数理解析研究所講究録 1253 巻 2002 年 172-186

172
Fidelity in aquantum teleportation model
東京理科大学理工学部情報科学科宮寺隆之 (Takayuki Miyadera)
大矢雅則 (Masanori Ohya)

Science University of Tokyo
Priedrich-Schiller University, Karl-Heinz Fichtner

1Introduction

Amodel of Quantum Teleportation has been first given by Bennett etal.[l, 2], in which Alice perfectly sends an unlatown state to Bob using the
EPR entangled state. In their model, every state is perfectly teleported.
The key to make aperfect teleportation scheme is to use amaximallyentangled state (EPR state) over Alice and Bob. It is known, however,that preparation of such amaximally entangled states is difficult to realze. Therefore it is important to consider schemes with partialy (notmaximally) entangled states. As having been pointed out[3], with suchan incompletely entangled state one can not obtain aperfect teleortation scheme. In $[4, 5]$ , protocols employing apartially entangled state
constructed by beam splitting technique were introduced to provide theexamples for both perfect and nonperfect teleportation.The scheme introduced in $[4, 5]$ generalzed that of Bennett et al.. In the protocol in
nonperfect realstic teleportation, Alice and Bob make tests on their own
systems and give up the experiments if the tests are not passed. If thetests are fortunately passed, the obtained state by Bob is shown to be perfectly same with the original one first possessed by Alice. We calculated
the probability to complete successful teleportation, which approaches
unity as the mean energy of the entangled state goes to infinity even in
the nonperfect model.
We, in the present paper, do not employ the protocol with tests $[4, 5]$ but
original naive protocol given in [3] with beam splittings. For fixing thenotations, let us review what the naive scheme is (See [3, 12]).

Step 0: Agirl named Alice has an unknown quantum state on an Ndimensional subspace of aHilbert spaceand she was asked to$\rho$$\mathcal{L}$$\mathcal{H}_{1}$173
teleport it to aboy named Bob.
,
and
Step 1: For this purpose, we need two other Hilbert spaces
is attached to Bob. Prearrange a
is attached to Alice andhaving certain correlations
s0-called entangled state on
and prepare an ensemble of the combined system in the state$\mathcal{H}_{3}$$\mathcal{H}_{2}$$\mathcal{H}_{3}$$\mathcal{H}_{2}$$\sigma$$\mathcal{H}_{2}\otimes \mathcal{H}_{3}$$\rho\otimes\sigma$on$\mathcal{H}_{1}\otimes \mathcal{H}_{2}\otimes \mathcal{H}_{3}$.

Step 2: One then fixes family of mutually orthogonal projections(& fi2 corresponding to an observable $F:=$
on the Hilbert space. To complete the set of projections, we define another$(F_{nm})_{n,m=1}^{N}$$\mathcal{H}_{1}$$\sum_{n,m}z_{n,m}F_{nm}$projection $F_{0}:=1$ $- \sum_{nm}F_{nm}$ . Alice performs ameasurement ofpart of the system
the observable $F$ , involving only theand 0. When
. Possible outcomes arein the state, according to the von Neumann rule, after Alice’s
Alice obtains
measurement, the state becomes$\mathcal{H}_{1}\otimes \mathcal{H}2$$\{z_{nm}\}’ \mathrm{s}$$\rho\otimes\sigma$$z_{nm}$$\rho_{nm}^{(123)}:=\frac{(F_{nm}\otimes 1)\rho\otimes\sigma(F_{nm}\otimes 1)}{\mathrm{t}\mathrm{r}_{123}(F_{nm}\otimes 1)\rho\otimes\sigma(F_{nm}\otimes 1)}$on the Hilbert spaceis the full
where
the other hand, when Alice obtains 0, the state becomes$\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}$$\mathrm{t}\mathrm{r}_{123}$$\rho_{0}^{(123)}$$\mathcal{H}_{1}\otimes \mathcal{H}_{2}\otimes \mathcal{H}_{3}$$.= \frac{(F_{0}\otimes 1)\rho\otimes\sigma(F_{0}\otimes 1)}{\mathrm{t}\mathrm{r}_{123}(F_{0}\otimes 1)\rho\otimes\sigma(F_{0}\otimes 1)}$

. On

.

Step 3: Bob is informed which outcome was obtained by Alice. This isorequivalent to transmit the information that the eigenvalue
0was detected. This information is transmitted from Alice to Bob
without disturbance and by means of classical tools.
$z_{nm}$Step 4: Having been informed an outcome of Alice’s measurement, Bob
performs acorresponding unitary operation onto his system. That
, Bob operates aunitary operator
is, if the outcome wasand change the state into$z_{nm}$$W_{nm}$.

