Introduction to Quantum Secret Sharing

Sharing Quantum SecretsTerm Paper - II

Arunabha SahaRoll No: 91/CSE/111033

Supervisors :

Dr. Guruprasad KarPhysics and Applied Mathematics Unit

Indian Statistical Institute, Kolkata

Dr. Pritha BanerjeeDept. of Computer Science & Engineering

University of Calcutta

June 13, 2014Arunabha Saha (CU) Sharing Quantum Secrets June 13, 2014 1 / 39

Outline

1 Demystifying Secrets

2 Entanglement

3 No Cloning Theorem

4 Classical Secret Sharing

5 Quantum Secret Sharing

6 Conclusion

7 References

8 Appendix

Arunabha Saha (CU) Sharing Quantum Secrets June 13, 2014 2 / 39

Demystifying Secrets


2 Entanglement




6 Conclusion

7 References

8 Appendix






What is a Secret??

simple..!!, which is not known publicly.

mathematically secret refers to limited access to a finite set ofmembers, say S = {s1, s2, ...., sr}.When we share completely, basically we are increasing the cardinalityof that set. Say to n members the secret is shared then thecardinality increase by n.



What is a Secret??





What is a Secret??





Problem with secret sharing

The complete sharing can be possible, iff

new shareholders as well as the existing ones are entirely trusted, inshort honest.

the medium is secure, through which sharing takes place.















But...Life is not so easy

Now what if the things are not in favour..!!??

shareholders are dishonest

medium is not completely secure

But we have to share, no alternative..!!
























Hope, that keep us alive

In such a hostile situation....

we still can assume (by heart or by probability) that there 1 or fewparties are honest..!!

we have to curb the complete access to everyone(we believe and don‘tbelieve!!)

we need a scheme or protocol which works fine with dishonest as wellas honest guys, and can ensure security.























Entanglement


2 Entanglement




6 Conclusion

7 References

8 Appendix


Entanglement

Bipartite System

Suppose we have two parties, Alice and Bob, each having a qubit.Alice‘s qubit: |ψ1〉 = α0 |0〉+ α1 |1〉Bob‘s qubit: |ψ2〉 = β0 |0〉+ β1 |1〉What will be the state of composite system of Alice and Bob ?

given as

|ψ12〉 = (α0 |0〉+ α1 |1〉)⊗ (β0 |0〉+ β1 |1〉)= α0β0 |00〉1 +α0β1 |01〉+ α1β0 |10〉+ α1β1 |11〉

The converse is also true. If we have the bipartite system(|ψ12〉) thenwe can factorise it to get the corresponding qubits involved.

1|0〉 ⊗ |0〉 ≡ |0〉 |0〉 ≡ |00〉Arunabha Saha (CU) Sharing Quantum Secrets June 13, 2014 10 / 39

Entanglement

Bipartite System


given as

|ψ12〉 = (α0 |0〉+ α1 |1〉)⊗ (β0 |0〉+ β1 |1〉)= α0β0 |00〉1 +α0β1 |01〉+ α1β0 |10〉+ α1β1 |11〉



Entanglement

Bipartite System


given as

|ψ12〉 = (α0 |0〉+ α1 |1〉)⊗ (β0 |0〉+ β1 |1〉)= α0β0 |00〉1 +α0β1 |01〉+ α1β0 |10〉+ α1β1 |11〉



Entanglement

Bipartite System


given as

|ψ12〉 = (α0 |0〉+ α1 |1〉)⊗ (β0 |0〉+ β1 |1〉)= α0β0 |00〉1 +α0β1 |01〉+ α1β0 |10〉+ α1β1 |11〉



Entanglement

Entangled State

A pure state of two systems is entangled if it cannot be written as aproduct of two states -

|ψAB〉 6= |ψA〉 ⊗ |ψB〉

We have four such states entangled states called Bell States[4] or EPRStates.

|φ+〉 = 1√2(|00〉+ |11〉)

|φ−〉 = 1√2(|00〉 − |11〉)

|ψ+〉 = 1√2(|01〉+ |10〉)

|ψ−〉 = 1√2(|01〉 − |10〉)

These 4 states are called Maximally Entangled States.Any state of the form a |00〉 ± b |11〉 or a a |01〉 ± b |10〉, where a 6= b, iscalled Pure Entangled State.


No Cloning Theorem


2 Entanglement




6 Conclusion

7 References

8 Appendix


No Cloning Theorem

No Cloning Theorem

No Cloning Theorem

In 1982, Wootters and Zurek showed in their paper that a single quantumcannot be cloned[2].

