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Insteadofthesingle-qubitstategeneration,theability
Bob touches some traps.
of single-qubit measurements is also enough for Alice: it
Recently,anew verificationprotocolthatdoesnotuse
wasshowninRef.[15]thatAlicewhocandoonlysingle-
the traptechnique was proposed[22] (see alsoRef. [23]).
qubitmeasurementscanperformblindquantumcomput-
In this protocol, Bob generates a graph state, and sends
ing. The idea is that Bob generates a graph state and
each qubit of it one by one to Alice. Alice directly veri-
sendseachqubitonebyonetoAlice. IfBobishonest,he
fies the correctness of the graph state (and therefore the
generatesthecorrectgraphstateandthereforeAlice can
correctness of the computation) sent from Bob by mea-
perform the correct measurement-based quantum com-
suringstabilizer operators. This verificationtechnique is
puting (correctness). If Bob is malicious, he might send
called the stabilizer test. Note that the stabilizer test is
a wrongstate to Alice, but whateverBobsends to Alice,
useful also in quantum interactive proof system [24–28].
Alice’s measurement angles, which contain information
aboutAlice’scomputation,cannotbetransmittedtoBob Althoughcomputingitselfisverifiablethroughthesta-
due to the no-signalingprinciple (blindness). The proto- bilizertest,theinputisnotifitisaquantumstatewhose
col is called the measurement-only protocol, since Alice classical efficient description is not known to Alice. For
needs only measurements. example, let us assume that Alice receives a state ψ
| i
A problem in all these protocols is the lack of the ver- fromCharlie,andshewantstoapplyaunitaryU on ψ .
| i
If she delegates the quantum computation to Bob in the
measurement-only style, a possible procedure is as fol-
lows. Alicefirstsends ψ toBob. Bobnextentangles ψ
| i | i
∗Electronicaddress: morimae@gunma-u.ac.jp tothegraphstate. Bobthensendseachqubitofthestate
2
one by one to Alice, and Alice does measurement-based In fact, if p 1 ǫ, we obtain
pass
≥ −
quantum computing on it. If Bob is honest, Alice can
realize U ψ . Furthermore, as is shown in Refs. [24], Al- k I +g
| i Tr j ρ 1 2ǫ.
ice can verify the correctness of the graphstate by using
the stabilizer test even if some states are coupled to the (cid:16) jY=1 2 (cid:17)≥ −
graph state. However, in the procedure, the correctness
Let
of the input state is not guaranteed, since Bob does not
necessarily couple ψ to the graph state, and Alice can- k
| i I+g j
not check the correctness of the input state. Bob might Λ .
≡ 2
discard ψ and entangles completely different state ψ′ jY=1
| i | i
to the graph state. In this case what Alice obtains is
not U ψ but U ψ′ . Can Alice verify that Bob honestly From the gentle measurement lemma [29],
| i | i
coupled her input state to the graph state?
1
In this paper, we introduce a new protocol of ρ ΛρΛ 1 Tr(Λρ)
1
2k − k ≤ −
measurement-only verifiable blind quantum computing p
where notonlythe computationitself but alsothe quan- 1 (1 2ǫ)
≤ − −
tum input are verifiable. Our strategy is to combine the = p√2ǫ.
traptechniqueandthestabilizertest. Thecorrectnessof
the computing is verified by the stabilizer test, and the Note that
correctness of the quantum input is verified by checking
ΛρΛ ΛρΛ
the trap qubits that are randomly hidden in the input g g =
j j
Tr(Λρ) Tr(Λρ)
state. When the traps are checked, the state can be iso-
lated from the graph state by measuring the connecting
foranyj,andtherefore,ΛρΛ/Tr(Λρ)isastabilizedstate.
qubits in Z basis. The main technical challenge in our
For any positive operator M,
proof is to show that the trap verification and the stabi-
lizer verification can coexist with each other. Tr(Mρ) Tr(ΛρΛ) √2ǫ,
− ≤
which means
II. STABILIZER TEST
ΛρΛ
Tr(Mρ) Tr M Tr(Λρ)+√2ǫ
We first review the stabilizer test. Let us consider an ≤ (cid:16) Tr(Λρ) (cid:17)
N-qubitstateρandasetg ≡{g 1,...,g n }ofgeneratorsof Tr M ΛρΛ +√2ǫ.
a stabilizer group. The stabilizer test is a following test: ≤ (cid:16) Tr(Λρ) (cid:17)
1. Randomly generate an n-bit string k And, for any positive operator M,
(k ,...,k ) 0,1 n. ≡
1 n
∈{ } Tr(ΛρΛ) Tr(Mρ) √2ǫ,
2. Measure the operator − ≤
which means
n
jkj.
s k ≡ g Tr(Mρ) Tr M ΛρΛ Tr(Λρ) √2ǫ
jY=1 ≥ (cid:16) Tr(Λρ) (cid:17) −
ΛρΛ
Note that this measurement can be done with Tr M (1 2ǫ) √2ǫ.
single-qubit measurements, since s k is a tensor ≥ (cid:16) Tr(Λρ) (cid:17) − −
product of Pauli operators.
3. If the result is +1 ( 1), the test passes (fails). III. OUR PROTOCOL