1 Introduction to Quantum Information Processing CS 667 / PH 767 / CO 681 / AM 871 Richard Cleve DC 2117 [email protected] Lecture 14 (2009)

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Introduction to Introduction to Quantum Information ProcessingQuantum Information Processing

CS 667 / PH 767 / CO 681 / AM 871CS 667 / PH 767 / CO 681 / AM 871

Richard Cleve DC [email protected]

Lecture 14 (2009)

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Bloch sphere for qubits

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Bloch sphere for qubits (1)Bloch sphere for qubits (1)Consider the set of all 2x2 density matrices

Note that the coefficient of I is ½, since X, Y, Y are traceless

They have a nice representation in terms of the Pauli matrices:

01

10σ Xx

0

0σ

i

iYy

10

01σ Zz

Note that these matrices—combined with I—form a basis for the vector space of all 2x2 matrices

We will express density matrices in this basis

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Bloch sphere for qubits (2)Bloch sphere for qubits (2)

2

ZcYcXcIρ zyx We will express

First consider the case of pure states , where, without

loss of generality, = cos()0 + e2isin()1 (, R)

θθe

θeθ

θθθe

θθeθρ

φi

φi

φi

φi

2cos12sin

2sin2cos1

2

1

sinsincos

sincoscos2

2

22

22

Therefore cz = cos(2), cx = cos(2)sin(2), cy = sin(2)sin(2)

These are polar coordinates of a unit vector (cx , cy , cz) R3

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Bloch sphere for qubits (3)Bloch sphere for qubits (3)

+

0

1

–

+i

–i+i = 0 + i1–i = 0 – i1

– = 0 – 1+ = 0 +1

Pure states are on the surface, and mixed states are inside (being weighted averages of pure states)

Note that orthogonal corresponds to antipodal here

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Distinguishing mixed states

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Distinguishing mixed states (1)Distinguishing mixed states (1)

10

01

2

12ρ

0 with prob. ½ 0 + 1 with prob. ½

0 with prob. ½ 1 with prob. ½

4121

21431 //

//ρ

0 with prob. cos2(/8) 1 with prob. sin2(/8)

0

+

0

1

0 with prob. ½ 1 with prob. ½

What’s the best distinguishing strategy between these two mixed states?

1 also arises from this orthogonal mixture: … as does 2 from:

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Distinguishing mixed states (2)Distinguishing mixed states (2)

8πsin0

08πcos2

2

2/

/ρ

0

+

0

1

10

01

2

11ρ

We’ve effectively found an orthonormal basis 0, 1 in which both density matrices are diagonal:

Rotating 0, 1 to 0, 1 the scenario can now be examined using classical probability theory:

Question: what do we do if we aren’t so lucky to get two density matrices that are simultaneously diagonalizable?

Distinguish between two classical coins, whose probabilities of “heads” are cos2(/8) and ½ respectively (details: exercise)

1

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General quantum operations

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General quantum operations (1)General quantum operations (1)

Example 1 (unitary op): applying U to yields U U†

Also known as: “quantum channels”“completely positive trace preserving maps”, “admissible operations”

Let A1, A2 , …, Am be matrices satisfying

Then the mapping

m

jjj AA

1

t

is a general quantum op

Note: A1, A2 , …, Am do not have to be square matrices

m

jjj IAA

1

t

m

jjj AρAρ

1

t

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General quantum operations (2)General quantum operations (2)

Example 2 (decoherence): let A0 = 00 and A1 = 11

This quantum op maps to 0000 + 1111

Corresponds to measuring “without looking at the outcome”

2

2

2

2

0

0

β

α

ββα

αβαFor ψ = 0 + 1,

After looking at the outcome, becomes 00 with prob. ||2

11 with prob. ||2

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General quantum operations (3)General quantum operations (3)

Example 3 (discarding the second of two qubits):

Let A0 = I0 and A1 = I1

0100

0001

1000

0010

States of the form (product states) become

State becomes 110011002

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12

12

1

Note 1: it’s the same density matrix as for ((½ , 0), (½ , 1))

10

01

2

1

Note 2: the operation is called the partial trace Tr2

More about the partial traceMore about the partial trace

If the 2nd register is discarded, state of the 1st register remains

Two quantum registers in states and (resp.) are independent when the combined system is in state =

In general, the state of a two-register system may not be of the form (it may contain entanglement or correlations)

The partial trace Tr2 , can also be characterized as the

unique linear operator satisfying the identity Tr2( ) =

For d-dimensional registers, Tr2 is defined with respect to the

operators Ak = Ik , where 0, 1, …, d1 can be any

orthonormal basis

The partial trace Tr2 gives the effective state of the first register

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Partial trace continuedPartial trace continued

1111101001110010

1101100001010000

1111101101110011

1110101001100010

1101100101010001

1100100001000000

2Tr,,,,

,,,,

,,,,

,,,,

,,,,

,,,,

ρρρρ

ρρρρ

ρρρρ

ρρρρ

ρρρρ

ρρρρ

For 2-qubit systems, the partial trace is explicitly

1111010110110001

1110010010100000

1111101101110011

1110101001100010

1101100101010001

1100100001000000

1Tr,,,,

,,,,

,,,,

,,,,

,,,,

,,,,

ρρρρ

ρρρρ

ρρρρ

ρρρρ

ρρρρ

ρρρρ

and

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General quantum operations (4)General quantum operations (4)Example 4 (adding an extra qubit):

Just one operator A0 = I0

00

10

00

01

States of the form become 00

More generally, to add a register in state , use the

operator A0 = I

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POVM measurements

(POVM = Positive Operator Valued Measre)

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POVM measurements (1)POVM measurements (1)

Let A1, A2 , …, Am be matrices satisfying IAA j

m

jj

1

t

Corresponding POVM measurement is a stochastic operation

on that, with probability , produces outcome:

j (classical information)

tjj AρATr

tt

jj

jj

AρA

AρA

Tr(the collapsed quantum state)

Example 1: Aj = jj (orthogonal projectors)

This reduces to our previously defined measurements …

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POVM measurements (2)POVM measurements (2)

Moreover,

tjj AρATr

jj

j

jjjj

jj

jj φφψφ

φφψψφφ

AρA

AρA 2Tr t

t

When Aj = jj are orthogonal projectors and = ,

= Trjjjj

= jjjj

= j2

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POVM measurements (3)POVM measurements (3)

Example 3 (trine state “measurent”):

Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20 3/21

Then IAAAAAA 221100ttt

If the input itself is an unknown trine state, kk, then the

probability that classical outcome is k is 2/3 = 0.6666…

00

01

3

2Define A0 = 2/300

A1= 2/311 A2= 2/322

62

232

4

1

62

232

4

1

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POVM measurements (4)POVM measurements (4)

Simplified definition for POVM measurements:

Let E1, E2 , …, Em be positive definite and such that IEm

jj

1

The probability of outcome j is jjjj AAAA tt TrTr

Often POVMs arise in contexts where we only care about the classical part of the outcome (not the residual quantum state)

The probability of outcome j is jETr

This is usually the way POVM measurements are defined

““Mother of all operations”Mother of all operations”Let A1,1, A1,2 , …, A1,m1

satisfy

A2,1, A2,2 , …, A2,m2

Ak,1, Ak,2 , …, Ak,mk

IAAk

jij

m

iij

j

1

,1

,t

Then there is a quantum operation that, on input , produces

with probability the state:

j (classical information)

(the collapsed quantum state)

jm

iijij AA

1,,Tr t

jm

iijij AA

1,,t

jm

ii,ji,j AρA

1

Tr t