We want to create an copying device for states(say) in C2. Let A be thesystem which state we want to copy into some other state of system B.The unitary transformation Uc that performs this action,

Uc(|0〉A ⊗ |0〉B) = |0〉A ⊗ |0〉BUc(|1〉A ⊗ |0〉B) = |1〉A ⊗ |1〉B

But then Uc act on |ψ〉 = a |0〉+ b |1〉 as

Uc((a |0〉A + b |1〉A)⊗ |0〉B) = a |0〉A ⊗ |0〉B + b |1〉A ⊗ |1〉B 6= |ψ〉 ⊗ |ψ〉,

this implies that Uc has failed to copy |ψ〉


Classical Secret Sharing


2 Entanglement




6 Conclusion

7 References

8 Appendix






Problem

Problem of Secret Sharing

Suppose that the president of a bank wants to give access to a vault tothree vice presidents who are not entirely trusted. Instead of giving thecombination to any one individual, it may be desirable to distributeinformation in such a way that no vice president alone has any knowledgeof the combination, but any two of them can jointly determine thecombination[7]



(k, n) threshold scheme

In 1979, Shamir[5] introduced an idea to share a secret.The secret data D is divided into n pieces D1, D2, ...., Dn such a way:

knowledge of any k or more Di pieces is sufficient to determine secretdata D

knowledge of any k − 1 or few Di pieces leaves secret data Dundetermined

the size of each piece does not exceed the size of the original data

Di can be changed kept k fixed and without changing the data D

this scheme based upon the polynomial interpolationpolynomial of the form : q(x) = a0 + a1x+ ......+ ak−1x

k−1










k−1










k−1










k−1










k−1










k−1


Quantum Secret Sharing


2 Entanglement




6 Conclusion

7 References

8 Appendix



Quantum Secret Sharing(QSS)



QSS scheme

In QSS scheme, it is possible to combine quantum cryptography withsecret sharing to detect whether a eavesdropper is active during the secretsharing protocol. To do this one have to apply quantum cryptographicprotocols to each of the bit strings which results from the secret sharing.By using the multipartite entanglement, it eliminates the classicalsecret-splitting procedure and can detect the eavesdropper altogether[6].



QSS using GHZ states

Hillery et al.[6] presented a QSS scheme for splitting a message into twoparts by using maximally entangled tripartite state or GHZ states. In thisscheme it allows Alice to send qubits to Bob and Charlie in such a waythat only by working together they can determine the original string. Thisscheme can be generalized for more than three parties.This is a ((2,2)) scheme.

The GHZ state ,

|ψ〉 = 1√2(|000〉+ |111〉)

Let Alice, Bob and Charlie each have one particle from the triplet.They each decide randomly in which direction they will measure anddeclare it publicly but not the result of their measurement.





The GHZ state ,

|ψ〉 = 1√2(|000〉+ |111〉)






The GHZ state ,

|ψ〉 = 1√2(|000〉+ |111〉)




QSS uisng GHZ states

Defining x and y eigenstates

|±x〉 =1√2

(|0〉 ± |1〉), |±y〉 =1√2

(|0〉 ± i |1〉) (1)

from eq. 1 we can express

|0〉 =1√2

(|+x〉+ |−x〉), |1〉 =1√2

(|+x〉 − |−x〉) (2)

By using these results we can write,

|ψ〉 =1

2√

2[(|+x〉a |+x〉b + |−x〉a |−x〉b)(|0〉c + |1〉c)+

(|+x〉a |−x〉b + |−x〉a |+x〉b)(|0〉c − |1〉c)](3)





|±x〉 =1√2

(|0〉 ± |1〉), |±y〉 =1√2

(|0〉 ± i |1〉) (1)


|0〉 =1√2

(|+x〉+ |−x〉), |1〉 =1√2

(|+x〉 − |−x〉) (2)


|ψ〉 =1

2√

2[(|+x〉a |+x〉b + |−x〉a |−x〉b)(|0〉c + |1〉c)+

(|+x〉a |−x〉b + |−x〉a |+x〉b)(|0〉c − |1〉c)](3)





|±x〉 =1√2

(|0〉 ± |1〉), |±y〉 =1√2

(|0〉 ± i |1〉) (1)


|0〉 =1√2

(|+x〉+ |−x〉), |1〉 =1√2

(|+x〉 − |−x〉) (2)


|ψ〉 =1

2√

2[(|+x〉a |+x〉b + |−x〉a |−x〉b)(|0〉c + |1〉c)+

(|+x〉a |−x〉b + |−x〉a |+x〉b)(|0〉c − |1〉c)](3)




This decomposition of |ψ〉 tells that if Alice and Bob get the sameresult on measurement along x direction then Charlie will have the

state(|0〉c)+|1〉c√

2; if they got different results, then Charlie will have the

state(|0〉c)−|1〉c√

2.

