# Controlled bidirectional quantum teleportation of superposed coherent state
using five qubit cluster-type entangled coherent state as a resource

###### Abstract

We propose a scheme of bi-directional quantum teleportation of coherent state
qubit between two distant partners Alice and Bob, with the consent of
controller, Charlie. We use five-mode cluster-type entangled coherent state as
the quantum resource to achieve this task. The scheme uses linear optical
devices such as beam splitter, phase shifters, and photon counters. It is
shown that for moderately large coherent amplitude, near perfect bi-
directional controlled teleportation can be obtained in terms of the average
fidelity of teleportation.
Keywords: Quantum Teleportation Bidirectional Quantum Teleportation Entangled
Coherent State Cluster State.

Ravi Kamal Pandey$\alpha$, Phool Singh Yadav$\beta$ Ranjana Prakash$\gamma$,
Hari Prakash$\delta$,

Physics Department, University of Allahabad, Allahabad- 211002, India.

E-mail: $\alpha$<EMAIL_ADDRESS>
$\beta$<EMAIL_ADDRESS>
$\gamma$<EMAIL_ADDRESS>
$\delta$<EMAIL_ADDRESS>
##  1 Introduction

Quantum teleportation allows a sender (Alice) to communicate to a remote
receiver (Bob) by encoding information in quantum state of a system [1]. Most
often, it is required that both Alice and Bob can communicate simultaneously
which (in most simplistic scenario) can be acheived using two quantum
teleportation protocol, one from Alice to Bob and vice-versa. This shall
require two Bell states being shared between Alice and Bob and both have to
make a Bell state measurement (BSM) on their location, subsequently, sharing
classically the BSM outcome so as to make appropriate unitary operation and
retrieve the original state. However, in such a protocol, if one of the
partner denies to make BSM, even then single way communication is still
possible. One can circumvent such issue by considering Zha et al.f [2] scheme
of controlled bi-directional quantum teleportation (CBQT). The scheme uses a
genuinely entangled five-qubit cluster entangled state, instead of tensor
product of two Bell state, as quantum resource to obtain two-way communication
between Alice and Bob with the consent of controller, Charlie. In fact, it has
been shown by Shukla et al. [3] that there is nothing special about five-qubit
cluster state, and that one can construct infinite number of five-qubit
genuinely entangled state which can be used to obtain CBQT. Following these,
numerous schemes of bidirectional teleportaion of single-qubit as well as
multi-qubits were proposed using genuinely entangled resource of five or more
qubits[4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18].

The coherent state of radiation field can be a promising physical system for
performing many important quantum information tasks[19]. Phase opposite
coherent state of equal amplitude has been used to encode single qubit
information which can be teleported using entangled coherent state (ECS) as
quantum resource[20]. One can even obtain teleportation fidelity almost unity
using only linear optics and photon number resolving detectors[21]. Variants
of quantum teleportation as well as other quantum information processing can
also be obtained using ECS and linear optics with almost perfect success [22,
23, 24, 25, 26, 27, 28, 29, 30, 31].

Recently, an attempt has been made to obtain bi-directional teleportation of
information encoded in superposition of phase opposite coherent state (SCS)
using 3-mode entangled coherent state as a resource[32]. The success
probability of the scheme was shown to be $1/2$. However, there are few lapses
in their proposed scheme. For example, the mixing of modes using a beam
splitter has been done for modes which are at different location, i.e, one
mode is with Alice and the other mode is with Bob. Such non-local mixing is
not allowed unless we have some other ancillary mode and using teleportation.
Furthermore, the two controlled SWAP gate-operations performed by Alice and
Bob are conditioned to the state of the remaining mode with them. However, no
corresponding change is considered by the authors in the state of the system
after making the measurement. Therefore, such a scheme of obtaining bi-
directional teleportation of SCS is not feasible.

We present a scheme of obtaining CQBT using five-mode cluster-type genuine ECS
in this paper. Our scheme requires linear optical devices, such as symmetric
beam splitter, phase shifter and photon resolving detectors for its
implementation. The average fidelity of teleportation becomes almost unity for
appreciable mean coherent amplitude of the involved modes.

##  2 BCQT scheme

We consider three remote partners, Alice, Bob and Charlie. Alice and Bob each
has a single qubit information encoded in phase opposite coherent states,
$|\alpha\rangle$, $|-\alpha\rangle$ in mode $0$ and $1$ given by

$\begin{gathered}|I^{A}\rangle_{0}=\epsilon_{+}|\alpha\rangle_{0}+\epsilon_{-}|-\alpha\rangle_{0}\\\
|I^{B}\rangle_{1}=\omega_{+}|\alpha\rangle_{1}+\omega_{-}|-\alpha\rangle_{1},\end{gathered}$
(1)

respectively. The normalization of the information states requires,
$|\epsilon|_{+}^{2}+|\epsilon|_{-}^{2}+2x^{2}Re(\epsilon_{+}^{\star}\epsilon_{-})=|\omega|_{+}^{2}+|\omega|_{-}^{2}+2x^{2}Re(\omega_{+}^{\star}\omega_{-})=1$.
The information states can also be written in orthogonal basis using even and
odd coherent states [33],

$\begin{gathered}|+,\alpha\rangle=[\sqrt{2(1+x^{2})}]^{-1}(|{\alpha}\rangle+|-{\alpha}\rangle)\\\
|-,\alpha\rangle=[\sqrt{2(1-x^{2})}]^{-1}(|{\alpha}\rangle-|-{\alpha}\rangle,\end{gathered}$
(2)

as,

$\begin{gathered}|I^{A}\rangle_{0}=A_{+}|+,\alpha\rangle_{0}+A_{-}|-,\alpha\rangle_{0}\\\
|I^{B}\rangle_{1}=B_{+}|+,\alpha\rangle_{1}+B_{-}|-,\alpha\rangle_{1}.\end{gathered}$
(3)

Without loss in generality, we can write $A_{+}=\cos\theta/2$,
$A_{-}=\exp(i\phi)\sin\theta/2$ and $B_{+}=\cos\theta^{{}^{\prime}}/2$,
$B_{-}=\exp(i\phi^{{}^{\prime}})\sin\theta^{{}^{\prime}}/2$ . In order to
fulfil the task of simultaneously teleporting the information to each other
with the consent of Charlie; Alice, Bob and Charlie share a five mode cluster-
type ECS given as,

$\displaystyle|E\rangle_{2,3,4,5,6}$ $\displaystyle=$
$\displaystyle[2\sqrt{1+2x^{6}+x^{8}}]^{-1}(|\alpha,\alpha,\alpha,\alpha,\alpha\rangle+|\alpha,\alpha,-\alpha,-\alpha,-\alpha\rangle+|-\alpha,-\alpha,\alpha,-\alpha,\alpha\rangle$
(4) $\displaystyle+|-\alpha,-\alpha,\alpha,-\alpha,\alpha\rangle)_{2,3,4,5,6}$

where, Alice has modes $2$ and $4$, Bob has modes $3$ and $6$, while Charlie
has mode $5$. At this point, the overall state of the system consisting of
seven modes can be written as,

