1

Richard Cleve

Lectures 10 ,11 and 12

DENSITY DENSITY MATRICES, traces, MATRICES, traces,

Operators and Operators and MeasurementsMeasurements

Michael A. Nielsen

Michele Mosca

Sources:

2

ReviewReview: : Density matrices Density matrices of pure statesof pure states

We have represented quantum states as vectors (e.g. ψ, and all such states are called pure states)

An alternative way of representing quantum states is in terms of density matrices (a.k.a. density operators)

The density matrix of a pure state ψ is the matrix = ψ ψ

Example: the density matrix of 0 + 1 is

2

2

ββα

αβαβα

β

αρ

3

Reminder:Reminder: Trace of a matrix Trace of a matrixThe 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

4

Example:Example: Notation of Density Notation of Density Matrices and tracesMatrices and traces

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

5

Review:Review: 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

6

Density matrices of mixed statesDensity matrices of mixed states

A probability distribution on pure states is called a mixed state:

( (ψ1, p1), (ψ2, p2), …, (ψd, pd))

The density matrix associated with such a mixed state is:

d

kkkkp

1

ψψρ

Example: the density matrix for ((0, ½ ), (1, ½ )) is:

10

01

2

1

10

00

2

1

00

01

2

1

Question: what is the density matrix of

((0 + 1, ½ ), (0 − 1, ½ )) ?

7

Density matrix of a mixed Density matrix of a mixed state (use of trace)state (use of trace)

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

i

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

ρ

φφ

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.

8

RecapRecap: operationally : operationally indistinguishable statesindistinguishable states

Since these are expressible in terms of density matrices alone (independent of any specific probabilistic mixtures), states with identical density matrices are operationally indistinguishable

9

Applying Unitary Operator to a Applying Unitary Operator to a Density Matrix of a pure stateDensity Matrix of a pure state

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

with density matrix

U

tt UUUU

10

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

Applying Unitary Operator to a Applying Unitary Operator to a Density Matrix of a Density Matrix of a mixedmixed state state

How do quantum operations work for these mixed states?

11

Operators on Density matrices of Operators on Density matrices of mixed mixed states.states.

Effect of a unitary operation on a density matrix:

applying U to still yields U U†

Effect of a measurement on a density matrix: measuring state with respect to the basis 1, 2,..., d,

still yields the k th outcome with probability k kWhy?

Thus this is true always

12

Effect of a measurement on a density matrix: measuring state with respect to the basis 1, 2,..., d, yields the k

th outcome with probability

k k

How do quantum operations How do quantum operations work using density matrices?work using density matrices?

(this is because k k = kψ ψk = kψ2 )

—and the state collapses to k k

13

More examples of density matricesMore examples of density matrices

The density matrix of the mixed state

((ψ1, p1), (ψ2, p2), …,(ψd, pd)) is:

d

kkkk ψψpρ

1

1. & 2. 0 + 1 and −0 − 1 both have

3. 0 with prob. ½ 1 with prob. ½

4. 0 + 1 with prob. ½ 0 − 1 with prob. ½

6. 0 with prob. ¼ 1 with prob. ¼ 0 + 1 with prob. ¼ 0 − 1 with prob. ¼

Examples (from previous lecture):

11

11

2

1ρ

10

01

2

1ρ

14

5. 0 with prob. ½ 0 + 1 with prob. ½

7. The first qubit of 01 − 10

Examples (continued):

4/12/1

2/14/3

2/12/1

2/12/1

2

1

00

01

2

1has:

...? (later)

More examples of density matricesMore examples of density matrices

15

To Remember:To Remember: Three Properties of Density Three Properties of Density MatricesMatrices

Three properties of :• Tr = 1 (Tr M = M11 + M22 + ... + Mdd )

• =† (i.e. is Hermitian)

• 0, for all states

d

kkkk ψψpρ

1

Moreover, for any matrix satisfying the above properties,

there exists a probabilistic mixture whose density matrix is

Exercise: show this

16

Use of Density Matrix and Trace to Use of Density Matrix and Trace to Calculate the probability of obtaining Calculate the probability of obtaining

state in measurementstate in measurement

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

Tr2 This is for a pure

state.

