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

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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 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

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

We will express

First consider the case of pure states , where, without loss of generality, =cos()0 +e2isin()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 +i1

–i = 0 –i1

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

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

1 also arises from this orthogonal mixture:

… as does 2 from:

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

1

1

+

0

Rotating 0,1to 0, 1the 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

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 yieldsUU†

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

11 with prob. ||2

Example 2 (decoherence):letA0 =00andA1 =11

This quantum op maps to 0000 + 1111

For ψ =0 +1,

Corresponds to measuring “without looking at the outcome”

Example 3 (discarding the second of two qubits):

LetA0 =I0andA1 =I1

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

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 =Ik, where0, 1, …, d1 can be any orthonormal basis

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

For 2-qubit systems, the partial trace is explicitly

and

Example 4 (adding an extra qubit):

Just one operatorA0 =I0

States of the form become 00

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)

LetA1, A2 , …, Am be matrices satisfying

Example 1:Aj = jj (orthogonal projectors)

This reduces to our previously defined measurements …

When Aj = jj are orthogonal projectors and = ,

= Trjjjj

= jjjj

= j2

Moreover,

DefineA0 =2/300

A1=2/311 A2=2/322

Example 3 (trine state“measurent”):

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

Then

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

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

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)