$|\psi\rangle_{0,1,2,3,4,5,6}=|I^{A}\rangle_{0}|I^{B}\rangle_{1}|E\rangle_{2,3,4,5,6}.$
(5)

Alice mixes modes $0$ and $2$ using a symmetric beam splitter (BS-I) which
outputs modes $7$ and $8$. Similarly, Bob also mixes modes $1$ and $6$ using
another symmetric beam splitter (BS-II) to become modes $9$ and $10$ (see Fig.
1). Both BS-I and BS-II are fitted with two phase shifters at its second input
and second output port. The phase shifter changes state $|\alpha\rangle$ to
$|-i\alpha\rangle$. This combination of a beam splitter with two phase
shifters transforms the input ports $x$ and $y$, to output modes $u$ and $v$
as,
$|{\alpha},{\beta}\rangle_{x,y}\rightarrow|\frac{{\alpha}+{\beta}}{\sqrt{2}},\frac{{\alpha}-{\beta}}{\sqrt{2}}\rangle_{u,v}$.
Then, the overall state of the system can be written as,

$\displaystyle|\psi\rangle_{7,8,4,5,3,10,9}$ $\displaystyle=$
$\displaystyle[2\sqrt{1+2x^{6}+x^{8}}]^{-1}[\epsilon_{+}\omega_{+}(|\sqrt{2}\alpha,0,\alpha,\alpha,\alpha,\sqrt{2}\alpha,0\rangle+|\sqrt{2}\alpha,0,-\alpha,-\alpha,\alpha,0,-\sqrt{2}\alpha\rangle$
(6)
$\displaystyle+|0,\sqrt{2}\alpha,\alpha,-\alpha,-\alpha,\sqrt{2}\alpha,0\rangle+|0,\sqrt{2}\alpha,-\alpha,\alpha,-\alpha,0,-\sqrt{2}\alpha\rangle)_{7,8,4,5,3,10,9}$
$\displaystyle+\epsilon_{+}\omega_{-}(|\sqrt{2}\alpha,0,\alpha,\alpha,\alpha,\sqrt{2}\alpha,0\rangle+|\sqrt{2}\alpha,0,-\alpha,-\alpha,\alpha,-\sqrt{2}\alpha,0\rangle$
$\displaystyle+|0,\sqrt{2}\alpha,\alpha,-\alpha,-\alpha,0,\sqrt{2}\alpha\rangle+|0,\sqrt{2}\alpha,-\alpha,\alpha,-\alpha,0,-\sqrt{2}\alpha\rangle)_{7,8,4,5,3,10,9}$
$\displaystyle+\epsilon_{-}\omega_{+}(|0,-\sqrt{2}\alpha,\alpha,\alpha,\alpha,\sqrt{2}\alpha,0\rangle+|0,-\sqrt{2}\alpha,0,-\alpha,-\alpha,\alpha,-\sqrt{2}\alpha,0\rangle$
$\displaystyle+|0,\sqrt{2}\alpha,\alpha,-\alpha,-\alpha,\sqrt{2}\alpha,0\rangle+|0,\sqrt{2}\alpha,-\alpha,\alpha,-\alpha,0,-\sqrt{2}\alpha\rangle)_{7,8,4,5,3,10,9}$
$\displaystyle+\epsilon_{-}\omega_{-}(|\sqrt{2}\alpha,0,\alpha,\alpha,\alpha,\sqrt{2}\alpha,0\rangle+|\sqrt{2}\alpha,0,-\alpha,-\alpha,\alpha,0,-\sqrt{2}\alpha\rangle$
$\displaystyle+|0,\sqrt{2}\alpha,\alpha,-\alpha,-\alpha,\sqrt{2}\alpha,0\rangle+|0,\sqrt{2}\alpha,-\alpha,\alpha,-\alpha,0,-\sqrt{2}\alpha\rangle)_{7,8,4,5,3,10,9}].$

Figure 1: Schematic of the proposed scheme for the BCQT of SCS using five mode
cluster-type ECS. Alice mixes the information state, $|I\rangle_{0}$ in mode 0
with mode 1 of entangled state $|E\rangle_{2,3,4,5,6}$ using symmetric beam
splitter (BS-I) and performs photon counting measurement on the output modes 7
and 8 using detectors $D_{1}$ and $D_{2}$, respectively. Bob also mixes his
information state, $|I\rangle_{1}$, with mode 6 using symmetric beam splitter
(BS-II) and perform PC on output modes 9 and 10 using detectors $D_{3}$ and
$D_{4}$. Both Alice and Bob communicates their PC outcome to each other using
classical channel, CC-1. The controller, Charlie, does a PC on mode 5 using
detector $D_{5}$, and communicaate the outcome to Alice and Bob using
classical channels CC-2 and CC-3, respectively. Depending on the classical
inputs, Alice and Bob does appropriate unitary operation, $U_{A}$ and $U_{B}$,
on modes 4 and 3 respectively, which completes the teleportation protocol.

Alice performs photon counting (PC) on modes $7$ and $8$ while Bob does PC on
modes $9$ and $10$. Following Prakash et al. [21], we suppose that the photon
counter is sensitive enough to distinguish between a vacuum, non-zero even
(NZE) and odd photons. We thus write,

$|\pm\sqrt{2}{\alpha}\rangle=x|0\rangle+\frac{1-x^{2}}{\sqrt{2}}|{NZE,\sqrt{2}\alpha}\rangle\pm\sqrt{\frac{1-x^{4}}{2}}|{-,\sqrt{2}\alpha}\rangle,$
(7)

where,
$|NZE,\sqrt{2}\alpha\rangle=[\sqrt{2}(1-x^{2})]^{-1}(|\sqrt{2}\alpha\rangle+|-\sqrt{2}\alpha\rangle-2x|0\rangle)$
represents the normalized state of non-zero even number of photons. At this
point, the role of Charlie becomes crucial. Charlie also performs PC on mode
$5$ with him so as to distinguish between an even or odd photon. Using Eq.7,
we expand modes 7, 8, 9 and 10 in orthogonal basis
{$|0\rangle$,$|NZE,\sqrt{2}\alpha\rangle$,$|-,\sqrt{2}\alpha\rangle\\}$, while
we expand rest of the modes 4, 5 and 3 in the basis
{$|+,\alpha\rangle,|-,\alpha\rangle$} using Eq.2 to rewrite state in Eq.6 as,