$(1 \otimes 1 \otimes W_{nm})\rho_{nm}^{(123)}(1 \otimes 1 \otimes W_{nm}^{*})=\frac{(F_{nm}\otimes W_{nm})\rho\otimes\sigma(F_{nm}\otimes W_{nm}^{*})}{\mathrm{t}\mathrm{r}_{123}(F_{nm}\otimes 1)\rho\otimes\sigma(F_{nm}\otimes 1})$

174
If the outcome was 0, Bob operates aunitary operator
state becomes$W_{0}$and the.

$(1 \otimes 1 \otimes W_{0})\rho_{0}^{(123)}(1 \otimes 1 \otimes W_{0})=\frac{(F_{0}\otimes W_{0})\rho\otimes\sigma(F_{0}\otimes W_{0}^{*})}{\mathrm{t}\mathrm{r}_{123}(F_{0}\otimes 1)\rho\otimes\sigma(F_{0}\otimes 1)}$

Step 5: Making only partial measurements on the third part on the system means that Bob
control astate given by the partialon. Thus the state obtained by Bob is$\mathrm{w}\mathrm{i}\mathbb{I}$$\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}$$\mathcal{H}_{1}\otimes \mathcal{H}_{2}$$\Gamma_{nm}^{*}(\rho)$$=\mathrm{t}\mathrm{r}_{12}(1 \otimes 1 @ W_{nm})\rho_{nm}^{(123)}(1 \otimes 1 \otimes W_{nm}^{*})$$= \mathrm{t}\mathrm{r}_{12}\frac{(F_{nm}\otimes W_{nm})\rho\otimes\sigma(F_{nm}\otimes W_{nm}^{*})}{\mathrm{t}\mathrm{r}_{123}(F_{nm}\otimes 1)\rho\otimes\sigma(F_{nm}\otimes 1)}$in case when the outcome is$\Gamma_{0}^{*}(\rho)$$z_{nm}\mathrm{a}\mathrm{n}\mathrm{d}$$=\mathrm{t}\mathrm{r}_{12}(1 \otimes 1 \otimes W_{0})\rho_{0}^{(123)}(1 \otimes 1 \otimes W_{0}^{*})$.

$= \mathrm{t}\mathrm{r}_{12}\frac{(F_{0}\otimes W_{0})\rho\otimes\sigma(F_{0}\otimes W_{0}^{*})}{\mathrm{t}\mathrm{r}_{123}(F_{0}\otimes 1)\rho\otimes\sigma(F_{0}\otimes 1)}$

if the outcome was 0. Thus the whole teleportation scheme givenby the family $(F_{nm})$ and the entangled state can be characterized
by the family
andof channels from the set of states on
into the set of states on
and the family
given
and
by
$\sigma$$\Gamma_{nm}$$\Gamma_{0}$$\mathcal{H}_{1}$$\{p_{nm}(\rho)\}$$\mathcal{H}_{3}$$p_{0}(\rho)$$p_{nm}(\rho):=\mathrm{t}\mathrm{r}_{123}(F_{nm}\otimes 1)\rho\otimes\sigma(F_{nm}\otimes 1)$$p\mathrm{o}(\rho):=\mathrm{t}\mathrm{r}_{123}(F_{0}\otimes 1)\rho\otimes\sigma(F_{0}\otimes 1)$of the probabilities that Alice’s measurement according to the observable $F$ will show the valueand 0.
$z_{nm}$Once knowing the result of Alice’s measurement, the channel becomes
nonlinear because of the probabilties
andwhich appear in
the denominator. We, however, do not know the result of Alice’s meaeurement before the experiment. Therefore it is also important to consider an$p_{nm}(\rho)$$p_{0}(\rho)$

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expected state which is obtained by mixing all possible states with multiplying their probabilities to occur. That is, the teleportation scheme canbe written by alinear channel (completely positive maP)$—*( \rho)=\sum_{nm}---*nm(\rho)+---*\mathrm{o}(\rho)$

(1)

,

where
$–nm-*(\rho)$$–0-*(\rho)$$:=p_{nm}(\rho)\Gamma_{nm}^{*}(\rho)=\mathrm{t}\mathrm{r}_{1,2}(F_{nm}\otimes W_{nm})\rho(F_{nm}\otimes W_{nm})^{*}$$:=p_{0}(\rho)\Gamma_{0}^{*}(\rho)=\mathrm{t}\mathrm{r}_{1,2}(F_{0}\otimes W_{0})\rho(F\circ\otimes W_{0})^{*}$

to the original state .
We investigate how close the obtained state
In the next section, we review some mathematical notions which are used
to construct rigorously ateleportation scheme by beam splittings. In
section 3we introduce anaive teleportation scheme and in section 4we
discuss how perfect the protocol is by use of aquantity, fidelity.
$—*(\rho)$$\rho$

2Basic Notions and NotationsFirst we collect some basic facts concerning the (symmetric) Fock space.
We will introduce the Fock space in away adapted to the language ofcounting measures. For details we refer to [6, 7, 8, 9, 10] and other papers
cited in [8].

