introduction to quantum information processing lecture 4 michele mosca

Introduction to Quantum Information Processing

Lecture 4

Michele Mosca

Overview

Von Neumann measurements General measurements Traces and density matrices and

partial traces

“Von Neumann measurement in the computational basis”

Suppose we have a universal set of quantum gates, and the ability to measure each qubit in the basis

If we measure we get with probability

}1,0{

2

bαb)10( 10

In section 2.2.5, this is described as follows

00P0 11P1

We have the projection operatorsand satisfying

We consider the projection operator or “observable”

Note that 0 and 1 are the eigenvalues When we measure this observable M, the

probability of getting the eigenvalue is and we

are in that case left with the state

IPP 10

110 PP1P0M

b2

ΦΦ)Pr( bbPb αbb

)b(p

P

b

bb

“Expected value” of an observable

b If we associate with outcome the

eigenvalue then the expected outcome is

ΦΦΦΦ

ΦΦΦΦ

)Pr(

MTrbPTr

bPPb

bb

bb

bb

bb

b

b

“Von Neumann measurement in the computational basis”

Suppose we have a universal set of quantum gates, and the ability to measure each qubit in the basis

Say we have the state If we measure all n qubits, then we

obtain with probability Notice that this means that probability of

measuring a in the first qubit equals

}1,0{x

n}1,0{xx

x 2x

0

1n}1,0{0x

2x

Partial measurements

(This is similar to Bayes Theorem)

xp1n}1,0{0x 0

x

0

1n}1,0{0x

2x0p

If we only measure the first qubit and leave the rest alone, then we still get with probability

The remaining n-1 qubits are then in the renormalized state

In section 2.2.5

This partial measurement corresponds to measuring the observable

1n1n I111I000M

Von Neumann Measurements

A Von Neumann measurement is a type of projective measurement. Given an orthonormal basis , if we perform a Von Neumann measurement with respect to of the state

then we measure with probability

}{ k

kk}{ k

k

kkkk

kk

2

k

2

k

TrTr

Von Neumann Measurements

E.x. Consider Von Neumann measurement of the state with respect to the orthonormal basis

Note that

2

10,

2

10

2

10

22

10

2

)10(

We therefore get with probability

2

10

2

2

Von Neumann Measurements

Note that22

10

22

10 **

22

10

2

10Tr

2

10

2

10

2

How do we implement Von Neumann

measurements?

If we have access to a universal set of gates and bit-wise measurements in the computational basis, we can implement Von Neumann measurements with respect to an arbitrary orthonormal basis

as follows.}{ k

How do we implement Von Neumann

measurements?

Construct a quantum network that implements the unitary transformation

kU k Then “conjugate” the measurement

operation with the operation U

kk U k2

kprob

1Uk

Another approach

kk U 1U

kk

000

kkk

000k000

kkk

kkk

2

kprob

Ex. Bell basis change

100101

Consider the orthonormal basis consisting of the “Bell” states

110000

110010 100111 Note that

xyx

y

H

Bell measurement We can “destructively” measure

Or non-destructively project

xyy,x

y,x x

y

H2

xyprob

xyy,x

y,x xyy,x

H

2

xyprob 00

H

Most general measurement

kk

000U

Trace of a matrix

The trace of a matrix is the sum of its diagonal elementse.g.

221100

222120

121110

020100

aaa

aaa

aaa

aaa

Tr

Some properties:

i

ii AATr

ATrUAUTr

CABTrABCTr

BATrABTr

ByTrAxTryBxATr

φφ

t

][][

Orthonormal basis { }iφ

Density Matrices

Notice that 0=0|, and 1

=1|.So the probability of getting 0 when measuring | is:

22

0 0)0( p

ρφφ

φφφφ

φφφφ

0000

0000

0000

TrTr

Tr

where = || is called the density matrix for the state |

10 10 ααφ

Mixture of pure states

A state described by a state vector | is called a pure state.

What if we have a qubit which is known to be in the pure state |1 with probability p1, and in |2 with probability p2 ? More generally, consider probabilistic mixtures of pure states (called mixed states):

... , ,, , 2211 pp φφφ

Density matrix of a mixed state

…then the probability of measuring 0 is given by conditional probability:

i

iipp state pure given 0 measuring of prob.)0(

ρφφ

φφ

00

00

00

Tr

pTr

Trp

iiii

iiii

where

i

iiip is the density matrix for the mixed state

Density matrices contain all the useful information about an arbitrary quantum state.