$\displaystyle|\psi\rangle_{7,8,4,5,3,10,9}=\frac{2x^{2}A_{+}B_{+}}{2(1+x^{2})\sqrt{1+2x^{6}+x^{8}}}|0,0,0,0\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}|+,\alpha\rangle_{4}|+,\alpha\rangle_{6}$
$\displaystyle+(1-x^{2})|-,\alpha\rangle_{5}|-,\alpha\rangle_{4}|+,\alpha\rangle_{6}]+\frac{x(1-x^{2})}{2\sqrt{(1+x^{2})(1+2x^{6}+x^{8})}}\\{B_{+}|NZE,0,0,0\rangle_{7,8,10,9}$
$\displaystyle[(1+x^{2})|+,\alpha\rangle_{5}|+,\alpha\rangle_{4}(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})+(1-x^{2})|-,\alpha\rangle_{5}|-,\alpha\rangle_{4}(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta A_{-}|+,\alpha\rangle_{6})]$
$\displaystyle+B_{+}|0,NZE,0,0\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}|+,\alpha\rangle_{4}(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})+(1-x^{2})|-,\alpha\rangle_{5}|-,\alpha\rangle_{4}$
$\displaystyle(-\gamma A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})]+A_{+}|0,0,NZE,0\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})$
$\displaystyle|+,\alpha\rangle_{6}+(1-x^{2})|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta
B_{-}|+,\alpha\rangle_{4})|-,\alpha\rangle_{6}]+A_{+}|0,0,0,NZE\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}$
$\displaystyle(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})|+,\alpha\rangle_{6}+(1-x^{2})|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta
B_{-}|+,\alpha\rangle_{4})|-,\alpha\rangle_{6}]\\}$
$\displaystyle+\frac{x\sqrt{(1-x^{2})}}{2\sqrt{(1+2x^{6}+x^{8})}}\\{B_{+}|ODD,0,0,0\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}|+,\alpha\rangle_{4}(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta A_{-}|+,\alpha\rangle_{6})$
$\displaystyle+(1-x^{2})|-,\alpha\rangle_{5}|-,\alpha\rangle_{4}(A_{+}|+,\alpha\rangle_{6}+\delta
A_{-}|-,\alpha\rangle_{6})]+B_{+}|0,ODD,0,0\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}|+,\alpha\rangle_{4}$
$\displaystyle(-\gamma A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})+(1-x^{2})|-,\alpha\rangle_{5}|-,\alpha\rangle_{4}(A_{+}|+,\alpha\rangle_{6}-\delta
A_{-}|-,\alpha\rangle_{6})]$
$\displaystyle+A_{+}|0,0,ODD,0\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta
B_{-}|+,\alpha\rangle_{4})|+,\alpha\rangle_{6}+(1-x^{2})|-,\alpha\rangle_{5}$
$\displaystyle(B_{+}|+,\alpha\rangle_{4}+\delta+B_{-}|-,\alpha\rangle_{4})|-,\alpha\rangle_{6}]+A_{+}|0,0,0,ODD\rangle_{7,8,10,9}[(1+x^{2})|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta B_{-}|+,\alpha\rangle_{4})$
$\displaystyle|+,\alpha\rangle_{6}+(1-x^{2})|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})|-,\alpha\rangle_{6}]\\}+\frac{(1-x^{2})^{2}}{4\sqrt{2(1+2x^{6}+x^{8})}}\\{|NZE,0,NZE,0\rangle_{7,8,10,9}$
$\displaystyle[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}$ $\displaystyle+\delta
B_{-}|+,\alpha\rangle_{4})(\gamma A_{+}|+,\alpha\rangle_{6}+\delta
A_{-}|-,\alpha\rangle_{6})]+|NZE,0,0,NZE\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}$
$\displaystyle-
B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta B_{-}|+,\alpha\rangle_{4})(\gamma
A_{+}|+,\alpha\rangle_{6}$ $\displaystyle+\delta
A_{-}|-,\alpha\rangle_{6})]+|0,NZE,NZE,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}$
$\displaystyle-
A_{-}|-,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta B_{-}|+,\alpha\rangle_{4})(\gamma
A_{+}|+,\alpha\rangle_{6}-\delta A_{-}|-,\alpha\rangle_{6})]$
$\displaystyle+|0,NZE,0,NZE\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})$
$\displaystyle+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta B_{-}|+,\alpha\rangle_{4})(\gamma
A_{+}|+,\alpha\rangle_{6}-\delta
A_{-}|-,\alpha\rangle_{6})]+\frac{(1-x^{4})}{4\sqrt{2(1+2x^{6}+x^{8})}}$
$\displaystyle\\{|ODD,0,ODD,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta B_{-}|+,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta A_{-}|+,\alpha\rangle_{6})$
$\displaystyle+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})]+|ODD,0,0,ODD\rangle_{7,8,10,9}$
$\displaystyle[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta B_{-}|+,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}$
$\displaystyle(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})]+|0,ODD,ODD,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}$ $\displaystyle+\delta
B_{-}|+,\alpha\rangle_{4})(\gamma A_{+}|-,\alpha\rangle_{6}-\delta
A_{-}|+,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})]$
$\displaystyle+|0,ODD,0,ODD\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta B_{-}|+,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}-\delta A_{-}|+,\alpha\rangle_{6})$
$\displaystyle+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})]+\frac{(1-x^{2}\sqrt{1-x^{4}})}{4\sqrt{2(1+2x^{6}+x^{8})}}$
$\displaystyle\\{|ODD,0,NZE,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta A_{-}|+,\alpha\rangle_{6})$
$\displaystyle+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta
B_{-}|+,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})]+|ODD,0,0,NZE\rangle_{7,8,10,9}$
$\displaystyle[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}$
$\displaystyle(\gamma B_{+}|-,\alpha\rangle_{4}-\delta
B_{-}|+,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})]+|0,ODD,NZE,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}$
$\displaystyle(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(-\gamma
A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(-A_{+}|+,\alpha\rangle_{4}+A_{-}|-,\alpha\rangle_{4})$
$\displaystyle(\gamma B_{+}|-,\alpha\rangle_{6}+\delta
B_{-}|+,\alpha\rangle_{6})]+|0,ODD,0,NZE\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}-\delta
B_{-}|-,\alpha\rangle_{4})$ $\displaystyle(-\gamma
A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(-\gamma
B_{+}|-,\alpha\rangle_{4}+\delta
B_{-}|+,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})]$
$\displaystyle+|NZE,0,ODD,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta
B_{-}|+,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})$
$\displaystyle+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})]+|NZE,0,0,ODD\rangle_{7,8,10,9}$
$\displaystyle[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}-\delta
B_{-}|+,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}+A_{-}|-,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}$
$\displaystyle(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})(\gamma
A_{+}|-,\alpha\rangle_{6}+\delta
A_{-}|+,\alpha\rangle_{6})]+|0,NZE,ODD,0\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}$
$\displaystyle(\gamma B_{+}|-,\alpha\rangle_{4}+\delta
B_{-}|+,\alpha\rangle_{4})(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}+B_{-}|-,\alpha\rangle_{4})$
$\displaystyle(\gamma A_{+}|-,\alpha\rangle_{6}-\delta
A_{-}|+,\alpha\rangle_{6})]+|0,NZE,0,ODD\rangle_{7,8,10,9}[\sqrt{(1+x^{2})}|+,\alpha\rangle_{5}(\gamma
B_{+}|-,\alpha\rangle_{4}+\delta B_{-}|+,\alpha\rangle_{4})$
$\displaystyle(A_{+}|+,\alpha\rangle_{6}-A_{-}|-,\alpha\rangle_{6})+\sqrt{1-x^{2}}|-,\alpha\rangle_{5}(B_{+}|+,\alpha\rangle_{4}-B_{-}|-,\alpha\rangle_{4})(-\gamma
A_{+}|-,\alpha\rangle_{6}+\delta A_{-}|+,\alpha\rangle_{6})]\\},$ (8)

where $\gamma=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$ and
$\delta=\sqrt{\frac{1+x^{2}}{1-x^{2}}}$. We shall investigate all possible PC
outcomes in the following section.

