# General teleportation channel in Fermionic Quantum Theory

Sanam Khan<EMAIL_ADDRESS>Homi Bhabha National Institute, Training
School Complex, Anushaktinagar, Mumbai 400094, India  Computer Division,
Indira Gandhi Centre for Atomic Research, Kalpakkam 603102,India  R.
Jehadeesan  Computer Division, Indira Gandhi Centre for Atomic Research,
Kalpakkam 603102,India  Sibasish Ghosh<EMAIL_ADDRESS>Optics & Quantum
Information Group, The Institute of Mathematical Sciences, C. I. T. Campus,
Taramani, Chennai 600113, India  Homi Bhabha National Institute, Training
School Complex, Anushaktinagar, Mumbai 400094, India

###### Abstract

Quantum Teleportation is a very useful scheme for transferring quantum
information. Given that the quantum information is encoded in a state of a
system of distinguishable particles, and given that the shared bi-partite
entangled state is also that of a system of distinguishable particles, the
optimal teleportation fidelity of the shared state is known to be
$(F_{max}d+1)/(d+1)$ with $F_{max}$ being the ‘maximal singlet fraction’ of
the shared state. In the present work, we address the question of optimal
teleportation fidelity given that the quantum information to be teleported is
encoded in Fermionic modes while a $2N$-mode state of a system of Fermions
(with maximum $2N$ no. of Fermions – in the second quantization language) is
shared between the sender and receiver with each party possessing $N$ modes of
the $2N$-mode state. Parity Superselection Rule (PSSR) in Fermionic Quantum
Theory (FQT) puts constraint on the allowed set of physical states and
operations, and thereby, leads to a different notion of Quantum Teleportation.
Due to PSSR, we introduce restricted Clifford twirl operations that constitute
the Unitary 2-design in case of FQT, and show that the structure of the
canonical form of Fermionic invariant shared state differs from that of the
isotropic state – the corresponding canonical invariant form for teleportation
in Standard Quantum Theory (SQT). We provide a lower bound on the optimal
teleportation fidelity in FQT and compare the result with teleportation in
SQT. Surprisingly, we find that, under separable measurements on a bipartite
Fermionic state, input and output states of the Fermionic teleportation
channel cannot be distinguished operationally, even if a particular kind of
resource state with ‘maximal singlet fraction’ being less than unity is used.

††preprint: APS/123-QED

##  I Introduction

Quantum Teleportation (QT) is a quantum information transmission task, first
introduced by Bennett et al. [1] where the sender and the receiver utilizes a
pre-shared EPR-correlated pairs [2] of particles. The first experimental
demonstration of quantum teleportation was due to Bouwmeester et. al. [3]
where the resource state used is entangled photon generated using parametric
down conversion. Apart from sending unknown quantum state of a particle, QT
also find its application in quantum computation. Gottesman and Chuang, in
their seminal paper [4], showed the essence of QT towards achieving fault-
tolerant quantum computing. Also, the success probability of quantum gate
operations in the linear optics quantum computation (LOQC) setup can be
increased using non-deterministically prepared entangled states and quantum
teleportation [5]. Recent advancement in QT [6] also finds it’s application
towards quantum networks [7].

The question of finding optimal QT fidelity was first answered by Horodecki et
al. [8] where the authors have shown that the optimal teleportation fidelity
is related to the maximal singlet fraction of the resource state (generally,
noisy) shared between sender and receiver. The authors utilized local twirling
operations to reduce any bi-partite state to the form of isotropic state which
is also physically isomorphic to a depolarizing channel in SQT – the
corresponding reduced channel under local twirling operations on the channel
operation. Such optimal teleportation has recently been used for benchmarking
the performance of NISQ devices [9].

In case, information, encoded in a system of indistinguishable particles,
needs to be teleported efficiently, one generally needs to consider an
‘entangled’ state of a bi-partite system consisting of a system of
indistinguishable particles (see, for example, ref. [10]). Entanglement of
indistinguishable particles has been at the centre of debate [11] since last
two decades. Based on the definition of subsytem accessible to local
observers, two different perspectives have been put forward – particle local
and mode local. In particle local picture, it was pointed out that due to the
symmetrized and antisymmetrized structures of the respective Hilbert spaces,
the von-neumann entropy of the reduced density matrix obtained through partial
trace, behaves differently for Bosons and Fermions [12]. However, this problem
can be overcome in the $C^{\ast}$-algebraic formalism by substituting the
partial trace operation with the restrictions of state to the corresponding
subalgebras, and applying GNS construction [13]. Analysis of mode entanglement
utilizes the second quantization method, and hence, the total number of
particles need not be fixed. In mode local picture, the entanglement is
defined relative to a given partioning of subsystems. This can be demonstrated
using Bogoliubov transformation [14]. Such mode local picture has also found
it’s application to quantify the quantum mutual information between orbitals
in quantum chemistry [15].

Fermionic entanglement has been extensively studied both in particle picture
[12, 16, 17, 18] as well as mode picture [19, 20, 21, 22]. In mode picture –
which, we are going to use in the present work – the set of allowed fermionic
states and operations must adhere to the Parity Superselection Rule (PSSR)
[23, 24, 25]. Fermionic entanglement as a resource has been utilised in
various quantum information processing tasks such as fermionic teleportation,
superdense coding, thermodynamic processes and quantum metrology [26, 27, 28,
22]. Experimental demonstration of fermionic teleportation has become a
reality using Solid-state system [29, 30]. Recently graphene nanoribbons has
been identfied as a quantum channel for teleportating Werner states with
average teleportation fidelity more than $80\%$ [31]. Apart from quantum
informational tasks, Fermionic Linear Optics(FLO) has been shown to be
advantageous over classical simulation for quantum computation when
initialized with proper input states [32].

In this work, we attempt to answer the question of finding out the ‘optimal
teleportation fidelity’ in FQT using the help of mode local picture and wedge
product formalism [24]. We follow the similar approach as that of finding the
optimal teleportation fidelity in SQT [8]. However, we adhere to the
operational notion of teleportation where one of the two $N$-modes subsystems
of an $N\times N$ -mode arbitrary bi-partite fermionic state is being
teleported, and the fidelity of this teleportation is measured operationally.
In this regard, we introduce the notion of ‘separable fidelity’ (to be defined
later) for fermionic system. The main results of this paper include the
structure of shared fermionic state under PSSR-respected local twirling
operations and lower bound on the separable fidelity – considering the shared
fermionic state ($N\times N$-mode) as a resource. We also compare this
fidelity bound with the well-known optimal teleportation fidelity in SQT with
the help of an example of a bi-partite state.

The rest of the paper is structured as follows. In Sec.II, we introduce the
mathematical structure of fermionic Hibert space and the associated Fock
space. We discuss the importance of PSSR in FQT, and the restrictions on
states and operations.We also discuss the operational notion of Entanglement
in FQT. In Sec.III, we prove the exact teleportation scheme for an $N\times N$
-mode resource state. In Sec.IV, we derive the structure of the ‘reduced’
fermionic state under local twirling operation by introducing a Haar measure
over a PSSR-respected Unitary group. In Sec.V, we construct an Unitary-2
design for this restricted Haar measure using PSSR-respected Clifford group
operations. We prove the fermionic State-Channel Isomorphism in Sec.VI and
introduce the notion of fermionic channel twirling operation. Finally, in
Sec.VII, we define the average separable fidelity for fermionic channel and
derive a lower bound for optimal fermionic teleportation scheme.

##  II PRELIMINARIES

###  II.1 Fermionic Hilbert Space

For an $N$-mode fermionic system, consider $f_{j}$ and $f_{j}^{\dagger}$ to be
the annihilation and creation operators respectively, satisfying the relations

$\\{f_{i},f_{j}\\}=0\ \ \ \\{f_{i}^{\dagger},f_{j}^{\dagger}\\}=0\ \ \
\\{f_{i},f_{j}^{\dagger}\\}=\delta_{ij}\mathbb{I}$

with $i,j\in\\{1,2,\ldots,N\\}\equiv[1,N]$.

In this work, we will utilize wedge-product Hilbert space instead of tensor-
product Hilbert space for a bipartite system. Details about the relevant
properties of wedge-product Hilbert spaces can be found in [24]. For example,
if $\mathcal{H_{PQ}}$ denotes the wedge-product Hilbert space of
$\mathcal{H_{P}}$ and $\mathcal{H_{Q}}$ then we write
$\mathcal{H_{PQ}}=\mathcal{H_{P}}\wedge\mathcal{H_{Q}}$.

Consider the Hilbert space $\mathcal{H_{A}}$ of an $N$-mode fermionic system A
with each mode spanning a 2-dimensional subspace. Let us assume that the basis
for each of these subspaces be given by $\\{|\Omega\rangle,|j\rangle\\}$ where

$|j\rangle=f_{j}^{\dagger}|\Omega\rangle$

and $|\Omega\rangle$ being vacuum state (for $j=1,2,\ldots,N$). Extending this
basis over two modes using wedge product of Hilbert spaces, we get
$\\{|\Omega\rangle,\ |i\rangle,\ |j\rangle,\ |i\rangle\wedge|j\rangle\\}$ as a
basis for $\mathcal{H}_{ij}=\mathcal{H}_{i}\wedge\mathcal{H}_{j}$. Here,
$\mathcal{H}_{i}$ and $\mathcal{H}_{j}$ correspond to the Hilbert spaces of
$i$th and $j$th mode respectively along with the constraint $i<j$.

Thus, the global Hilbert space of an $N$-mode fermionic system is given by

$\mathcal{H_{A}}=\mathcal{H}_{1}\wedge\mathcal{H}_{2}\wedge\dots\wedge\mathcal{H}_{N}$

i.e. the wedge-product of all single mode subspaces. One can define a basis
for $\mathcal{H_{A}}$ [24] as

$\displaystyle\mathcal{B_{A}}=\\{|j_{1}\rangle\wedge|j_{2}\rangle$
$\displaystyle\wedge\dots\wedge|j_{k}\rangle$ (1) $\displaystyle\ |\
j_{1}<j_{2}<\dots<j_{k},\ k\in[0,N]\\}.$

Each $j_{1},j_{2},\dots,j_{k}$ in Eq. (1) is associated with a particular
mode. Notice, $k=0$ gives us the state $|\Omega{\rangle}$. In terms of
fermionic creation operators, the aforesaid basis elements can also be written
as

$|j_{1}\rangle\wedge|j_{2}\rangle\wedge\dots\wedge|j_{k}\rangle=f_{j_{1}}^{\dagger}f_{j_{2}}^{\dagger}\dots
f_{j_{k}}^{\dagger}|\Omega\rangle.$

Dimension of this $N$-mode fermionic Hilbert space $\mathcal{H}$ is $2^{N}$.

As per Pauli exclusion principle, each mode can be occupied by maximum one
fermion. Basis elements generated by even number of fermions are denoted by
$|E_{i}\rangle$ while elements generated by odd number of fermions are denoted
by $|O_{i}\rangle$ with $i$ ranging from $0$ to $2^{N-1}$. In this new
notation, the basis $\mathcal{B_{A}}$ can be rewritten as

$\mathcal{B_{A}^{P}}=\\{|E_{1}\rangle,|E_{2}\rangle,\dots,|E_{2^{N-1}}\rangle,|O_{1}\rangle,|O_{2}\rangle,\dots,|O_{2^{N-1}}\rangle\\}$
(2)

Here, $|E_{1}\rangle\equiv|\Omega\rangle$, i.e. the ground state.

###  II.2 Parity superselection rule and physical states

Parity superselection rule (PSSR) imposes constraint on the form of a physical
fermionic state. It states that, coherent superposition of fermionic state
with even number of fermions and fermionic state with odd number of fermions
is prohibited. For example, states $f_{i}^{\dagger}|\Omega\rangle$ \+
$|\Omega\rangle$ is not allowed due to having superposition of odd and even
number of particles respectively. PSSR for spinor field was initially
discussed in the context of special relativity [33]. Recently, PSSR has been
imposed on the states of fermionic systems to remove the problem of non-
symmetric bipartitions in terms of eigenvalues [23] using the spin-statistics
connection. Howerver it was argued that the necessity of imposing PSSR is
profoundly related to the no-signalling principle [34].

PSSR has also been derived in the framework of operational probabilistic
theories for fermionic quantum systems following a set of assumptions [25].

Now consider a mixed state $\rho$ of an $N$-mode fermionic system. Due to
PSSR, the most general form of $\rho$ in the basis $\mathcal{B_{A}^{P}}$ is
given by

$\rho=\sum_{i=1,j=1}^{i=2^{N-1},j=2^{N-1}}A_{ij}|E_{i}\rangle\langle
E_{j}|+B_{ij}|O_{i}\rangle\langle O_{j}|$ (3)

with suitable on A, B so that $\rho$ satisfies positivity and the trace
condition ${\rm Tr}{\rho}=1$.

To describe a physical state of a $N\times M$ -mode bipartite fermionic
system, where $N$ corresponds to local fermionic modes of subsystem A and $M$
corresponds to local fermionic modes of subsystem B, we extend the Basis
$\mathcal{B_{A}^{P}}$ over the Hilbert space
$\mathcal{H_{A}}\wedge\mathcal{H_{B}}$. Accordingly, the extended basis
$\mathcal{B_{AB}^{P}}$ will have the form

$\displaystyle\mathcal{B_{AB}^{P}}=$
$\displaystyle\bigg{\\{}|E_{i}^{A}\rangle\wedge|E_{j}^{B}\rangle,|E_{i}^{A}\rangle\wedge|O_{j}^{B}\rangle,|O_{i}^{A}\rangle\wedge|E_{j}^{B}\rangle,$
(4) $\displaystyle|O_{i}^{A}\rangle\wedge|O_{j}^{B}\rangle\ \bigg{|}\
|E_{i(j)}^{A(B)}\rangle,|O_{i(j)}^{A(B)}\rangle\in\mathcal{B_{A(B)}^{P}}\bigg{\\}}$

.

with $i$ ranging from $0$ to $2^{N-1}$ and $j$ ranging from $0$ to $2^{M-1}$.
Using this extended basis, one can then express a generic PSSR-respected
density matrix $W$ as

$\displaystyle W=$ $\displaystyle\sum_{ijkl}a_{ijkl}|E_{i}\rangle\langle
E_{k}|\wedge|E_{j}\rangle\langle E_{l}|+b_{ijkl}|E_{i}\rangle\langle
O_{k}|\wedge|E_{j}\rangle\langle O_{l}|+c_{ijkl}|O_{i}\rangle\langle
E_{k}|\wedge|O_{j}\rangle\langle E_{k}|+d_{ijkl}|O_{i}\rangle\langle
O_{k}|\wedge|O_{j}\rangle\langle O_{l}|$ (5)
$\displaystyle+A_{ijkl}|E_{i}\rangle\langle E_{k}|\wedge|O_{j}\rangle\langle
O_{l}|+B_{ijkl}|E_{i}\rangle\langle O_{k}|\wedge|O_{j}\rangle\langle
E_{l}|+C_{ijkl}|O_{i}\rangle\langle E_{k}|\wedge|E_{j}\rangle\langle
O_{l}|+D_{ijkl}|O_{i}\rangle\langle O_{k}|\wedge|E_{j}\rangle\langle E_{l}|$

with suitable conditions on the coefficients $a,b,c,d,A,B,C$ and $D$ such that
$W>0$ and $tr(W)=1$.

###  II.3 Operations on fermionic states

As described in [24], the most generic PSSR-respected unitary operation
assumes block-diagonal form $U_{BD}$ in the basis
$B^{\prime}=\\{B_{e},B_{o}\\}$ due to no-signalling principle where, $B_{e}$
($B_{o}$) is the basis of even-parity (odd-parity) states of the fermionic
system. However, one can always embed any anti-block diagonal unitary
$U_{ABD}$ into the form of a block-diagonal unitary using an ancilla system
[26]. We explicitly prove this fact in the following lemma for dimension
$2^{N}$.

Lemma 1. Any anti-block diagonal unitary in the basis
$\mathcal{B_{A}^{P}}\equiv\\{B_{e},B_{o}\\}$ can always be embedded into the
form of a block-diagonal unitary using a single ancilla system.

Proof. Consider a single mode fermionic ancilla system $C$, with creation and
annihilation operators described by $f_{c}^{\dagger}$ and $f_{c}$
respectively. Hence the basis can be written as
$\\{|\Omega_{c}\rangle,f_{c}^{\dagger}|\Omega_{c}\rangle\\}\equiv\\{|E_{c}\rangle,|O_{c}\rangle\\}$.
Let the structure of $U_{BD}$ and $U_{ABD}$ in the basis $\mathcal{B_{A}^{P}}$
be given by

$\displaystyle U_{BD}=$ $\displaystyle\sum_{ij}u_{ij}|E_{i}\rangle\langle
E_{j}|+v_{ij}|O_{i}\rangle\langle O_{j}|$ (6) $\displaystyle U_{ABD}=$
$\displaystyle\sum_{ij}u^{\prime}_{ij}|E_{i}\rangle\langle
O_{j}|+v^{\prime}_{ij}|O_{i}\rangle\langle E_{j}|$ (7)

Consider the following unitary $U_{c}=(f_{c}^{\dagger}+f_{c})$ on the
ancillary mode. Then, any unitary of the form $U_{C}\wedge U_{ABD}$ will have
the form of $U_{BD}$. $\blacksquare$

###  II.4 Fermionic Observables

In this work, we consider the physically allowed fermionic observable to be
PSSR-respected. For a bipartite system $\mathcal{AB}$, let $\hat{O}_{P}^{A}$
denote the set of all physical observables for system $A$ and
$\hat{O}_{P}^{B}$ denote the set of all physical observables for system $B$.
Then any product observables for system $\mathcal{AB}$ denoted by
$\hat{O}_{P}^{A}\wedge\hat{O}_{P}^{B}$ respects local PSSR. Here, local PSSR
is defined as the PSSR on the subsystems of a given fermionic bi=partite
state.

In general, any observable for system $A$ can be expressed in the basis
$\mathcal{B_{A}^{P}}$ as

$\hat{O}_{P}^{A}=\sum_{ij}\alpha_{ij}|E_{i}^{A}\rangle\langle
E_{j}^{A}|+\beta_{ij}|O_{i}^{A}\rangle\langle O_{j}^{A}|$ (8)

with suitable conditions on the coefficients $\alpha$ and $\beta$.

###  II.5 Entanglement in indistinguishable fermions

Entanglement for the system of indistinguishable particles has been a topic of
debate since the last decade [11]. Moreover, for indistinguishable fermions
there exist different notions of entanglement [20]. Following the discussions
of [24], we define the operational fermionic product state as follows

Definition 1. A fermionic state $\rho^{AB}$ of a bipartite system
$\mathcal{AB}$ is said be a product state, if for any PSSR-respected
observable $\hat{O}_{P}^{A}$ and $\hat{O}_{P}^{B}$ of system $A$ and system
$B$ respectively, there exist $\rho^{A(B)}$ such that the following condition

$Tr(\rho^{AB}(\hat{O}_{P}^{A}\wedge\hat{O}_{P}^{B}))=Tr(\rho^{A}\hat{O}_{P}^{A})\
Tr(\rho^{B}\hat{O}_{P}^{B})$ (9)

satisfied, where $\rho^{A(B)}$ is a legitimate PSSR-respected density matrix
for system $A(B)$. Notice that, in general, for any operational product state
${\rho}^{AB}$, we have $\rho^{AB}\neq\rho^{A}\wedge\rho^{B}$ due to the
absence of local discriminability [25, 35]. However, the following is always
true for any local PSSR-respected product state.