Density Matrix

If we apply the unitary operation U to the resulting state is

with density matrix

U

tt UUUU

Density Matrix

If we apply the unitary operation U to the resulting state is with density matrix

kkq ψ,

kk Uq ψ,

t

t

t

UU

UqU

UUq

kk

kk

kk

kk

ρ

ψψ

ψψ

Density Matrix

If we perform a Von Neumann measurement of the state wrt a basis containing , the probability of obtaining is

Tr2

Density MatrixIf we perform a Von Neumann

measurement of the state wrt a basis containing the

probability of obtaining is

kkq ψ,

φφρ

φφψψ

φφψψφψ

Tr

qTr

Trqq

kkkk

kkkk

kkk

2

Density Matrix

In other words, the density matrix contains all the information necessary to compute the probability of any outcome in any future measurement.

Spectral decomposition

Often it is convenient to rewrite the density matrix as a mixture of its eigenvectors

Recall that eigenvectors with distinct eigenvalues are orthogonal; for the subspace of eigenvectors with a common eigenvalue (“degeneracies”), we can select an orthonormal basis

Spectral decomposition

In other words, we can always “diagonalize” a density matrix so that it is written as

kk

kkp φφρ

where is an eigenvector with eigenvalue and forms an orthonormal basis

kφ

kp kφ

Partial Trace

How can we compute probabilities for a partial system?

E.g.

yxp

p

yx

yx

y x y

xyy

y xxy

yxxy

,

Partial Trace

If the 2nd system is taken away and never again (directly or indirectly) interacts with the 1st system, then we can treat the first system as the following mixture

E.g.

ρρα

ρα

22,2 Trx

pp

yxp

p

x y

xyy

Trace

y x y

xyy

Partial Trace

ρρα

ρα

22,2 Trx

pp

yxp

p

x y

xyy

Trace

y x y

xyy

yyy

ypTr ΦΦ2 ρ x y

xyy x

p

αΦ

Why?

the probability of measuring e.g. in the first register depends only on

ρ

αα

2

2

2

ΦΦ

ΦΦ

TrwwTr

pwwTr

wwTrp

pp

yyy

y

yyy

y

y y y

wyywy

w

ρ2Tr

Partial Trace

Notice that it doesn’t matter in which orthonormal basis we “trace out” the 2nd system, e.g.

1100110022

2 βαβα Tr

In a different basis

1

2

10

2

110

2

11100 βαβα

1

2

10

2

110

2

1βα

Partial Trace

1100

10102

1

10102

1

22

**

**2

βα

βαβα

βαβα

Tr

1

2

10

2

110

2

1βα

1

2

10

2

110

2

1βα

Distant transformations don’t change the local density

matrix

Notice that the previous observation implies that a unitary transformation on the system that is traced out does not affect the result of the partial trace

I.e.

ρρ

ρ

22,2 Φ

Φ

Trp

UIyUp

yy

Trace

yyy

Distant transformations don’t change the local density

matrix

In fact, any legal quantum transformation on the traced out system, including measurement (without communicating back the answer) does not affect the partial trace

I.e. ρρ 22,

2 Φ

Φ,

Trp

yp

yy

Trace

yy

Why??

Operations on the 2nd system should not affect the statistics of any outcomes of measurements on the first system

Otherwise a party in control of the 2nd system could instantaneously communicate information to a party controlling the 1st system.

Principle of implicit measurement

If some qubits in a computation are never used again, you can assume (if you like) that they have been measured (and the result ignored)

The “reduced density matrix” of the remaining qubits is the same

Partial Trace

This is a linear map that takes bipartite states to single system states.

We can also trace out the first system

We can compute the partial trace directly from the density matrix description

kijljlki

ljTrkiljkiTr

2

Partial Trace using matrices

Tracing out the 2nd system

33223120

13021100

3332

2322

3130

2120

1312

0302

1110

0100

33323130

23222120

13121110

03020100

2

aaaa

aaaa

aa

aaTr

aa

aaTr

aa

aaTr

aa

aaTr

aaaa

aaaa

aaaa

aaaa

Tr

Most general measurement

000U

0000002 Tr