##  3 Investigation of PC cases

From Eq.2, we observe that one of the mode with Alice (7 or 8) as well as Bob
(9 or 10) always gives a zero count while the other mode can give a zero,
$NZE$ or odd counts. All of these cases are tabulated (see Table 1 in the
Appendix). There are a total of 50 mutually exclusive PC outcomes that can be
obtained from our scheme. We find that from a total of 50 cases, 32 cases
amounts to bi-directional teleportation with fidelity either unity or close to
unity (depending upon mean photon amplitude). Whereas, 16 cases lead to uni-
directional teleportation and for remaining two cases, teleportation fails. It
shall be useful for the discussions what follow to denote the state teleported
to Bob as,

$\begin{gathered}|B^{0}\rangle\sim A_{+}|+,\alpha\rangle\\\ |B^{1}\rangle\sim
A_{+}|-,\alpha\rangle\\\ |B^{2}\rangle=A_{+}|+,\alpha\rangle-
A_{-}|-,\alpha\rangle\\\ |B^{3}\rangle\sim\gamma A_{+}|-,\alpha\rangle+\delta
A_{-}|+,\alpha\rangle\\\ |B^{4}\rangle\sim\gamma A_{+}|-,\alpha\rangle-\delta
A_{-}|+,\alpha\rangle,\end{gathered}$ (9)

while the state teleported to Alice as,

$\begin{gathered}|A^{0}\rangle\sim B_{+}|+,\alpha\rangle\\\ |A^{1}\rangle\sim
B_{+}|-,\alpha\rangle\\\ |A^{2}\rangle=B_{+}|+,\alpha\rangle-
B_{-}|-,\alpha\rangle\\\ |A^{3}\rangle\sim\gamma B_{+}|-,\alpha\rangle+\delta
B_{-}|+,\alpha\rangle\\\ |A^{4}\rangle\sim\gamma B_{+}|-,\alpha\rangle-\delta
B_{-}|+,\alpha\rangle,\end{gathered}$ (10)

(where $\sim$ denotes un-normalized state) and the unitary operations,

$\begin{gathered}U_{1}=|+,\alpha\rangle\langle-,\alpha|+|-,\alpha\rangle\langle+,\alpha|\\\
U_{2}=|+,\alpha\rangle\langle-,\alpha|-|-,\alpha\rangle\langle+,\alpha|\\\
U_{3}=|+,\alpha\rangle\langle+,\alpha|-|-,\alpha\rangle\langle-,\alpha|.\end{gathered}$
(11)

For information state $|\psi^{I}\rangle$ and teleported state
$|\psi^{T}\rangle$, the teleportation fidelity is given by,

$F=|\langle\psi^{I}|\psi^{T}\rangle|^{2}.$ (12)

Since, there are two teleportations involved, we have two fidelities as well.
For teleportation from Alice to Bob, the fidelity obtained will be one of the
following,

$F^{AB}_{1}=|A_{+}|^{2};F^{AB}_{2}=|A_{-}|^{2};F^{AB}_{3}=1-\frac{x^{4}\sin^{2}\theta}{1+x^{4}-2x^{2}\cos\theta}.$
(13)

On the other hand, for teleportation from Bob to Alice, the fidelity can be,

$F^{BA}_{1}=|B_{+}|^{2};F^{BA}_{2}=|B_{-}|^{2};F^{BA}_{3}=1-\frac{x^{4}\sin^{2}\theta^{{}^{\prime}}}{1+x^{4}-2x^{2}\cos\theta^{{}^{\prime}}}.$
(14)

We shall now discuss the PC outcomes.

###  3.1 Cases of failure

These cases correspond to situations when both the output of BS-I and BS-II
give vacuum while Charlie obtain either even or odd counts in mode 5. For case
1 and 2, the teleported with {Alice,Bob} are given by
{$|A^{0}\rangle,|B^{0}\rangle$} and {$|A^{1}\rangle,|B^{1}\rangle$},
respectively, amounting to respective fidelity $\\{F_{1}^{BA},F_{1}^{AB}\\}$
and $\\{F_{2}^{BA},F_{2}^{AB}\\}$. For both of these cases, the teleported
states can not be transformed to the information states using any unitary
transformation. Hence, for these cases, we say that the teleportation fails
for both sides. The probability of obtaining such a case are given by

$P_{I,+}=\frac{2x^{4}(1+x^{2})|A_{+}|^{2}||B_{+}|^{2}}{1+2x^{6}+x^{8}}$ (15)

for case 1, and

$P_{I,-}=\frac{2x^{4}(1-x^{2})^{3}|A_{+}|^{2}||B_{+}|^{2}}{(1+x^{2})^{2}(1+2x^{6}+x^{8})}$
(16)

for case 2. Both $P_{I,+}$ and $P_{I,-}$ depend on information parameters
$\theta$ and $\theta^{{}^{\prime}}$ as well as $|\alpha|^{2}$. However, both
these expressions contain the product of $x^{4}$ with
$|A_{+}|^{2}|B_{+}|^{2}$, where the latter is always less than or equal to
one, while $x^{4}\simeq 3.3\times 10^{-4}$ for $|\alpha|^{2}=2$. Hence, for
appreciable coherent amplitude ($|\alpha|^{2}\geq 2$) both these probabilities
become vanishingly small implying little chance of occurrence of these cases.
The probabilities are plotted for the special case when
$|A_{+}|^{2}|B_{+}|^{2}=1$, which is the upper bound for probability obtained
for any other value of $A_{+}$ and $B_{+}$. Both $P^{max}_{I,+}$ and
$P^{max}_{I,-}$ become almost zero for $|\alpha|^{2}\geq 2$ (Fig.2).

Figure 2: Red and blue curve shows the variation of $P^{max}_{I,+}$ and
$P^{max}_{I,-}$ respectively with respect to mean photons in the coherent
state, $|\alpha|^{2}$. Both rapidly decreases with increasing $|\alpha|^{2}$
to become almost zero for $|\alpha|^{2}\geqslant 2$.

###  3.2 Cases of unidirectional teleportation

####  3.2.1 3 to 10

Cases 3 to 10 occur for collective PC outcomes (NZE,0,0,0,even),
(NZE,0,0,0,odd), (0,NZE,0,0,even), (0,NZE,0,0,odd), (odd,0,0,0,even),
(odd,0,0,0,odd), (0,odd,0,0,even), (0,odd,0,0,odd) in modes (7,8,9,10,5),
respectively. For these cases, the teleportation is unidirectional in a sense
that Bob can get the information of Alice (with fidelity either unity or
$F_{3}^{AB}$) but Alice’s state is either an even or odd coherent state
depending upon the outcome of the controller Charlie.

