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shuffling algorithm in polynomial time is not known and a quantum gate sequence x(U), decide whether the quan- |
it is unlikely. Instead, we introduce a concept of a tum gate sequence of C(cid:48) is in g(G(cid:48)) ≡ g(V V ⊗I) or |
U 0 L R |
sufficiently-random gate shuffling algorithm. To ensure g(G(cid:48))≡g((V ⊗I)C (V ⊗I)) where U (cid:54)=I. |
1 L U R |
informational indistinguishability of the applied order of We investigate the action of the algorithm A on |
V onH control, shufflingalgorithmsarerequiredthattwo a sufficiently-random shuffled gate sequence z(cid:48)(C U(cid:48) ) ∈ |
sets of quantum gate sequences obtained by shuffling D(z(C(cid:48) )). SinceEq.(2)hastobealsosatisfiedforN key |
U |
lz al p= pedx s( iV gn)z ifi(C caI n) tla yn .d Thz ur s= wez d( eC fiI n) ex a(V su) ffish cio eu nl td ly-b re ano dv oe mr- s {t 0a ,t 1e }s N|Φ ai nN d(cid:105) l= ineV aR ri† ty⊗ o. f.. th⊗ eV aR lg† o| ri iN th(cid:105) mfo ,r Ei qN .(= 2)j h1 o· ld·· sj eN ve∈ |
n |
gateshufflingalgorithmasthefollowing: Agateshuffling if we replace the key state by any mixture of key states, |
algorithm S in polynomial time is said to be sufficiently- namely, |Φ (cid:105)(cid:104)Φ |→(cid:80) p |Φ (cid:105)(cid:104)Φ | where {p } |
random if for a restricted quantum gate sequence z(C I) isanarbitri aN rypri oN babilityiN disi tN ribui tN ion. FiN orthecaseoiN fa |
constructedfromarbitraryx(I)∈g p(n)(I)andagatese- unitary operation C U(cid:48) , the key space H does not con- |
key |
quence x(V) of V on H control, there exists a polynomial tainthedummyspace. Thus,wecanreplacethemixture |
function q(n) for the number of qubit n such that by a completely mixed state I/2N. This means that Eve |
hasnoadvantagefromcollectingextrakeys, instead, she |
|D(x(V)z(C ))∩D(z(C )x(V))| 1 |
I I =O( ), (4) justneedstoprepareI/2N byherself. Thus,wecanomit |
|D(x(V)z(C ))∪D(z(C )x(V))| q(n) |
I I |Φ (cid:105)inEq.(2)withoutlossofgenerality,andtheaction |
iN |
of the algorithm A can be simplified to |
where D(•) is a distribution obtained by applying S on |
a quantum gate sequence • in polynomial time. A(cid:48) :(z(cid:48)(C(cid:48) ), |φ (cid:105))(cid:55)→(y(F(cid:48)), |φ (cid:105)⊗|ψy (cid:105)). (5) |
U i i z(cid:48)i |
Notethatasufficiently-randomgateshufflingdoesnot |
Further, Eq. (5) can be represented by a CPTP map |
require a completely-random shuffling from the follow- |
Λ (|φ (cid:105)(cid:104)φ |) = |φ (cid:105)(cid:104)φ | ⊗ (cid:80) py |y(cid:105)(cid:104)y| ⊗ |ψy (cid:105)(cid:104)ψy |, |
i gng r (e Ia )so isns a. suA bses tet ofo gf z(C I) (Ic )o an nst dru nc ot ted unf ir foo rm m.x( TI h) us∈ wz h(cid:48) erei i 1i andi {0y ,1}z ∗(cid:48)i abbrevz(cid:48) ii ationz(cid:48)i |
, (cid:80) py = y ∈ is an of |
p(n) p(n+1) y x(cid:48)i |
D(x(V)z(C )) and D(z(C )x(V)) are not necessarily re- thequantumgatesequencey(F(cid:48))foraunitaryoperation |
I I |
quired to be an uniform distribution of g (I). Fur- F(cid:48) satisfying Eq. (3). Using the Steinspring representa- |
p(n+1) |
ther,q(n)ofEq.(4)maydependonz(C )andx(V),since tion[8],Λ canbesimulatedbyaunitaryoperationW |
I z(cid:48) z(cid:48) |
4 |
byaddinganappropriatedimensionalancilla|¯0(cid:105)=|0...0(cid:105) given quantum gate sequence x(U), belongs to g(G(cid:48)) |
0 |
as or g(G(cid:48)) by observing the difference in probabilities of |
1 |
above two cases by repeating the processes many (but |
(cid:88)(cid:113) |
W (|φ (cid:105)⊗|¯0(cid:105))=|φ (cid:105)⊗ py |y(cid:105)⊗|ψy (cid:105)⊗|yz(cid:48)i(cid:105),(6) polynomial) times. Then it is possible to check that the |
z(cid:48) i i z(cid:48)i z(cid:48)i |
givenquantumgatesequencex(U)isanidentityornotin |
