# Introduction to Quantum Information Processing CS 667 / PH 767 / CO 681 / AM 871 - PowerPoint PPT Presentation

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Introduction to Quantum Information Processing CS 667 / PH 767 / CO 681 / AM 871. Lecture 14 (2009). Richard Cleve DC 2117 cleve@cs.uwaterloo.ca. Bloch sphere for qubits. Bloch sphere for qubits (1). Consider the set of all 2x2 density matrices .

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Introduction to Quantum Information Processing CS 667 / PH 767 / CO 681 / AM 871

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## Introduction to Quantum Information ProcessingCS 667 / PH 767 / CO 681 / AM 871

Lecture 14 (2009)

Richard Cleve

DC 2117

cleve@cs.uwaterloo.ca

Bloch sphere for qubits

### Bloch sphere for qubits (1)

Consider the set of all 2x2 density matrices 

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

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

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

### Bloch sphere for qubits (2)

We will express

First consider the case of pure states , where, without loss of generality,  =cos()0 +e2isin()1 (,R)

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

0

–i

+

–

+i

1

+ = 0 +1

– = 0 – 1

+i = 0 +i1

–i = 0 –i1

### Bloch sphere for qubits (3)

Note that orthogonal corresponds to antipodal here

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

Distinguishing mixed states

0 with prob. ½

1 with prob. ½

0 with prob. cos2(/8)

1 with prob. sin2(/8)

0 with prob. ½

1 with prob. ½

1

+

0

0 with prob. ½

0 +1 with prob. ½

0

### Distinguishing mixed states (1)

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

1 also arises from this orthogonal mixture:

… as does 2 from:

### Distinguishing mixed states (2)

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

1

1

+

0

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

0

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

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

General quantum operations

### General quantum operations (1)

Also known as:

“quantum channels”

“completely positive trace preserving maps”, “admissible operations”

LetA1, A2 , …, Am be matrices satisfying

Then the mapping

is a general quantum op

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

Example 1 (unitary op): applying U to yieldsUU†

After looking at the outcome,  becomes 00 with prob. ||2

11 with prob. ||2

### General quantum operations (2)

Example 2 (decoherence):letA0 =00andA1 =11

This quantum op maps  to 0000 + 1111

For ψ =0 +1,

Corresponds to measuring  “without looking at the outcome”

### General quantum operations (3)

Example 3 (discarding the second of two qubits):

LetA0 =I0andA1 =I1

States of the form (product states) become 

State becomes

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

Note 2: the operation is called the partial traceTr2 

### More about the partial trace

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

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

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

The partial trace Tr2gives the effective state of the first register

For d-dimensional registers, Tr2 is defined with respect to the operators Ak =Ik, where0, 1, …, d1 can be any orthonormal basis

The partial trace Tr2 ,can also be characterized as the unique linear operator satisfying the identity Tr2( )= 

### Partial trace continued

For 2-qubit systems, the partial trace is explicitly

and

### General quantum operations (4)

Example 4 (adding an extra qubit):

Just one operatorA0 =I0

States of the form become 00

More generally, to add a register in state , use the

operator A0=I

POVM measurements

(POVM = Positive Operator Valued Measre)

Corresponding POVM measurement is a stochastic operation on  that, with probability , produces outcome:

j (classical information)

(the collapsed quantum state)

### POVM measurements (1)

LetA1, A2 , …, Am be matrices satisfying

Example 1:Aj = jj (orthogonal projectors)

This reduces to our previously defined measurements …

### POVM measurements (2)

When Aj = jj are orthogonal projectors and = ,

= Trjjjj

= jjjj

= j2

Moreover,

DefineA0 =2/300

A1=2/311 A2=2/322

### POVM measurements (3)

Example 3 (trine state“measurent”):

Let 0 = 0, 1 =1/20 + 3/21, 2 =1/20 3/21

Then

If the input itself is an unknown trine state, kk, then the probability that classical outcome is k is 2/3 = 0.6666…

### POVM measurements (4)

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

Simplified definition for POVM measurements:

LetE1, E2 , …, Em be positive definite and such that

The probability of outcome j is

This is usually the way POVM measurements are defined

### “Mother of all operations”

LetA1,1, A1,2 , …, A1,m1 satisfy

A2,1, A2,2 , …, A2,m2

Ak,1, Ak,2 , …, Ak,mk

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

with probability the state:

j (classical information)

(the collapsed quantum state)