For cases 3 to 6, the received states with {Alice,Bob} are given by,
$\\{|A^{0}\rangle,|I^{B}\rangle\\}$, $\\{|A^{1}\rangle,|B^{3}\rangle\\}$,
$\\{|A^{0}\rangle,|B^{2}\rangle\\}$, $\\{|A^{1}\rangle,-|B^{4}\rangle\\}$,
respectively, requiring unitary operations, $\\{I,I\\}$, $\\{U_{1},U_{1}\\}$,
$\\{I,U_{3}\\}$ and $\\{U_{1},-U_{2}\\}$, respectively (where $I$ is the
identity operation). Although, performing non-identity operation by Alice will
not help her to get to any close to Bob’s information, however, it does later
help us to increase the average fidelity of our scheme. The fidelity of
teleported state with Alice, $F^{BA}$, becomes $F^{BA}_{1}$ for cases 3 to 6,
while the fidelity of Bob, $F^{AB}$, is unity for cases 3 and 5, and
$F^{AB}_{3}$ for cases 4 and 6. It can be observed by Eq. 2 that the
teleported state, the required unitary operation and corresponding fidelities
for cases 7, 8, 9 and 10 are same as that for cases 4, 3, 6 and 5
respectively.

The probability of obtaining case 3 (same for case 5) is given by

$P_{II,+}=\frac{x^{2}(1-x^{2})^{2}(1+x^{2})|B_{+}|^{2}}{4(1+2x^{6}+x^{8})}$
(17)

while the probability of obtaining case 4 (same for case 6) is given by

$P_{II,-}=\frac{x^{2}(1-x^{2})^{4}(\gamma^{2}|A_{+}|^{2}+\delta^{2}|A_{-}|^{2})|B_{+}|^{2}}{4(1+2x^{6}+x^{8})}.$
(18)

Similarly, the probability of obtaining case 7 (same for case 9) is given by,

$P_{III,+}=\frac{x^{2}(1+x^{2})^{2}(1-x^{2})(\gamma^{2}|A_{+}|^{2}+\delta^{2}|A_{-}|^{2})|B_{+}|^{2}}{4(1+2x^{6}+x^{8})},$
(19)

while for case 8 (same for case 10) is given by

$P_{III,-}=\frac{x^{2}(1-x^{2})^{3}|B_{+}|^{2}}{4(1+2x^{6}+x^{8})}.$ (20)

The probabilities $P_{III,-}$ and $P_{IV,+}$ are function of $x$ and
information parameters $\theta$ and $\theta^{{}^{\prime}}$ while $P_{III,+}$
and $P_{IV,-}$ only depend on $x$ and $\theta^{{}^{\prime}}$. We have plotted
these probabilities in units of $|B_{+}|^{2}$ in Fig. 4 from which it is
evident that the probabilities decrease with increasing coherent amplitude to
become almost zero. Hence, one can infer that these cases also have little
chance of occurrence at appreciably large coherent amplitude
($|\alpha|^{2}\geq 3$).

Figure 3: (a), (b), (c) and (d) show the variation of probability of
occurrence in units of $|B_{+}|^{2}$ for cases 3, 4, 7 and 8 with respect to
$|\alpha|^{2}$ and information parameter $\theta$. The probabilities becomes
almost zero for $|\alpha|^{2}\geq 2$.

####  3.2.2 11 to 18

Cases 11 to 18 occur for collective PC outcomes (0,0,NZE,0,even),
(0,0,NZE,0,odd), (0,0,0,NZE,even), (0,0,0,NZE,odd), (0,0,odd,0,even),
(0,0,odd,0,odd), (0,0,0,odd,even), (0,0,0,odd,odd) in modes (7,8,9,10,5),
respectively. These cases can be obtained by inter-changing the roles of Alice
and Bob in cases 3 to 10, respectively. Therefore, the received state with
Alice (Bob) for cases 11 to 18 will be that of Bob (Alice) for cases 3 to 10,
with $B_{\pm}(A_{\pm})$ replaced with $A_{\pm}(B_{\pm})$, respectively. Also,
the expression for probability for these cases can be obtained using eq. 16 to
19, by the same replacement of $A(B)$ to $B(A)$. Explicitly, if $P_{IV,+}$,
$P_{IV,-}$, $P_{V,+}$ and $P_{V,-}$ represent the probability of obataining
case 11 (same for case 13), case 12 (same for case 14), case 15 (same for 17)
and case 16 (same for case 18), respectively, we have,

$P_{IV,+}=\frac{x^{2}(1-x^{2})^{2}(1+x^{2})|A_{+}|^{2}}{4(1+2x^{6}+x^{8})},$
(21)
$P_{IV,-}=\frac{x^{2}(1-x^{2})^{4}(\gamma^{2}|B_{+}|^{2}+\delta^{2}|B_{-}|^{2})|A_{+}|^{2}}{4(1+2x^{6}+x^{8})},$
(22)
$P_{V,+}=\frac{x^{2}(1+x^{2})^{2}(1-x^{2})(\gamma^{2}|B_{+}|^{2}+\delta^{2}|B_{-}|^{2})|A_{+}|^{2}}{4(1+2x^{6}+x^{8})},$
(23) $P_{V,-}=\frac{x^{2}(1-x^{2})^{3}|A_{+}|^{2}}{4(1+2x^{6}+x^{8})}.$ (24)

If we plot these probabilities in units of $|A_{+}|^{2}$, we would have
similar variation as we have already obtained for expression 16 to 19 as shown
in Fig. 2.

