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quantum

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The probability of passing the stabilizer test is
Now we explain our protocol and analyze it, which is
1 I+s the main result of the present paper. Let us consider
k
p = Tr ρ .
pass 2n 2 the following situation: Alice possesses an m-qubit state
k∈X {0,1}n (cid:16) (cid:17) ψ , but she neither knows its classical description nor a
| i
classical description of a quantum circuit that efficiently
We canshowthatif the probabilityofpassingthe test is
generates the state. (For example, she just receives ψ
high,p pass 1 ǫ,thenρis“close”toacertainstabilized | i
≥ − from her friend Charlie, etc.) She wants to perform a
state σ in the sense of
polynomial-sizequantum computing U onthe input ψ ,
| i
Tr(Mσ)(1 2ǫ) √2ǫ Tr(Mρ) Tr(Mσ)+√2ǫ (1) but she cannot do it by herself. She therefore asks Bob,
− − ≤ ≤ who is very powerful but not trusted, to perform her
for any POVM element M. quantum computing. We show that Alice can delegate
3
herquantumcomputingtoBobwithoutrevealing ψ and basis. If the Z-basis measurement result on
| i
U,andshecanverifythe correctnessofthe computation the nearest-neighbour of jth vertex in V is
2
and input. 1, Alice applies Z on the jth vertex in V for
2
For simplicity, we assume that Alice wants to solve a j = 1,...,3m. If Bob was honest, the state of
decision problem L. Alice measures the output qubit of V is now
2
U ψ in the computational basis, and accepts (rejects)
| i 3m
if the result is 1 (0). As usual, we assume that for any Ψ′ = XxjZzj P Ψ .
yes instance x, i.e., x L, the acceptance probability is | i (cid:16)Oj=1 j j (cid:17) | i
∈
larger than a, and for any no instance x, i.e., x / L, the
acceptance probability is smaller than b, where∈ a b AlicefurtherappliesP†( 3m XxjZzj)onV .
− ≥ j=1 j j 2
1/poly(x). If Bob was honest, theNstate of V is now
2
Our p| ro| tocol runs as follows: Ψ = ψ 0 ⊗m + ⊗m. Then Alice
| i | i ⊗ | i ⊗ | i
performs the projection measurement Λ
0
1. Alicerandomlychoosesa3m-qubitpermutationP, 0 0⊗m + +⊗m,Λ = I⊗2m Λ{ , o≡ n
1 0
and applies it on | thi eh la| st 2⊗ m| quih bit| s of V . If she ge− ts Λ} , she
2 0
accepts. Otherwise, she rejects.
Ψ ψ 0 ⊗m + ⊗m
| i≡| i⊗| i ⊗| i
to generate P Ψ . Alice further chooses a random
6m-bit string| (xi 1,...,x 3m,z 1,...,z 3m) 0,1 6m, G
and applies 3m XxjZzj on P Ψ to g∈ en{ erate}
j=1 j j | i
N
3m
Ψ′ XxjZzj P Ψ . '
| i≡(cid:16)Oj=1 j j (cid:17) | i (cid:1)
She sends Ψ′ to Bob. (Or, it is reasonable to as- FIG. 1: The state |G Ψ′i. The state in thered dotted box is
sume that| Chi arlie givesAlice Ψ′ andinformation |Gi and that in the blue dotted box is |Ψ′ i. E connect is the
| i
of P and (x ,...,x ,z ,...,z ) in stead of giving set of edges that connects qubits in the red dotted box and
1 3m 1 3m
ψ .) those in the bluedotted box.
| i
2. If Bob is honest, he generates the (3m+N)-qubit Nowletusanalyzetheprotocol. First,theblindnessis
state obvious, since what she sends to Bob is the completely-
mixed state from Bob’s view point, and due to the no-
G Ψ′ CZ e (Ψ′ G ), (2) signalingprinciple,Alice’soperationsonstatessentfrom
| i≡(cid:16) e∈EO (cid:17) | i⊗| i Bob do not transmit any information to Bob.
connect
Second, let us consider the case of x L. In this
andsendseachqubitofitonebyonetoAlice,where case, honest Bob generates the correct∈ state, G Ψ′ ,
CZ eistheCZgateontheverticesoftheedgee,and Eq. (2), and therefore Alice cando correctcomput| ationi ,
E connect is the set of edges that connects qubits in if she chooses the measurement-basedquantum comput-
G and Ψ′ (Fig. 1). If Bobis malicious, he sends ing, passes the stabilizer test with probability 1, if she
| i | i
any (3m+N)-qubit state ρ to Alice. choosesthestabilizertest,andpassestheinput-statetest
with probability 1, if she chooses the input-state test.
3. 3-a. With probability q, which is specified later,
Therefore, the acceptance probability, px∈L, is
Alice does the measurement-based quantum acc
computing on qubits sent from Bob. If the px∈L qa+ 1 −q 1+ 1 −q 1
computation result is accept (reject), she ac- acc ≥ 2 × 2 ×
cepts (rejects). (During the computation, Al- = qa+(1 q) α.
− ≡
ice of course corrects the initial randomPauli
operatorandpermutation( 3m XxjZzj)P.) Finally, let us consider the case of x ∈/ L. In this case,
j=1 j j Bob might be malicious, and can send any (3m+N)-
N
3-b. Withprobability(1 q)/2,Alicedoesthesta- qubit state ρ. Let p and p be the probability
Gpass ψpass
−
bilizer test, and if she passes (fails) the test, of passing the stabilizer test and the initial-state test,
she accepts (rejects). respectively.
Let ǫ = 1 . It is easy to see that the acceptance
3-c. With probability(1 q)/2,Alice doesthe fol- poly(|x|)
lowingtest,whichw− ecalltheinput-statetest: probability, px∈/L, is given as follows:
acc