Lemma 2. For any local PSSR-respected fermionic state $\rho^{AB}$ of a
bipartite system,

$\displaystyle Tr(\rho^{AB}(\hat{O}_{P}^{A}\wedge\hat{O}_{P}^{B}))=$
$\displaystyle Tr(\rho^{A}\hat{O}_{P}^{A})\ Tr(\rho^{B}\hat{O}_{P}^{B})$
$\displaystyle\Longleftrightarrow$ $\displaystyle\rho^{AB}=$
$\displaystyle\rho^{A}\wedge\rho^{B}\ \ \ \ \forall\hat{O}_{P}^{A(B)}$ (10)

with $\rho^{A(B)}$ being legitimate PSSR-respected density matrix for system
$A(B)$.

The proof of the Lemma 2 is given in Appendix A.

A separable state of fermionic system $\mathcal{AB}$ is the one which can be
represented as a convex combination of product fermionic states
$\\{\rho_{i}^{AB}\\}$:

$\rho_{sep}^{AB}=\sum_{i}p_{i}\rho_{i}^{AB}$ (11)

One can write the most general form of fermionic separable state as

$\displaystyle\rho_{sep}^{AB}$
$\displaystyle=\sum_{i}p_{i}\rho_{i}^{A}\wedge\rho_{i}^{B}+\sum_{abcd}A^{\prime}_{abcd}|E_{a}^{A}\rangle\langle
O_{c}^{A}|\wedge|E_{b}^{B}\rangle\langle O_{d}^{B}|$ (12)
$\displaystyle+(A^{\prime}_{abcd})^{\ast}|O_{c}^{A}\rangle\langle
E_{a}^{A}|\wedge|O_{d}^{B}\rangle\langle
E_{b}^{B}|+C^{\prime}_{abcd}|E_{a}^{A}\rangle\langle O_{c}^{A}|$
$\displaystyle\wedge|O_{b}^{B}\rangle\langle
E_{d}^{B}|+(C^{\prime}_{abcd})^{\ast}|O_{c}^{A}\rangle\langle
E_{a}^{A}|\wedge|E_{d}^{B}\rangle\langle O_{b}^{B}|$

with ${()}^{\ast}$ denoting the complex conjugate. Any state which is not
separable, is said to be an entangled state of the fermionic system
$\mathcal{AB}$.

Lemma 3. Transformation of a separable fermionic state into an entangled
fermionic state is forbidden under the operation of local anti-block diagonal
unitary. In other words,

$(U_{ABD}\wedge U^{\prime}_{ABD(BD)})\ \rho_{sep}^{AB}\
(U_{ABD}^{\dagger}\wedge{U^{\prime}_{ABD(BD)}}^{\dagger})\in SEP$

where $SEP$ denotes the set of all separable fermionic states. Notice that,
the case of $(U_{BD}\wedge U_{BD})$ is trivial.

Proof of Lemma 3 is given in Appendix A.

##  III Fermionic teleportation for N$\times$N mode systems

Two different notions of fermionic teleportation have been discussed in [26]:

  1. 1.

Fermionic single-mode exact teleportation is associated with the procedure in
which Alice starts with a bipartite fermionic state
$|\psi_{\tilde{A}\tilde{A}^{\prime}}\rangle$ (possibly entangled) and the
information about the reduced state of Alice’s side $\rho_{\tilde{A}}$ is
transferred to Bob’s side such that the final state becomes
$|\psi_{B\tilde{A}^{\prime}}\rangle$. This process utilizes a single copy of a
maximally entangled two-mode fermionic state $|\psi_{AB}\rangle$.

  2. 2.

Fermionic two-mode teleportation is associated with the procedure to transfer
one qubit of information encoded in terms of complex amplitude $\alpha$ and
$\beta$ ($|\alpha|^{2}+|\beta|^{2}=1$). However, to encode the information of
$\alpha$ and $\beta$ using fermionic state, one need minimum two number of
modes. Hence, one has to perform the Fermionic single-mode exact teleportation
twice to transmit the information encoded in terms of $\alpha$ and $\beta$.

In this paper, we generalize the concept of Fermionic single-mode
teleportation for N$\times$N mode system where we assume each fermionic system
$\tilde{A},\tilde{A}^{\prime},A$ and $B$ consists of $N$ local fermionic modes
(LFM). But before delve into the fermionic teleportation, let us briefly
discuss the teleportation protocol in d-dimensional Hilbert space as described
in [36] in the context of teleporting subsystem.

###  III.1 Teleportation in d-dimensional Hilbert space in SQT

In this quantum teleportation protocol, four quantum systems are used.
Consider the teleportation channel being applied to a part of the bipartite
state $\rho_{\tilde{A}^{\prime}\tilde{A}}$ . Also, consider a pre-shared
maximally entangled state $|\omega\rangle\langle\omega|_{AB}$ between Alice
and Bob. Hence the initial state can be written as

$\rho_{\tilde{A}^{\prime}\tilde{A}}\otimes|\omega\rangle\langle\omega|_{AB}$

Then a successful teleportation is represented by the equation

$\displaystyle\sum_{x\in
X}\text{tr}((\rho_{\tilde{A}^{\prime}\tilde{A}}\otimes|\omega\rangle\langle\omega|_{AB})$
$\displaystyle(\hat{O}_{\tilde{A}^{\prime}}\otimes M_{x}\otimes
N_{x}(\hat{O}_{B})))$ (13)
$\displaystyle=\text{tr}(\rho_{\tilde{A}^{\prime}B}\hat{O}_{\tilde{A}^{\prime}}\otimes\hat{O}_{B})$

with $M_{x}=|\Psi_{x}\rangle\langle\Psi_{x}|_{\tilde{A}A}$ and
$\langle\Psi_{x}|\Psi_{y}\rangle=\delta_{xy}$. Here, $\Psi_{x}$ constitutes
the basis of maximally entangled states and $X$ denotes a set of $d^{2}$
elements. Also,

$N_{x}(\hat{O}_{B})=U_{x}^{\dagger}(\hat{O}_{B})U_{x}$

and

$|\Psi_{x}\rangle\langle\Psi_{x}|=(U_{x}\otimes\mathbb{I})|\omega\rangle\langle\omega|(U_{x}^{\dagger}\otimes\mathbb{I})$

defines the $U_{x}$ for each x. Hence, $U_{x}$ also forms the bases of unitary
operators. To prove Eq. (13), let us proceed, as given in [36], by choosing
$\rho_{\tilde{A}^{\prime}\tilde{A}}=|\phi_{2}\rangle_{\tilde{A}^{\prime}}\langle\phi_{4}|\otimes|\phi_{1}\rangle_{\tilde{A}}\langle\phi_{3}|$,
$\hat{O}_{B}=|\psi_{1}\rangle_{B}\langle\psi_{3}|$ and
$\hat{O}_{\tilde{A}^{\prime}}=|\psi_{2}\rangle_{\tilde{A}^{\prime}}\langle\psi_{4}|$.
Then, considering the $x^{th}$ term in Eq. (13) we get

$\displaystyle Term_{x}=\langle\phi_{4}\otimes$
$\displaystyle\phi_{3}\otimes\omega|\psi_{2}\otimes\Psi_{x}\otimes
U_{x}^{\dagger}\psi_{1}\rangle\times$ (14)
$\displaystyle\langle\psi_{4}\otimes\Psi_{x}\otimes
U_{x}^{\dagger}\psi_{3}|\phi_{2}\otimes\phi_{1}\otimes\omega\rangle$

Now substituting $|\Psi_{x}\rangle=(U_{x}\otimes\mathbb{I})|\omega\rangle$ and
utilizing the inner product definition of tensor product Hilbert space we get

$Term_{x}=d^{-2}\langle\phi_{3}|\psi_{1}\rangle\langle\psi_{3}|\phi_{1}\rangle\langle\phi_{4}|\psi_{2}\rangle\langle\psi_{4}|\phi_{2}\rangle$

Hence, summing over $x$ we get the desired result.

Notice, SQT satisfies local distinguishability of states [37]. Hence,
observable of type separable on the bipartite system is sufficient to
distinguish two different bipartite states.

###  III.2 Teleportation in d-dimensional Hilbert space in FQT

In $2^{N}$-dim, teleportation of complex amplitudes of even-parity and odd-
parity vectors at the same time is prohibited. This makes a difference between
the teleportation protocol in SQT and FQT. However, incoherent mixtures of
even-parity vectors and odd parity vectors are allowed. Fermionic partial
trace of a bi-partite fermionic state may result in a reduced density matrix
with incoherent mixtures of even-parity vectors and odd parity vectors. In
this work, we have demonstrated the teleportation in the sense of subsystem
swapping. To see this, consider a fermionic state
$\rho_{\tilde{A}^{\prime}\tilde{A}}^{f}$ of an $N\times N$ bipartite system
where $N$ denotes LFMs. Our objective is to faithfully teleport the
information about the state of subsystem $\tilde{A}$ to Bob using a pre-shared
resource state $\omega_{AB}^{f}$ between Alice and Bob as shown in Fig. 1.

Figure 1:  Teleportation in FQT. In this scheme $\omega_{AB}^{f}$ is the
resource state used to to teleport one half of the initial state
$\rho_{\tilde{A}^{\prime}\tilde{A}}^{f}(initial)$ and produces the final state
$\rho_{\tilde{A}^{\prime}B}^{f}(final)$.

We define the resource state to be used for this teleportation as,

$\omega_{AB}^{f}=\text{loc-p}(|\omega\rangle\langle\omega|)+\Upsilon\overline{\text{loc-p}}(|\omega\rangle\langle\omega|)$
(15)

where, $\Upsilon$ is real and loc-p() indicates the local-parity preserving
part, while $\overline{\text{loc-p()}}$ represents local-parity violating
part. We now describe the first result of the paper.

Theorem 1. A successful fermionic teleportation of a subsystem of a bipartite
fermionic state can be described by the equation

$\displaystyle\sum_{x\in
X}\text{tr}((\rho_{\tilde{A}^{\prime}\tilde{A}}^{f}\wedge\omega_{AB}^{f})$
$\displaystyle(\hat{O}_{\tilde{A}^{\prime}}\wedge M_{x}\wedge
N_{x}(\hat{O}_{B})))$ (16)
$\displaystyle=\text{tr}(\rho_{\tilde{A}^{\prime}B}^{f}\hat{O}_{\tilde{A}^{\prime}}\wedge\hat{O}_{B})$

provided the local fermionic observable respects local-PSSR. The operational
meaning of Eq. (16) is that the input and output fermionic states of the above
mentioned teleportation channel cannot be locally distinguishable.

Proof of Theorem 1 is given in Appendix B.

##  IV $U\wedge U$ invariant fermionic state

In this section we derive the family of fermionic states invariant under the
$U\wedge U$ operations with $U$ satisfying Eq. (6) and Eq. (7). We follow the
procedure mentioned in [38, 39]. Consider an operator $W$ on fermionic wedge
product space of $N\times N$ mode system, given by, Eq. (5). At first, let us
impose the constraint

$(U_{BD}\wedge U_{BD})\ W\ (U_{BD}^{\dagger}\wedge U_{BD}^{\dagger})=W$ (17)

This gives us the first reduced form of $W$ with nonzero coefficients of type
$a_{iiii}$, $a_{ijji}$, $a_{ijij}$, $d_{iiii}$, $d_{ijji}$, $d_{ijij}$,
$A_{iiii}$, $A_{ijij}$, $B_{iiii}$, $B_{ijji}$, $C_{iiii}$, $C_{ijji}$,
$D_{iiii}$ and $D_{ijij}$. For example, equating the coefficients of the basis
$|E_{\alpha}\rangle\wedge|E_{\gamma}\rangle\langle O_{q}|\wedge\langle
O_{s}|$, we get the constraint equation $b_{\alpha\gamma
qs}=u_{\alpha\beta}u_{\gamma\delta}b_{\beta\delta
pr}v_{qp}^{\ast}v_{sr}^{\ast}$. Now let us assume that, for a given unitary
$U_{BD}$,

$\displaystyle u_{\alpha\beta}=0\ \ \text{if}\ \ \alpha\neq\beta\ \ \text{and
}\ \ u_{\alpha\beta}=1\ \text{or}\ -1\ \ \text{if}\ \ \alpha=\beta$
$\displaystyle v_{\alpha\beta}=0\ \ \text{if}\ \ \alpha\neq\beta\ \ \text{and
}\ \ v_{\alpha\beta}=1\ \text{or}\ -1\ \ \text{if}\ \ \alpha=\beta$

Hence, the only possible non-zero elements are $b_{\alpha\alpha qq}$. But this
can also be equated to zero by considering one of $u_{\alpha\alpha}$ or
$v_{qq}$ equal to $i$. Similarly, one can show that $c_{\alpha\gamma qs}=0$
for all $\alpha,\gamma,q$ and $s$.

Now, equating the coefficients of the basis
$|E_{\alpha}\rangle\wedge|O_{\gamma}\rangle\langle E_{q}|\wedge\langle
O_{s}|$, we get the constraint equation $A_{\alpha\gamma
qs}=u_{\alpha\beta}v_{\gamma\delta}A_{\beta\delta
pr}u_{qp}^{\ast}v_{sr}^{\ast}$. With suitable choice of $U_{BD}$, one can show
that $A_{\alpha qq\alpha}=0$ for all $\alpha$ and $q$. Proceeding similarly,
one can show that, $D_{\alpha qq\alpha}=0$, $B_{\alpha q\alpha q}=0$ and
$C_{\alpha q\alpha q}=0$ for all $\alpha,\gamma,q$ and $s$.

Next, consider the permutation of basis in the even and odd parity sectors
separately. This parity-respected permutation of basis can be obtained using
block diagonal unitary operation $U_{BD}$. With this parity-respected
permutation of basis in the even and odd sectors, along with the constraint
equation (17) one can show that all the elements within the sets
$\\{a_{ijij}|i\neq j\\}$, $\\{a_{ijji}|i\neq j\\}$, $\\{a_{iiii}\\}$,
$\\{d_{ijij}|i\neq j\\}$, $\\{d_{ijji}|i\neq j\\}$, $\\{d_{iiii}\\}$,
$\\{A_{ijij}\\}$, $\\{B_{ijji}\\}$, $\\{C_{ijji}\\}$ and $\\{D_{ijij}\\}$ are
equal. Thus $W$ depends on maximum 10 independent parameters. Now, consider
the constraint equation

$(U_{ABD}\wedge U_{ABD})\ W\ (U_{ABD}^{\dagger}\wedge U_{ABD}^{\dagger})=W$
(18)

This constraint equation (18) gives us further reduction of $W$ in terms of
number of independent parameters as follows: $\\{a_{ijij}|i\neq
j\\}=\\{d_{ijij}|i\neq j\\}=\\{\tilde{a}^{E}\\}$, $\\{a_{ijji}|i\neq
j\\}=\\{d_{ijji}|i\neq j\\}=\\{\tilde{b}^{E}\\}$,
$\\{a_{iiii}\\}=\\{d_{iiii}\\}=\\{A^{E}\\}$,
$\\{A_{ijij}\\}=\\{D_{ijij}\\}=\\{a^{O}\\}$ and
$\\{B_{ijji}\\}=\\{C_{ijji}\\}=\\{b^{O}\\}$ where, $\\{\tilde{a}^{E}\\}$,
$\\{\tilde{b}^{E}\\}$, $\\{A^{E}\\}$, $\\{a^{O}\\}$ and $\\{b^{O}\\}$
represents singleton set. Hence, one can write down $W$, invariant under
twirling operation as,

$\displaystyle W=$ $\displaystyle\tilde{a}^{E}\sum_{ij,i\neq
j}|E_{i}\rangle\wedge|E_{j}\rangle\langle E_{i}|\wedge\langle
E_{j}|+|O_{i}\rangle\wedge|O_{j}\rangle\langle O_{i}|\wedge\langle
O_{j}|+\tilde{b}^{E}\sum_{ij,i\neq j}|E_{i}\rangle\wedge|E_{j}\rangle\langle
E_{j}|\wedge\langle E_{i}|+|O_{i}\rangle\wedge|O_{j}\rangle\langle
O_{j}|\wedge\langle O_{i}|$ (19)
$\displaystyle+A^{E}\sum_{i}|E_{i}\rangle\wedge|E_{i}\rangle\langle
E_{i}|\wedge\langle E_{i}|+|O_{i}\rangle\wedge|O_{i}\rangle\langle
O_{i}|\wedge\langle O_{i}|+a^{O}\sum_{ij}|E_{i}\rangle\wedge\
|O_{j}\rangle\langle E_{i}|\wedge\langle
O_{j}|+|O_{i}\rangle\wedge|E_{j}\rangle\langle O_{i}|\wedge\langle E_{j}|$
$\displaystyle+b^{O}\sum_{ij}|E_{i}\rangle\wedge|O_{j}\rangle\langle
O_{j}|\wedge\langle E_{i}|+|O_{i}\rangle\wedge|E_{j}\rangle\wedge\langle
E_{j}|\wedge\langle O_{i}|$

However, the sums with coefficients $A^{E},\tilde{a}^{E}$ and $\tilde{b}^{E}$
in the R.H.S of Eq. (19) are not invariant under $U_{BD}\wedge U_{BD}$
twirling individually. Hence, $A^{E}$ must depend on $\tilde{a}^{E}$ and
$\tilde{b}^{E}$. We define,

$\displaystyle\mathbb{I}^{|E\rangle\langle E|}=$
$\displaystyle\sum_{ij}|E_{i}\rangle\wedge|E_{j}\rangle\langle
E_{i}|\wedge|\langle E_{j}|+|O_{i}\rangle\wedge|O_{j}\rangle\langle
O_{i}|\wedge|\langle O_{j}|$ $\displaystyle\mathbb{I}^{|O\rangle\langle O|}=$
$\displaystyle\sum_{ij}|E_{i}\rangle\wedge|O_{j}\rangle\langle
E_{i}|\wedge|\langle O_{j}|+|O_{i}\rangle\wedge|E_{j}\rangle\langle
O_{i}|\wedge|\langle E_{j}|$ $\displaystyle H_{N}^{|E\rangle\langle E|}=$
$\displaystyle\sum_{ij}|E_{i}\rangle\wedge|E_{j}\rangle\langle
E_{j}|\wedge|\langle E_{i}|+|O_{i}\rangle\wedge|O_{j}\rangle\langle
O_{j}|\wedge|\langle O_{i}|$ $\displaystyle H_{N}^{|O\rangle\langle O|}=$
$\displaystyle\sum_{ij}|E_{i}\rangle\wedge|O_{j}\rangle\langle
O_{j}|\wedge|\langle E_{i}|+|O_{i}\rangle\wedge|E_{j}\rangle\langle
E_{j}|\wedge|\langle O_{i}|$

Thus, $\mathbb{I}^{|E\rangle\langle E|}+\mathbb{I}^{|O\rangle\langle
O|}=\mathbb{I}$ and $H_{N}^{|E\rangle\langle E|}+H_{N}^{|O\rangle\langle
O|}=H_{N}$

Lemma 4. Any fermionic states invariant under the fermionic unitary operation
$U\wedge U$ is of the form

$W_{werner}^{f}=a^{E}\mathbb{I}^{|E\rangle\langle
E|}+a^{O}\mathbb{I}^{|O\rangle\langle O|}+b^{E}H_{N}^{|E\rangle\langle
E|}+b^{O}H_{N}^{|O\rangle\langle O|}$ (20)

Proof. At first, we prove that $W$ is invariant under the fermionic operation
$U\wedge U$. For this purpose, notice

$\displaystyle(U_{BD}\wedge U_{BD})(\mathbb{I}^{|E\rangle\langle
E|})(U_{BD}^{\dagger}\wedge U_{BD}^{\dagger})$
$\displaystyle=\Bigg{(}\sum_{efgh}p_{ef}p_{gh}|E_{e}^{A}\rangle\wedge|E_{g}^{B}\rangle\langle
E_{f}^{A}|\wedge\langle
E_{h}^{B}|+p_{ef}q_{gh}|E_{e}^{A}\rangle\wedge|O_{g}^{B}\rangle\langle
E_{f}^{A}|\wedge\langle
O_{h}^{B}|+q_{ef}p_{gh}|O_{e}^{A}\rangle\wedge|E_{g}^{B}\rangle\langle
O_{f}^{A}|\wedge\langle E_{h}^{B}|\Bigg{.}$
$\displaystyle+\Bigg{.}q_{ef}q_{gh}|O_{e}^{A}\rangle\wedge|O_{g}^{B}\rangle\langle
O_{f}^{A}|\wedge\langle O_{h}^{B}|\Bigg{)}\Bigg{(}\mathbb{I}^{|E\rangle\langle
E|}\Bigg{)}\Bigg{(}\sum_{ijkl}(p_{ij})^{\ast}(p_{kl})^{\ast}|E_{j}^{A}\rangle\wedge|E_{l}^{B}\rangle\langle
E_{i}^{A}|\wedge\langle
E_{k}^{B}|+(p_{ij})^{\ast}(q_{kl})^{\ast}|E_{j}^{A}\rangle\wedge|O_{l}^{B}\rangle\Bigg{.}$
$\displaystyle\langle E_{i}^{A}|\wedge\langle
O_{k}^{B}|+\Bigg{.}(q_{ij})^{\ast}(p_{kl})^{\ast}|O_{j}^{A}\rangle\wedge|E_{l}^{B}\rangle\langle
O_{i}^{A}|\wedge\langle
E_{k}^{B}|+(q_{ij})^{\ast}(q_{kl})^{\ast}|O_{j}^{A}\rangle\wedge|O_{l}^{B}\rangle\langle
O_{i}^{A}|\wedge\langle O_{k}^{B}|\Bigg{)}$
$\displaystyle=\mathbb{I}^{|E\rangle\langle E|}$

where we have used the relation $\sum_{f}p_{ef}(p_{if})^{\ast}=\delta_{ei}$
and $\sum_{f}q_{ef}(q_{ei})^{\ast}=\delta_{ei}$. Similarly one can show that,
$(U_{BD}\wedge U_{BD})(\mathbb{I}^{|O\rangle\langle
O|})(U_{BD}^{\dagger}\wedge U_{BD}^{\dagger})=\mathbb{I}^{|O\rangle\langle
O|}$. Same results can also be obtained for $U_{ABD}\wedge U_{ABD}$ operation.