###  3.3 Cases of bi-directional teleportation

####  3.3.1 Cases 19 to 34

Cases 19 to 34 correspond to collective PC outcome (NZE,0,NZE,0,even),
(NZE,0,NZE,0,odd), (NZE,0,0, NZE,even), (NZE,0,0,NZE,odd), (0,NZE,NZE,0,even),
(0,NZE,NZE,0,odd), (0,NZE,0,NZE,even), (0,NZE, 0,NZE,odd), (odd,0,odd,0,even),
(odd,0,odd,0,odd), (odd,0,0,odd,even), (odd,0,0,odd,odd), (0,odd,odd,0 ,even),
(0,odd,odd,0,odd), (0,odd,0,odd,even), (0,odd,0,odd,odd) in modes
(7,8,9,10,5), respectively. We have for these cases, the received state with
{Alice,Bob} as, $\\{|I^{B}\rangle,|I^{A}\rangle\\}$,
$\\{|A^{3}\rangle,|B^{3}\rangle\\}$, $\\{|A^{2}\rangle,|I^{A}\rangle\\}$,
$\\{|A^{4}\rangle,|B^{3}\rangle\\}$, $\\{|I^{B}\rangle,|B^{2}\rangle\\}$,
$\\{|A^{4}\rangle,|B^{4}\rangle\\}$, $\\{|A^{2}\rangle,|B^{2}\rangle\\}$,
$\\{|A^{4}\rangle,|B^{4}\rangle\\}$, $\\{|A^{3}\rangle,|B^{3}\rangle\\}$,
$\\{|I^{B}\rangle,|I^{A}\rangle\\}$, $\\{|A^{3}\rangle,|B^{3}\rangle\\}$,
$\\{|A^{2}\rangle,|I^{A}\rangle\\}$, $\\{|A^{3}\rangle,|B^{4}\rangle\\}$,
$\\{|I^{B}\rangle,|B^{2}\rangle\\}$, $\\{|A^{4}\rangle,|B^{4}\rangle\\}$,
$\\{|A^{2}\rangle,|B^{2}\rangle\\}$, respectively, requiring unitary
operations, $\\{I,I\\}$, $\\{U_{1},U_{1}\\}$, $\\{U_{3},I\\}$ and
$\\{U_{2},U_{1}\\}$, $\\{I,U_{3}\\}$, $\\{U_{2},U_{2}\\}$,
$\\{U_{3},U_{3}\\}$, $\\{U_{2},U_{2}\\}$, $\\{U_{1},U_{1}\\}$, $\\{I,I\\}$,
$\\{U_{1},U_{1}\\}$, $\\{U_{3},I\\}$, $\\{U_{1},U_{2}\\}$, $\\{I,U_{2}\\}$,
$\\{U_{2},U_{2}\\}$, $\\{U_{3},U_{3}\\}$, respectively. For all these cases,
there is either perfect teleportation (in both way) with unit fidelity or with
fidelity almost unity at appreciable coherent amplitude. The fidelity of state
teleported to Alice (Bob) then becomes unity for cases 19, 21, 23, 25, 28, 30,
32, 34 and $F_{3}^{BA}$($F_{3}^{AB}$) for cases 20, 22, 24, 26, 27, 29, 31 and
33.

Representing $P_{VI,+}$ and $P_{VI,-}$ the probability of obtaining case 19
(same for case 21, 23, 25) and case 20 (same for 22,24 and 26), respectively,
are given by

$P_{VI,+}=\frac{(1-x^{2})^{4}(1+x^{2})}{32(1+2x^{6}+x^{8})}$ (25)
$P_{VI,-}=\frac{(1-x^{2})^{5}(\gamma^{2}|A_{+}|^{2}+\delta^{2}|A_{-}|^{2})(\gamma^{2}|B_{+}|^{2}+\delta^{2}|B_{-}|^{2})}{32(1+2x^{6}+x^{8})}.$
(26)

Similarly, if $P_{VII,+}$ and $P_{VII,-}$ represent the probability of
obtaining case 27 (same for case 29, 31, 33) and case 28 (same for 30, 32 and
34), respectively, then we have

$P_{VII,+}=\frac{(1-x^{2})^{2}(1+x^{2})^{3}(\gamma^{2}|A_{+}|^{2}+\delta^{2}|A_{-}|^{2})(\gamma^{2}|B_{+}|^{2}+\delta^{2}|B_{-}|^{2})}{32(1+2x^{6}+x^{8})}$
(27) $P_{VII,-}=\frac{(1-x^{2})^{3}(1+x^{2})^{2}}{32(1+2x^{6}+x^{8})}.$ (28)

For $|\alpha|^{2}\geq 2$, $x\simeq 0$, implying, $\gamma,\delta\rightarrow 1$.
In this particular limit, we have $P_{VI,\pm}$ and $P_{VII,\pm}$
asymptotically converges to a constant value of 1/32.

####  3.3.2 35 to 50

Cases 35 to 42 correspond to collective PC outcome (NZE,0,odd,0,even),
(NZE,0,odd,0,odd), (NZE,0,0, odd,even), (NZE,0,0,odd,odd), (0,NZE,odd,0,even),
(0,NZE,odd,0,odd), (0,NZE,0, odd,even), (0,NZE,0, odd,odd),
(odd,0,NZE,0,even), (odd,0,NZE,0,odd), (odd,0,0,NZE,even), (odd,0,0,NZE,odd),
(0,odd,NZE, 0,even), (0,odd,NZE,0,odd), (0,odd,0,NZE,even), (0,odd,0,NZE,odd)
in modes (7,8,9,10,5), respectively. For these cases, the received state with
{Alice,Bob} are given by $\\{|A^{3}\rangle,|I^{A}\rangle\\}$,
$\\{|I^{B}\rangle,|B^{3}\rangle\\}$, $\\{|A^{4}\rangle,|I^{A}\rangle\\}$,
$\\{|I^{B}\rangle,|B^{3}\rangle\\}$, $\\{|A^{3}\rangle,|B^{2}\rangle\\}$,
$\\{|I^{B}\rangle,|B^{4}\rangle\\}$, $\\{|A^{3}\rangle,|B^{2}\rangle\\}$,
$\\{|A^{2}\rangle,-|B^{4}\rangle\\}$ , $\\{|I^{B}\rangle,|B^{3}\rangle\\}$,
$\\{|A^{3}\rangle,|I^{A}\rangle\\}$, $\\{|A^{2}\rangle,|B^{3}\rangle\\}$,
$\\{|A^{4}\rangle,|I^{A}\rangle\\}$, $\\{|I^{B}\rangle,-|B^{4}\rangle\\}$,
$\\{|A^{3}\rangle,-|B^{2}\rangle\\}$, $\\{|A^{2}\rangle,-|B^{4}\rangle\\}$,
$\\{-|A^{4}\rangle,|B^{2}\rangle\\}$, respectively, requiring unitary
operations, $\\{U_{1},I\\}$, $\\{I,U_{1}\\}$, $\\{U_{2},I\\}$,
$\\{I,U_{1}\\}$, $\\{U_{1},U_{2}\\}$, $\\{I,U_{2}\\}$, $\\{U_{1},U_{3}\\}$,
$\\{U_{3},-U_{2}\\}$, $\\{I,U_{1}\\}$, $\\{U_{1},I\\}$, $\\{U_{3},U_{1}\\}$,
$\\{U_{2},I\\}$, $\\{I,-U_{2}\\}$, $\\{U_{1},-U_{3}\\}$, $\\{U_{3},-U_{2}\\}$,
$\\{-U_{2},U_{3}\\}$, respectively. The fidelity of state teleported to Alice
and Bob becomes 1 and $F_{3}^{AB}$, respectively, for cases 36, 38, 40, 42,
43, 45, 47 and 49. For rest of the cases, 37, 39, 41, 44, 46, 48 and 50, the
fidelity becomes $F_{3}^{BA}$ for teleported state with Alice, and unity for
state teleported to Bob.