y |
polynomial time. However, the exact non-identity check |
where (cid:104)yz(cid:48)i|y(cid:48)z(cid:48)i(cid:105)=δ yy(cid:48). problem has been shown to be NQP-complete and it is |
For U (cid:54)=I, note that hard to solve in polynomial time without using the wit- |
ness state. Therefore, Eve’s cracking strategy A, ana- |
(cid:113) |
(cid:88) py py(cid:48) (cid:104)y|y(cid:48)(cid:105)(cid:104)ψy |ψy(cid:48) (cid:105)(cid:104)yz(cid:48)0|y(cid:48)z(cid:48)1(cid:105)=0. (7) lyzing the quantum gate sequence x(cid:48)(G(cid:48)) obfuscated by |
z(cid:48)0 z(cid:48)1 z(cid:48)0 z(cid:48)1 |
thesufficiently-randomshufflingalgorithmS,isshownto |
y,y(cid:48) |
be a computationally hard problem even using quantum |
Applying W on |φ (cid:105) ⊗ |¯0(cid:105), where |φ (cid:105) = V†(|0(cid:105) + computers. |
√ z(cid:48) + + R |
|1(cid:105))/ 2, and tracing out the ancilla qubits, we obtain In this letter, we propose authorized quantum com- |
Γ (|φ (cid:105)(cid:104)φ |) = tr [W (|φ (cid:105)(cid:104)φ |⊗|¯0(cid:105)(cid:104)¯0|)W†] = I/2. putation, where only a user with a non-cloneable quan- |
Tz h(cid:48) us,+ for al+ l V anda V z(cid:48) and+ all z+ (cid:48)(C )∈g(G(cid:48))z ,(cid:48) we have tum authorization key can perform a unitary operation |
L R U 1 genuinely created by a programmer. In our scheme, the |
(cid:104)φ |Γ (|φ (cid:105)(cid:104)φ |)|φ (cid:105)=δ , (8) unitary operation is encrypted into another unitary op- |
i z(cid:48) j j i ij |
eration acting on a larger Hilbert space in the form of |
(cid:104)φ |Γ (|φ (cid:105)(cid:104)φ |)|φ (cid:105)=1/2. (9) |
+ z(cid:48) + + + programmable quantum arrays proposed by Nielsen and |
For U =I, note that z(C(cid:48))∈g(G(cid:48))=g(V V ⊗I)= Chuang. Further, the quantum gate sequence of the en- |
g(V (cid:48)V (cid:48) ⊗ I) for V (cid:54)= I V (cid:48) and0 V (cid:54)= L VR (cid:48) where cryptedunitaryoperationisobfuscatedbyasufficiently- |
L R L L R R |
V V = V (cid:48)V (cid:48). For O(1/q(n+1)) of the sufficiently- random shuffling algorithm and then, it is authenticated |
L R L R |
random shuffled quantum gate sequences of S(z(C(cid:48))), andpubliclyannounced. Toperformtheoriginalunitary |
I operation, the user needs to obtain an quantum autho- |
we cannot determine which V is taken. This property |
R |
rizationkey,whichisprovidedbytheprogrammertothe |
leads a contradiction if we assume that we cannot per- |
authorized user, and then performs the obfuscated gate |
formtheexactnon-identitycheckprobleminpolynomial |
sequence together with the key. |
time without using a witness state. |
Undertheimpossibilityoftheexactnon-identitycheck The security of our authorized quantum computation |
problem, the two probabilities given by Eqs. (8) and is based on the quantum computational complexity of |
(9) for U = I should not be different more than forging the quantum authorization key from the obfus- |
O(1/poly). By taking V = V = H, where H de- cated quantum gate sequence. Under the assumption |
R L |
notes a Hadamard operation, we have the probabilities of the existence of a sufficiently-random shuffling algo- |
(cid:104)+|Γ (|+(cid:105)(cid:104)+|)|+(cid:105) = 1 and (cid:104)0|Γ (|0(cid:105)(cid:104)0|)|0(cid:105) = 1/2. rithm, we have shown that the problem is NQP-hard by |
z(cid:48) z(cid:48) |
However,undertheexistenceofsufficiently-randomshuf- reducing it to a NQP-Complete problem, the exact non- |
fling, we can also take V(cid:48) = V(cid:48) = I. To satisfy the im- identity check problem of large quantum gate sequences. |
R L |
possibility of the exact non-identity check problem, the Therefore, our authorized quantum computation can be |
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