Let be an arbitrary complete separable metric space. Further, letbe alocally finite diffuse measure on , i.e. $\mu(B)<+\infty$ for boundedordermeasurable subsets of and $\mu(\{x\})=0$ for all singletons $x\in G$ .
to describe the teleportation of states on afinite dimensional Hilbert space, especially we are concerned withthrough the dimensional space
the case
$G$

$\mu$

$G$

$\ln$

$G$

$k$

$\mathrm{R}^{k}$

$G$

$=$

$\mu=$$\mathrm{R}^{k}\cross\{1, \ldots, N\}$$l\cross\#$

where 1is the -dimensional Lebesgue measure anding measure on {1,N}.
$k$\ldots ,$\#$denotes the count-

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Now by $M=M(G)$ we denote the set of

on

$G$

finite counting measures. Since $\varphi\in M$ can be written in the form
for some$\mathrm{a}\mathbb{I}$$\varphi=\sum_{j=1}^{n}\delta_{x_{\mathrm{j}}}$$n=0,1,2$ ,and $Xj\in G$ (where
denotes the Dirac measures corresponding to $\in G$ ) the elements of $M$ can be interpreted as finite(symmetric) point configurations in $G$ . We equip $M$ with its canonical(cf. [6], [7]) and we consider the measure $F$ by setting$\delta_{x}$$\ldots$$x$$\sigma-\mathrm{a}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}\mathrm{r}\mathrm{a}\mathfrak{M}$$F( \mathrm{Y}):=\mathcal{X}_{\mathrm{Y}}(O)+\sum_{n\geq 1}\frac{1}{n!}\int_{G^{*}}.\mathcal{X}_{\mathrm{Y}}(\sum_{j=1}^{n}\delta_{x_{\mathrm{j}}})\mu^{n}(d[x_{1}, \ldots, x_{n}])(\mathrm{Y}\in \mathfrak{B})$

Hereby,denotes the indicator function of a set and represents
the empty configuration, . , $O(G)=0$ . Observe that $F$ is aa-finite$O$$\mathrm{Y}$$\mathcal{X}_{\mathrm{Y}}$$\mathrm{i}$$\mathrm{e}.$measure.
Since was assumed to be diffuse one easily checks that $F$ is concentrated on the set of asimple configurations (i.e., without multiple points)$\mu$$\hat{M}:=${$\varphi\in M|\varphi(\{x\})\leq 1$

for ffi $x\in G$ }$=\mathcal{M}(G):=L^{2}(M, W, F)$ is called the (symDEFINITION 2.1
metric) Fock space over .
$\mathcal{M}$$G$

In [6] it was proved thatand the Boson Fock space
usual definition are isomorphic.
For eachwith
we denote bynormalzed vector$\mathcal{M}$$\Phi$$\in \mathcal{M}$$\Phi$$\neq 0$$\Gamma(L^{2}(G))$$|\Phi>\mathrm{t}\mathrm{h}\mathrm{e}$in thecorresponding$| \Phi>:=\frac{\Phi}{||\Phi||}$

Further, $|\Phi><\Phi|$ denotes the corresponding one–dimensional projection, describing the pure state given by the normalzed vector $|\Phi>$ . Now,
for each $n\geq 1$ let
be the -fold tensor product of the Hilbert space. Obviously,
can be identified with $L^{2}(M^{n}, F^{n})$ .
$\mathcal{M}^{Qn}$$\mathcal{M}$$n$$\mathcal{M}^{\Phi n}$

DEFINITION 2.2 For a given function g : G$arrow \mathbb{C}$the function

177$\exp(g)$:$Marrow \mathbb{C}$defined by$\exp(g)(\varphi):=\{\begin{array}{l}1\mathrm{i}\mathrm{f} \varphi=0\prod_{x\in G,\varphi(\{x\})>0}g(x)otherwise\end{array}$is called exponential vector generated by .
$g$

Observe that $\exp(g)\in \mathcal{M}$ if and only if $g\in L^{2}(G)$ and one has in thiscaseand $|\exp(g)>=e^{-\frac{1}{2}||g||^{2}}\exp(g)$ . The projection$|\exp(g)><\exp(g)|$ is called the coherent state corresponding to $g\in$$L^{2}(G)$ . In the special case
we get the vacuum state$||\exp(g)||^{2}=e^{||g||^{2}}$$g\equiv 0$$|\exp(0)>=\mathcal{X}_{\{0\}}$.

