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Quantum Computation

Howard A. Blair (Syracuse University) QuantumComputing 11/04/14 2 / 33


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Nondeterministic classical states work like this:

Someone puts either a black marble in the box with probability p,

or a white marble in the box with probability 1 p.The state of the box with regard to the color of the marble it

contains is either black with probability p, or white probability

(1 p).You look in the box tomeasurethe state of the box in regard to the

color of the marble. Suppose you see a black marble. Then there

was a marble there along and it was black.



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Quantum States:

A simple example: One bit. We physically represent bits many

ways. One way: circularly polarized photons.

3D glasses at the movies.

Clockwize polarization can be regarded as representing bit value

0 and counterclockwize polarization regarded as representing 1.



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Spin:



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Quantum States:



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Leibniz gave us two things important for computer

science

The Principle ofInertia: Everything remains in the state in which it

is, unlesscausedto change state.The Principle of theIdentity of Indiscernables: Two things are

the actually the same thing if, and only if, they bear exactly the

sameessentialrelationships to everything. (HUGE)

Essential?



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And then came Mach

Ernst Machs positivist/empiricist view: Stick to observables. If

with the help of the atomic hypothesis one could actually establish

a connection between several observable properties which

without it would remain isolated, then I should say that this

hypothesis was an economical one. (i.e: The atom concept is,maybe, a useful hack.)

Machs positivist/empiricist view, which remains at the heart of

contemporary physics, leads right back into metaphysics. What

we observe hasprofound metaphysical implications: Forsomething to exist it must have properties observable in

principle. (Theres the rub!)



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Superposition:

The circular polarization property of a photon is oneofimplementing aqubit: its state, i.e. its polarization direction, is a

quantum state: a0 + b1, called awave function.

a, bare square roots of probabilities withphase- called

amplitudes. Wave functions can with combine constructively and

destructively.

a0 + b1 = ei

p0 + ei(1 p)1

The squares of the absolute values ofaandbare probabilities.

The absolute value operation removes phase. e.g.

eip = p



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The rules (of projective measurement)

Observe a qubit thats in state a0 + b1.

The result of your observation will be either 0or1.

The probability of getting 0 is |a|2 and of getting 1 is |b|2.(Wave functioncollapse.) After the observation the photon iseither

in state0or in state1

depending on whether0or1was observed.

Important Question: How do we know the state really wasnt 0 or1, like the marble in the box, before we observed it?



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2 qubits: Entanglement

a00 + b01 + c10 + d11Suppose we have two photons, the first in state a0 + b1and thesecond in statec0 + d1.

The combined system is in theseparablestate

(a0 + b1) (c0 + d1) = ac00 + ad01 + bc10 + bd11

What about states like

1

2 00 + 1

2 11The quantum states of the separate qubits arent there. Therefore

those qubits do not separately exist? (Quantum Leibniz)



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Quantum Teleportation

The Setup 1: On an apparatus in her lab in New York, Alice has a

qubit in some quantum statea0 + b1. She does not know thequantum state of her qubit. She wants to send the qubit to Bob

who is in Paris. Specifically, she wants to cause the quantum state

of the qubit storage apparatus in Bobs lab in Paris to become

a0 + b1.

The Setup 2: Alice and Bob have prepared a pair of entangled

qubits in (BellorEPR) state

12

00 + 1

211

on a pair of single qubit storage devices, one of which they each

took to their labs.

We have suggested there are no separate qubit states stored in

their qubits from the entangled pair.



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Initially, there are three qubit storage devices in play: (1) holdsAlices qubit that she wants to teleport; (2) Alices qubit

participating in the entangled pair; (3) Bobs qubit participating in

the entangled pair.

AToBeSent AEntangled BEntangled

The state of the three devices is partially entangled like this:

(a0 + b1)

(

1

200 +

1

211)

=a

2000 +

a2

011 + b

2100 +

b2

111



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Alice pretends that the state of her qubit part of the entangled pair

really exists and sends her qubit state that she wants to teleportand her qubit state from the entangled pair through a

controlled-notquantum gate. The state of the three devices

becomes:

a

2 000 + a

2 011 + b

2 110 + b

2 101Classical controlled-not gate:

0 0

0

1 1

0

Control



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Quantum controlled-not gate

Qubit values0and1, respectively:

1

0

0

1

The quantum controlled-not gate must map pairs of inputs to pairs

of outputs in the same that the classical controlled-not gate does.

Pairs of quantum states on separate, distinguishable components

result in tensor products of the separate states. In matrix form wetake Kronecker products.



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Distinguishable multi-partite quantum systems

Suppose we have two quantum bits

a0 + b1 c0 + d1

In matrix notation

a0 + b1 = a

1

0

+ b

0

1

=

a

b

c0 + d1 = c

10+ d

01

=

cd



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The combined system of two qubits is in state

(a0 + b1) (c0 + d1) =

acad

bc

bd



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For the controlled-not gate we have thefollowing inputs

0 0 =

1

0

0

0

0 1 =

0

1

0

0

1 0 =

0

0

1

0

1 1 =

0

0

0

1

and corresponding outputs

0 0 = 1

00

0

0 1 =

0

10

0

1 1 =

0

00

1

1 0 =

0

01

0



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In matrix notation thisunitaryoperation is

1 0 0 0

0 1 0 0

0 0 0 1

0 0 1 0

Classical (sequential) computation: A sequence of small

changes in thecomputational state; i.e. a state is given by the

values of all of the variables involved. State change consists of

changing the value of one of the variables.

