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COS 433: Cryptography

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COS 433: Cryptography

Princeton University Fall 2005

Boaz Barak

Disclaimer

Lecture 12: Idiot’s Guide to Quantum Computing & Crypto

"Do not take the lecture too seriously . . . just relax and enjoy it. I am going to tell you what nature behaves like. If you will simply admit that maybe she does behave like this, you will find her a delightful, entrancing thing. Do not keep saying to yourself "But how can it be like that?" because you will get . . . into a blind alley from which nobody has yet escaped. Nobody knows how it can be like that."

Richard Feynmann on Quantum Mechanics.

Strange aspects of quantum mechanics:

- Superposition– object doesn’t have definite properties (location, speed) but has probabilities over them.

- Interference– probabilities can be negative.

- Entanglement– properties of many particles can be correlated.

- Measurement– object’s properties collapse to definite value when measured, collapsing also properties of other entangled objects.

How does electron passing thru top slit know to avoid mid point if bottom slit is open?

We can never catch an electron “red-handed” behaving bizarrely

If we place detector then pattern turns to be as expected.

In fact, to make the math work nicely, assume:

- p,q can be arbitrary complex numbers.
- p2+q2=1 (prob of system measuring to |0> is p2=|p|2)b

Consider object/system that can be in one of two states.

State |1> - electron hit mid point

b

State |0> - electron did not hit mid point.

Deterministic view:

System is either in state |0> or state |1>

Probabilistic view:

System is in state |0> w.prob p and state |1> w.prob q with p+q=1

Quantum view:

System is in state p|0>+q|1> with |p|+|q|=1

(p,q can be negative!)

b

b1

b2

Consider object/system that can be in one of two states.

State |1> - electron hit mid point

State |0> - electron did not hit mid point.

Quantum view:

System is in state p|0>+q|1> with |p|2+|q|2=1

(p,q complex)

Suppose system consists of two bits – has four possible states: |00> , |01> , |10> , |11>

Quantum view:

System is in state p1|00>+p2|01>+p3|10>+p4|11>

where |p1|2+|p2|2+|p3|2+|p4|2=1

When measured, system will collapse to ithstate w.prob |pi|2.

Note: Need 2n numbers to keep track of state of n-bit system.

Democritos Newton Einstein:

Underlying everything are small particles interacting locally using simple well-defined rules (“billiard balls”).

Quantum Mechanics:

Nature has a secret HUGE piece of paper containing >210000000000000000 complex numbers, keeping track of a superposition of all particles in the world, but allows us only to make some specific measurements of these numbers.

“Corollary”: We do not know how to simulate quantum system of n particles for t time units in time poly(n.t).

Rephrase: There are some computations performed by quantum systems of n particles and t time units that we don’t know to perform in a classical computer in time poly(n,t)

Maybe can use quantum system to solve hard computational problems??

- There is a mathematical model for computing devices exploiting quantum mechanics – “quantum computers”.

- Many technical difficulties (and maybe fundamental difficulties?) in building such machines.

- (Unsurprisingly) there is no proof that quantum computers are more powerful than classical computers/Boolean circuits/Turing machines.

- There are polynomial algorithms for quantum computers solving problems unknown to be solvable classically in poly-time:
- Simulation of quantum system
- Factoring integers and discrete logs.

- There are hard problems with no quantum poly-time algorithms:
- SAT, 3COL and all the NP-complete problems.
- Inverting many candidate one-way functions and permutations, private key encryption and signature schemes.
- Problems on lattices (can be used for public-key encryption).

- If quantum computers can be built, then many popular encryption and signature schemes can be broken (RSA,Diffie-Hellman)

- However, there are still other candidates for encryption schemes not known to be broken. This is especially true for private key cryptography and signature schemes.

- Many (but not all) of the proofs of security in crypto carry over from the classical model to the quantum model, as long as the underlying hard problem is assumed hard for quantum computers.

- Exciting possibilities of using quantum mechanics to obtain perfectly unconditionally secure cryptography. Does not require full fledged quantum computers – prototype systems already being built.

Quantum Key Distribution (QKD)

b1

b2

Transfer qubit b2

|00>+|11>

Consider system of two bits initialized to 1/p2 |00> + 1/p2 |11>

Give b1to Alice and b2 to Bob.