, so thatis dense in
The linear span of the exponential vectors ofbounded operators and certain unbounded operators can be characterized
by their actions on exponential vectors.
$\mathcal{M}$$\mathcal{M}$be a linear operator on $L^{2}(G)$ with $||T||\leq 1$ .
called second quantization of is the (uniquely
Then the operator
fulfillingdetermined) bounded operator on

DEFINITION 2.3 Let

$T$

$T$

$\Gamma(T)$$\mathcal{M}$$\Gamma(T)\exp(g)=\exp(Tg)$$(g\in L^{2}(G))$

Clearly, it holds

It follows that
operator on $L^{2}(G)$ .

$\Gamma(T)$

LEMMA 2.4 Let$\Gamma(T_{1})\Gamma(T_{2})$$=$$\Gamma(T_{1}T_{2})$$\Gamma(T^{*})$$=$$\Gamma(T^{*})$

is an unitary operator on$K_{1}$,$K_{2}$(2)$\mathcal{M}$

if

$T$

is an unitary

be linear operators on $L^{2}(G)$ with property$K_{1}^{*}K_{1}+K_{2}^{*}K_{2}=1$

Then there exists exactly one isometrywith$\nu_{K_{1},K_{2}}$

.

from

$\nu_{K_{1},K_{2}}\exp(g)=\exp(K_{1}g)\otimes\exp(K_{2}g)$(3)$\mathcal{M}$to$\mathcal{M}^{\otimes 2}=\mathcal{M}\otimes \mathcal{M}$$(g\in L^{2}(G))$

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178
Further it holds(5)$\nu_{K_{1},K_{2}}=(\Gamma(K_{1})\otimes\Gamma(K_{2}))D$(at least on
The adjoint$\nu_{K_{1},K_{2}}^{*}$

but one has the unique extension).
is characterized byof$\mathrm{d}\mathrm{o}\mathrm{m}(D)$$\nu_{K_{1},K_{2}}^{*}$$\nu_{K_{1},K_{2}}$( $\exp$ (h)&exp (g)) $=\exp(K_{1}^{*}h+K_{2}^{*}g)$

REMARK 2.5 }$vm$$f$$K_{1}$, usingas a certain splitting (cf.

$\mathcal{M}$$arrow \mathcal{M}$$\otimes \mathcal{M}$,$K_{2}$$\nu_{K_{1},K_{2}}$(g, h $\in L^{2}(G))$ .

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we get a transition expectation:and the lifting
may be interpreted$\xi_{K_{1}K_{2}}$$\xi_{K_{1}K_{2}}^{*}$).

$[\mathit{9}J$

Here we explain fundamental scheme of beam splitting [8]. We define an
isometric operator
for coherent vectors such that
$V_{\alpha,\beta}$$V_{\alpha,\beta}|\exp(g)\rangle=|\exp(\alpha g)\rangle\otimes|\exp(\beta g)\rangle$

with $|\alpha|^{2}+|\beta|^{2}=1$ . This beam splitting is amathematicalexpression for optical communication and quantum measurements [9].
$\mathrm{u}\mathrm{s}\mathrm{e}\mathrm{f}\mathrm{i}4$

EXAMPLE 2.6 ( $\alpha=\beta=1/\sqrt{2}$ above) Let $K_{1}=K_{2}$ be the following
operator of multiplication on $L^{2}(G)$$K_{1}g= \frac{1}{\sqrt{2}}g=K_{2}g$$(g\in L^{2}(G))$

We put$\nu$$:=\nu_{K_{1},K_{2}}$and obtain$\nu$$\exp(g)$ $=\exp$

EXAMPLE 2.7 Let
subspaces.
and$( \frac{1}{\sqrt{2}}g)\otimes\exp(\frac{1}{\sqrt{2}}g)$$(g \in L^{2}(G))$

be the orthogonal sum of thedenote the corresponding projections.

$L^{2}(G)=\mathcal{H}_{1}\oplus \mathcal{H}_{2}$$\mathcal{H}_{1},\mathcal{H}_{2}$$K_{1}$$K_{2}$

We
use Example 2.6 in order to describe ateleportation model
where Bob performs his experiments on the same ensemble of the system$\mathrm{w}\mathrm{i}\mathbb{I}$179
like Alice.