Following the same steps one can show that, for $U=\\{U_{BD},U_{ABD}\\}$

$\displaystyle(U\wedge U)(H_{N}^{|E\rangle\langle E|})(U^{\dagger}\wedge
U^{\dagger})=H_{N}^{|E\rangle\langle E|}\ \ \text{and}\ \ (U\wedge
U)(H_{N}^{|O\rangle\langle O|})(U^{\dagger}\wedge
U^{\dagger})=H_{N}^{|O\rangle\langle O|}$ (21)

Hence,$W_{werner}^{f}$ in Eq. (20) is an invariant fermionic state under the
fermionic operation $(U\wedge U)$. Notice, for
$A^{E}=\tilde{a}^{E}+\tilde{b}^{E}$, the $W$ in Eq. (19) becomes equal to
$W_{werner}^{f}$. $\blacksquare$

We now show that, there exist a unique Haar measure on the restricted set
$U_{res}\equiv\\{U_{BD},U_{ABD}\\}$. In this regard, we provide the following
lemma.

Lemma 5. There exist a unique left invariat Haar measure on the restricted set
$U_{res}\equiv\\{U_{BD},U_{ABD}\\}$

Proof of the lemma is given in Appendix A. Then one can easily show that the
integral

$\int_{U_{res}}dU^{\prime}(U^{\prime}\wedge
U^{\prime})\rho(U^{\prime\dagger}\wedge U^{\prime\dagger})$

is invariant under the operation $U\wedge U$. Hence the local twirling
operation reduces any state to the invariant form given in Lemma 4.

Lemma 6. The Overlap of an $N\times N$-mode fermionic state $\rho^{f}$ with
the operators $\mathbb{I}^{|E\rangle\langle E|}+\mathbb{I}^{|O\rangle\langle
O|}$, $\mathbb{I}^{|E\rangle\langle E|}-\mathbb{I}^{|O\rangle\langle O|}$,
$H^{|E\rangle\langle E|}_{N}$ and $H^{|O\rangle\langle O|}_{N}$ remain
invariant under the restricted Unitary group twirling operation on fermionic
state.

Proof of Lemma 6. is given in Appendix C.

##  V Reduction of $N\times N$ mode fermionic states under twirling

Let us first introduce the fermionic Pauli operator acting on a single
fermionic mode as follows

$\displaystyle\sigma_{z}^{f}=$
$\displaystyle|\Omega\rangle\langle\Omega|-f^{\dagger}|\Omega\rangle\langle\Omega|f\equiv|E\rangle\langle
E|-|O\rangle\langle O|$ $\displaystyle\sigma_{x}^{f}=$
$\displaystyle|\Omega\rangle\langle\Omega|f+f^{\dagger}|\Omega\rangle\langle\Omega|\equiv|E\rangle\langle
O|+|O\rangle\langle E|$ $\displaystyle\sigma_{y}^{f}=$
$\displaystyle-i|\Omega\rangle\langle\Omega|f+if^{\dagger}|\Omega\rangle\langle\Omega|\equiv-i|E\rangle\langle
E|+i|O\rangle\langle O|$

We define the fermionic Pauli group [40] on 1-mode system as

$P_{1}\equiv\\{\pm\mathbb{I},\pm\sigma_{x}^{f},\pm\sigma_{y}^{f},\pm\sigma_{z}^{f},\pm
i\mathbb{I},\pm i\sigma_{x}^{f},\pm\ i\sigma_{y}^{f},\pm i\sigma_{z}^{f}\\}$

For $N$-mode system, the fermionic Pauli group $P_{N}$ can be written by
taking the $N$-fold wedge product of the fermionic Pauli operators along with
identity element and multiplicative factor $\\{\pm 1,\pm i\\}$. We now define
the Clifford group $C_{N}$ as the normalizer of the fermionic Pauli group as
mentioned in [41]. However, all the Clifford group elements are not legitimate
fermionic operator due to it’s non-block/non-antiblock diagonal structure in
the basis $\mathcal{B_{A}^{P}}$.

Consider, $C^{res}_{N}$ as the restricted set with elements from Clifford
group such that

$\displaystyle C^{res}_{N}$ $\displaystyle=\\{C\in
C_{N}|\Gamma^{BD}\in\Xi^{BD}\implies C\Gamma^{BD}C^{\dagger}\in\Xi^{BD}\ $
(22) $\displaystyle\text{and}\ \Gamma^{ABD}\in\Xi^{ABD}\implies\
C\Gamma^{ABD}C^{\dagger}\in\Xi^{ABD}\\}$

where the set $\Xi^{BD}$ consists of block-diagonal elements of the fermionic
Pauli group $P_{N}$ and the set $\Xi^{ABD}$ consists of antiblock-diagonal
elements of the fermionic Pauli group $P_{N}$ in the basis
$\mathcal{B_{A}^{P}}$. Using this restricted set $C^{res}_{N}$ we define a
restricted Clifford twirl operation as follows

$\mathcal{T}_{C^{res}_{N}}[\rho^{f}]=\frac{1}{|C^{res}_{N}|}\sum_{c\in
C^{res}_{N}}(c\wedge c)\rho^{f}(c^{\dagger}\wedge c^{\dagger})$ (23)

where $\rho^{f}$ is any fermionic state of a $N\times N$-mode system.

Lemma 7. All the elements of $C^{res}_{N}$ are legitimate unitary operator in
FQT.

The proof of Lemma 7 is given in Appendix C.

As in [41], we will denote $P_{N}^{H}$ as the hermitian subset of $P_{N}$. We
will now prove that restricted Clifford twirling operation reduces any
$N\times N$-mode fermionic state into the form of Eq. (20).

Theorem 2. The following statements are true:

  1. 1.

$\mathcal{T}_{C^{res}_{N}}$=$\mathcal{T}_{C^{res}_{N}}^{\dagger}$

  2. 2.

$\mathcal{T}_{C^{res}_{N}}[\mathbb{I}_{N}]=\mathbb{I}_{N}$ and
$\mathcal{T}_{C^{res}_{N}}[(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{A}\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{B}]=(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{A}\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{B}$

  3. 3.

$\mathcal{T}_{C^{res}_{N}}[H^{|E\rangle\langle E|}_{N}]=H^{|E\rangle\langle
E|}_{N}$ and $\mathcal{T}_{C^{res}_{N}}[H^{|O\rangle\langle
O|}_{N}]=H^{|O\rangle\langle O|}_{N}$

  4. 4.

For any $N\times N$-mode fermionic state $\rho^{f}$,
$\mathcal{T}_{C^{res}_{N}}[\rho^{f}]$ becomes the linear combination of
$\mathbb{I}^{|E\rangle\langle E|}$, $\mathbb{I}^{|O\rangle\langle O|}$,
$H^{|E\rangle\langle E|}_{N}$ and $H^{|O\rangle\langle O|}_{N}$

where, $\mathcal{T}_{C^{res}_{N}}^{\dagger}=\frac{1}{|C^{res}_{N}|}\sum_{c\in
C^{res}_{N}}(c^{\dagger}\wedge c^{\dagger})\rho^{f}(c\wedge c)$.

Proof of Theorem 2 is given in Appendix C.

Corollary 1. The Overlap of an $N\times N$-mode fermionic state $\rho^{f}$
with the operators $\mathbb{I}_{N}\wedge\mathbb{I}_{N},\
(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}),\
H^{|E\rangle\langle E|}_{N}$ and $H^{|O\rangle\langle O|}_{N}$ remains
invariant under the restricted Clifford twirling operation on fermionic state.

Hence, comparing Eq. (20) and Theorem 2 we get that the restricted Clifford
twirling operation reduces any fermionic state to the invariant form under
$U\wedge U$ operation as given in Lemma 4.

For any $N\times N$-mode fermionic state $\rho^{f}$ which is invariant under
$U\wedge U$, $(\mathbb{I}_{N}\wedge\mathbb{T}_{f})(\rho^{f})$ is invariant
under the operation $U\wedge U^{\ast}$ with $U^{\ast}$ denotes the complex
conjugation of $U$. Here, $\mathbb{T}_{f}$ denotes the partial transpose
operator (see Appendix C).

Corollary 2. The following statements are true:

  1. 1.

$\mathcal{T}^{\ast}_{C^{res}_{N}}$=$(\mathcal{T}^{\ast}_{C^{res}_{N}})^{\dagger}$

  2. 2.

$\mathcal{T}^{\ast}_{C^{res}_{N}}[\mathbb{I}_{N}]=\mathbb{I}_{N}$ and
$\mathcal{T}^{\ast}_{C^{res}_{N}}[(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{A}\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{B}]=(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{A}\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{B}$

  3. 3.

$\mathcal{T}^{\ast}_{C^{res}_{N}}[\Psi^{|E\rangle\langle
E|}_{N}]=\Psi^{|E\rangle\langle E|}_{N}$ and
$\mathcal{T}^{\ast}_{C^{res}_{N}}[\Psi^{|E\rangle\langle
O|}_{N}]=\Psi^{|E\rangle\langle O|}_{N}$

  4. 4.

For any $N\times N$-mode fermionic state $\rho^{f}$,
$\mathcal{T}^{\ast}_{C^{res}_{N}}[\rho^{f}]$ becomes a linear combination of
$\mathbb{I}^{|E\rangle\langle E|}$, $\mathbb{I}^{|O\rangle\langle O|}$,
$\Psi^{|E\rangle\langle E|}_{N}$ and $\Psi^{|E\rangle\langle O|}_{N}$

where we define

$\displaystyle\mathcal{T}^{\ast}_{C^{res}_{N}}[\rho^{f}]$
$\displaystyle=\frac{1}{|C^{res}_{N}|}\sum_{c\in C^{res}_{N}}(c\wedge
c^{\ast})\rho^{f}(c^{\dagger}\wedge(c^{\ast})^{\dagger})$
$\displaystyle\Psi^{|E\rangle\langle E|}_{N}$
$\displaystyle=(\mathbb{I}_{N}\wedge\mathbb{T}_{f})H^{|E\rangle\langle
E|}_{N}$ $\displaystyle\Psi^{|E\rangle\langle O|}_{N}$
$\displaystyle=(\mathbb{I}_{N}\wedge\mathbb{T}_{f})H^{|O\rangle\langle
O|}_{N}$

Proof. For a generic $N\times N$-mode fermionic state $W$given by Eq. (5), it
can be easily proved that

$\displaystyle(U\wedge
U^{\ast})\\{(\mathbb{I}_{N}\wedge\mathbb{T}_{f})(W)\\}(U^{\dagger}\wedge(U^{\ast})^{\dagger})$
$\displaystyle=(\mathbb{I}_{N}\wedge\mathbb{T}_{f})\\{(U\wedge
U)(W)(U^{\dagger}\wedge U^{\dagger})\\}$

where, $U\in\\{U_{BD},U_{ABD}\\}$. Therefore, for any $\rho^{f}$

$\displaystyle\mathcal{T}^{\ast}_{C^{res}_{N}}[\rho^{f}]=(\mathbb{I}_{N}\wedge\mathbb{T}_{f})\mathcal{T}_{C^{res}_{N}}[(\mathbb{I}_{N}\wedge\mathbb{T}_{f})\rho^{f}]$
(24)

Hence statement (1)-(4) can be proved using the results of Theorem 2.
$\blacksquare$

##  VI State-Channel isomorphism in FQT

A Quantum Operations for FQT in it’s most general form or Fermionic Channel
has already been introduced in [24] where the equivalence of operator-sum
representation and completely Positive map has been proved. In this section,
following the procedure mentioned in [8], we will establish an isomorphism
given by

$\rho^{f}_{\mathcal{E}}=(\mathbb{I}\wedge\mathcal{E})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$
(25)

Here, $|\psi^{+}\rangle\langle\psi^{+}|_{f}=\Psi^{|E\rangle\langle
E|}_{N}+\Psi^{|E\rangle\langle O|}_{N}$, represented in basis
$\mathcal{B_{AB}^{P}}$ and the first reduction of $\rho^{f}_{\mathcal{E}}$
following fermionic partial tracing procedure gives $\mathbb{I}/d$ with
$\mathcal{E}$ representing a fermionic channel. To prove the surjective
property of this isomorphism. let us assume that the spectral decomposition of
$\rho^{f}_{\mathcal{E}}$ is given by

$\rho^{f}_{\mathcal{E}}=\sum_{m}p_{m}^{E}|\psi^{E}_{m}\rangle\langle\psi^{E}_{m}|+\sum_{m}p_{m}^{O}|\psi^{O}_{m}\rangle\langle\psi^{O}_{m}|$

where $|\psi^{E}_{m}\rangle$ and $|\psi^{O}_{m}\rangle$ represents
eigenvectors of $\rho^{f}_{\mathcal{E}}$ for all $m$. Let,

$\displaystyle|\psi^{E}_{1}\rangle=$
$\displaystyle\sum_{ij}A_{ij}^{EE}|E_{i}\rangle\wedge|E_{j}\rangle+A_{ij}^{OO}|O_{i}\rangle\wedge|O_{j}\rangle$
$\displaystyle|\psi^{O}_{1}\rangle=$
$\displaystyle\sum_{ij}A_{ij}^{EO}|E_{i}\rangle\wedge|O_{j}\rangle+A_{ij}^{OE}|O_{i}\rangle\wedge|E_{j}\rangle$

One can then represent $|\psi^{E}_{1}\rangle$ in terms of a fermionic operator
$K_{1}^{BD}$as

$|\psi^{E}_{1}\rangle=(\mathbb{I}\wedge
K_{1}^{BD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$

with, $K_{1}^{BD}$ being block diagonal in the basis $\mathcal{B^{P}_{A}}$. To
see this, let us assume

$K_{1}^{BD}=\sum_{ij}S_{ij}|E_{i}\rangle\langle
E_{j}|+T_{ij}|O_{i}\rangle\langle O_{j}|$

Then, $(\mathbb{I}\wedge K_{1}^{BD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$ is
given by

$=\frac{1}{\sqrt{d}}\sum_{ij}S_{ij}|E_{j}\rangle\wedge|E_{i}\rangle+T_{ij}|O_{j}\rangle\wedge|O_{i}\rangle$

Fixing $\langle E_{i}|K_{1}^{BD}|E_{j}\rangle=\sqrt{d}A_{ij}^{EE}$ and
$\langle O_{i}|K_{1}^{BD}|O_{j}\rangle=\sqrt{d}A_{ij}^{OO}$ we get
$|\psi^{E}_{1}\rangle=(\mathbb{I}\wedge
K_{1}^{BD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$. Similarly one can find
anti-block diagonal $K_{1}^{ABD}$ and show that
$|\psi^{O}_{1}\rangle=(\mathbb{I}\wedge
K_{1}^{ABD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$. Hence,
$\rho^{f}_{\mathcal{E}}$ can be written as

$\displaystyle\rho^{f}_{\mathcal{E}}=$
$\displaystyle\sum_{m}p_{m}^{E}(\mathbb{I}\wedge
K_{m}^{BD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})(\mathbb{I}\wedge{K_{m}^{BD}}^{\dagger})+$
(26) $\displaystyle p_{m}^{O}(\mathbb{I}\wedge
K_{m}^{ABD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})(\mathbb{I}\wedge{K_{m}^{ABD}}^{\dagger})$
$\displaystyle=(\mathbb{I}\wedge\mathcal{E})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$

Then $(\mathbb{I}\wedge\mathcal{E})$ is a positive map with
$\\{K_{m}^{BD},K_{m}^{ABD}|\ \forall m\\}$ representing the Kraus operator for
the channel $\mathcal{E}$.

