Short course on quantum computing Andris Ambainis University of Latvia.

Short course on quantum computing

Andris AmbainisUniversity of Latvia

Lecture 3

Recent results in quantum cryptography

Quantum cryptography

Unconditional secure key distribution.Unconditional security for other tasks?

Setting

QKD: two honest parties, connected by insecure channel. Protection from eavesdropping.Two (or more) parties, some of them might be dishonest. Honest parties need to be protected from dishonest ones.

Bit commitment

Alice has a bit a. She wants to commit it to Bob so that Bob does not learn a, Alice cannot change it.

Coin flipping

Alice and Bob want to flip a coin so that neither of them controls the outcome.If both honest, 0 (1) with probability 1/2.If one honest, 0 (1) with probability at most 1/2+.

Oblivious transfer

Alice has two bits x0, x1. Bob wants to learn xb so that:

Alice does not learn b. Alice is guaranteed that Bob gets only

one bit.

Multiparty computation

Alice has x, Bob has y. They want to compute f(x, y) so that: Alice learns nothing about y except

f(x, y). Bob learns nothing about x except f(x,

y).

Generalizes to more than two parties.

Coin flipping

Alice and Bob want to flip a coin so that neither of them controls the outcome.If both honest, 0 (1) with probability 1/2.If one honest, 0 (1) with probability at most 1/2+.

Classical coin flipping

If hard functions are available,Information-theoretically (unlimited computational power), one party can always force one outcome with probability 1.

nc

1

Quantum coin flipping

Protocol with =1/4 [A, 2000].Lower bound of 1/2+ 1/2 [Kitaev, 2001].Better protocols with weaker definition [A, RS, 2002].

Classical coin flipping

a{0, 1} b{0, 1}Commit (a)

b

Reveal (a)

Result: (a+b) mod 2.

Why is this secure?

Bob is honest, Alice cheating.Alice’s bit a does not depend on b because Alice has to commit a before seeing b.Bob picks 0/1 with probability ½.The result is a or (a+1) mod 2 with probability ½.

Quantum coin flipping

a, x{0, 1} b{0, 1}

b

a,x

Result: (a+b) mod 2.

ax

General quantum states

k-dimensional quantum system.Basis |1>, |2>, …, |k>.General state

1|1>+2|2>+…+k|k>,

|1|^2+…+ |k|^2=12k dimensional system can be constructed as a tensor product of k quantum bits.

Measurements

Measuring 1|1>+2|2>+…+k|k>

in the basis |1>, |2>, …, |k> gives |i> with probability |i|2.

Any orthogonal basis can be used.

Quantum coin flipping

a, x{0, 1} b{0, 1}

b

a,x

Result: (a+b) mod 2.

ax

States

12|2

10|

2

1

0,12|2

10|

2

1

1,01|2

10|

2

1

01|2

10|

2

1

xa

xa

xa

xa

ax

Security result

Theorem. Alice (Bob) cannot achieve 0 (1) with probability more than 3/4.

Cheating Bob

Bob could measure the state in basis |0>, |1>, |2>.If a=0, he gets |0> or |1> with probabilities 1/2.If a=1, |0> or |2> with probabilities 1/2.Learns a with probability 1/2, no information otherwise.

Mixed states

If a=0, Alice sends |0>|1> with probabilities 1/2. If a=1, Alice sends |0>|2> with probabilities 1/2. How well can Bob distinguish these two?

Mixed states

Probabilistic combinations of quantum states.(|0> with probability 1/2 and |1> with probability 1/2) not the same as |0>+|1>.

|1>

|0>

|0> +|1>|0> -|1>

Equivalent mixed states

Let 0 be |0> or |1> with probabilities 1/2.Let 1 be |0>|1> with probabilities 1/2.Any measurement on 0 produces the same probability distribution as on 1.

Bra-ket notation

kaa k |...1|1

ka

a

a

...2

1

**2

*1 ... kaaa

Bra-ket notation

i

ii

k

k ba

b

b

b

aaa *2

1

**2

*1 ...

...

Inner product

Density matrix

Consider the mixed state that is |i> with probabilities pi.

The density matrix is ,i

iiip

**

2

*1

*21

*11

*11

...

.........

...

...

ikikiik

ikiii

ii

ii

ii

aaaa

aaaa

aa

aa

Density matrix

Let 1|2

10|

2

1

000

02

1

2

1

02

1

2

1

02

1

2

1

02

12

1

Cheating Bob

Alice sends 0, 1.