Further we will use aspecial case of Example 2.7 in order to describe
ateleportation model where Bob and Alice are spatially separated (cf.
section 5).
REMARK 2.8 The property (3) implies$||K_{1}g||^{2}+||K_{2}g||^{2}=||g||^{2}$(g $\in L^{2}(G))$(7)

V be unitary operators on $L^{2}(G)$ . If operators
REMARK 2.9 Let
satisfy (3), then the pair $\hat{K}_{1}=UK_{1},\hat{K}_{2}=VK_{2}$ fulfill (3).
,$U_{f}$$K_{1}$$K_{2}$

3Anaive teleportation scheme
In this section we define anaive version of the teleportation scheme bybeam splitting $[4, 5]$ . We fix an ONS $\{g_{1}, \ldots,g_{N}\}\subseteq L^{2}(G)$ , operators,
on $L^{2}(G)$ with (3), an unitary operator $T$ on $L^{2}(G)$ , and $d>0$ .
$K_{1}$$K_{2}$

We assume$(k=1, \ldots, N)$ ,(8)$(k\neq j;k,j=1\ldots, N)$ ,(9)$TK_{1}g_{k}=K_{2}g_{k}$$\langle K_{1}g_{k}, K_{1}g_{j}\rangle=0$

Using (7) and (8) we get$||K_{1}g_{k}||^{2}=||K_{2}g_{k}||^{2}= \frac{1}{2}$

.

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Prom (8) and (9) we get$\langle K_{2}g_{k}, K_{2}g_{j}\rangle=0$$(k\neq j;k,j=1, \ldots, N)$ .

(11)

The state of Alice asked to teleport is of the tyPe$\rho=\sum_{s=1}^{N}\lambda_{s}|\Phi_{s}\rangle\langle\Phi_{s}|$

,

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$| \Phi_{s}\rangle=\sum_{j=1}^{N}c_{sj}|\exp(aK_{1}g_{j})-\exp(0)\rangle$$( \sum_{j}|c_{sj}|^{2}=1$; s $=1$ ,\ldots ,N)(13)

and $|a|^{2}=d$ . One easily chedcs that $(|\exp(aK_{1}g_{j})-\exp(0)\rangle)_{j=1}^{N}$ and$(|\exp aK_{2}g_{\dot{J}})-\exp(0)\rangle)_{j=1}^{N}$ arein. That is, the state of Aliceasked to teleport lives in an -dimensional subspace of the Fock spacespanned by the ONS.
$\mathrm{O}\mathrm{N}\mathrm{S}$$\mathcal{M}$$\mathrm{N}$In order to achieve that$(|\Phi_{s}\rangle)_{-1}^{N}$$\sum_{j=1}^{N}\overline{c}_{sj}c_{kj}=0$

Denote$c_{s}=[c_{s1},\ldots,c_{sN}]\in \mathbb{C}^{N}$

Now let$(b_{n})_{n=1}^{N}$is still an ONS in$\mathcal{M}$$(j\neq k;j, k=1, \ldots, N)$, thenbe asequence in$(c_{s})_{s=1}^{N}$$\mathbb{C}^{N}$

we assume.

is an CONS in(14)$\mathbb{C}^{N}$.

,$b_{n}=[b_{n1},\ldots,b_{nN}]$

with properties$|b_{nk}|=1$$\langle b_{n}, b_{j}\rangle=0$$(n, k=1, \ldots, N)$ ,(n$\neq j;$(15)

n,j $=1,$\ldots ,N).

(16)

Then Alice’s measurements are performed with projection$F_{nm}=|\xi_{nm}\rangle\langle\xi_{nm}|$(n,m $=1,$\ldots ,N)(17)given by$| \xi_{nm}\rangle=\frac{1}{\sqrt{N}}\sum_{j=1}^{N}b_{nj}|\exp(aK_{1}g_{j})-\exp(0)\rangle\otimes|\exp(aK_{1}g_{j\oplus m})-\exp(0)\rangle$