To see the trace-preservion condition of the channel $\mathcal{E}$, we need to
show that the operator

$Z^{f}=\sum_{m}p_{m}^{E}{K_{m}^{BD}}^{\dagger}K_{m}^{BD}+p_{m}^{O}{K_{m}^{ABD}}^{\dagger}K_{m}^{ABD}=\mathbb{I}$

. For this, recall that the consistency condition for partial trace in FQT
[21] is given by

$\text{tr}(\mathcal{O}_{A}\rho^{f}_{A})=\text{tr}((\mathcal{O}_{A}\wedge\mathbb{I}_{B})\rho^{f}_{AB})$

for any bi-partite fermionic states $\rho^{f}_{AB}$ with reduced state
$\rho^{f}_{A}$. For $\rho^{f}_{A}=\mathbb{I}/d$ and letting
$\rho^{f}_{AB}=\rho^{f}_{\mathcal{E}}$, we get

$\displaystyle\text{tr}(\mathcal{O}_{A})=d\text{tr}((\mathcal{O}_{A}\wedge\mathbb{I}_{B})\rho^{f}_{AB})=$
$\displaystyle
d\sum_{m}p_{m}^{E}\text{tr}((\mathcal{O}_{A}\wedge\mathbb{I}_{B})(\mathbb{I}\wedge
K_{m}^{BD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})(\mathbb{I}\wedge{K_{m}^{BD}}^{\dagger}))$
$\displaystyle+p_{m}^{O}\text{tr}((\mathcal{O}_{A}\wedge\mathbb{I}_{B})(\mathbb{I}\wedge
K_{m}^{ABD})(|\psi^{+}\rangle\langle\psi^{+}|_{f})(\mathbb{I}\wedge{K_{m}^{ABD}}^{\dagger}))$
$\displaystyle=d\text{tr}((|\psi^{+}\rangle\langle\psi^{+}|_{f})(\mathcal{O}_{A}\wedge
Z^{f}))$
$\displaystyle=d\text{tr}((|\psi^{+}\rangle\langle\psi^{+}|_{f})(\mathcal{O}_{A}(Z^{f})^{T}\wedge\mathbb{I}_{B}))$
$\displaystyle=\text{tr}(\mathcal{O}_{A}(Z^{f})^{T})\ \ \ \ \
\forall\mathcal{O}_{A}$

where we have used the properties $(\mathbb{I}\wedge
Z^{f})|\psi^{+}\rangle_{f}=((Z^{f})^{T}\wedge\mathbb{I})|\psi^{+}\rangle_{f}$
and partial trace of $|\psi^{+}\rangle\langle\psi^{+}|_{f}$ is given by
$\mathbb{I}/d$. Hence, $(Z^{f})^{T}=Z^{f}=\mathbb{I}$. This completes the
proof of surjectivity.

To show the injectivity property of the isomorphism, let there exists two
channels $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ such that

$\displaystyle(\mathbb{I}\wedge\mathcal{E}_{1})(|\psi^{+}\rangle\langle\psi^{+}|_{f})=(\mathbb{I}\wedge\mathcal{E}_{2})(|\psi^{+}\rangle\langle\psi^{+}|_{f})$
$\displaystyle\implies(\mathbb{I}\wedge(\mathcal{E}_{1}-\mathcal{E}_{2}))(|\psi^{+}\rangle\langle\psi^{+}|_{f})=0$
(27)

Let $(\mathcal{E}_{1}-\mathcal{E}_{2})=\mathcal{E}^{\prime}$. Notice, that

$\displaystyle(\mathbb{I}\wedge\mathcal{E}^{\prime})(|\psi^{+}\rangle\langle\psi^{+}|_{f})=\frac{1}{d}\sum_{ij}|E_{i}\rangle\langle
E_{j}|\wedge\mathcal{E}^{\prime}(|E_{i}\rangle\langle E_{j}|)$
$\displaystyle+|O_{i}\rangle\langle
O_{j}|\wedge\mathcal{E}^{\prime}(|O_{i}\rangle\langle
O_{j}|)+|E_{i}\rangle\langle
O_{j}|\wedge\mathcal{E}^{\prime}(|E_{i}\rangle\langle O_{j}|)+$
$\displaystyle|O_{i}\rangle\langle
E_{j}|\wedge\mathcal{E}^{\prime}(|O_{i}\rangle\langle E_{j}|)$ (28)

Notice, under the action of the trace annhilating map $\mathcal{E}^{\prime}$
on the basis $|E_{i}\rangle\langle E_{j}|$ is given by
$\gamma_{ijkl}|E_{k}\rangle\langle E_{l}|$ or
$\gamma_{ijkl}|O_{k}\rangle\langle O_{l}|$ due to PSSR restriction on most
general operations. Similarly, one can find the action of
$\mathcal{E}^{\prime}$ on other basis. Comparing, Eq. (27) and Eq. (28) one
gets $\mathcal{E}^{\prime}=0$. This proves the injectivity.

As in the case of SQT, in FQT also we define the restricted twirling operation
on fermionic channel as follows: Alice applies unitary which is an element of
restricted clifford set on one half of the fermionic bi-partite state and send
it through the channel $\mathcal{E}$ to Bob. Bob upon receiving the state,
applies the inverse Unitary.

Lemma 8. For any fermionic channel $\mathcal{E}$ we have the relation

$\rho_{\mathcal{T}^{\ast}_{C^{res}_{N}}[\mathcal{E}]}^{f}=\mathcal{T}^{\ast}_{C^{res}_{N}}[\rho_{\mathcal{E}}^{f}]$

where,
$\rho_{\mathcal{T}_{C^{res}_{N}}[\mathcal{E}]}^{f}=(\mathbb{I}\wedge\mathcal{T}^{\ast}_{C^{res}_{N}}[\mathcal{E}])(|\psi^{+}\rangle\langle\psi^{+}|_{f})$
with $\mathcal{T}^{\ast}_{C^{res}_{N}}[\mathcal{E}]$ representing restricted
twirling operation on fermionic channel.

Proof. Given for FQT,
$\rho_{\mathcal{T}^{\ast}_{C^{res}_{N}}[\mathcal{E}]}^{f}$

$\displaystyle=(\mathbb{I}\wedge\mathcal{T}_{C^{res}_{N}}[\mathcal{E}])(|\psi^{+}\rangle\langle\psi^{+}|_{f})$
$\displaystyle=\frac{1}{|C^{res}_{N}|}\sum_{c\in C^{res}_{N}}(\mathbb{I}\wedge
c^{\dagger})[(\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c)|\psi^{+}\rangle\langle\psi^{+}|_{f}$ $\displaystyle(\mathbb{I}\wedge
c^{\dagger}))](\mathbb{I}\wedge c)$
$\displaystyle=\frac{1}{|C^{res}_{N}|}\sum_{c\in C^{res}_{N}}(\mathbb{I}\wedge
c^{\dagger})[(\mathbb{I}\wedge\mathcal{E})((c^{T}\wedge\mathbb{I})|\psi^{+}\rangle\langle\psi^{+}|_{f}$
$\displaystyle(c^{\ast}\wedge\mathbb{I}))](\mathbb{I}\wedge c)$
$\displaystyle=\frac{1}{|C^{res}_{N}|}\sum_{c\in C^{res}_{N}}(c^{T}\wedge
c^{\dagger})\rho_{\mathcal{E}}^{f}(c^{\ast}\wedge c)$
$\displaystyle=\mathcal{T}^{\ast}_{C^{res}_{N}}[\rho_{\mathcal{E}}^{f}]$ (29)

where we used the fact that, for all $c\in C^{res}_{N}$ we also have,
$c^{\dagger}\in C^{res}_{N}$, thus the sum remain invariant under hermitian
conjugate. $\blacksquare$

##  VII Lower Bound on Optimal teleportation fidelity

In the previous section we found that under restricted clifford twirling
operation, any $N\times N$-mode fermionic state can be reduced to the
canonical form of

$\displaystyle\omega_{noise}^{f}=\alpha
d^{2}\Bigg{(}\frac{\mathbb{I}_{N}\wedge\mathbb{I}_{N}}{d^{2}}+\Upsilon_{1}\frac{\mathbb{S}_{z}\wedge\mathbb{S}_{z}}{d^{2}}\Bigg{)}+\beta
d(\omega_{AB}^{f})$ (30)

where, $\mathbb{S}_{z}=(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})$,
$\alpha,\ \beta,\ \Upsilon_{1}$ are real and $\omega_{AB}^{f}$ is given by Eq.
(15). Also, $\alpha,\ \beta,\ \Upsilon_{1}$ can be expressed in terms of the
twirl invariant overlap (see Corollary 1).

In case of any fermionic channel $\mathcal{E}$ in FQT we define the average
separable fidelity of the channel as

$\displaystyle F(\mathbb{I}\wedge\mathcal{E})=$
$\displaystyle\int\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(|\psi_{E}\rangle\langle\psi_{E}|E_{m}^{A}\wedge
E_{m}^{B})}$ (31)
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge\mathcal{E}(|\psi_{E}\rangle\langle\psi_{E}|))E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}d\psi_{E}$
$\displaystyle+\int\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(|\psi_{o}\rangle\langle\psi_{o}|E_{m}^{A}\wedge
E_{m}^{B})}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge\mathcal{E}(|\psi_{o}\rangle\langle\psi_{o}|))E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}d\psi_{o}$

where $|\psi_{O}\rangle\langle\psi_{O}|$ represents fermionic state with
Parity of $|\psi_{O}\rangle$ being odd and $|\psi_{E}\rangle\langle\psi_{E}|$
represents fermionic state with Parity of $|\psi_{E}\rangle$ being even. Also,
the the integrals are calculated considering uniform measure over all even and
odd states respectively. As the measurement is restricted to be separable, we
consider here the separable POVM elements $\\{E_{m}^{A}\wedge E_{m}^{B}\\}$.
Notice, this average separable fidelity of the fermionic channel is invariant
under the restricted twirling operation on fermionic channel as introduced in
section VI (see Appendix D).

Now we consider the fermionic teleportation scheme explained in section III,
with the shared resource $\omega_{noise}^{f}$ (given in Eq. (30)).

Lemma 9. For any fermionic state
$|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}$, the output state of
the fermionic teleportation channel with the teleportation scheme mentioned in
Theorem 1 and shared resource $\omega_{noise}^{f}$ is given by

$\displaystyle\rho^{f}_{\tilde{A}^{\prime}B}=$ $\displaystyle\alpha
d^{2}\Bigg{(}\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d}\pm\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d}\Bigg{)}$
$\displaystyle+\beta d\
\text{loc-p}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}B})+\Upsilon\beta d\
\overline{\text{loc-p}}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}B})$
$\displaystyle\equiv\alpha d^{2}(\rho^{f}_{\tilde{A}^{\prime}B})_{noisy}+\beta
d(\rho^{f}_{\tilde{A}^{\prime}B})_{exact}$

with $\pm$ depending upon the parity of the input state
$|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}$. Here,
$\text{loc-p}(|\psi\rangle\langle\psi|)$ denotes the local-parity respecting
part of $|\psi\rangle\langle\psi|$ and
$\overline{\text{loc-p}}(|\psi\rangle\langle\psi|)$ denotes the local-parity
violating part of $|\psi\rangle\langle\psi|$. Here, we define

$\displaystyle(\rho^{f}_{\tilde{A}^{\prime}B})_{noisy}=$
$\displaystyle\Bigg{(}\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d}\pm\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d}\Bigg{)}$
$\displaystyle(\rho^{f}_{\tilde{A}^{\prime}B})_{exact}=$
$\displaystyle\text{loc-p}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}B})+\Upsilon\
\overline{\text{loc-p}}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}B})$

The proof of Lemma 9 is given in Appendix E.

Notice, in FQT also, the mathematical state-channel isomorphism between the
noisy fermionic resource state and the corresponding teleportation channel is
physical. Indeed, applying the fermionic teleportation channel (whose action
is given by Lemma 9) on one half of the state
$|\psi^{+}\rangle\langle\psi^{+}|_{f}$ outputs noisy fermionic resource state
given by Eq. (LABEL:eq:25). Hence using Corollary 2 and Lemma 8 one can show
that any Fermionic channel can be reduced to the form mentioned in Lemma 9.

To calculate the teleportation fidelity of this teleportation channel (with
the scheme mentioned in Theorem 1), one can easily show that for the exact
case, $F(\mathbb{I}\wedge\mathcal{E}_{exact})=1$ (see Appendix E).

For the noisy part, we get, according to Appendix E,

$\displaystyle F(\mathbb{I}\wedge\mathcal{E}_{noisy})\geq(1+\Upsilon_{1})\int
dU_{res}\frac{\text{tr}((\rho^{f}_{\tilde{A}^{\prime}}(U_{res}))^{2})}{d}$

where $U_{res}$ denotes the Harr measure on the set of bipartite
unitary$\\{U_{BD},U_{ABD}\\}$ and the even-parity (odd-parity) states
$|{\psi}_{E}{\rangle}$ of $\tilde{A}^{\prime}\tilde{A}$ are generated by the
action of $U_{BD}$’s ($U_{ABD}$’s) on a fixed state $|{\psi}_{0}{\rangle}$ (of
given parity) of $\tilde{A}^{\prime}\tilde{A}$.

Hence, the overall lower bound on the average separable fidelity for
teleportation is given by

$\displaystyle F(\mathbb{I}\wedge\mathcal{E})\geq\alpha
d^{2}(1+\Upsilon_{1})\int
dU_{res}\frac{\text{tr}((\rho^{f}_{\tilde{A}^{\prime}}(U_{res}))^{2})}{d}+\beta
d$

We compare this result with the well known optimal teleportation fidelity in
SQT with an example state (see Appendix F).

##  VIII Conclusion

In this work, we characterized general teleportation channel in FQT by
introducing a fidelity measure based on separable measurements (also called as
‘product’ measurements). The motivation behind the use of this measure is the
physical scenario where Alice and Bob are far away from each other. Although,
due to local discriminability in SQT [37], product measurements result in
different probablities for different state, the lack of local discriminability
in FQT may result in the same probablity for different states when measured
with product measurements. Hence, in general, Alice and Bob may not be able to
distinguish the bi-partite state before and after the teleportation
operationally even though the mathematical structures of the initial and final
bi-partite states are different. We considered this situation as the exact
teleportation channel with respect to the separable fidelity measure.

At first, we derived the fermionic invariant state under local twirling
operation and found that the structure of this invariant state differ from
that of Werner state - the corresponding invariant state in SQT. We introduced
a Haar measure on the restricted Unitary group $U_{res}$ and showed that any
state can be reduced to the invariant form using restricted Unitary group
twirling. We constitute the Unitary-2 design of the restricted Unitary group
twirling by introducing a restricted Clifford set. Introducing the partial
transpose operation on fermionic modes, we derived the fermionic isotropic
state - a linear combination of noise in even subspace, noise in odd subspace
and the fermionic resource state equivalent to the exact teleportation
channel. We introduced the state-channel isomorphism and channel twirling
operation in FQT, and provided a lower bound on the optimal teleportation
fidelity. The lower bound depends only on the parameters invariant under the
local-twirling operations. Finally, we compared this lower bound with the
optimal teleportation fidelity in SQT. However, the difference in the
operational teleportation fidelity in SQT and FQT can be considered to be the
consequence of local discriminability. While utilizing the operational
definition of teleportation fidelity in FQT, Fermionic channel twirling
operation need careful attention due to the non-linearity involved in the
definition of the separable fidelity.

Even though we have provided the lower bound on the optimal teleportation
fidelity calculated operationally, finding the exact expression for the
optimal teleportation fidelity still remains open. Also, the effect of
filtering on the twirling invariant parameters in FQT is yet to be figured
out. In this paper, we establish the fermionic state-channel isomorphism by
applying the channel on one half of the resource state with $\Upsilon$ being
unity. In this regard, possibility to extend the fermionic state-channel
isomorphism with $\omega_{AB}^{f}$ still remains open. Apart from FQT, one may
also try to find the optimal teleportation fidelity for quantum information
encoded in Bosonic modes. Also, general quantum teleportation channel may find
its application in Fermionic Quantum Computation in the NISQ-like settings.

###### Acknowledgements.

We wish to thank Dr. Aravinda S for fruitful discussions on Quantum
Information of system of indistinguishable particles.

##  Appendix A Fermionic Hilbert space

Proof of Lemma 2. Let us assume,

$\hat{O}_{P}^{A}=\sum_{ij}p_{ij}^{A}|E_{i}^{A}\rangle\langle
E_{j}^{A}|+q_{ij}^{A}|O_{i}^{A}\rangle\langle O_{j}^{A}|$

denotes an observable of system $A$ and

$\hat{O}_{P}^{B}=\sum_{ij}p_{ij}^{B}|E_{i}^{B}\rangle\langle
E_{j}^{B}|+q_{ij}^{B}|O_{i}^{B}\rangle\langle O_{j}^{B}|$

denotes an observable of system $B$. We denote a local PSSR-respected
fermionic state $\rho_{local}^{AB}$ as

$\displaystyle\rho_{local}^{AB}=$
$\displaystyle\sum_{abcd}A^{\prime}_{abcd}|E_{a}^{A}\rangle\wedge|E_{b}^{B}\rangle\langle
E_{c}^{A}|\wedge\langle
E_{d}^{B}|+B^{\prime}_{abcd}|E_{a}^{A}\rangle\wedge|O_{b}^{B}\rangle\langle
E_{c}^{A}|\wedge\langle O_{d}^{B}|+$ $\displaystyle
C^{\prime}_{abcd}|O_{a}^{A}\rangle\wedge|E_{b}^{B}\rangle\langle
O_{c}^{A}|\wedge\langle
E_{d}^{B}|+D^{\prime}_{abcd}|O_{a}^{A}\rangle\wedge|O_{b}^{B}\rangle\langle
O_{c}^{A}|\wedge\langle O_{d}^{B}|$

with suitable conditions on the coefficients
$A^{\prime},B^{\prime},C^{\prime}$ and $D^{\prime}$ such that
$\rho_{local}^{AB}>0$ and $tr(\rho_{local}^{AB})=1$.