The probability of obtaining case 35 (same for case 37, 39, 41) becomes

$P_{VIII,+}=\frac{(1-x^{2})^{3}(1+x^{2})^{2}(\gamma^{2}|B_{+}|^{2}+\delta^{2}|B_{-}|^{2})}{32(1+2x^{6}+x^{8})},$
(29)

while probability of obtaining case 36 (same for case 38, 40, 42) is

$P_{VIII,-}=\frac{(1-x^{2})^{4}(1+x^{2})(\gamma^{2}|A_{+}|^{2}+\delta^{2}|A_{-}|^{2})}{32(1+2x^{6}+x^{8})}$
(30)

Similarly, the probability of obtaining case 43 (same for case 45, 47, 49) is
given by

$P_{IX,+}=\frac{(1-x^{2})^{3}(1+x^{2})^{2}(\gamma^{2}|A_{+}|^{2}+\delta^{2}|A_{-}|^{2})}{32(1+2x^{6}+x^{8})}$
(31)

and the probability of obtaining case 44 (same for case 46, 48, 50) becomes as

$P_{IX,-}=\frac{(1-x^{2})^{4}(1+x^{2})(\gamma^{2}|B_{+}|^{2}+\delta^{2}|B_{-}|^{2})}{32(1+2x^{6}+x^{8})}.$
(32)

The probabilities $P_{VIII,\pm}$ and $P_{IX,\pm}$ also converge asymptotically
to 1/32 for $|\alpha|^{2}\geq 2$.

##  4 Average fidelity and minimum assured fidelity of teleportation

Figure 4: (a) Variation of average fidelity of teleportation with respect to
information parameter $\theta$ and mean photons in the coherent state,
$|\alpha|^{2}$. (b) Variation of minimum average fidelity of teleportation
with respect to mean photons in the coherent state, $|\alpha|^{2}$. Both
increase monotonically to become almost unity for $|\alpha|^{2}\geq 2$.

In view of previous section, it is evident that the individual fidelities
$F^{AB}$ and $F^{BA}$ for many cases depend on information parameter, $\theta$
and $\theta^{{}^{\prime}}$ as well as on mean coherent amplitude
$|\alpha|^{2}$. In such a scenario to effectively quantify the overall quality
of teleportation, we find the average fidelity of teleportation ($F_{avg.}$)
which is the sum of product of individual fidelity with its probability of
occurrence. For individual cases, we calculate the minimum assured fidelity
(MAF) which is the minimum value of fidelity for every possible information
state [34].

As there are two teleportations involved, we need to find average fidelity for
both teleportation from Alice to Bob, denoted by $F_{avg.}^{AB}$ and
$F^{BA}_{avg.}$ for teleportation from Bob to Alice. Although, $F^{AB}\neq
F^{BA}$ for several cases, it turns out that $F_{avg.}^{AB}(F^{BA}_{avg.})$
have similar functional dependence on information parameters as well as $x$,

$F^{AB(BA)}_{avg.}\sum_{i=1}^{50}P_{i}F_{i}=1-\frac{x^{2}(1-x^{2}+2x^{4})\sin^{2}\theta(\theta^{{}^{\prime}})}{2(1-x^{2}+x^{4}+x^{6})}.$
(33)

$F^{AB(BA)}_{avg.}$ minimizes for $\theta(\theta^{{}^{\prime}})=\pi/2$, while
its maxima is obtained for $\theta(\theta^{{}^{\prime}})=\\{0,\pi\\}$ equal to
1. The average fidelity $F^{AB}_{avg.}$ as well as $F^{AB}_{avg.min.}$ is
plotted with respect to information parameter $\theta$ and $|\alpha|^{2}$ in
Fig.4. It is evident that both becomes almost unity for $|\alpha|^{2}\geq 2$.

We shall now discuss MAF for all PC cases. Let us first consider those cases
for which the fidelity of state teleported to either Bob or Alice is one of
$F_{1}^{AB},F_{2}^{AB}$, $F_{1}^{BA},F_{2}^{BA}$. For all such cases, the
fidelity is proportional to either $|A_{\pm}|$ or $|B_{\pm}|$, which is zero
if information states are $\theta=\pi/2,0$ or $\theta^{{}^{\prime}}=\pi/2,0$,
respectively, and for all allowed value of $\phi$ and $\phi^{{}^{\prime}}$.
Hence for these cases, MAF is 0. For those cases in which the teleported state
with Alice or Bob gives fidelity, $F_{3}^{AB}$ or $F_{3}^{BA}$, then a minimum
is obtained for $\theta^{(^{\prime})}=\cos^{-1}x^{2}$, therefore, MAF for such
cases becomes, $(1-x^{4})$. For rest of the cases, fidelity is unity and we
have ideal teleportation.

##  5 Conclusion

We may conclude that CBQT of information encoded in phase opposite state
coherent state can be obtained using five-mode cluster-type ECS as a resource
with average fidelity of almost unity at moderate coherent amplitude. We note
that complete failure of CBQT is obtained for cases 1 and 2, while uni-
directional teleportation is obtained for cases 3 to 18. However, the
probability of occurrence of all such cases becomes negligibly small as we
increase mean photon in coherent state. For each of Cases 19 to 50, we obtain
almost perfect success for CBQT in terms of fidelity of teleportation. The
probability of occurrence of each of these cases becomes almost equal
converging asymptotically to a constant value of 1/32 for $|\alpha|^{2}\geq
2$. We only require linear optical devices, such as, symmetric beam splitters,
phase shifters, and photon counters for its practical realization.

## Acknowledgement

We dedicate this paper in the memory of Prof. Ranjana Prakash and Prof. Hari
Prakash. May their soul rest in peace. RKP is thankful to UGC for providing
financial assistance under CSIR-UGC SRF fellowship. Discussions with Dr. D K
Mishra, Dr. M K Mishra, Dr. V Verma and Ms. S Javed are gratefully
acknowledged.