(18)

where
. One easily checks thatis an$(n, m =1, 2, \cdots N)$ doaae not form acomin. Because
pletley orthonormal system of, we introduce another projection$j\oplus m:=j+m(\mathrm{m}\mathrm{o}\mathrm{d} N)$$\mathrm{O}\mathrm{N}\mathrm{S}$$\mathcal{M}^{\otimes 2}$$(|\xi_{nm}\rangle)_{n,m=1}^{N}$$|\xi_{nm}\rangle$$\mathcal{M}$$\otimes \mathcal{M}$,181
operator $F_{0}:=1$ $- \sum_{nm}F_{nm}$ . Thus the measurement of the observable$F$ distingishes
corresponds to the case an outand , where$F_{0}$$F_{0}$$\{F_{n}m\}’ \mathrm{s}$some 1s zero.
Further, the state vector$|\xi\rangle$of the entangled state$\sigma=|\xi\rangle$$| \xi\rangle=\frac{\gamma}{\sqrt{N}}\sum_{k}|\exp(aK_{1}g_{k})\rangle\otimes|\exp(aK_{2}g_{k})\rangle$$\langle$$\xi|$

is given by

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,

which is naturally prepared by use of Beam splitting technique, whereis determined by anormalization condition. However, the physical naturalness requires asacrifice. That is, the state is not maximally entangledstate any longer.
,, $N$ we have
As for unitary operation of Bob, for each , $m=1$ ,given byon$\gamma$$U_{m}$$\cdots$$n$$B_{n}$$\mathcal{M}$$B_{n}|\exp(aK_{1}g_{j})-\exp(0)\rangle$$=$$b_{nj}|\exp(aK_{1}g_{j})-\exp(0)\rangle$$B_{n}|\exp(0)\rangle$$=$$|\exp(0)\rangle$$U_{m}|\exp(aK_{1}g_{j})-\exp(0)\rangle$$=$$|\exp(aK_{1}g_{j\oplus m})-\exp(0)\rangle$$U_{m}|\exp(0)\rangle$$=$$|\exp(0)\rangle$

where$j\oplus m:=j+m(\mathrm{m}\mathrm{o}\mathrm{d} N)$$(j=1, \ldots, N)$(20)$(j=1, \ldots, N)$(21)

and define$W_{nm}:=B_{n}U_{m}^{*}\Gamma(T)$

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’

In addition we have some arbitrary unitary operator
not specify yet.

$W_{0}$, which we do

4Fidelity
We need some proper quantity (for e.g., [12]) to measure how close twostates are. In this paper we take up fidelity $[13, 14]$ . The notion of fidelity
is frequently used in the context of quantum information, quantum optics
and so on. The fidelity of astate with respect to another state isdefined by$\sigma$$\rho$$F(\rho, \sigma):=\mathrm{t}\mathrm{r}[\sqrt{\sigma^{1/2}\rho\sigma^{1/2}}]$

,

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182
which possesses some nice properties.
$0\leq F(\rho, \sigma)$$F(\rho, \sigma)=1$$F(\rho, \sigma)$$\leq$$\Leftrightarrow$

$=$

1

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$\rho=\sigma$(25)$F(\sigma, \rho)$(26)

Thus we can say two states and are close when the fidelity between
them is close to unity. Moreover it satisfies akind of concavity relationas$\sigma$$\rho$$F( \sum_{\dot{1}}p:\rho:, \sum_{\dot{1}}q\dot{.}\sigma:)\geq\sum.\cdot\sqrt{p_{\dot{1}}q_{\dot{1}}}F(_{\beta:}, \sigma_{\dot{1}})$

,

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where :’s and ’s are states andand ’ are nonnegative numbers
satisfying $\sum_{:}p:=\sum_{:}q\dot{.}=1$ . In particular putting $p_{j}=1$ , one gets$\rho$$p_{\dot{1}}’ \mathrm{s}$$\sigma.\cdot$$q.\cdot$$\mathrm{s}$(28)$F( \rho, \sum\dot{.}q.\cdot\sigma)\geq\sqrt{q_{j}}F(\rho, \sigma_{j})$for $j=1$ , 2,.
To estimate $F(\rho,---*(\rho))$ we begin with acalculation of $—*( \rho)=\sum_{nm^{-}nm}^{-*}-(\beta)+$$\cdots$$–0-*(\rho)$.