According to QM until Alice measures b1, it is completely random, but once she measures it system collapses to either |00> or |11>

Thus Bob will measure the same value as Alice.

First idea for key exchange using QM:

Alice

Eve

Bob

b1b2= |00>+|11>

Measure b1

Measure b2

Transfer qubit b2

First idea for key exchange using QM:

Alice

Eve

Bob

b1b2= |00>+|11>

Measure b1

Measure b2

Problem: What if Eve measures b2on the way and learns it?

We can’t stop Eve from doing so, but we need a way for Bob to find out.

Problem can be solved but we need:

- Learn more about operations allowed in QM.

- Assume Bob and Alice can exchange authenticated but not secret classical messages.

b

Consider system of one bit.

Classically, there are not many operations we can perform on it – keep it the same or invert it.

In QM, system’s state is described as p|0>+q|1> - i.e., vector (p,q)2C2

According to QM, we can perform any operation A on system that is:

- Linear:A(p+p’,q+q’) = A(p,q) + A(p’,q’)

- Norm-preserving: If ||(p,q)||=pp2+q2 =1 then ||A(p,q)||=1

- Orthogonal:A(1,0)=A|0> is perpendicular to A(0,1)=A|1>

(p,q) ? (p’,q’) if pp’+qq’=0

Example:

H

H|0> = 1/p2 |0> + 1/p2 |1> ~ |0> + |1> = (1,1)

H|1> = 1/p2 |0> - 1/p2 |1> ~ |0> - |1> = (1,-1)

Transfer qubit b2

“I received the bit”

If applied H send “YES”

w.p. ½ send b2

Key exchange using QM:

Alice

Eve

Bob

b1b2= |00>+|11>

With prob ½, apply H to b1

If “YES” apply H to b2

Measure b2

Measure b1.If b1 b2 abort protocol.

Lemma 1: If Eve did not measure b2 then b1=b2 with prob 1.

Proof: If they did not apply H then clearly b1=b2

If both Alice and Bob apply H we get that b1b2 is transformed to

HH|00>+|11>= (|0>+|1>)(|0>+|1>)+(|0>-|1>)(|0>-|1>) =|00>+|10>+|01>+|11>+|00>-|10>-|01>+|11>=|00>+|11>

H

H|0> = 1/p2 |0> + 1/p2 |1> ~ |0> + |1> = (1,1)

H|1> = 1/p2 |0> - 1/p2 |1> ~ |0> - |1> = (1,-1)

Transfer qubit b2

“I received the bit”

If applied H send “YES”

w.p. ½ send b2

Key exchange using QM:

Alice

Eve

Bob

b1b2= |00>+|11>

With prob ½, apply H to b1

If “YES” apply H to b2

Measure b2

Measure b1.If b1 b2 abort protocol.

Lemma 2: If Eve did measure b2 then b1 b2 with prob ¸1/4.

Proof: As example, assume that Eve measured b2 and collapsed b1b2to |11>

If both Alice and Bob apply H we get that b1b2 is transformed to

HH|11>= (|0>-|1>)(|0>-|1>) = |00>-|10>-|01>+|11>

w.p. ½ this system collapses to either |10> or |01> and hence b1b2

H

H|0> = 1/p2 |0> + 1/p2 |1> ~ |0> + |1> = (1,1)

H|1> = 1/p2 |0> - 1/p2 |1> ~ |0> - |1> = (1,-1)

Transfer qubit b2

“I received the bit”

If applied H send “YES”

w.p. ½ send b2

Key exchange using QM:

Alice

Eve

Bob

b1b2= |00>+|11>

With prob ½, apply H to b1

If “YES” apply H to b2

Measure b2

Measure b1.If b1 b2 abort protocol.

Lemma 1: If Eve did not measure b2 then b1=b2 with prob 1.

Lemma 2: If Eve did measure b2 then b1 b2 with prob ¸1/4.

Idea: Continue this for 2n steps, and discard all bits that were made public. If did not abort, Alice and Bob can be almost certain Eve did not measure and has no information about undiscarded bits.

Proof generalized to case that Eve applies arbitrary unitary transformation.