##  Appendix A Appendix

Table 1: Various possible PC measurement results in modes 7, 8 with Alice, modes 9, 10 with Bob, and mode 5 with controller Charlie. For each case, the corresponding probability of occurrence, required unitary operation, teleported state and fidelity is given.  Cases | PC result obtained by |  Probability |  Unitary operation |  Teleported state | Fidelity
---|---|---|---|---|---
Alice (7,8) |  Bob (9,10) |  Charlie (5) | Alice | Bob | Alice | Bob | Alice | Bob
1 | (0,0) | (0,0) | even | $P_{I,+}$ | $I$ | $I$ | $|A^{0}\rangle$ | $|B^{0}\rangle$ | $F_{1}^{AB}$ | $F_{1}^{BA}$
2 | (0,0) | (0,0) | odd | $P_{I,-}$ | $I$ | $I$ | $|A^{1}\rangle$ | $|B^{1}\rangle$ | $F_{2}^{AB}$ | $F_{2}^{BA}$
3 | (NZE,0) | (0,0) | even | $P_{II,+}$ | $I$ | $I$ | $|A^{0}\rangle$ | $|I^{B}\rangle$ | $F_{1}^{AB}$ | 1
4 | (NZE,0) | (0,0) | odd | $P_{II,-}$ | $U_{1}$ | $U_{1}$ | $|A^{0}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{1}^{AB}$ | $F_{3}^{BA}$
5 | (0,NZE) | (0,0) | even | $P_{II,+}$ | $I$ | $U_{3}$ | $|A^{0}\rangle$ | $|I^{A}\rangle$ | $F_{1}^{AB}$ | 1
6 | (0,NZE) | (0,0) | odd | $P_{II,-}$ | $U_{1}$ | $U_{2}$ | $|A^{0}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{1}^{AB}$ | $F_{3}^{BA}$
7 | (odd,0) | (0,0) | even | $P_{III,+}$ | $I$ | $U_{1}$ | $|A^{0}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{1}^{AB}$ | $F_{3}^{BA}$
8 | (odd,0) | (0,0) | odd | $P_{III,-}$ | $U_{1}$ | $I$ | $|A^{0}\rangle$ | $|I^{A}\rangle$ | $F_{1}^{AB}$ | 1
9 | (0,odd) | (0,0) | even | $P_{III,+}$ | $I$ | $U_{2}$ | $|A^{0}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{1}^{AB}$ | $F_{3}^{BA}$
10 | (0,odd) | (0,0) | odd | $P_{III,-}$ | $U_{1}$ | $U{3}$ | $|A^{0}\rangle$ | $|I^{A}\rangle$ | $F_{1}^{AB}$ | $F_{3}^{BA}$
11 | (0,0) | (NZE,0) | even | $P_{IV,+}$ | $\hat{I}$ | $\hat{I}$ | $|I^{B}\rangle$ | $|B^{0}\rangle$ | 1 | $F_{1}^{BA}$
12 | (0,0) | (NZE,0) | odd | $P_{IV,-}$ | $U_{1}$ | $U_{1}$ | $|I^{{}^{\prime}B}\rangle$ | $|B^{0}\rangle$ | $F_{3}^{AB}$ | $F_{1}^{BA}$
13 | (0,0) | (0,NZE) | even | $P_{IV,+}$ | $U_{3}$ | $I$ | $|I^{B}\rangle$ | $|B^{0}\rangle$ | 1 | $F_{1}^{BA}$
14 | (0,0) | (0,NZE) | odd | $P_{IV,-}$ | $U_{2}$ | $U_{1}$ | $|I^{{}^{\prime}B}\rangle$ | $|B^{0}\rangle$ | $F_{3}^{AB}$ | $F_{1}^{BA}$
15 | (0,0) | (odd,0) | even | $P_{V,+}$ | $U_{1}$ | $\hat{I}$ | $|I^{{}^{\prime}B}\rangle$ | $B^{0}\rangle$ | $F_{3}^{AB}$ | $F_{1}^{BA}$
16 | (0,0) | (odd,0) | odd | $P_{V,-}$ | $I$ | $U_{1}$ | $|I^{B}\rangle$ | $|B^{0}\rangle$ | 1 | $F_{1}^{BA}$
17 | (0,0) | (0,odd) | even | $P_{V,+}$ | $U_{2}$ | $\hat{I}$ | $|I^{{}^{\prime}B}\rangle$ | $B^{0}\rangle$ | $F_{3}^{AB}$ | $F_{1}^{BA}$
18 | (0,0) | (0,odd) | odd | $P_{V,-}$ | $U_{3}$ | $U_{1}$ | $|I^{B}\rangle$ | $|B^{0}\rangle$ | 1 | $F_{1}^{BA}$
19 | (NZE,0) | (NZE,0) | even | $P_{VI,+}$ | $I$ | $I$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
20 | (NZE,0) | (NZE,0) | odd | $P_{VI,-}$ | $U_{1}$ | $U_{1}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
21 | (NZE,0) | (0,NZE) | even | $P_{VI,+}$ | $U_{3}$ | $U_{3}$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
22 | (NZE,0) | (0,NZE) | odd | $P_{VI,-}$ | $U_{2}$ | $U_{2}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
23 | (0,NZE) | (NZE,0) | even | $P_{VI,+}$ | $I$ | $U_{3}$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
24 | (0,NZE) | (NZE,0) | odd | $P_{VI,-}$ | $U_{1}$ | $U_{2}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
25 | (0,NZE) | (0,NZE) | even | $P_{VI,+}$ | $U_{3}$ | $U_{3}$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
26 | (0,NZE) | (0,NZE) | odd | $P_{VI,-}$ | $U_{2}$ | $U_{2}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
27 | (odd,0) | (odd,0) | even | $P_{VII,+}$ | $U_{1}$ | $U_{1}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
28 | (odd,0) | (odd,0) | odd | $P_{VII,-}$ | $I$ | $I$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
29 | (odd,0) | (0,odd) | even | $P_{VII,+}$ | $U_{1}$ | $U_{1}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
30 | (odd,0) | (0,odd) | odd | $P_{VII,-}$ | $U_{3}$ | $I$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
31 | (0,odd) | (odd,0) | even | $P_{VII,+}$ | $U_{1}$ | $U_{2}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
32 | (0,odd) | (odd,0) | odd | $P_{VII,-}$ | $I$ | $U_{3}$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
33 | (0,odd) | (0,odd) | even | $P_{VII,+}$ | $U_{2}$ | $U_{2}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | $F_{3}^{AB}$ | $F_{3}^{BA}$
34 | (0,odd) | (0,odd) | odd | $P_{VII,-}$ | $U_{3}$ | $U_{3}$ | $|I^{B}\rangle$ | $|I^{A}\rangle$ | 1 | 1
35 | (NZE,0) | (odd,0) | even | $P_{VIII,+}$ | $U_{1}$ | $I$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
36 | (NZE,0) | (odd,0) | odd | $P_{VIII,-}$ | $I$ | $U_{1}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
37 | (NZE,0) | (0,odd) | even | $P_{VIII,+}$ | $U_{2}$ | $I$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
38 | (NZE,0) | (0,odd) | odd | $P_{VIII,-}$ | $I$ | $U_{1}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
39 | (0,NZE) | (odd,0) | even | $P_{VIII,+}$ | $U_{1}$ | $U_{3}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
40 | (0,NZE) | (odd,0) | odd | $P_{VIII,-}$ | $I$ | $U_{2}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
41 | (0,NZE) | (0,odd) | even | $P_{VIII,+}$ | $U_{1}$ | $U_{3}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
42 | (0,NZE) | (0,odd) | odd | $P_{VIII,-}$ | $U_{3}$ | $-U_{2}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
43 | (odd,0) | (NZE,0) | even | $P_{IX,+}$ | $I$ | $U_{1}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
44 | (odd,0) | (NZE,0) | odd | $P_{IX,+}$ | $U_{1}$ | $I$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
45 | (odd,0) | (0,NZE) | even | $P_{IX,+}$ | $U_{3}$ | $U_{1}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
46 | (odd,0) | (0,NZE) | odd | $P_{IX,+}$ | $U_{2}$ | $I$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
47 | (0,odd) | (NZE,0) | even | $P_{IX,+}$ | $I$ | $-U_{2}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
48 | (0,odd) | (NZE,0) | odd | $P_{IX,+}$ | $U_{1}$ | $-U_{3}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1
49 | (0,odd) | (0,NZE) | even | $P_{IX,+}$ | $U_{3}$ | $-U_{2}$ | $|I^{B}\rangle$ | $|I^{{}^{\prime}A}\rangle$ | 1 | $F_{3}^{BA}$
50 | (0,odd) | (0,NZE) | odd | $P_{IX,+}$ | -$U_{2}$ | $U_{3}$ | $|I^{{}^{\prime}B}\rangle$ | $|I^{A}\rangle$ | $F_{3}^{AB}$ | 1

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