Using the partial tracing procedure outlined in [24] we obtain the reduced
density matrix for system $A$ as

$\displaystyle\rho_{local}^{A}=$
$\displaystyle\sum_{abc}A^{\prime}_{abcb}|E_{a}^{A}\rangle\langle
E_{c}^{A}|+B^{\prime}_{abcb}|E_{a}^{A}\rangle\langle
E_{c}^{A}|+C^{\prime}_{abcb}|O_{a}^{A}\rangle\langle
O_{c}^{A}|+D^{\prime}_{abcb}|O_{a}^{A}\rangle\langle O_{c}^{A}|$

and reduced density matrix for system $B$ as

$\displaystyle\rho_{local}^{B}=$
$\displaystyle\sum_{abd}A^{\prime}_{abad}|E_{b}^{B}\rangle\langle
E_{d}^{B}|+C^{\prime}_{abad}|E_{b}^{B}\rangle\langle
E_{d}^{B}|+B^{\prime}_{abad}|O_{b}^{B}\rangle\langle
O_{d}^{B}|+D^{\prime}_{abad}|O_{b}^{B}\rangle\langle O_{d}^{B}|$

Then imposing the trace condition of product state (see Eq. (9)) and comparing
coefficients of $p_{ca}^{A}\ p_{db}^{B},\ p_{ca}^{A}\ q_{db}^{B},\ q_{ca}^{A}\
p_{db}^{B}$ and $q_{ca}^{A}\ q_{db}^{B}$ we get

$\rho_{local}^{AB}=\rho_{local}^{A}\wedge\rho_{local}^{B}$

Reverse implication is trivially satisfied. $\blacksquare$

Proof of Lemma 3. Let us assume that the anti block-diagonal and block-
diagonal unitary matrices acting on system $A$ and $B$ respectively is given
by

$U_{ABD}^{A}=\sum_{ij}p_{ij}|E_{i}^{A}\rangle\langle
O_{j}^{A}|+q_{ij}|O_{i}^{A}\rangle\langle E_{j}^{A}|\ \ \ \text{and}\ \ \
{U^{\prime}_{BD}}^{B}=\sum_{kl}r_{kl}|E_{k}^{B}\rangle\langle
E_{l}^{B}|+s_{kl}|O_{k}^{A}\rangle\langle O_{l}^{B}|$

Now consider a separable fermionic state $\rho_{sep}^{AB}$ given by Eq. (12)
with

$\rho_{i}^{A}=\sum_{s}|\psi_{s}^{i}\rangle\langle\psi_{s}^{i}|\ \ \
\text{and}\ \ \ \rho_{i}^{B}=\sum_{t}|\phi_{t}^{i}\rangle\langle\phi_{t}^{i}|$

with $|\psi_{s}^{i}\rangle,|\phi_{s}^{i}\rangle$ unnormalized. Without loss of
generality, consider $|\psi_{s}^{i}\rangle,|\phi_{s}^{i}\rangle$ to be even
and odd fermionic state respectively. Hence, expanding in the basis
$\mathcal{B_{A(B)}^{P}}$ we get

$\displaystyle|\psi_{s}^{i}\rangle\langle\psi_{s}^{i}|=\sum_{y_{s}z_{s}}{\alpha_{y_{s}}^{i}}{(\alpha_{z_{s}}^{i})}^{\ast}|E_{y_{s}}^{A}\rangle\langle
E_{z_{s}}^{A}|\ \ \ \text{and}\ \ \
|\phi_{t}^{i}\rangle\langle\phi_{t}^{i}|=\sum_{m_{t}n_{t}}{\beta_{m_{t}}^{i}}{(\beta_{n_{t}}^{i})}^{\ast}|O_{m_{t}}^{B}\rangle\langle
O_{n_{t}}^{B}|$ (33)

Then

$\displaystyle(U^{A}_{ABD}\wedge{U^{\prime}}^{B}_{BD})\ \rho_{sep}^{AB}\
({U^{A}_{ABD}}^{\dagger}\wedge{{U^{\prime}}^{B}_{BD}}^{\dagger})$
$\displaystyle=\Bigg{(}\sum_{efgh}p_{ef}r_{gh}|E_{e}^{A}\rangle\wedge|E_{g}^{B}\rangle\langle
O_{f}^{A}|\wedge\langle
E_{h}^{B}|+p_{ef}s_{gh}|E_{e}^{A}\rangle\wedge|O_{g}^{B}\rangle\langle
O_{f}^{A}|\wedge\langle
O_{h}^{B}|+q_{ef}r_{gh}|O_{e}^{A}\rangle\wedge|E_{g}^{B}\rangle\langle
E_{f}^{A}|\wedge\langle E_{h}^{B}|\Bigg{.}$
$\displaystyle+\Bigg{.}q_{ef}s_{gh}|O_{e}^{A}\rangle\wedge|O_{g}^{B}\rangle\langle
E_{f}^{A}|\wedge\langle
O_{h}^{B}|\Bigg{)}\rho_{sep}^{AB}\Bigg{(}\sum_{ijkl}(p_{ij})^{\ast}(r_{kl})^{\ast}|O_{j}^{A}\rangle\wedge|E_{l}^{B}\rangle\langle
E_{i}^{A}|\wedge\langle
E_{k}^{B}|+(p_{ij})^{\ast}(s_{kl})^{\ast}|O_{j}^{A}\rangle\wedge|O_{l}^{B}\rangle\langle
E_{i}^{A}|\wedge\langle O_{k}^{B}|\Bigg{.}$
$\displaystyle+\Bigg{.}(q_{ij})^{\ast}(r_{kl})^{\ast}|E_{j}^{A}\rangle\wedge|E_{l}^{B}\rangle\langle
O_{i}^{A}|\wedge\langle
E_{k}^{B}|+(q_{ij})^{\ast}(s_{kl})^{\ast}|E_{j}^{A}\rangle\wedge|O_{l}^{B}\rangle\langle
O_{i}^{A}|\wedge\langle O_{k}^{B}|\Bigg{)}$ Considering the $i^{th}$ term of
$\rho_{sep}^{AB}$ and dropping the index $i$ in Eq. (33) we get
$\displaystyle=\sum_{st}\Bigg{(}\sum_{ei}\sum_{y_{s}z_{s}}(q_{ey_{s}}\alpha_{y_{s}})(q_{iz_{s}}\alpha_{z_{s}})^{\ast}|O_{e}^{A}\rangle\langle
O_{i}^{A}|\Bigg{)}\wedge\Bigg{(}\sum_{gk}\sum_{m_{t}n_{t}}(s_{gm_{t}}\beta_{m_{t}})(s_{kn_{t}}\beta_{n_{t}})^{\ast}|O_{g}^{B}\rangle\langle
O_{k}^{B}|\Bigg{)}$ $\displaystyle+\ \ \text{terms with non-local parity}$
$\displaystyle=\Sigma^{A}\wedge\Sigma^{B}+\ \ \text{terms with local-PSSR
violation}\ \ \ \in SEP$ (34)

One can then obtain the desired result by taking convex combination over index
$i$. Similar result can be obtained by replacing block-diagonal unitary with
anti block-diagonal for system $B$. $\blacksquare$

Proof of Lemma 5. It is well known that, given G be a locally compact group
then, there exists a left Haar measure on G [42, 43]. Also, for G be a locally
compact group, and given $\mu$, $\mu^{\prime}$ be two left Haar measures on G.
Then, $\mu=\alpha\mu^{\prime}$ for some $\alpha\in\mathbb{R}^{+}$. We also
know that every closed subgroup of a locally compact group is a locally
compact group [44]. We also know that the Unitary group (closed and bounded).
Hence it is compact due to Heine-borel theorem. Hence, we only need to show
that $U_{res}$ is a closed subgroup of the Unitary group. The closure property
of $U_{res}$ can be proved using the property that

$U_{BD(ABD)}^{n}=\\{U\in\mathbb{M}_{BD(ABD)}(n,\mathbb{C})|U^{\dagger}U=UU^{\dagger}=\mathbb{I}\\}$

where $\mathbb{M}_{BD(ABD)}(n,\mathbb{C})$ represents the block(anti-
block)-diagonal subspace of $\mathbb{M}_{BD(ABD)}(n,\mathbb{C})$. To see that,
$U_{res}$ constitutes a subgroup of the Unitary gorup, notice that the
identity element $\mathbb{I}_{N}$ is also an element of $U_{res}$. Also,
$U_{BD(ABD)}^{-1}=U_{BD(ABD)}^{\dagger}$, hence an element in $U_{res}$. Also,
note that matrix multiplication of $U_{BD(ABD)}$ with $U_{BD(ABD)}$ is also an
element of $U_{res}$. Hence, $U_{res}$ is a closed subgroup of the Unitary
group. Hence, there exist a unique left Haar measure on the set $U_{res}$.
$\blacksquare$

##  Appendix B Exact Fermionic Teleportation

Proof of Theorem 1. We will consider the even and odd fermionic state
separately. Let us first consider the even fermionic bipartite state of the
system $\tilde{A}^{\prime}\tilde{A}$ given by

$\displaystyle\rho_{\tilde{A}^{\prime}\tilde{A}}^{f}=|\phi^{e}\rangle\langle\phi^{e}|=$
$\displaystyle\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(\tilde{A},\nu_{1})}^{\dagger}f_{(\tilde{A},\nu_{2})}^{\dagger}\dots
f_{(\tilde{A},\nu_{h})}^{\dagger}|\Omega\rangle$ (35)
$\displaystyle\langle\Omega|f_{(\tilde{A},\nu_{h^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\nu_{2}^{\prime})}f_{(\tilde{A},\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}$

with $P(\vec{\lambda})=P(\vec{\nu})$ and
$P(\vec{\lambda}^{\prime})=P(\vec{\nu}^{\prime})$. Without loss of generality,
we consider $Term_{x}$ in Eq. (16) with $U_{x}$ to be block-diagonal unitary
and defined as follows

$\displaystyle
U_{BD}^{x}=\sum_{\vec{z},\vec{y}}R_{\vec{z},\vec{y}}^{x}f_{z_{1}}^{\dagger}f_{z_{2}}^{\dagger}\dots
f_{z_{s}}^{\dagger}|\Omega\rangle\langle\Omega|f_{y_{t}}\dots
f_{y_{2}}f_{y_{1}}$ (36)

with $P(\vec{z})=P(\vec{y})$. Using this $U_{BD}^{x}$ one can express $M_{x}$
in fermionic system as

$\displaystyle
M_{x}=(U_{BD}^{x}\wedge\mathbb{I})|\omega\rangle\langle\omega|({U_{BD}^{x}}^{\dagger}\wedge\mathbb{I})$
(37)

Expressing $\omega_{AB}^{f}$ in terms of fermionic creation and annihilation
operator, we get

$\displaystyle\omega_{AB}^{f}$
$\displaystyle=\frac{1}{d}\sum_{\vec{\sigma},\vec{\sigma}^{\prime},P(\vec{\sigma})=P(\vec{\sigma}^{\prime})}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}\dots
f_{(A,\sigma_{j})}^{\dagger}f_{(B,\sigma_{1})}^{\dagger}f_{(B,\sigma_{2})}^{\dagger}\dots
f_{(B,\sigma_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\sigma_{r}^{\prime})}\dots
f_{(B,\sigma_{2}^{\prime})}f_{(B,\sigma_{1}^{\prime})}f_{(A,\sigma_{r}^{\prime})}\dots$
(38) $\displaystyle
f_{(A,\sigma_{2}^{\prime})}f_{(A,\sigma_{1}^{\prime})}+\frac{\Upsilon}{d}\sum_{\vec{\sigma},\vec{\sigma}^{\prime},P(\vec{\sigma})\neq
P(\vec{\sigma}^{\prime})}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}\dots
f_{(A,\sigma_{j})}^{\dagger}f_{(B,\sigma_{1})}^{\dagger}f_{(B,\sigma_{2})}^{\dagger}\dots
f_{(B,\sigma_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\sigma_{r}^{\prime})}\dots
f_{(B,\sigma_{2}^{\prime})}f_{(B,\sigma_{1}^{\prime})}$ $\displaystyle
f_{(A,\sigma_{r}^{\prime})}\dots
f_{(A,\sigma_{2}^{\prime})}f_{(A,\sigma_{1}^{\prime})}$

Here $P(.)$ denotes the parity of the vector.

Now the fermionic singlet state $|\omega\rangle\langle\omega|$ can be
expressed in terms of fermionic creation and annihilation operator using Eq.
(38) and setting the value of $\Upsilon=1$.Then $(M_{x})_{\tilde{A}A}$ can be
expressed as

$\displaystyle(M_{x})_{\tilde{A}A}=$
$\displaystyle\frac{1}{d}\sum_{\vec{z}^{\prime},\vec{y}^{\prime}}\sum_{\vec{\pi},\vec{\pi}^{\prime}}\sum_{\vec{z},\vec{y}}R_{\vec{z},\vec{y}}^{x}(R_{\vec{z}^{\prime},\vec{y}^{\prime}}^{x})^{\ast}f_{(\tilde{A},z_{1})}^{\dagger}f_{(\tilde{A},z_{2})}^{\dagger}\dots
f_{(\tilde{A},z_{s})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(\tilde{A},y_{t})}\dots
f_{(\tilde{A},y_{2})}f_{(\tilde{A},y_{1})}f_{(\tilde{A},\pi_{1})}^{\dagger}f_{(\tilde{A},\pi_{2})}^{\dagger}\dots
f_{(\tilde{A},\pi_{q})}^{\dagger}$ $\displaystyle
f_{(A,\pi_{1})}^{\dagger}f_{(A,\pi_{2})}^{\dagger}\dots
f_{(A,\pi_{q})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(A,\pi_{q^{\prime}}^{\prime})}\dots
f_{(A,\pi_{2}^{\prime})}f_{(A,\pi_{1}^{\prime})}f_{(\tilde{A},\pi_{q^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\pi_{2}^{\prime})}f_{(\tilde{A},\pi_{1}^{\prime})}f_{(\tilde{A},y_{1}^{\prime})}^{\dagger}f_{(\tilde{A},y_{2}^{\prime})}^{\dagger}\dots
f_{(\tilde{A},y_{t^{\prime}}^{\prime})}^{\dagger}|\Omega\rangle$
$\displaystyle\langle\Omega|f_{(\tilde{A},z_{s^{\prime}}^{\prime})}\dots
f_{(\tilde{A},z_{2}^{\prime})}f_{(\tilde{A},z_{1}^{\prime})}$ $\displaystyle=$
$\displaystyle\sum_{\vec{z}^{\prime},\vec{\pi}^{\prime}}\sum_{\vec{z},\vec{\pi}}R_{\vec{z},\vec{\pi}}^{x}(R_{\vec{z}^{\prime},\vec{\pi}^{\prime}}^{x})^{\ast}f_{(\tilde{A},z_{1})}^{\dagger}f_{(\tilde{A},z_{2})}^{\dagger}\dots
f_{(\tilde{A},z_{s})}^{\dagger}f_{(A,\pi_{1})}^{\dagger}f_{(A,\pi_{2})}^{\dagger}\dots
f_{(A,\pi_{q})}|\Omega\rangle\langle\Omega|f_{(A,\pi_{q^{\prime}}^{\prime})}\dots
f_{(A,\pi_{2}^{\prime})}f_{(A,\pi_{1}^{\prime})}f_{(\tilde{A},z_{s^{\prime}}^{\prime})}$
(39) $\displaystyle\dots
f_{(\tilde{A},z_{2}^{\prime})}f_{(\tilde{A},z_{1}^{\prime})}$

Here we have used the scalar product definition of the wedge product Hilbert
space. Without loss of generality, the local PSSR-respected observable
$\hat{O}_{B}$ and $\hat{O}_{\tilde{A}^{\prime}}$ can be written as

$\displaystyle\hat{O}_{B}=$
$\displaystyle|\psi_{1}\rangle\langle\psi_{3}|=\sum_{\vec{b},\vec{c}}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}f_{(B,b_{1})}^{\dagger}f_{(B,b_{2})}^{\dagger}\dots
f_{(B,b_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,c_{k})}\dots
f_{(B,c_{2})}f_{(B,c_{1})}$ $\displaystyle\hat{O}_{\tilde{A}^{\prime}}=$
$\displaystyle|\psi_{2}\rangle\langle\psi_{4}|=\sum_{\vec{d},\vec{e}}\xi_{\vec{d}}(\gamma_{\vec{e}})^{\ast}f_{(\tilde{A}^{\prime},d_{1})}^{\dagger}f_{(\tilde{A}^{\prime},d_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},d_{j^{\prime}})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(\tilde{A}^{\prime},e_{k^{\prime}})}\dots
f_{(\tilde{A}^{\prime},e_{2})}f_{(\tilde{A}^{\prime},e_{1})}$ (40)

with $P(\vec{b})=P(\vec{c})$ and $P(\vec{d})=P(\vec{e})$ to ensure local PSSR-
respected observable. Hence,

$\displaystyle
N_{x}(\hat{O}_{B})={U_{BD}^{x}}^{\dagger}(\hat{O}_{B}){U_{BD}^{x}}=$
$\displaystyle\sum_{\vec{z}^{\prime},\vec{y}^{\prime}}\sum_{\vec{z},\vec{y}}\sum_{\vec{b},\vec{c}}R_{\vec{z}^{\prime},\vec{y}^{\prime}}^{x}(R_{\vec{z},\vec{y}}^{x})^{\ast}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}f_{(B,y_{1})}^{\dagger}f_{(B,y_{2})}^{\dagger}\dots
f_{(B,y_{t})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,z_{s})}\dots
f_{(B,z_{2})}f_{(B,z_{1})}$ $\displaystyle
f_{(B,b_{1})}^{\dagger}f_{(B,b_{2})}^{\dagger}\dots
f_{(B,b_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,c_{k})}\dots
f_{(B,c_{2})}f_{(B,c_{1})}f_{(B,z_{1}^{\prime})}^{\dagger}f_{(B,z_{2}^{\prime})}^{\dagger}\dots
f_{(B,z_{s^{\prime}}^{\prime})}^{\dagger}|\Omega\rangle$
$\displaystyle\langle\Omega|f_{(B,y_{t^{\prime}}^{\prime})}\dots
f_{(B,y_{2}^{\prime})}f_{(B,y_{1}^{\prime})}$

Again using the inner product definition of the wedge product Hilbert space we
get

$\displaystyle
N_{x}(\hat{O}_{B})=\sum_{\vec{y},\vec{y}^{\prime}}\sum_{\vec{b},\vec{c}}R_{\vec{c},\vec{y}^{\prime}}^{x}(R_{\vec{b},\vec{y}}^{x})^{\ast}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}f_{(B,y_{1})}^{\dagger}f_{(B,y_{2})}^{\dagger}\dots
f_{(B,y_{t})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,y_{t^{\prime}}^{\prime})}\dots
f_{(B,y_{2}^{\prime})}f_{(B,y_{1}^{\prime})}$ (41)

with $P(\vec{y})=P(\vec{y}^{\prime})$. Substituting Eq. (38), Eq. (35), Eq.
(39), Eq. (40) and Eq. (41) into the $x_{th}$ term in Eq. (16), L.H.S of Eq.
(16) can be written as

$\displaystyle Term_{x}=$
$\displaystyle\text{tr}\Bigg{[}\frac{1}{d^{2}}\Bigg{(}\sum_{\vec{\sigma},\vec{\sigma}^{\prime},P(\vec{\sigma})=P(\vec{\sigma}^{\prime})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(\tilde{A},\nu_{1})}^{\dagger}f_{(\tilde{A},\nu_{2})}^{\dagger}\dots
f_{(\tilde{A},\nu_{h})}^{\dagger}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}\dots$
(42) $\displaystyle
f_{(A,\sigma_{j})}^{\dagger}f_{(B,\sigma_{1})}^{\dagger}f_{(B,\sigma_{2})}^{\dagger}\dots
f_{(B,\sigma_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\sigma_{r}^{\prime})}\dots
f_{(B,\sigma_{2}^{\prime})}f_{(B,\sigma_{1}^{\prime})}f_{(A,\sigma_{r}^{\prime})}\dots
f_{(A,\sigma_{2}^{\prime})}f_{(A,\sigma_{1}^{\prime})}f_{(\tilde{A},\nu_{h^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\nu_{2}^{\prime})}f_{(\tilde{A},\nu_{1}^{\prime})}$
$\displaystyle f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon\sum_{\vec{\sigma},\vec{\sigma}^{\prime},P(\vec{\sigma})\neq
P(\vec{\sigma}^{\prime})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(\tilde{A},\nu_{1})}^{\dagger}f_{(\tilde{A},\nu_{2})}^{\dagger}\dots$
$\displaystyle
f_{(\tilde{A},\nu_{h})}^{\dagger}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}\dots
f_{(A,\sigma_{j})}^{\dagger}f_{(B,\sigma_{1})}^{\dagger}f_{(B,\sigma_{2})}^{\dagger}\dots
f_{(B,\sigma_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\sigma_{r}^{\prime})}\dots
f_{(B,\sigma_{2}^{\prime})}f_{(B,\sigma_{1}^{\prime})}f_{(A,\sigma_{r}^{\prime})}\dots
f_{(A,\sigma_{2}^{\prime})}f_{(A,\sigma_{1}^{\prime})}$ $\displaystyle
f_{(\tilde{A},\nu_{h^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\nu_{2}^{\prime})}f_{(\tilde{A},\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{)}\Bigg{(}\sum_{\vec{y},\vec{y}^{\prime}}\sum_{\vec{b},\vec{c}}\sum_{\vec{d},\vec{e}}\sum_{\vec{z}^{\prime},\vec{\pi}^{\prime}}\sum_{\vec{z},\vec{\pi}}R_{\vec{c},\vec{y}^{\prime}}^{x}(R_{\vec{b},\vec{y}}^{x})^{\ast}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}\xi_{\vec{d}}(\gamma_{\vec{e}})^{\ast}R_{\vec{z},\vec{\pi}}^{x}$
$\displaystyle(R_{\vec{z}^{\prime},\vec{\pi}^{\prime}}^{x})^{\ast}f_{(\tilde{A}^{\prime},d_{1})}^{\dagger}f_{(\tilde{A}^{\prime},d_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},d_{j^{\prime}})}^{\dagger}f_{(\tilde{A},z_{1})}^{\dagger}f_{(\tilde{A},z_{2})}^{\dagger}\dots
f_{(\tilde{A},z_{s})}^{\dagger}f_{(A,\pi_{1})}^{\dagger}f_{(A,\pi_{2})}^{\dagger}\dots
f_{(A,\pi_{q})}f_{(B,y_{1})}^{\dagger}f_{(B,y_{2})}^{\dagger}\dots
f_{(B,y_{t})}^{\dagger}|\Omega\rangle$
$\displaystyle\langle\Omega|f_{(B,y_{t^{\prime}}^{\prime})}\dots
f_{(B,y_{2}^{\prime})}f_{(B,y_{1}^{\prime})}f_{(A,\pi_{q^{\prime}}^{\prime})}\dots
f_{(A,\pi_{2}^{\prime})}f_{(A,\pi_{1}^{\prime})}f_{(\tilde{A},z_{s^{\prime}}^{\prime})}f_{(\tilde{A}^{\prime},e_{k^{\prime}})}\dots
f_{(\tilde{A}^{\prime},e_{2})}f_{(\tilde{A}^{\prime},e_{1})}\Bigg{)}\Bigg{]}$