Effect of Alice and Bob measurement on Charlie’s state given in thetable

Alice→Bob↓

+x −x +y −y+x |0〉+ |1〉 |0〉 − |1〉 |0〉 − i |1〉 |0〉+ i |1〉+x |0〉 − |1〉 |0〉+ |1〉 |0〉+ i |1〉 |0〉 − i |1〉+y |0〉 − i |1〉 |0〉+ i |1〉 |0〉 − |1〉 |0〉+ |1〉−y |0〉+ i |1〉 |0〉 − i |1〉 |0〉+ |1〉 |0〉 − |1〉




This decomposition of |ψ〉 tells that if Alice and Bob get the sameresult on measurement along x direction then Charlie will have the

state(|0〉c)+|1〉c√

2; if they got different results, then Charlie will have the

state(|0〉c)−|1〉c√

2.

Effect of Alice and Bob measurement on Charlie’s state given in thetable

Alice→Bob↓

+x −x +y −y+x |0〉+ |1〉 |0〉 − |1〉 |0〉 − i |1〉 |0〉+ i |1〉+x |0〉 − |1〉 |0〉+ |1〉 |0〉+ i |1〉 |0〉 − i |1〉+y |0〉 − i |1〉 |0〉+ i |1〉 |0〉 − |1〉 |0〉+ |1〉−y |0〉+ i |1〉 |0〉 − i |1〉 |0〉+ |1〉 |0〉 − |1〉




From the table Charlie knows about the measurements of Alice andBob(x or y) and also can determine that their result is same or notbut cannot know the results explicitly. Similarly, Bob also unable togather information about the Alice measurement without the help ofCharlie.



Sharing in presence of Eve

The problem of Eve is she has no knowledge about the measurement basishave been used and if she measure in the wrong basis then introduce someerror and get caught.But Eve able to entangle an ancilla with the tripartitestate that Alice, Bob and Charlie are using. The combined state of ancillaand the GHZ triplet given by,

Ψ =

1∑j,k,n=0

|jkn〉3 |Rjkn〉ξ (4)

where |jkn〉3 is the state of the three particles and |Rjkn〉ξ is anunnormalized ancilla state.




Let Alice, Bob, Charlie all measure in x basis. If no error occur then onemust have,

p(C = +x|A = ±x,B = ±x) = 1,

p(C = −x|A = ±x,B = ∓x) = 1(5)

where p(C = +x|A = +x,B = +x) is the probability that Charliemeasures +x given that both Alice and Bob measure +x. These implies,

P (+x,+x,−x) |Ψ〉 = 0, P (−x,−x,−x) |Ψ〉 = 0,

P (+x,−x,+x) |Ψ〉 = 0, P (−x,+x,+x) |Ψ〉 = 0(6)

where P (+x,+x,−x) projection onto the subspace of thethree-particle-ancilla Hilbert space in which Alice particle is in +x, Bob’sone is in +x and Charlie’s particle is in −x direction.




Using the conditions in eqn. 6,

1√2

(|000〉3 − |111〉3),1√2

(|100〉3 − |011〉3),

1√2

(|010〉3 − |101〉3),1√2

(|110〉3 − |001〉3)(7)

if these projection operator act on Ψ then it results 0.Now say, alice measure her particle in x basis and Bob and Charliemeasure in y basis. For no error,

p(C = −y|A = ±x,B = ±y) = 1,

p(C = +y|A = ±x,B = ∓y) = 1,(8)




which implies,

P (+x,+y,+y) |Ψ〉 = 0, P (−x,−y,+y) |Ψ〉 = 0,

P (+x,−y,−y) |Ψ〉 = 0, P (−x,+y,−y) |Ψ〉 = 0(9)

expressing conditions of eqn. 9 in z basis,

1√2

(|000〉3 − |111〉3),1√2

(|100〉3 − |011〉3),

1√2

(|010〉3 + |101〉3),1√2

(|110〉3 + |001〉3)(10)

will annihilate |Ψ〉.




So far we have six vectors to which three-particle part of |Ψ〉 must beorthogonal. A seventh 1√

2(|100〉3 + |011〉3) occure when we demand no

error occurrence while Alice measure her particle in y direction, Bobmeasure in x direction and Charlie in y direction. These implies that |Ψ〉must be of form,

|ψ〉 =1√2

(|000〉3 + |111〉3) |R〉ξ (11)

It is in the product form of GHZ state and the ancilla, so there is noeavesdropping. If there some eavesdropping, there will be some error.