Quantum computation: State is multipartite: some componentsof the state a quantum states, others classical states. State

change is consists of classical changes to classical components,

orunitarychanges to quantum components, or measurements of

quantum components.



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We left Alices and Bobs three qubit multipartite system in state

a2

000 + a

2011 +

b2

110 + b

2101

Alice then sends her qubit (the one to be teleported) through a

Hadamard gate:

0 ( 12

0 + 1

21) 1 ( 1

20 1

21)

Substituting on the first qubit, the state of the three partially

entangled devices is now:

a2(

12

0 + 1

21)00 +

a2(

12

0 + 1

21)11

+ b

2(

12

0 12

1)10 + b

2(

12

0 12

1)01



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which simplifies to

a

2000 +

a

2100 +

a

2011 +

a

2111

+b

2010 b

2110 +

b

2001 b

2101

(and now for more quantum wierdness):

=1

200(a0+b1)+1

201(a1+b0)+1

210(a0b1)+1

211(a1b0)



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Alice performs a projective measurement on both of her qubits

which will cause a wave function partial collapse of the entirethree-qubit system shared by Alice and Bob.

Suppose for example that Alices qubit to sent is measured to be 1

and the qubit originally entangled with Bobs is measured to be0.

Then the quantum state of the system collapses to

10(a0 b1)

Bobs qubit is now in statea0 b1. He doesnt knowaandb.Alice emails Bob to tell him the result of her measurement. Bobthen knows that if he applies a phase-flip operation (i.e. 0 0and1 1) he will have Alices qubit (value) in the device in his lab

a0 + b1



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Factoring pure quantum states

Let Fn2 be the set of alln-tuples of bits.

A pure quantum state ofndistinguishable qubits has the form

=xFn

2

axx

where xFn

2

|ax|2 = 1

Consider theax arranged in a 2 2 . . . 2 array, i.e. anndimensional hypermatrixT. Then is factorable as a tensorproduct of individual qubit states if, and only if, the determinant of

every 2 2 submatrix ofT vanishes.



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Entangled?a

2000 +

a

2100 +

a

2011 +

a

2111

+b

2010 b

2110 +

b

2001 b

2101



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a/2 000 a/2 100

b/2 001 -b/2 101

b/2 010 -b/2 110

a/2 011 a/2 001



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SQIL

qubitp,q,r; // All qubits are assumed to be randomly

// initialized in some quantum state.

bitu,v,w; //u,v,w are initialized to 0.

measureqinw; // After execution the state of q and v// is either0and 0, or1and 1.

if(w = 1) // Notice that the state transition performed by

// this branching command is not unitary.

then

NOT(q); // The state of q after this branching command is// is0. NOT(x0+ y1) = y0+ x1



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SQIL

measurerinw;

if(w = 1)then

NOT(r); // At this point the state of the qubit pair (q,r) is

//00. The combined state of all three qubits is

// a 000+ b 100

BELL(q,r); // BELL(x,y) can be any 2-qubit unitary that maps

//00to the Bell state

// (1/sqrt2) 00+ (1/sqrt2) 11.

// To see the combined state of all three qubits,// multiply (a 0+ b textbf1)((1/sqrt2) 00+ (1/sqrt2) 11)

CNOT(p,q); // CNOT(00) =00. CNOT(01) =01.

// CNOT(10) =11. CNOT(11) =10.



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SQIL

Hadamard(p);measurepinu;

measureqinv; // Four possibilities for u, v

if(u = 0)and(v = 1)

then

NOT(r)else if(u = 1)and(v = 0)

then

PhaseFlipPi(r) //PhaseFlipPi(x 0+ y 1) = x 0- y 1

else

{PhaseFlipPi(r);NOT(r)

}


?


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How do we know quantum states are not classical?

Answer: Kochen-Specker, Bell/CSHS inequality, ...

Simon Kochen, Ernst Specker, 1968

J.F. Clauser, M.A. Horne, A. Shimony, R.A. Holt, 1969


B ll


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Bell

Alice and Bob are given a large collection of entangled pairs of

qubits, where each pair is entangled in a phase-flipped Bell state

12

01 12

10

For each pair, Alice has one of the qubits, Bob the other.

Alice prepares two apparatuses to perform a projective

measurement on each of her qubits that will return +1

corresponding to bit-value of 0 and -1 corresponding to a bit-value

of 1. Specifically, Alice prepares the observables

Z=

1 0

0 1

X=

0 1

1 0


B ll


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Bell

Bob prepares similar but more elaborateobservables

S= 1

2(Z X) T= 1

2(Z X)

Bob leaves for our Martian colony with his qubits and apparatuses.

At a prearranged time when Mars and Earth are about 20

light-minutes apart, Alice and Bob each measure their qubits

using their prepared observables. We assume their actions take

much less than 20 minutes and are therefore relativistically

causally separated.? What is the expected value of the random variable

V= ZASB+ XASB+ XATB ZATB


B ll


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Bell

If we assume that Alices and Bobs qubits had definite

probabilistically determined 0 or 1 valued states prior tomeasurement, then we obtain, independently of the probability

distribution, the firstBell inequality:

E(V) = E(ZASB+ XASB+ XATB

ZATB)

2

If we calculate the expected values of the quantum mechanical

observables on theentangledqubit pairs, we obtain for each of

the component observables

E(ZASB) = E(XASB) = E(XATB) = E(ZATB) = 12

Hence

E(V) = 2

2


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