Using the relation $P(\vec{y})=P(\vec{y}^{\prime})$, the inner product
definition in wedge product Hilbert space will compel the parity of
$\vec{\sigma}$ to be equal to the parity of $\vec{\sigma}^{\prime}$. Thus, we
get

$\displaystyle Term_{x}=$
$\displaystyle\frac{1}{d^{2}}\Bigg{(}\sum_{\vec{\sigma},\vec{\sigma}^{\prime},P(\vec{\sigma})=P(\vec{\sigma}^{\prime})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\sum_{\vec{b},\vec{c}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}R_{\vec{c},\vec{\sigma}}^{x}(R_{\vec{b},\vec{\sigma}^{\prime}}^{x})^{\ast}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}R_{\vec{\nu}^{\prime},\vec{\sigma}^{\prime}}^{x}(R_{\vec{\nu},\vec{\sigma}}^{x})^{\ast}\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{1}{d^{2}}\Bigg{[}\sum_{\vec{\lambda}_{even}}\sum_{\vec{\lambda}^{\prime}_{even}}\sum_{\vec{\nu}_{even}}\sum_{\vec{\nu}^{\prime}_{even}}\sum_{\vec{b}_{even}}\sum_{\vec{c}_{even}}\Bigg{(}\sum_{\vec{\sigma}_{even}}R_{\vec{c},\vec{\sigma}}^{x}(R_{\vec{\nu},\vec{\sigma}}^{x})^{\ast}\Bigg{)}\Bigg{(}\sum_{\vec{\sigma}^{\prime}_{even}}(R_{\vec{b},\vec{\sigma}^{\prime}}^{x})^{\ast}R_{\vec{\nu}^{\prime},\vec{\sigma}^{\prime}}^{x}\Bigg{)}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}\Bigg{]}$
$\displaystyle+\frac{1}{d^{2}}\Bigg{[}\sum_{\vec{\lambda}_{odd}}\sum_{\vec{\lambda}^{\prime}_{odd}}\sum_{\vec{\nu}_{odd}}\sum_{\vec{\nu}^{\prime}_{odd}}\sum_{\vec{b}_{odd}}\sum_{\vec{c}_{odd}}\Bigg{(}\sum_{\vec{\sigma}_{odd}}R_{\vec{c},\vec{\sigma}}^{x}(R_{\vec{\nu},\vec{\sigma}}^{x})^{\ast}\Bigg{)}\Bigg{(}\sum_{\vec{\sigma}^{\prime}_{odd}}(R_{\vec{b},\vec{\sigma}^{\prime}}^{x})^{\ast}R_{\vec{\nu}^{\prime},\vec{\sigma}^{\prime}}^{x}\Bigg{)}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}\Bigg{]}$
$\displaystyle=$
$\displaystyle\frac{1}{d^{2}}\Bigg{[}\sum_{\vec{\lambda}_{even}}\sum_{\vec{\lambda}^{\prime}_{even}}\sum_{\vec{\nu}_{even}}\sum_{\vec{\nu}^{\prime}_{even}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}+\sum_{\vec{\lambda}_{odd}}\sum_{\vec{\lambda}^{\prime}_{odd}}\sum_{\vec{\nu}_{odd}}\sum_{\vec{\nu}^{\prime}_{odd}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}\Bigg{]}$
(43)

Next, consider $Term_{y}$ $(y\neq x)$ in Eq. (16) with $U_{y}$ to be anti
block-diagonal unitary and defined as follows

$\displaystyle
U_{ABD}^{y}=\sum_{\vec{z},\vec{y}}R_{\vec{z},\vec{y}}^{x}f_{z_{1}}^{\dagger}f_{z_{2}}^{\dagger}\dots
f_{z_{s}}^{\dagger}|\Omega\rangle\langle\Omega|f_{y_{t}}\dots
f_{y_{2}}f_{y_{1}}$ (44)

with $P(\vec{z})\neq P(\vec{y})$. In this case also, we get

$\displaystyle
Term_{y}=\frac{1}{d^{2}}\Bigg{[}\sum_{\vec{\lambda}_{even}}\sum_{\vec{\lambda}^{\prime}_{even}}\sum_{\vec{\nu}_{even}}\sum_{\vec{\nu}^{\prime}_{even}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}+\sum_{\vec{\lambda}_{odd}}\sum_{\vec{\lambda}^{\prime}_{odd}}\sum_{\vec{\nu}_{odd}}\sum_{\vec{\nu}^{\prime}_{odd}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}\Bigg{]}$

Summing over all outcome, we get L.H.S of Eq. (16)

$\displaystyle=\sum_{\vec{\lambda}_{even}}\sum_{\vec{\lambda}^{\prime}_{even}}\sum_{\vec{\nu}_{even}}\sum_{\vec{\nu}^{\prime}_{even}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}+\sum_{\vec{\lambda}_{odd}}\sum_{\vec{\lambda}^{\prime}_{odd}}\sum_{\vec{\nu}_{odd}}\sum_{\vec{\nu}^{\prime}_{odd}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}$
(45)

Using Eq. (35) and Eq. (40) we get

$\displaystyle\text{tr}(\rho_{\tilde{A}^{\prime}B}^{f}\hat{O}_{\tilde{A}^{\prime}}\wedge\hat{O}_{B})$
$\displaystyle=\text{tr}\Bigg{[}\Bigg{(}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,\nu_{1})}^{\dagger}f_{(B,\nu_{2})}^{\dagger}\dots
f_{(B,\nu_{h})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\nu_{h^{\prime}}^{\prime})}$
(46) $\displaystyle\dots
f_{(B,\nu_{2}^{\prime})}f_{(B,\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{)}\Bigg{(}\sum_{\vec{d},\vec{e}}\sum_{\vec{b},\vec{c}}\Xi_{\vec{b}}(\Gamma_{\vec{c}})^{\ast}\xi_{\vec{d}}(\gamma_{\vec{e}})^{\ast}f_{(\tilde{A}^{\prime},d_{1})}^{\dagger}f_{(\tilde{A}^{\prime},d_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},d_{j^{\prime}})}^{\dagger}$ $\displaystyle
f_{(B,b_{1})}^{\dagger}f_{(B,b_{2})}^{\dagger}\dots
f_{(B,b_{j})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,c_{k})}\dots
f_{(B,c_{2})}f_{(B,c_{1})}f_{(\tilde{A}^{\prime},e_{k^{\prime}})}\dots
f_{(\tilde{A}^{\prime},e_{2})}f_{(\tilde{A}^{\prime},e_{1})}\Bigg{)}\Bigg{]}$
$\displaystyle=\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}\Xi_{\vec{\nu}^{\prime}}(\Gamma_{\vec{\nu}})^{\ast}\xi_{\vec{\lambda}^{\prime}}(\gamma_{\vec{\lambda}})^{\ast}$

with
$P(\vec{\lambda})=P(\vec{\nu})=P(\vec{c})=P(\vec{b})=P(\vec{\nu}^{\prime})=P(\vec{\lambda}^{\prime})$.
Therefore, L.H.S= R.H.S of Eq. (16). Similar result can also be proved for odd
state with suitable choice of observable. $\blacksquare$

##  Appendix C Reduction under Restricted Unitary group and Restricted
Clifford Twirling

Proof of Lemma 6. To check the ovarlap of a given state $\rho^{f}$ under
restricted Unitary group twirling, notice that

$\displaystyle\text{tr}\Bigg{[}H_{N}^{|E\rangle\langle
E|}\int_{U_{res}}dU(U\wedge U)\rho(U^{\dagger}\wedge
U^{\dagger})\Bigg{]}=\int_{U_{res}}dU\text{tr}[H_{N}^{|E\rangle\langle
E|}\rho]=\text{tr}[H_{N}^{|E\rangle\langle E|}\rho]$

where we have used Eq. (21). Similarly, one can show that overlap of
$\rho^{f}$ with the operators $\mathbb{I}^{|E\rangle\langle
E|}+\mathbb{I}^{|O\rangle\langle O|}$, $\mathbb{I}^{|E\rangle\langle
E|}-\mathbb{I}^{|O\rangle\langle O|}$ and $H^{|O\rangle\langle O|}_{N}$
remains invariant under the restricted Unitary group twirling operation.
$\blacksquare$

Proof of Lemma 7. We prove it by contradiction. Let $U\in C^{res}_{N}$ and $U$
is given by

$\sum_{ij}u_{ij}|E_{i}\rangle\langle E_{j}|+v_{ij}|O_{i}\rangle\langle
O_{j}|+u^{\prime}_{ij}|E_{i}\rangle\langle
O_{j}|+v^{\prime}_{ij}|O_{i}\rangle\langle E_{j}|$

Also, from Eq. (22) we get, for

$\Gamma^{ABD}\in\Xi^{ABD}\implies U\Gamma^{ABD}U^{\dagger}\in\Xi^{ABD}$

with

$\Gamma^{ABD}=\sum_{mn}A_{mn}|E_{m}\rangle\langle
O_{n}|+B_{mn}|O_{m}\rangle\langle E_{n}|$

Hence substituting the $U$ and $\Gamma^{ABD}$ we get the constraint for all
$i,p$

$\displaystyle\sum_{mq}u^{\prime}_{im}B_{mq}(u_{pq})^{\ast}=0\ \ \ \ $
$\displaystyle\sum_{mq}v^{\prime}_{im}A_{mq}(v_{pq})^{\ast}=0$
$\displaystyle\sum_{mq}u_{im}A_{mq}(u^{\prime}_{pq})^{\ast}=0\ \ \ \ $
$\displaystyle\sum_{mq}v_{im}B_{mq}(v^{\prime}_{pq})^{\ast}=0$

Let us assume that for a given $a,b,c$ and $d$

$u^{\prime}_{ab}\neq 0\ \text{and}\ u_{cd}\neq 0$ (47)

Now consider a matrix $M_{f}=|E_{d}\rangle\langle O_{b}|+|O_{b}\rangle\langle
E_{d}|$. Notice that the set $\Xi^{ABD}$ constitute the basis set (apart from
the multiplicative phase factor) for all complex anti- block diagonal matrices
of dimension $2^{n}\times 2^{n}$. Hence, $M_{f}$ can be expanded as

$M_{f}=\sum_{i}\Gamma^{ABD}_{i}\ \ \ \text{with}\ \ \
\Gamma^{ABD}\in\Xi^{ABD}$

Hence,

$UM_{f}U^{\dagger}=\sum_{i}U\Gamma^{ABD}_{i}U^{\dagger}\ \in\ \Xi^{ABD}$

But notice, $UM_{f}U^{\dagger}$ has a nonzero element in the basis
$|E_{a}\rangle\langle E_{c}|$ due to the condition in Eq. (47) - a
contradiction. $\blacksquare$

Proof of Theorem 2. Statement (1) can be proven by considering the adjoint of
Eq. (23). To prove Statement (2), notice that
$\mathcal{T}_{C^{res}_{N}}[\mathbb{I}_{N}]=\mathbb{I}_{N}$ is obvious from Eq.
(23). Also, notice that,

$\displaystyle
U_{BD}\Bigg{(}\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\Bigg{)}U_{BD}^{\dagger}=$
$\displaystyle\Bigg{(}\sum_{ij}u_{ij}|E_{i}\rangle\langle
E_{j}|+v_{ij}|O_{i}\rangle\langle
O_{j}|\Bigg{)}\Bigg{(}\sum_{k}|E_{k}\rangle\langle E_{k}|-|O_{k}\rangle\langle
O_{k}|\Bigg{)}$
$\displaystyle\Bigg{(}\sum_{pq}(u_{pq})^{\ast}|E_{q}\rangle\langle
E_{p}|+(v_{pq})^{\ast}|O_{q}\rangle\langle O_{p}|\Bigg{)}$ $\displaystyle=$
$\displaystyle\Bigg{(}\sum_{ik}u_{ik}|E_{i}\rangle\langle
E_{k}|-v_{ik}|O_{i}\rangle\langle
O_{k}|\Bigg{)}\Bigg{(}\sum_{pq}(u_{pq})^{\ast}|E_{q}\rangle\langle
E_{p}|+(v_{pq})^{\ast}|O_{q}\rangle\langle O_{p}|\Bigg{)}$ $\displaystyle=$
$\displaystyle\sum_{ipk}u_{ik}(u_{pk})^{\ast}|E_{i}\rangle\langle
E_{p}|-v_{ik}(v_{pk})^{\ast}|O_{i}\rangle\langle
O_{p}|=\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}$

where we have utilised the fact that

$\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}=\sum_{k}|E_{k}\rangle\langle
E_{k}|-|O_{k}\rangle\langle O_{k}$ (48)

by rewriting the wedge product of fermionic single mode basis in terms of the
elements of $\mathcal{B_{A}^{P}}$. Similarly, one can show that
$U_{ABD}\Bigg{(}\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\Bigg{)}U_{ABD}^{\dagger}=-\Bigg{(}\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\Bigg{)}$.
Hence for all $c\in C^{res}_{N}$

$c(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{A}c^{\dagger}\wedge
c(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{B}c^{\dagger}=(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{A}\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})^{B})$

One can then prove the statement (2) using Eq. (23). Before proving statement
(3), notice that $H_{N}$ can be written as $H_{N}=K_{N}\sum_{P\in
P_{N}^{H}}P\wedge P$ with $K_{N}$ being a constant, value of which depends on
the number of modes $N$ of the fermionic system. To see this, consider a
$1\times 1$-mode fermionic state given by

$\displaystyle|\psi^{+}\rangle\langle\psi^{+}|_{f}$
$\displaystyle=\frac{1}{2}[|E\rangle\wedge|E\rangle\langle E|\wedge\langle
E|+|O\rangle\wedge|O\rangle\langle O|\wedge\langle
O|+|E\rangle\wedge|E\rangle\langle O|\wedge\langle
O|+|O\rangle\wedge|O\rangle\langle E|\wedge\langle E|$

Define, partial transpose operator $(\mathbb{I}\wedge\mathbb{T}_{f})$ on the
Fermionic Hilbert Space. Hence,

$\displaystyle(\mathbb{I}\wedge\mathbb{T}_{f})(|\psi^{+}\rangle\langle\psi^{+}|_{f})=\frac{1}{2}\Bigg{[}|E\rangle\wedge|E\rangle\langle
E|\wedge\langle E|+|O\rangle\wedge|O\rangle\langle O|\wedge\langle
O|+|E\rangle\wedge|O\rangle\langle O|\wedge\langle
E|+|O\rangle\wedge|E\rangle\langle E|\wedge\langle O|\Bigg{]}$ (49)

Taking, N-fold wedge product of Eq. (49) and rewriting in terms of the
elements of the basis $\mathcal{B_{A}^{P}}$,we get

$\displaystyle[(\mathbb{I}\wedge\mathbb{T}_{f})(|\psi^{+}\rangle\langle\psi^{+}|_{f})]^{\wedge
N}=$
$\displaystyle\frac{1}{2^{N}}\Bigg{[}\sum_{ij}|E_{i}\rangle\wedge|E_{j}\rangle\langle
E_{j}|\wedge\langle E_{i}|+|O_{i}\rangle\wedge|O_{j}\rangle\langle
O_{j}|\wedge\langle O_{i}|$ (50)
$\displaystyle+|E_{i}\rangle\wedge|O_{j}\rangle\langle O_{j}|\wedge\langle
E_{i}|+|O_{i}\rangle\wedge|E_{j}\rangle\langle E_{j}|\wedge\langle
O_{i}|\Bigg{]}=\frac{1}{2^{N}}H_{N}$

Also, notice that,

$\displaystyle I\wedge
I+\sigma_{x}^{f}\wedge\sigma_{x}^{f}+\sigma_{z}^{f}\wedge\sigma_{z}^{f}+\sigma_{y}^{f}\wedge\sigma_{y}^{f}=$
$\displaystyle 2\Bigg{[}|E\rangle\wedge|E\rangle\langle E|\wedge\langle E|+$
(51) $\displaystyle|O\rangle\wedge|O\rangle\langle O|\wedge\langle
O|+|E\rangle\wedge|O\rangle\langle O|\wedge\langle
E|+|O\rangle\wedge|E\rangle\langle E|\wedge\langle O|\Bigg{]}$
$\displaystyle=4[(\mathbb{I}\wedge\mathbb{T}_{f})(|\psi^{+}\rangle\langle\psi^{+}|_{f})]$

Therefore, one can express $H_{N}$ as

$\displaystyle H_{N}=$ $\displaystyle
2^{N}[(\mathbb{I}\wedge\mathbb{T}_{f})(|\psi^{+}\rangle\langle\psi^{+}|_{f})]^{\wedge
N}\frac{2^{N}}{4^{N}}[I\wedge
I+\sigma_{x}^{f}\wedge\sigma_{x}^{f}+\sigma_{z}^{f}\wedge\sigma_{z}^{f}+\sigma_{y}^{f}\wedge\sigma_{y}^{f}]^{\wedge
N}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\frac{1}{2^{N}}\sum_{P\in
P_{N}^{H}}P\wedge P$ (52)

Now, $H^{|E\rangle\langle E|}_{N}$ is defined as the even subspace projection
of $H_{N}$ while $H^{|O\rangle\langle O|}_{N}$ is defined as the odd subspace
projection of $H_{N}$. Thus, $H^{|E\rangle\langle E|}_{N}$ and
$H^{|O\rangle\langle O|}_{N}$ can be expressed as

$\displaystyle H^{|E\rangle\langle
E|}_{N}=\frac{1}{2}\frac{1}{2^{N}}\sum_{\Gamma\in\Xi_{BD}^{H}}\Gamma\wedge\Gamma\
\ \ \text{and}\ \ \ H^{|O\rangle\langle
O|}_{N}=\frac{1}{2}\frac{1}{2^{N}}\sum_{\Gamma\in\Xi_{ABD}^{H}}\Gamma\wedge\Gamma$
(53)

where, $\Xi_{BD}^{H}$ and $\Xi_{ABD}^{H}$ consist of block-diagonal and anti-
block diagonal elements of $P_{N}^{H}$ respectively. Then,

$\displaystyle\mathcal{T}_{C^{res}_{N}}[H^{|E\rangle\langle E|}_{N}]=$
$\displaystyle\frac{1}{|C^{res}_{N}|}\sum_{c\in C^{res}_{N}}(c\wedge
c)H^{|E\rangle\langle E|}_{N}(c^{\dagger}\wedge
c^{\dagger})=\frac{1}{2^{N+1}}\frac{1}{|C^{res}_{N}|}\sum_{c\in
C^{res}_{N}}\sum_{\Gamma\in\Xi_{BD}^{H}}(c\Gamma c^{\dagger})\wedge(c\Gamma
c^{\dagger})$ (54) $\displaystyle=H^{|E\rangle\langle E|}_{N}$

as $c\in C^{res}_{N}$ permutes the elements of $\Xi_{BD}^{H}$. Similarly, for
$H^{|O\rangle\langle O|}_{N}$, one can prove that
$\mathcal{T}_{C^{res}_{N}}[H^{|O\rangle\langle O|}_{N}]=H^{|O\rangle\langle
O|}_{N}$ using the fact that $c\in C^{res}_{N}$ permutes the elements of
$\Xi_{ABD}^{H}$. This proves the statement (3).

