$. Such state will
make the three photons in the output modes D1D5D3 or D4D2D6. Finally, with the
probability of (1-F1)(1-F3), it is in the state
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{2}^{+}\rangle_{ABC}$. Such state will
make the three photons in the output modes D1D5D6 or D4D2D3. In this way, by
selecting the cases D1D2D3 or D4D5D6, they can ultimately obtain the
polarization state $|\Phi_{0}^{+}\rangle_{ABC}$. Interestingly, if the mixed
state is described as shown in Eq.(32), the bit-flip error can be completely
purified. The second case
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{0}^{+}\rangle_{ABC}$ will lead the
photons in D1D2D6 or D4D5D3 and such state will become
$|\Phi_{1}^{+}\rangle_{ABC}$. They can add a bit-flip operation on the first
photon and convert it to $|\Phi_{0}^{+}\rangle_{ABC}$ deterministically. On
the other hand, the second case
$|\Phi_{0}^{+}\rangle_{ABC}\otimes|\phi_{2}^{+}\rangle_{ABC}$ will lead the
photons in the spatial modes D1D5D3 or D4D2D6 and become
$|\Phi_{2}^{+}\rangle_{ABC}$. They can also add the bit-flip operation on the
second photon and convert it to $|\Phi_{0}^{+}\rangle_{ABC}$
deterministically. The final case
$|\Phi_{1}^{+}\rangle_{ABC}\otimes|\phi_{2}^{+}\rangle_{ABC}$ will also become
$|\Phi_{1}^{+}\rangle_{ABC}$ and can be converted to
$|\Phi_{0}^{+}\rangle_{ABC}$ deterministically. In this way, they can obtain
the maximally pure entangled state $|\Phi_{0}^{+}\rangle_{ABC}$ with the
probability of 100$\%$. For a general mixed state with bit-flip error, the
polarization part and spatial-mode part can be written as

$\displaystyle\rho^{\prime\prime P}_{ABC}$ $\displaystyle=$ $\displaystyle
F_{1}|\Phi_{0}^{+}\rangle_{ABC}\langle\Phi_{0}^{+}|+F_{2}|\Phi_{1}^{+}\rangle_{ABC}\langle\Phi_{1}^{+}|$
(33) $\displaystyle+$ $\displaystyle
F_{3}|\Phi_{2}^{+}\rangle_{ABC}\langle\Phi_{2}^{+}|+F_{4}|\Phi_{3}^{+}\rangle_{ABC}\langle\Phi_{3}^{+}|,$

and

$\displaystyle\rho^{\prime\prime S}_{ABC}$ $\displaystyle=$ $\displaystyle
F_{4}|\phi_{0}^{+}\rangle_{ABC}\langle\phi_{0}^{+}|+F_{5}|\phi_{1}^{+}\rangle_{ABC}\langle\phi_{1}^{+}|$
(34) $\displaystyle+$ $\displaystyle
F_{6}|\phi_{2}^{+}\rangle_{ABC}\langle\phi_{2}^{+}|+F_{7}|\phi_{3}^{+}\rangle_{ABC}\langle\phi_{3}^{+}|.$

Here $F_{1}+F_{2}+F_{3}+F_{4}=1$ and $F_{5}+F_{6}+F_{7}+F_{8}=1$. Similarly,
by selecting the output modes D1D2D3 or D4D5D6, they can obtain a new mixed
state as

$\displaystyle\rho_{ABC}^{\prime\prime}$ $\displaystyle=$ $\displaystyle
F^{\prime}_{1}|\Phi_{0}^{+}\rangle_{ABC}\langle\Phi_{0}^{+}|+F^{\prime}_{2}|\Phi_{1}^{+}\rangle_{ABC}\langle\Phi_{1}^{+}|$
(35) $\displaystyle+$ $\displaystyle
F^{\prime}_{3}|\Phi_{2}^{+}\rangle_{ABC}\langle\Phi_{2}^{+}|+F^{\prime}_{4}|\Phi_{3}^{+}\rangle_{ABC}\langle\Phi_{3}^{+}|.$

Here

$\displaystyle
F^{\prime}_{1}=\frac{F_{1}F_{5}}{F_{1}F_{5}+F_{2}F_{6}+F_{3}F_{7}+F_{4}F_{8}},$
$\displaystyle
F^{\prime}_{2}=\frac{F_{2}F_{6}}{F_{1}F_{5}+F_{2}F_{6}+F_{3}F_{7}+F_{4}F_{8}},$
$\displaystyle
F^{\prime}_{3}=\frac{F_{3}F_{7}}{F_{1}F_{5}+F_{2}F_{6}+F_{3}F_{7}+F_{4}F_{8}},$
$\displaystyle
F^{\prime}_{4}=\frac{F_{4}F_{8}}{F_{1}F_{5}+F_{2}F_{6}+F_{3}F_{7}+F_{4}F_{8}}.$
(36)

If $F_{1}>\frac{1}{2}$ and $F_{5}>\frac{1}{2}$, we can also obtain
$F^{\prime}_{1}>F_{1}$ and $F^{\prime}_{1}>F_{5}$. In this way, we can realize
the general purification.

