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−⊗ | i k −⊗ | ik | − i∼| − i
Since Alice, and the third qubit to Bob. Note that Alice’s
second qubit and Bob’s third qubit are from the
I Mϕ ϕ =I Q ϕ ϕ I Q ϕ ϕ
+ + + +
⊗ | −i ⊗ | −i− ⊗ −| −i entangledBellstate.Second,Aliceappliesacontrol-
= −|ϕ +ϕ −i, NOT gate to her qubits in |ϕ and obtains
0
i
I ⊗M |ϕ −ϕ + i= |ϕ −ϕ + i, |ϕ 1 i= √1 2[α 0 |0 i( |00 i+ |11 i)+α 1 |1 i( |10 i+ |01 i)].
that is, the post-measurement state ϕ ϕ (or
+ Third,sheappliesaHadamardgatetothefirstqubit
ϕ ϕ ) is the eigenvector of I M co| rrespo− ni ding
+ in ϕ and gets
t| o− eigeni value 1 (or +1), perfor⊗ ming measurement | 1 i
on I ⊗M in t− he post-measurement state must al- |ϕ 2 i= 1 2[α 0( |0 i+ |1 i)( |00 i+ |11 i)
ways yield measurement result 1 (or +1). Thus,
+α (0 1 )(10 + 01 )].
− 1
the measurement results of M on the two qubits of | i−| i | i | i
We regroup the terms of ϕ and rewrite it as fol-
ψ are always opposite to each other. 2
| i
| i lows:
3.3.2 Quantum teleportation Quantum teleporta-
tion is a process by which we can transfer the state |ϕ 2 i= 21[ |00 i(α 0 |0 i+α 1 |1 i)+ |01 i(α 0 |1 i+α 1 |0 i)
of a qubit from one location to another, without
+ 10 (α 0 α 1 )
0 1
transmitting it through the intervening space. We | i | i− | i
+ 11 (α 1 α 0 )].
illustrate the phenomenon as follows. Alice and Bob 0 1
| i | i− | i
together generated a Bell state long ago. Each took The new expression has four terms, and each term
onequbitoftheBellstatewhentheysplit.Nowthey has Alice’s qubits in one of four possible states 00 ,
arefarawayfromeachother.ThemissionforAliceis 01 , 10 and 11 , and Bob’s qubit is in the s| tati e
to deliver a qubit ψ to Bob,whileheis hiding,and | relai te| d i to the| orii ginal state ψ . If Alice performs
she can only send| ci lassical information to Bob but a measurement on her qubit| s ai nd informs Bob of
does not know the state of the qubit ψ . Quantum the measurement result, then his post-measurement
| i
teleportation is a way that Alice utilizes the entan- state is completely determined. For example, the
gled Bell state to send a qubit of unknown state to first term has Alice’s qubits in the state 00 and
Bob, with only a small overhead of classical com- | i
Bob’s qubit in state ψ . Therefore, if Alice’s mea-
munication. Recently a breakthrough in quantum | i
surementresultonherqubitsis00,thenBob’squbit
teleportation has been made by successfully trans-
willbeinstate ψ .BelowisalistofBob’sfourpost-
ferringcomplex quantumdata instantaneously from | i
measurement states corresponding to the results of
one place to another, paving the way for real-world
Alice’s measurements:
applications of quantum communications (Lee et al.
00 α 0 +α 1 , 01 α 1 +α 0 ,
(2011)). 0 1 0 1
→ | i | i → | i | i
Here is how it works. Alice interacts the qubit ψ 10 α 0 α 1 , 11 α 1 α 0 .
to be teleported with her half of the Bell state, a| ndi → 0 | i− 1 | i → 0 | i− 1 | i
As Alice’s measurement outcome on her qubits is
then performs a measurement on the two interacted
one of 00,01,10 and 11, depending on her measure-
qubits to obtain one of four possible two-classical-
ment outcome Bob’s qubit will be one of the above
bit results: 00,01,10 and 11. She sends the two-bit
four possible states. Once Alice sends to Bob her
informationviaclassicalcommunicationtoBob.De-
two-classical-bit measurement outcome through
pending on Alice’s classical message, Bob performs
a classical channel, he applies appropriate quantum
one of four operations on his half of the Bell state.
gates to his state and recovers ψ . For example, if
Surprisingly, the described procedure allows Bob to
| i
her measurement is 00, Bob’s state is ψ , and he
recover the original state ψ .
| i
| i does not need to apply any quantum gate. If her
Specifically assumethatthestate tobeteleported
measurement is 01, then Bob needs to apply a σ
is ψ =α 0 +α 1 ,whereα andα areunknown x
0 1 0 1
| i | i | i gate to his state α 1 +α 0 and yields ψ . If her
amplitudes. First, consider a three-qubit state 0 1
| i | i | i
measurement is 10, then applying a σ gate to his
z
00 + 11
|ϕ 0 i= |ψ i| i √2| i s mta et ne tα is0 | 10 1i ,− tα h1 e| n1 i BB oo bb cr ae nco five xrs up|ψ hi. isIf sh taer tem αeas 1ure-
0
| i−
1 α 0 to recover ψ by applying first a σ gate and
1 x
= [α 0 (00 + 11 )+α 1 (00 + 11 )], | i | i
√2 0 | i | i | i 1 | i | i | i then a σ z gate. Here the σ x and σ z gates are de-
10
Y.WANG
fined by Pauli matrices σ and σ given by (7). Suppose X , i = 1,2,3,4, are four random vari-
x z i
In summary, according to Alice’s measurement out- ables taking values 1. Consider an ordinary exper-
±