To prove statement (4), notice that, any generic fermionic density matrix
(PSSR-respected) of $N\times N$-mode bipartite fermionic system can be
expressed as

$\rho^{f}=\sum_{\Gamma_{i},\Gamma_{j}\in\Xi_{BD}^{H}}A_{ij}\Gamma_{i}\wedge\Gamma_{j}+\sum_{\Gamma_{i},\Gamma_{j}\in\Xi_{ABD}^{H}}B_{ij}\Gamma_{i}\wedge\Gamma_{j}$
(55)

Now, following the procedure mentioned in [41], choose a basis
$\Gamma_{1}\wedge\Gamma_{2}\in\Xi_{BD}^{H}\wedge\Xi_{BD}^{H}$ such that
$\Gamma_{1}\neq\Gamma_{2}$. Also, let us assume that
$\Gamma_{1}\neq\mathbb{I}_{N}$ and
$\Gamma_{1}\neq\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}$. Consider an
element $c^{\prime}\in C^{res}_{N}$ such that
$c^{\prime}\Gamma_{1}c^{\prime\dagger}=\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I}$
and
$c^{\prime}\Gamma_{2}c^{\prime\dagger}=\gamma=\mathbb{I}\wedge\sigma_{z}^{f}\wedge\dots\wedge\mathbb{I}$
or
$\sigma_{x}^{f}\wedge\sigma_{x}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I}$
or $\mathbb{I}_{N}$ or $\pm(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})$
depending upon the condition $[\Gamma_{1},\Gamma_{2}]=0$ or
$\\{\Gamma_{1},\Gamma_{2}\\}=0$ or $\Gamma_{2}=\mathbb{I}_{N}$ or
$\Gamma_{2}=\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}$. Hence,

$\displaystyle\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{2}]=\frac{1}{|C^{res}_{N}|}\sum_{c\in
C^{res}_{N}}(c\Gamma_{1}c^{\dagger})\wedge(c\Gamma_{2}c^{\dagger})=\frac{1}{|C^{res}_{N}|}\sum_{c\in
C^{res}_{N}}\Bigg{[}(cc^{\prime\dagger}(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})c^{\prime}c^{\dagger})\wedge(cc^{\prime\dagger}\gamma
c^{\prime}c^{\dagger})\Bigg{]}$

Notice that due to the closure property of the group $C_{N}$,
$c.c^{\prime\dagger}\in C_{N}$. Also the set of all block-diagonal and anti-
block diagonal matrices in the basis $\mathcal{B^{P}_{A}}$ is closed under
matrix multiplication. Hence, for any $c,c^{\prime}\in C^{res}_{N}$ and
$cc^{\prime\dagger}\in C_{N}$, we have $Cc^{\prime\dagger}\in C^{res}_{N}$.
Let, $cc^{\prime\dagger}=\tilde{c}$, then

$\displaystyle\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{2}]=\frac{1}{|C^{res}_{N}|}\sum_{\tilde{c}\in
C^{res}_{N}}\Bigg{[}$
$\displaystyle(\tilde{c}(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})\tilde{c}^{\dagger})\wedge(\tilde{c}\gamma\tilde{c}^{\dagger})\Bigg{]}$

Notice, for any $\tilde{c}$ we can always find a $\tilde{c}^{\prime}$ such
that

$\tilde{c}^{\prime}(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})\tilde{c}^{\prime\dagger}=-\tilde{c}(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})\tilde{c}^{\dagger}$

and
$\tilde{c}^{\prime}\gamma\tilde{c}^{\prime\dagger}=\tilde{c}\gamma\tilde{c}^{\dagger}$.
Hence $\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{2}]=0$. For,
$\Gamma_{1}=\Gamma_{2}$

$\displaystyle\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{1}]=$
$\displaystyle\frac{1}{|C^{res}_{N}|}\sum_{c\in
C^{res}_{N}}c(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})c^{\dagger}\wedge\
c(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})c^{\dagger}$ (56)

Now for $c\in C^{res}_{N}$,
$c(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})c^{\dagger}$
corresponds to all the elements in
$\Xi_{BD}^{H}-\\{I,\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\\}$. Also,
irrespective of the element $c\in C^{res}_{N}$, number of occurrence of each
elements of
$\Xi_{BD}^{H}-\\{I,\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\\}$ in the
sum will be same. This can be argued by noticing that all the elements $c\in
C_{N}$ that takes
$c(\sigma_{z}^{f}\wedge\mathbb{I}\wedge\dots\wedge\mathbb{I})c^{\dagger}$ to
the elements in
$\Xi_{BD}^{H}-\\{I,\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\\}$ are also
part of $c\in C^{res}_{N}$. Hence, if elements of $C_{N}$ outputs the same
occurrence of elements in
$\Xi_{BD}^{H}-\\{I,\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\\}$, then so
does the elements of $C^{res}_{N}$. Hence we have,

$\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{1}]=\frac{1}{|\Xi_{BD}^{H}|-2}\sum_{P\in\Xi_{BD}^{H}-\\{I,\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}\\}}P\wedge
P$

Similarly, choosing a basis
$\Gamma_{1}\wedge\Gamma_{2}\in\Xi_{ABD}^{H}\wedge\Xi_{ABD}^{H}$ it can be
proved that for $\Gamma_{1}\neq\Gamma_{2}$, we get
$\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{2}]=0$ and

$\mathcal{T}_{C^{res}_{N}}[\Gamma_{1}\wedge\Gamma_{1}]=\frac{1}{|\Xi_{ABD}^{H}|}\sum_{P\in\Xi_{ABD}^{H}}P\wedge
P$

Hence, $\mathcal{T}_{C^{res}_{N}}[\rho^{f}]$ can be expressed as the linear
combinantion of
$\mathbb{I}_{N}\wedge\mathbb{I}_{N},(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f}),H^{|E\rangle\langle
E|}_{N}$ and $H^{|O\rangle\langle O|}_{N}$ Now notice that
$\mathbb{I}^{|E\rangle\langle E|}+\mathbb{I}^{|O\rangle\langle
O|}=\mathbb{I}$. Using Eq. (48), it can be shown that

$\mathbb{I}^{|E\rangle\langle E|}-\mathbb{I}^{|O\rangle\langle
O|}=(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})\wedge(\sigma_{z}^{f}\wedge\dots\wedge\sigma_{z}^{f})$

Hence, under the twirling $\mathcal{T}_{C^{res}_{N}}$, any $N\times N$-mode
fermionic state $\rho^{f}$ can be reduced to the linear combination of
$\mathbb{I}^{|E\rangle\langle E|},\ \mathbb{I}^{|O\rangle\langle O|},\
H^{|E\rangle\langle E|}_{N}$ and $H^{|O\rangle\langle O|}_{N}$. $\blacksquare$

##  Appendix D Invariance of Optimal Teleportation Fidelity

Notice that,
$F(\mathbb{I}\wedge\mathcal{T}^{\ast}_{C^{res}_{N}}[\mathcal{E}])$ can be
written as

$\displaystyle=$
$\displaystyle\frac{1}{|C^{res}_{N}|}\int_{\psi_{even}}d\psi_{E}\sum_{c\in
C^{res}_{N}}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}((\mathbb{I}\wedge
c^{\dagger})[(\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c)(|\psi_{E}\rangle\langle\psi_{E}|)(\mathbb{I}\wedge
c^{\dagger}))](\mathbb{I}\wedge c)(E_{m}^{A}\wedge E_{m}^{B}))}$
$\displaystyle\sqrt{\text{tr}(|\psi_{E}\rangle\langle\psi_{E}|E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}+\frac{1}{|C^{res}_{N}|}\int_{\psi_{odd}}d\psi_{o}\sum_{c\in
C^{res}_{N}}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(|\psi_{o}\rangle\langle\psi_{o}|E_{m}^{A}\wedge
E_{m}^{B})}$ $\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge
c^{\dagger})[(\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c)(|\psi_{O}\rangle\langle\psi_{O}|)(\mathbb{I}\wedge
c^{\dagger}))](\mathbb{I}\wedge c)(E_{m}^{A}\wedge E_{m}^{B}))}\Bigg{)}^{2}$
$\displaystyle=$
$\displaystyle\frac{1}{|C^{res}_{N}|}\int_{\psi_{even}}d\psi_{E}\sum_{c\in
C^{res}_{N}}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}((\mathbb{I}\wedge
c^{\dagger})[(\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c)(|\psi_{E}\rangle\langle\psi_{E}|)(\mathbb{I}\wedge
c^{\dagger}))](\mathbb{I}\wedge c)(E_{m}^{A}\wedge E_{m}^{B}))}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge c^{\dagger})(\mathbb{I}\wedge
c)|\psi_{E}\rangle\langle\psi_{E}|(\mathbb{I}\wedge
c^{\dagger})(\mathbb{I}\wedge c)(E_{m}^{A}\wedge E_{m}^{B}))}\Bigg{)}^{2}$
$\displaystyle+\frac{1}{|C^{res}_{N}|}\int_{\psi_{odd}}d\psi_{o}\sum_{c\in
C^{res}_{N}}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}((\mathbb{I}\wedge
c^{\dagger})(\mathbb{I}\wedge
c)|\psi_{O}\rangle\langle\psi_{O}|(\mathbb{I}\wedge
c^{\dagger})(\mathbb{I}\wedge c)(E_{m}^{A}\wedge E_{m}^{B}))}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge
c^{\dagger})[(\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c)(|\psi_{O}\rangle\langle\psi_{O}|)(\mathbb{I}\wedge
c^{\dagger}))](\mathbb{I}\wedge c)(E_{m}^{A}\wedge E_{m}^{B}))}\Bigg{)}^{2}$

In general, the measures $d\psi_{E}$ and $d\psi_{O}$ is not invariant under
the change of variable $\psi_{O}^{\prime}=c^{ABD}\psi_{E}$ and
$\psi_{E}^{\prime}=c^{ABD}\psi_{O}$ respectively. Hence seprating the action
due to $c^{BD}$ and $c^{ABD}$ we get

$\displaystyle=$ $\displaystyle\frac{1}{|C^{res}_{N}|}\sum_{c^{BD}\in
C^{res}_{N}}\int_{\psi_{even}}d\psi_{E}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}([(\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c^{BD})(|\psi_{E}\rangle\langle\psi_{E}|)(\mathbb{I}\wedge{c^{BD}}^{\dagger}))](E_{m}^{A}\wedge
c^{BD}E_{m}^{B}{c^{BD}}^{\dagger}))}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge
c^{BD})|\psi_{E}\rangle\langle\psi_{E}|(\mathbb{I}\wedge{c^{BD}}^{\dagger})(E_{m}^{A}\wedge
c^{BD}E_{m}^{B}{c^{BD}}^{\dagger}))}\Bigg{)}^{2}$
$\displaystyle+\frac{1}{|C^{res}_{N}|}\sum_{c^{ABD}\in
C^{res}_{N}}\int_{\psi_{even}}d\psi_{E}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}((\mathbb{I}\wedge
c^{ABD})|\psi_{E}\rangle\langle\psi_{E}|(\mathbb{I}\wedge{c^{ABD}}^{\dagger})(E_{m}^{A}\wedge
c^{ABD}E_{m}^{B}{c^{ABD}}^{\dagger}))}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c^{ABD})(|\psi_{E}\rangle\langle\psi_{E}|)(\mathbb{I}\wedge{c^{ABD}}^{\dagger}))(E_{m}^{A}\wedge
c^{ABD}E_{m}^{B}{c^{ABD}}^{\dagger}))}\Bigg{)}^{2}$
$\displaystyle+\frac{1}{|C^{res}_{N}|}\sum_{c^{BD}\in
C^{res}_{N}}\int_{\psi_{odd}}d\psi_{O}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}((\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c^{BD})(|\psi_{O}\rangle\langle\psi_{O}|)(\mathbb{I}\wedge{c^{BD}}^{\dagger}))(E_{m}^{A}\wedge
c^{BD}E_{m}^{B}{c^{BD}}^{\dagger}))}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge
c^{BD})|\psi_{E}\rangle\langle\psi_{E}|(\mathbb{I}\wedge{c^{BD}}^{\dagger})(E_{m}^{A}\wedge
c^{BD}E_{m}^{B}{c^{BD}}^{\dagger}))}\Bigg{)}^{2}$
$\displaystyle+\frac{1}{|C^{res}_{N}|}\sum_{c^{ABD}\in
C^{res}_{N}}\int_{\psi_{odd}}d\psi_{O}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}((\mathbb{I}\wedge
c^{ABD})|\psi_{O}\rangle\langle\psi_{O}|(\mathbb{I}\wedge{c^{ABD}}^{\dagger})(E_{m}^{A}\wedge
c^{BD}E_{m}^{B}{c^{BD}}^{\dagger}))}$
$\displaystyle\sqrt{\text{tr}((\mathbb{I}\wedge\mathcal{E})((\mathbb{I}\wedge
c^{ABD})(|\psi_{O}\rangle\langle\psi_{O}|)(\mathbb{I}\wedge{c^{ABD}}^{\dagger}))(\mathbb{I}\wedge
c^{ABD})(E_{m}^{A}\wedge c^{BD}E_{m}^{B}{c^{BD}}^{\dagger}))}\Bigg{)}^{2}$
$\displaystyle=$ $\displaystyle F(\mathbb{I}\wedge\mathcal{E})$ (58)

where we utilized the changed of variables in the equations and corresponding
change in the measure for each element of the restricted set of clifford group
$C^{res}_{N}$.

##  Appendix E General Teleportation Channel in FQT

Proof of Lemma 9. Let $W^{f}_{AB}$ be any arbitrary bi-partite $N\times
N$-mode Fermionic state shared between Alice and Bob. Initially Alice prepare
a Fermionic state $|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}$
locally and wish to teleport part of the state to Bob using the canonical form
of the state $W^{f}_{AB}$ as resource. In this case, Initial state can be
written as,
$(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge(\omega_{noise}^{f})_{AB})$
where $(\omega_{noise}^{f})_{AB}=\mathcal{T}_{C^{res}_{N}}[W^{f}_{AB}]$ with
$\alpha,\ \beta,\ \Upsilon_{1}$ and $\omega_{AB}^{f}$ depends on $W^{f}_{AB}$.
Upon performing Bell basis measurement by Alice with
$M_{x}=|\Psi_{x}\rangle\langle\Psi_{x}|_{\tilde{A}A}$, the $x^{th}$ term of
the initial state can be written as

$\displaystyle\text{tr}_{\tilde{A}A}\Bigg{[}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge(\omega_{noise}^{f})_{AB})(\mathbb{I}_{\tilde{A}^{\prime}}\wedge
M_{x}\wedge\mathbb{I}_{B})\Bigg{]}$ $\displaystyle=\alpha
d^{2}\text{tr}_{\tilde{A}A}\Bigg{[}\Bigg{(}|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge\Bigg{(}\frac{\mathbb{I}_{N}\wedge\mathbb{I}_{N}}{d^{2}}+\Upsilon_{1}\frac{\mathbb{S}_{z}\wedge\mathbb{S}_{z}}{d^{2}}\Bigg{)}\Bigg{)}\Bigg{(}\mathbb{I}_{\tilde{A}^{\prime}}\wedge
M_{x}\wedge\mathbb{I}_{B}\Bigg{)}\Bigg{]}+\beta
d\text{tr}_{\tilde{A}A}\Bigg{[}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge(\omega^{f})_{AB})$
$\displaystyle(\mathbb{I}_{\tilde{A}^{\prime}}\wedge
M_{x}\wedge\mathbb{I}_{B})\Bigg{]}$ (59)

Following the partial tracing procedure mentioned in [21] that utilizes the
physical consistency condition to eliminate the possibility of any sign
ambiguity, we get

$\displaystyle\text{tr}_{\tilde{A}A}\Bigg{[}\Bigg{(}|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge\Bigg{(}\frac{\mathbb{I}_{N}\wedge\mathbb{I}_{N}}{d^{2}}+\Upsilon_{1}\frac{\mathbb{S}_{z}\wedge\mathbb{S}_{z}}{d^{2}}\Bigg{)}\Bigg{)}\Bigg{(}\mathbb{I}_{\tilde{A}^{\prime}}\wedge
M_{x}\wedge\mathbb{I}_{B}\Bigg{)}\Bigg{]}$
$\displaystyle=\text{tr}_{\tilde{A}A}\Bigg{[}\frac{1}{d^{3}}\Bigg{(}\sum_{\vec{\sigma},\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(\tilde{A},\nu_{1})}^{\dagger}f_{(\tilde{A},\nu_{2})}^{\dagger}\dots
f_{(\tilde{A},\nu_{h})}^{\dagger}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}\dots
f_{(A,\sigma_{j})}^{\dagger}f_{(B,d_{1})}^{\dagger}$ $\displaystyle
f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(A,\sigma_{j})}\dots
f_{(A,\sigma_{2})}f_{(A,\sigma_{1})}f_{(\tilde{A},\nu_{h^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\nu_{2}^{\prime})}f_{(\tilde{A},\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}$ $\displaystyle
f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon_{1}\sum_{\vec{\sigma},\vec{d},P(\vec{\sigma})=P(\vec{d})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(\tilde{A},\nu_{1})}^{\dagger}f_{(\tilde{A},\nu_{2})}^{\dagger}\dots
f_{(\tilde{A},\nu_{h})}^{\dagger}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}\dots
f_{(A,\sigma_{j})}^{\dagger}$ $\displaystyle
f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(A,\sigma_{j})}\dots
f_{(A,\sigma_{2})}f_{(A,\sigma_{1})}f_{(\tilde{A},\nu_{h^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\nu_{2}^{\prime})}f_{(\tilde{A},\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}\dots$
$\displaystyle
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}-\Upsilon_{1}\sum_{\vec{\sigma},\vec{d},P(\vec{\sigma})\neq
P(\vec{d})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(\tilde{A},\nu_{1})}^{\dagger}f_{(\tilde{A},\nu_{2})}^{\dagger}\dots
f_{(\tilde{A},\nu_{h})}^{\dagger}f_{(A,\sigma_{1})}^{\dagger}f_{(A,\sigma_{2})}^{\dagger}$
$\displaystyle\dots
f_{(A,\sigma_{j})}^{\dagger}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(A,\sigma_{j})}\dots
f_{(A,\sigma_{2})}f_{(A,\sigma_{1})}f_{(\tilde{A},\nu_{h^{\prime}}^{\prime})}\dots
f_{(\tilde{A},\nu_{2}^{\prime})}f_{(\tilde{A},\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{l^{\prime}}^{\prime})}$
$\displaystyle\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{)}\Bigg{(}\sum_{\vec{b},\vec{c}}\sum_{\vec{z}^{\prime},\vec{\pi}^{\prime}}\sum_{\vec{z},\vec{\pi}}R_{\vec{z},\vec{\pi}}^{x}(R_{\vec{z}^{\prime},\vec{\pi}^{\prime}}^{x})^{\ast}f_{(\tilde{A}^{\prime},b_{1})}^{\dagger}f_{(\tilde{A}^{\prime},b_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},b_{j^{\prime}})}^{\dagger}f_{(\tilde{A},z_{1})}^{\dagger}f_{(\tilde{A},z_{2})}^{\dagger}\dots
f_{(\tilde{A},z_{s})}^{\dagger}f_{(A,\pi_{1})}^{\dagger}f_{(A,\pi_{2})}^{\dagger}\dots$
$\displaystyle
f_{(A,\pi_{q})}f_{(B,c_{1})}^{\dagger}f_{(B,c_{2})}^{\dagger}\dots
f_{(B,c_{t})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,c_{t})}\dots
f_{(B,c_{2})}f_{(B,c_{1})}f_{(A,\pi_{q^{\prime}}^{\prime})}\dots
f_{(A,\pi_{2}^{\prime})}f_{(A,\pi_{1}^{\prime})}f_{(\tilde{A},z_{s^{\prime}}^{\prime})}f_{(\tilde{A}^{\prime},b_{j^{\prime}})}\dots
f_{(\tilde{A}^{\prime},b_{2})}f_{(\tilde{A}^{\prime},b_{1})}\Bigg{)}\Bigg{]}$