LEMMA 4.1 141 For each $n,m$ ,$(F_{nm}\otimes 1)$$(|\Phi_{s}\rangle\otimes|\tilde{\xi}\rangle)$

$=$

$s$

$(=1, \ldots, N)$, it holds$\frac{\gamma}{N}(1-e^{-\frac{d}{2}})|\xi_{nm}\rangle\otimes(\Gamma(T)U_{m}B_{n}^{*}|\Phi_{s}\rangle)$$+ \frac{\gamma}{N}(\frac{e^{\frac{d}{2}}-1}{e^{d}})^{\frac{1}{2}}\langle b_{n},c_{s}\rangle_{\mathrm{C}^{N}}\xi_{nm}\otimes|\exp(0)\rangle$$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}$$\alpha_{k_{\dot{\theta}},r}$For ffi $k,j$ , $r=1$ ,$:=$$\ldots$, $N$, we get$\langle|\exp(aK_{1}g_{r})-\exp(0)\rangle\otimes||\exp(aK_{1}g_{\Phi m},)-\exp(0)\rangle$,$|\exp(aK_{1}g_{j})-\exp(0)\rangle\otimes|\exp(aK_{1}g_{k})\rangle\rangle$$=$$\{.0(\frac{e^{a}\tau^{2}-1}{e^{a}\tau^{2}})$$\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{i}\mathrm{f}r=j\mathrm{a}\mathrm{n}\mathrm{d}k=r\oplus m$and$|\exp(aK_{2}g_{j\oplus m})\rangle=e^{-\frac{a^{2}}{2}}(e^{\frac{a^{2}}{2}}-1)^{\frac{1}{2}}|\exp(aK_{2}g_{j\oplus m})-\exp(0)\rangle+e^{-\frac{a^{2}}{2}}|\exp(0$183
On the other hand, we have$(F_{nm} \otimes 1)(|\Phi_{s}\rangle\otimes|\tilde{\xi}\rangle)=\frac{\gamma}{N}\sum_{k}\sum_{j}\sum_{r}c_{sj}\overline{b}_{nr}\alpha_{k,j,\prime}\xi_{nm}\otimes|\exp(aK_{2}g_{k})\rangle$

It follows with $a^{2}=d$$(F_{nm}\otimes 1)$$(\Phi_{s}\otimes\tilde{\xi})$$=$$\frac{\gamma}{N}(e^{\frac{d}{2}}-1)e^{-\frac{d}{2}}\xi_{nm}$$\otimes(\sum_{j}c_{sj}\overline{b}_{nj}|\exp(aK_{2}g_{j\oplus m})-\exp(0)\rangle)$$+ \frac{\gamma}{N}(e^{\frac{d}{2}}-1)^{\frac{1}{2}}e^{-\frac{d}{2}}\sum_{j}c_{sj}\overline{b}_{nj}\xi_{nm}\otimes|\exp(0)\rangle$$=$$\frac{\gamma}{N}$$(1-e^{-\frac{d}{2}}$)$\xi_{nm}\otimes(\Gamma(T)U_{m}B_{n}^{*}\Phi_{s})$$+ \frac{\gamma}{N}(\frac{e^{\frac{d}{2}}-1}{e^{d}})\frac{1}{2}\langle b_{n}, c_{s}\rangle_{\mathbb{C}^{N}}\xi_{nm}\otimes|\exp(0)\rangle$.

$\blacksquare$

The following lemma follows immediattiely.

LEMMA 4.2 For a general state$—*nm(\rho)$$\rho=\sum_{s}\lambda_{s}|\Phi_{s}\rangle\langle$ $\Phi_{s}|$, it holds$=$$\frac{\gamma^{2}}{N^{2}}(1 -e^{-d/2})^{2}\rho+\frac{\gamma^{2}}{N^{2}}(\frac{e^{d/2}-1}{e^{d}})\sum_{s}\lambda_{s}|\langle b_{n}, c_{s}\rangle|^{2}|\exp(0)\rangle\langle\exp(0)|$$+$$\frac{\gamma^{2}}{N^{2}}(\frac{e^{d/2}-1}{e^{d}})^{1/2}(1 -e^{-d/2})(\sum_{s}\langle b_{n}, c_{s}\rangle^{*}\lambda_{s}|\Phi_{s}\rangle\langle\exp(0)|$$+$$\sum_{s}\langle b_{n}, c_{s}\rangle\lambda_{s}|\exp(0)\rangle\langle\Phi_{s}|$Next we investigate an expression of

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).
$–*-\mathrm{o}(\rho)$.

LEMMA 4.3 It holds$(F_{0}\otimes 1)$$(|\Phi_{s}\rangle\otimes|\tilde{\xi}\rangle)$$=$$| \Phi_{s}\rangle\otimes|\exp(0)\rangle\otimes \mathrm{e}^{-|a|^{2}/4}\frac{\gamma}{\sqrt{N}}\sum_{k=1}^{N}|\exp(aK_{2}g_{k})\rangle$

asubsapce spanned by an(0) $(k=1, \cdots, N)$ , $\sum_{nm}F_{n}m$ is aprojection onto
we obtain$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}$If we let$\mathcal{L}$$\mathrm{O}\mathrm{N}\mathrm{S}\{|\exp(aK_{1}g_{k})-$$\mathcal{L}\otimes \mathcal{L}.\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}$$)$$(F_{0}\otimes 1)(\Phi_{s}\otimes\tilde{\xi})=(1 \otimes|\exp(0)\rangle\langle\exp)(0)|\otimes 1)(\Phi_{s}\otimes\tilde{\xi})$

.