000

02

10

002

1

0

2

100

000

002

1

1

How well can Bob distinguish these two?

Cheating Bob

Theorem: The best probability with which Bob can guess i, given i, is

For matrices in our protocol, ||0-1||t=1, probability 3/4.

,42

1 10 t

AATrA T

t

Cheating Alice.

Fidelity of two density matrices.Bounds how one state can be transformed into another.Probability that Alice can convince Bob that a=0 is F(, 0).

Probability that Alice can convince Bob that a=1 is F(, 1).

Quantum coin flipping

a, x{0, 1} b{0, 1}

b

a,x

Result: (a+b) mod 2.

ax

Better bit commitment

Quantum bit commitment => Quantum coin flipping.Better commitment?Bob can’t guess a at all, but Alice can’t change it?

Impossibility theorem

Theorem [Mayers, 1996]. Perfect quantum bit commitment is impossible. If Bob’s state contains no information about Alice’s bit, Alice can change commitment perfectly.Note: there was a “provably secure” protocol before Mayers’ proof.

Delayed measurements

Any measurement can be delayed till end of protocol. Any classical random variable can be replaced by a quantum state.E.g. 0/1 random bit can be replaced by.1

2

10

2

1

State after commitment

By delayed measurement, pure state |>.Let |0> be the state if Alice commits 0, |1> be the state if Alice commits 1.How well Bob can distinguish |0> and |1>?

Tracing out

Imagine that Alice measures her part. Then, Bob is left with mixed state.

.112

100

2

1

|0> |1>

Distinguishability

If Bob cannot access Alice’s part, distinguishing |0> and |1> is equivalent to distinguishing 0 and 1.

Bob can guess commitment with probability

Perfectly secure if ||0-1 ||t=0, i.e. 0=1.

.42

1 10 t

Transformability

Theorem. If 0=1, then there is a unitary U on Alice’s part such that U|0>= |1>.

Perfectly hiding commitments are completely non-binding.Almost perfecly hiding commitments?

Fidelity

F(0, 1)=max |<0 | 1>|2, over all | 0>, | 1> that give 0, 1 if Alice’s part is traced out.Any test that accepts | 0> with certainty, accepts | 1> with probability at least |<0 | 1>|2.

Fidelity

Theorem. For any | 0>, |1> Alice can transform | 0> into a state that is accepted as |1> with probability F(0, 1).

Theorem [Ullman, 1972] TrF ),(

Trace distance vs. fidelity

Theorem [Fuchs, van de Graaf, 1997]

Tradeoff between Alice’s and Bob’s cheating probabilities.

1),(2 10

10

Ft

Summary on bit commitment

In any protocol, either Alice or Bob is capable of cheating with a constant success probability.Protocols in which both parties can’t cheat perfectly, exist.

Coin flipping

Trace distance vs. fidelity gives some lower bounds for coin flipping.Based on one-round commitment [A,RS, 2001]: 3/4.Based on multi-round commitment: 9/16 [Nayak,Shor,2002].Not based on commitment?

Different protocol [Salvail, 2000]

Alice generate two copies of

sends second qubits to Bob.Bob randomly chooses one and verifies it.Alice and Bob measure the other pair.

,112

100

2

1

Security

Theorem [Salvail, 2000] No party can achieve 0 (1) with probability more than 3/4.

Lower bound [Kitaev, 2002]

Theorem. In any protocol, one party can force 0 (1) with probability at least 1/.Proof. Write a semidefinite program for max probability achieved by Alice/ Bob. Look at the dual program.Combine the dual programs.

Weak CF

Assume that Alice can achieve 0 with probability 1 and Bob can achieve 1 with probability 1.Would the protocol be useful?

Yes, if Alice wants 1 and Bob wants 0.Still allowed by Kitaev’s theorem.

Weak CF

Only interested in probability of Alice achieving 1 and Bob achieving 0.Kitaev’s lower bound allows 1/2+.Theorem [A, Rudolph-Spekkens, 2002] There is a protocol with probability 1/2.

Protocol

11|00| Alice prepares

12|11|00| Bob maps

11|00| |12>

Bob wins,Alice verifies

Alice wins,Bob verifies

CF summary

Strong Weak

3/4 1/2

1/2 >0

Protocol

Lower bound

CF open problems

Better protocols/lower bounds.Coin flipping with penalty for cheating. Party caught cheating loses k coins instead of 1.Best result achievable by cheater?The tradeoff between successful cheating vs. being caught.

Open problems

Other cryptographic primitives.Quantum zero knowledge?Multiparty computation. Composing the primitives.