Using the scaler product definition in wedge-product space we can proceed with
the proof as follows

$\displaystyle\text{tr}_{\tilde{A}A}\Bigg{[}\Bigg{(}|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge\Bigg{(}\frac{\mathbb{I}_{N}\wedge\mathbb{I}_{N}}{d^{2}}+\Upsilon_{1}\frac{\mathbb{S}_{z}\wedge\mathbb{S}_{z}}{d^{2}}\Bigg{)}\Bigg{)}\Bigg{(}\mathbb{I}_{\tilde{A}^{\prime}}\wedge
M_{x}\wedge\mathbb{I}_{B}\Bigg{)}\Bigg{]}$
$\displaystyle=\frac{1}{d^{3}}\Bigg{[}\sum_{\vec{\sigma},\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}R_{\vec{\nu}^{\prime},\vec{\sigma}}^{x}(R_{\vec{\nu},\vec{\sigma}}^{x})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}$ $\displaystyle
f_{(B,d_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon_{1}\sum_{\vec{\sigma},\vec{d},P(\vec{\sigma})=P(\vec{d})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}R_{\vec{\nu}^{\prime},\vec{\sigma}}^{x}(R_{\vec{\nu},\vec{\sigma}}^{x})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,d_{1})}^{\dagger}$
$\displaystyle f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}-\Upsilon_{1}\sum_{\vec{\sigma},\vec{d},P(\vec{\sigma})\neq
P(\vec{d})}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu},\vec{\nu}^{\prime}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}^{\prime}})^{\ast}R_{\vec{\nu}^{\prime},\vec{\sigma}}^{x}$
$\displaystyle(R_{\vec{\nu},\vec{\sigma}}^{x})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{]}$

Notice that $R^{x}$ represents PSSR-respected unitary. Hence, using properties
of PSSR-respected unitary matrices, we get

$\displaystyle\text{tr}_{\tilde{A}A}\Bigg{[}\Bigg{(}|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}\tilde{A}}\wedge\Bigg{(}\frac{\mathbb{I}_{N}\wedge\mathbb{I}_{N}}{d^{2}}+\Upsilon_{1}\frac{\mathbb{S}_{z}\wedge\mathbb{S}_{z}}{d^{2}}\Bigg{)}\Bigg{)}\Bigg{(}\mathbb{I}_{\tilde{A}^{\prime}}\wedge
M_{x}\wedge\mathbb{I}_{B}\Bigg{)}\Bigg{]}$
$\displaystyle=\frac{1}{d^{3}}\Bigg{[}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}$
$\displaystyle\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon_{1}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots$
$\displaystyle
f_{(B,d_{2})}f_{(B,d_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}-\Upsilon_{1}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}$
$\displaystyle|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{]}$

with $P(\vec{\lambda})=P(\vec{\lambda}^{\prime})=P(\vec{d})$ in the second
term and $P(\vec{\lambda})=P(\vec{\lambda}^{\prime})\neq P(\vec{d})$ in the
third term, given the corresponding $x$ represents block-diagonal unitary. For
$x$ representing anti block-diagonal unitary,
$P(\vec{\lambda})=P(\vec{\lambda}^{\prime})\neq P(\vec{d})$ satisfies in the
second term and $P(\vec{\lambda})=P(\vec{\lambda}^{\prime})=P(\vec{d})$ in the
third term.

Based on the outcome $x$, Bob then applies the corresponding Unitary
operation. Hence, the output state due to noisy part of the resource is given
by

$\displaystyle(\rho^{f}_{\tilde{A}^{\prime}B})_{noisy}$
$\displaystyle=\frac{1}{d^{3}}\Bigg{[}\sum_{\vec{p},\vec{q}}\sum_{\vec{p}^{\prime},\vec{q}^{\prime}}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}R_{\vec{p},\vec{q}}^{x}(R_{\vec{p}^{\prime},\vec{q}^{\prime}}^{x})^{\ast}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,p_{1})}^{\dagger}f_{(B,p_{2})}^{\dagger}\dots
f_{(B,p_{m})}^{\dagger}|\Omega\rangle\langle\Omega|$ $\displaystyle
f_{(B,q_{n})}\dots
f_{(B,q_{2})}f_{(B,q_{1})}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(B,q_{1}^{\prime})}^{\dagger}f_{(B,q_{2}^{\prime})}^{\dagger}\dots
f_{(B,q_{n^{\prime}}^{\prime})}^{\dagger}|\Omega\rangle\langle\Omega|$
$\displaystyle f_{(B,p_{m^{\prime}}^{\prime})}\dots
f_{(B,p_{2}^{\prime})}f_{(B,p_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon_{1}\sum_{\vec{p},\vec{q}}\sum_{\vec{p}^{\prime},\vec{q}^{\prime}}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}R_{\vec{p},\vec{q}}^{x}(R_{\vec{p}^{\prime},\vec{q}^{\prime}}^{x})^{\ast}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}$
$\displaystyle f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,p_{1})}^{\dagger}f_{(B,p_{2})}^{\dagger}\dots
f_{(B,p_{m})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,q_{n})}\dots
f_{(B,q_{2})}f_{(B,q_{1})}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|$ $\displaystyle
f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(B,q_{1}^{\prime})}^{\dagger}f_{(B,q_{2}^{\prime})}^{\dagger}\dots
f_{(B,q_{n^{\prime}}^{\prime})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,p_{m^{\prime}}^{\prime})}\dots
f_{(B,p_{2}^{\prime})}f_{(B,p_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}$
$\displaystyle-\Upsilon_{1}\sum_{\vec{p},\vec{q}}\sum_{\vec{p}^{\prime},\vec{q}^{\prime}}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}R_{\vec{p},\vec{q}}^{x}(R_{\vec{p}^{\prime},\vec{q}^{\prime}}^{x})^{\ast}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,p_{1})}^{\dagger}f_{(B,p_{2})}^{\dagger}\dots
f_{(B,p_{m})}^{\dagger}|\Omega\rangle\langle\Omega|$ $\displaystyle
f_{(B,q_{n})}\dots
f_{(B,q_{2})}f_{(B,q_{1})}f_{(B,d_{1})}^{\dagger}f_{(B,d_{2})}^{\dagger}\dots
f_{(B,d_{g})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,d_{g})}\dots
f_{(B,d_{2})}f_{(B,d_{1})}f_{(B,q_{1}^{\prime})}^{\dagger}f_{(B,q_{2}^{\prime})}^{\dagger}\dots
f_{(B,q_{n^{\prime}}^{\prime})}^{\dagger}|\Omega\rangle$
$\displaystyle\langle\Omega|f_{(B,p_{m^{\prime}}^{\prime})}\dots
f_{(B,p_{2}^{\prime})}f_{(B,p_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{]}$
$\displaystyle=\frac{1}{d^{3}}\Bigg{[}\sum_{\vec{p}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,p_{1})}^{\dagger}f_{(B,p_{2})}^{\dagger}\dots
f_{(B,p_{m})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,p_{m})}\dots
f_{(B,p_{2})}$ $\displaystyle
f_{(B,p_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon_{1}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,p_{1})}^{\dagger}f_{(B,p_{2})}^{\dagger}\dots$
$\displaystyle
f_{(B,p_{m})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,p_{m})}\dots
f_{(B,p_{2})}f_{(B,p_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}-\Upsilon_{1}\sum_{\vec{d}}\sum_{\vec{\lambda},\vec{\lambda}^{\prime}}\sum_{\vec{\nu}}\eta_{\vec{\lambda},\vec{\nu}}(\eta_{\vec{\lambda}^{\prime},\vec{\nu}})^{\ast}f_{(\tilde{A}^{\prime},\lambda_{1})}^{\dagger}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}$
$\displaystyle\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,p_{1})}^{\dagger}f_{(B,p_{2})}^{\dagger}\dots
f_{(B,p_{m})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,p_{m})}\dots
f_{(B,p_{2})}f_{(B,p_{1})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}\Bigg{]}$

with $P(\vec{\lambda})=P(\vec{\lambda}^{\prime})=P(\vec{p})$ in the second
term and $P(\vec{\lambda})=P(\vec{\lambda}^{\prime})\neq P(\vec{p})$ in the
third term irrespective of $x$. Hence,

$\displaystyle(\rho^{f}_{\tilde{A}^{\prime}B})_{noisy}=\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d^{3}}+\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d^{3}}$

Summing over all $x$ we get the noisy part of Eq. (LABEL:eq:25) for even input
states. For odd input state, notice that,
$P(\vec{\lambda})=P(\vec{\lambda}^{\prime})\neq P(\vec{p})$ in the second term
and $P(\vec{\lambda})=P(\vec{\lambda}^{\prime})=P(\vec{p})$ in the third term
irrespective of $x$. Hence,

$\displaystyle(\rho^{f}_{\tilde{A}^{\prime}B})_{noisy}=\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d^{3}}-\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d^{3}}$

In this case also, summing over all $x$ we get the noisy part of Eq.
(LABEL:eq:25) for odd input states. Proceeding similarly, the exact part of
the output state can be written as

$\displaystyle(\rho^{f}_{\tilde{A}^{\prime}B})_{exact}=$
$\displaystyle\sum_{\vec{\lambda},\vec{\lambda}^{\prime},\vec{\nu},\vec{\nu}^{\prime},P(\vec{\nu})=P(\vec{\nu}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,\nu_{1})}^{\dagger}f_{(B,\nu_{2})}^{\dagger}\dots
f_{(B,\nu_{h})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\nu_{m}^{\prime})}\dots
f_{(B,\nu_{2}^{\prime})}f_{(B,\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots$
$\displaystyle
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}+\Upsilon\sum_{\vec{\lambda},\vec{\lambda}^{\prime},\vec{\nu},\vec{\nu}^{\prime},P(\vec{\nu})\neq
P(\vec{\nu}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{2})}^{\dagger}\dots
f_{(\tilde{A}^{\prime},\lambda_{l})}^{\dagger}f_{(B,\nu_{1})}^{\dagger}f_{(B,\nu_{2})}^{\dagger}\dots
f_{(B,\nu_{h})}^{\dagger}|\Omega\rangle\langle\Omega|f_{(B,\nu_{m}^{\prime})}\dots$
$\displaystyle
f_{(B,\nu_{2}^{\prime})}f_{(B,\nu_{1}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{j^{\prime}}^{\prime})}\dots
f_{(\tilde{A}^{\prime},\lambda_{2}^{\prime})}f_{(\tilde{A}^{\prime},\lambda_{1}^{\prime})}$
$\displaystyle=$
$\displaystyle\text{loc-p}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}B})+\Upsilon\
\overline{\text{loc-p}}(|\psi\rangle\langle\psi|_{\tilde{A}^{\prime}B})$ (62)

Hence, the output state of the teleportation channel can be expressed as in
Eq. (LABEL:eq:25). $\blacksquare$

To find the teleportation fidelity of the give teleportation channel for the
exact case i.e. to prove $F(\mathbb{I}\wedge\mathcal{E})_{exact}=1$, notice,

$\displaystyle F(\mathbb{I}\wedge\mathcal{E})=$ $\displaystyle\int
d\psi_{E}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(|\psi_{E}\rangle\langle\psi_{E}|E_{m}^{A}\wedge
E_{m}^{B})\text{tr}((\mathbb{I}\wedge\mathcal{E}_{tel}(|\psi_{E}\rangle\langle\psi_{E}|))E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}$ $\displaystyle+\int
d\psi_{o}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(|\psi_{o}\rangle\langle\psi_{o}|E_{m}^{A}\wedge
E_{m}^{B})\text{tr}((\mathbb{I}\wedge\mathcal{E}_{tel}(|\psi_{o}\rangle\langle\psi_{o}|))E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}$ $\displaystyle=$ $\displaystyle\int
d\psi_{E}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(\text{loc-p}(|\psi_{E}\rangle\langle\psi_{E}|)E_{m}^{A}\wedge
E_{m}^{B})\text{tr}(\text{loc-p}(|\psi_{E}\rangle\langle\psi_{E}|)E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}$ $\displaystyle+\int
d\psi_{o}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}(\text{loc-p}(|\psi_{o}\rangle\langle\psi_{o})|E_{m}^{A}\wedge
E_{m}^{B})\text{tr}((\text{loc-p}(|\psi_{o}\rangle\langle\psi_{o}|))E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}$ $\displaystyle=$ $\displaystyle\int
d\psi_{E}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\text{tr}(\text{loc-p}(|\psi_{E}\rangle\langle\psi_{E}|)E_{m}^{A}\wedge
E_{m}^{B})\Bigg{)}^{2}+\int d\psi_{o}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\text{tr}(\text{loc-p}(|\psi_{o}\rangle\langle\psi_{o})|E_{m}^{A}\wedge
E_{m}^{B})\Bigg{)}^{2}$ $\displaystyle=$ $\displaystyle\int d\psi_{E}+\int
d\psi_{O}=1$

where we have used the relation $\sum_{m}E_{m}^{A}\wedge E_{m}^{B}=\mathbb{I}$
and $\text{tr}(\text{loc-p}(|\psi_{E(O)}\rangle\langle\psi_{E(O)}|))=1$.

For the noisy part of the channel, the lower bound can be derived as follows.
Notice that,

$\displaystyle F(\mathbb{I}\wedge\mathcal{E})=$ $\displaystyle\int
d\psi_{E}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}\Bigg{(}\Bigg{(}\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d}+\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d}\Bigg{)}\Bigg{(}E_{m}^{A}\wedge
E_{m}^{B}\Bigg{)}\text{tr}(|\psi_{E}\rangle\langle\psi_{E}|E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}$ $\displaystyle+\int
d\psi_{O}\Bigg{(}\min_{\\{E_{m}^{A}\wedge
E_{m}^{B}\\}}\sum_{m}\sqrt{\text{tr}\Bigg{(}\Bigg{(}\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d}-\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d}\Bigg{)}\Bigg{(}E_{m}^{A}\wedge
E_{m}^{B}\Bigg{)})\text{tr}(|\psi_{o}\rangle\langle\psi_{o}|E_{m}^{A}\wedge
E_{m}^{B})}\Bigg{)}^{2}$

Minimum value of the integrand in the first(second) integral [40] is achieved
if $\\{E_{m}^{A}\wedge E_{m}^{B}\\}$ is chosen to be equal to
$|\psi_{E(O)}\rangle\langle\psi_{E(O)}|$. However, this is not a legitimate
separable POVM if $|\psi_{E(O)}\rangle\langle\psi_{E(O)}|$ is not separable.
Hence, we get

$\displaystyle F(\mathbb{I}\wedge\mathcal{E})_{noisy}\geq$ $\displaystyle\int
d\psi_{E}\Bigg{(}\sqrt{\left\langle\psi_{E}\left|\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d}+\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d}\right|\psi_{E}\right\rangle}\Bigg{)}^{2}$
$\displaystyle+\int
d\psi_{O}\Bigg{(}\sqrt{\left\langle\psi_{O}\left|\frac{\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N}}{d}-\Upsilon_{1}\frac{(\rho^{f}_{\tilde{A}^{\prime}}\wedge\mathbb{I}_{N})(\mathbb{S}_{z}\wedge\mathbb{S}_{z})}{d}\right|\psi_{O}\right\rangle}\Bigg{)}^{2}$
$\displaystyle=$ $\displaystyle\int
d\psi_{E}\frac{\text{tr}({\rho^{f}_{\tilde{A}^{\prime}}}^{2})}{d}(1+\Upsilon_{1})+\int
d\psi_{O}\frac{\text{tr}({\rho^{f}_{\tilde{A}^{\prime}}}^{2})}{d}(1+\Upsilon_{1})$
$\displaystyle=$ $\displaystyle(1+\Upsilon_{1})\int
dU_{res}\frac{\text{tr}({\rho^{f}_{\tilde{A}^{\prime}}}^{2}(U_{res}))}{d}$
(63)

##  Appendix F Comparison with SQT based teleportation

Let us consider a $4\times 4$ bi-partite state of distinguishable particles,
given by

$\rho_{AB}=\setcounter{MaxMatrixCols}{16}\begin{pmatrix}1/4&0&0&1/4&a/4&0&0&a/4&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
1/4&0&0&1/4&a/4&0&0&a/4&0&0&0&0&0&0&0&0\\\
a/4&0&0&a/4&1/4&0&0&1/4&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ a/4&0&0&a/4&1/4&0&0&1/4&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
\end{pmatrix}$

The eigenvalues of this state are given by $0.5a+0.5$ with repetition 1,
$0.5a-0.5$ with repetition 1 and zero with repetition 14. Hence, for $-1\leq
a\leq 1$, $\rho_{AB}$ becomes a legitimate density matrix. We want to
calculate the separable fidelity which is operational in SQT using $\rho_{AB}$
as a resource state.

Notice, the negativity [45] of $\rho_{AB}$, given by

$N(\rho_{AB})=0.5+|a|$

depends on $a$, so does the upper bound on input-output fidelity.

Using the monotonicity of fidelity we have,

$F(\sigma_{AB},\mathcal{E}(\sigma_{AB}))\leq
F(\text{tr}_{A}(\sigma_{AB}),\text{tr}_{A}(\mathcal{E}(\sigma_{AB})))$

Using the operational definition of partial-trace one can see that separable
fidelity is upper bounded by the input-output fidelity. Now the fidelity
$F(\text{tr}_{A}(\sigma_{AB}),\text{tr}_{A}(\mathcal{E}(\sigma_{AB})))<1$ for
$a<1$. Hence, the separable fidelity is also less than $1$. However, for a
resource state $\rho_{AB}^{f}$ (in FQT) in the basis $\mathcal{B_{AB}^{P}}$
given by the same matrix as of $\rho_{AB}$, we know the separable
teleportation fidelity is equal to 1 (Lemma 7.)

## References

  * Bennett  _et al._ [1993] C. H. Bennett, G. Brassard, C. Crépeau, R. Jozsa, A. Peres, and W. K. Wootters, Phys. Rev. Lett. 70, 1895 (1993).