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184
Here we used afact that$|\exp(0)\rangle$

is orthogonal to$\{|\exp(aK_{1}g_{k})-\exp(0)\rangle\}’ \mathrm{s}$.

$\blacksquare$

LEMMA 4.4 For a general state$\rho=\sum_{s}\lambda_{s}|\Phi_{s}\rangle\langle$$\Phi_{s}|$, it holds(31)$—* \mathrm{o}(\rho)=e^{-|a|^{2}/2}\frac{\gamma^{2}}{N}\sum_{k=1}^{N}\sum_{l=1}^{N}W_{0}|\exp(aK_{2}g_{k})\rangle\langle\exp(aK_{2}g_{l})|W_{0}^{*}$

Now let us estimate the fidelity between
compute$\rho^{1/2-*}--(\rho)\rho^{1/2}$$=$$—*(\rho)$and . We must first$\rho$$\sum_{nm}\rho^{1/2}---*nm(\rho)\rho^{1/2}+\rho^{1/2-}--0\rho^{1/2}$$=\gamma^{2}(1-e^{-d/2})^{2}\rho^{1/2}\rho\rho^{1/2}$$+ \rho^{1/2}e^{-|a|^{2}/2}\frac{\gamma^{2}}{N}\sum_{k=1}^{N}\sum_{l=1}^{N}W_{0}|\exp(aK_{2}g_{k})\rangle\langle\exp(aK_{2}g_{l})|W_{0}^{*}\rho^{1/2}\mathrm{J}$where we used the relation. Becauseis positive
operator,
becomes astate and we can rearrange the expression of fidelity as$\langle\exp(0)|\Phi_{s}\rangle=0$$—*\mathrm{o}(\rho)$$–0-*(\rho)/\mathrm{t}\mathrm{r}[_{-0}^{-*}-(\rho)]$$F(\rho, ---*(\rho))=F(\rho,\gamma^{2}(1-e^{-d/2})^{2}\rho+\mathrm{t}\mathrm{r}[_{-0}^{-*}-(\rho)]_{-0}^{-*}-(\rho)/\mathrm{t}\mathrm{r}[_{-0}^{-*}-(\rho)])$

.

(32)

Thanks to the concavity property (28) of fidelity, we obtain$F(\rho, ---*(\rho))\geq\gamma(1-e^{-d/2})F(\rho,\rho)=\gamma(1-e^{-d/2})$

.

(33)

Thus we obtain the folowing theorem.

THEOREM 4.5 For any imput state , it holds$\rho$$F(\rho,---*(\rho))\geq\sqrt{\frac{(1-e^{-d/2})^{2}}{1+(N-1)e^{-d}}}$

.

(34)

Therefore the naive teleportation protocol approaches perfect one as theparameter
goes to infinity.
With some additional condition, one can strengthen the above estimate
to an equalty.
$|a|$

185
PROPOSITION 4.6 Let
the subspaces
and $W_{0}=1$ .

$\mathcal{H}_{1}$,$\mathcal{H}_{2}$.

and$K_{1}$

be the orthogonal sum of
denote the corresponding projections$L^{2}(G)=\mathcal{H}_{1}\oplus \mathcal{H}_{2}$$K_{2}$(35)$F(\rho,---*(\rho))=$holds for any imput state .
$\rho$$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}$Because,$\langle\exp(aK_{1}g_{k})-\exp(0)|\exp(aK_{2}g_{l})\rangle=0\mathrm{h}\mathrm{o}\mathrm{l}\grave{\mathrm{d}}\mathrm{s}$$\gamma^{2}(1-e^{-d/2})^{2}\rho^{2}\mathrm{f}\mathrm{o}\mathbb{I}\mathrm{o}\mathrm{w}\mathrm{s}$.

$\rho^{1/2}---*(\rho)\rho^{1/2}=$$\blacksquare$

We have cosidered the fidelity of naive teleportation scheme with beam
goes to infinity the fisplittings. We showed that as the parameter
delity approaches unity and the naive teleporation scheme also approaches
aperfect scheme as the teleportation scheme with tests does. In fact the
fidelity can be bounded from below by square route of probability tocomplete successful teleportation with tests.
$|a|$

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