It is interesting to calculate the purification efficiency in a practical
environment. As shown in Fig. 1, the N-photon GHZ state was distributed to $N$
parties. The transmission efficiency is $\eta_{t}=e^{-\frac{L}{L_{0}}}$. The
detector efficiency is $\eta_{d}$. The $\eta_{c}$ is the probability of
coupling a photon to the single-photon detector. $L_{0}$ is the attenuation
length of the channel (25 km for commercial fibre) munro . $L$ is transmission
distance. The success probability is $p_{1}=F_{1}F_{2}+(1-F_{1})(1-F_{2})$.
For N-photon purification, the total purification efficiency can be calculated
as

$\displaystyle P^{N}_{one}=p_{1}\eta^{N}_{t}\eta^{N}_{d}\eta^{N}_{c}.$ (37)

In existing multipartite EPPs multipurification1 ; multipurification3 ;
multipurification4 ; multipurification7 , they exploit two pairs of N-photon
GHZ states to perform the purification. Therefore, for linear optical system,
the total purification efficiency can be calculated as

$\displaystyle
P^{N}_{two}=\frac{1}{4}p_{1}\eta^{2N}_{t}\eta^{2N}_{d}\eta^{2N}_{c}.$ (38)

The ratio of $P^{N}_{one}$ and $P^{N}_{two}$ can be calculated as

$\displaystyle
R=\frac{P^{N}_{one}}{P^{N}_{two}}=\frac{4}{\eta^{N}_{t}\eta^{N}_{d}\eta^{N}_{c}}=\frac{4}{(e^{-\frac{L}{L_{0}}})^{N}\eta^{N}_{d}\eta^{N}_{c}}.$
(39)

If we let $\eta_{d}=0.9$, $\eta_{c}=0.95$ munro .

Figure 4: The ratio $R$ of entanglement purification efficiency plotted
against length of entanglement distribution. We let the photon number of GHZ
state as $N=3$ and $N=6$, respectively.  Figure 5: The ratio $R$ of
entanglement purification efficiency plotted against photon number of GHZ
state. We let $L=L_{0}=25km$.

Fig. 4 shows the relationship between the coefficient $L$ and $R$. Here we let
$N=3$ and $N=6$ respectively. We change the distance $L$ from 20km to 100km.
The ratio $R$ increases rapidly. The $R$ can reach more than $10^{10}$ when
$N=6$ and $L=100$km. In Fig. 5, we also calculated the $R$ altered with $N$.
We let $L=L_{0}=25$km. We also showed that the $R$ increases rapidly with the
photon number $N$. On the other hand, in existing multipartite EPPs
multipurification1 ; multipurification3 ; multipurification4 ;
multipurification7 , after each party performing the CNOT or similar
operation, they should measure the target particles to judge that the
purification is successful or not. In this EPP, the parties are not required
to measure the particles and they can judge whether the purification is
successful or not according to the output modes of the photons. In this way,
this EPP is more economical and practical in future application.

Finally, let us briefly discuss the possible realization. This protocol mainly
exploits the common linear optics, such as PBS, BS, BD, HWP. Meanwhile, this
protocol require the multi-partite hyperentanglement. Such hyperentanglement
was also realized in experiment source , which show that this protocol is
feasible in current experiment condition.

##  VI Conclusion

In conclusion, we have proposed the EPP for multipartite entanglement
purification using hyperentanglement. After performing the EPP, the spatial
entanglement can be used to purify the polarization entanglement. Different
from the previous works, this EPP has several advantages. Firstly, with the
same purification success probability, this EPP only requires one pair of
multipartite GHZ states, while existing EPPs usually require two pairs of
multipartite GHZ state. Secondly, if consider the practical transmission and
detector efficiency, this EPP may be extremely useful for the ratio of
purification efficiency increases rapidly with both the number of photons and
the transmission distance. Thirdly, this protocol requires linear optics and
does not add additional measurement operations, so that it is feasible for
experiment. All these advantages will make this protocol have potential
application for future quantum information processing.

## ACKNOWLEDGEMENTS

This work was supported by the National Natural Science Foundation of China
(No. 11974189).

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