Conclusion


2 Entanglement




6 Conclusion

7 References

8 Appendix


Conclusion

Conclusion

QSS also experimentally verified for single qubit[8] and four qubitsystems[9].The main beauty of this scheme is that it can communicate securelyamong two or more parties even in the presence of an Eve.These scheme deals with teleportation and prohibit Eve by using no-cloningtheorem. By using GHZ states the scheme can be extended for four party.


References


2 Entanglement




6 Conclusion

7 References

8 Appendix


References

References

[1] Teleporting an Unknown Quantum State via Dual Classical andEinstein-Podolsky-Rosen ChannelsC. H. Bennett, G. Brassard, C. Crepeau, R. Jozsa, A. Peres and W. K. WoottersPhys. Rev. Lett. vol. 70, pp 1895-1899 (1993)

[2] A single quantum cannot be clonedW. K. Wootters & W. H. ZurekNature(London) 299,802(1982)

[3]Can Quantum-Mechanical Description of Physical Reality Be Considered Complete?Einstein, A. and Podolsky, B. and Rosen, N.Phys. Rev. 47, 777-780 (1935)

[4] Quantum Computation and Quantum Information, p-25Nielson, Michael A., and Chuang, Issac L.

[5]How to Share SecretAdi ShamirCommun. ACM 22, 612 (1979)

[6] Quantum secret sharingM. Hillery, V. Buzek, A. BerthiaumePhys. Rev. A 59, 1829 (1999)


http://prl.aps.org/abstract/PRL/v70/i13/p1895_1



http://www.nature.com/nature/journal/v299/n5886/abs/299802a0.html

http://www.nature.com/nature/journal/v299/n5886/abs/299802a0.html

http://link.aps.org/doi/10.1103/PhysRev.47.777

http://link.aps.org/doi/10.1103/PhysRev.47.777

References

References

[7] How to share a quantum secretR. Cleve, D. Gottesman, H. K. LoPhys. Rev. Lett. 83, 648 (1999)

[8]Experimental single qubit quantum secret sharingC. Schmid, P. Trojek, M. Bourennane,C. Kurtsiefer, M. Zukowski, H. WeinfurterPhys.Rev. Lett. 95, 230505 (2005)

[9]Experimental demonstration of four-party quantum secret sharingS. Gaertner, C. Kurtsiefer, M. Bourennane, H. WeinfurterPhys. Rev. Lett. 98, 020503 (2007)


References

Thank You


Appendix

Appendix A: Tenson Product

Tensor product is a way of putting vector spaces together to form largervector spaces. Denoted by |ψ〉 ⊗ |φ〉.If |ψ〉 is of m dimensional and |φ〉 is of n dimensional, then |ψ〉 ⊗ |φ〉 is ofmn dimensional.

[12

]⊗[23

]=

1 × 21 × 32 × 22 × 3

=

2346


Appendix

Appendix B: Quantum Gates

Pauli matrices are named after the physicist Wolfgang Pauli.

These are a set of 2 x 2 complex matrices which are Hermitian andunitary.

They look like

σ0 ≡ I ≡[1 00 1

], σ1 ≡ σx ≡ X ≡

[0 11 0

]σ2 ≡ σy ≡ Y ≡

[0 − ii 0

], σ3 ≡ σz ≡ Z ≡

[1 00 − 1

]


Appendix




They look like

σ0 ≡ I ≡[1 00 1

], σ1 ≡ σx ≡ X ≡

[0 11 0

]σ2 ≡ σy ≡ Y ≡

[0 − ii 0

], σ3 ≡ σz ≡ Z ≡

[1 00 − 1

]


Appendix




They look like

σ0 ≡ I ≡[1 00 1

], σ1 ≡ σx ≡ X ≡

[0 11 0

]σ2 ≡ σy ≡ Y ≡

[0 − ii 0

], σ3 ≡ σz ≡ Z ≡

[1 00 − 1

]


Appendix

Appendix C: CNOT Gate

The CNOT gate flips the second qubit (the target qubit) if and only if thefirst qubit (the control qubit) is 1.


Appendix

Appendix D: Hadamard Gate

The Hadamard gate acts on a single qubit.

It maps the basis state |0〉 to |0〉+|1〉√2

and |1〉 to |0〉−|1〉√2

Hadamard

Matrix, H = 1√2

[1 11 − 1

]