How it would be for a mixed state?

17

Density Matrix

If 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

Use of Density Matrix and Trace to Calculate the Use of Density Matrix and Trace to Calculate the probability of obtaining state in measurement (now probability of obtaining state in measurement (now

for measuring a mixed state)for measuring a mixed state)

The same state

18

Conclusion: Conclusion: Density Matrix HasDensity Matrix Has Complete InformationComplete Information

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

19

Spectral decomposition can be used to Spectral decomposition can be used to represent a represent a useful form of density matrixuseful form of density matrix

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

20

Continue Continue - Spectral decomposition used - Spectral decomposition used to diagonalize the density matrixto diagonalize the density matrix

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

21

Taxonomy of Taxonomy of various normal various normal

matricesmatrices

22

Normal matricesNormal matricesDefinition: A matrix M is normal if M†M = MM†

Theorem: M is normal iff there exists a unitary U such that

M = U†DU, where D is diagonal (i.e. unitarily diagonalizable)

Examples of abnormal matrices:

10

11 is not even diagonalizable

20

11 is diagonalizable, but not unitarily

eigenvectors:

dλ

λ

λ

D

00

00

00

2

1

23

Unitary and Hermitian matricesUnitary and Hermitian matrices

dλ

λ

λ

M

00

00

00

2

1 with respect to some orthonormal basis

Normal:

Unitary: M†M = I which implies |k |2 = 1, for all k

Hermitian: M = M† which implies k R, for all k

Question: which matrices are both unitary and Hermitian?

Answer: reflections (k {+1,1}, for all k)

24

Positive semidefinite matricesPositive semidefinite matrices

Positive semidefinite: Hermitian and k 0, for all k

Theorem: M is positive semidefinite iff M is Hermitian and,

for all , M 0

(Positive definite: k > 0, for all k)

25

Projectors and density matricesProjectors and density matrices

Projector: Hermitian and M 2 = M, which implies that M is

positive semidefinite and k {0,1}, for all k

Density matrix: positive semidefinite and Tr M = 1, so 11

d

kkλ

Question: which matrices are both projectors and density matrices?

Answer: rank-one projectors (k = 1 if k = k0 and k = 0 if k k0 )

26

Taxonomy of normal matricesTaxonomy of normal matrices

normal

unitary Hermitian

reflectionpositive

semidefinite

projectordensitymatrix

rank oneprojector

If Hermitian then If Hermitian then normalnormal

27

Review:Review: Bloch sphere for qubits Bloch sphere for qubits

Consider the set of all 2x2 density matrices

Note that the coefficient of I is ½, since X, Y, Z have trace zero

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

28

Bloch sphere for qubits: Bloch sphere for qubits: polar polar coordinatescoordinates

2

ZcYcXcIρ zyx We will express

First consider the case of pure states , where, without

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

θ2cos1θ2sin

θ2sinθ2cos1

2

1

θsinθsinθcos

θsinθcosθcosρ

φ2

φ2

2φ2

φ22

i

i

i

i

e

e

e

e

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

29

Bloch sphere for qubits: location of pure Bloch sphere for qubits: location of pure and mixed statesand mixed states

+

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

30

General General quantum quantum

operationsoperationsDecoherence, partial traces, measurements.

31

General quantum operations (I)General quantum operations (I)

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

General quantum operationsGeneral quantum operations are also called “completely positive trace preserving maps”, or “admissible operations”

IAA j

m

jj

1

t

Then the mapping

m

jjj AA

1

t is a general quantum operator

Let A1, A2 , …, Am be matrices satisfying

conditioncondition

32

General quantum operations: General quantum operations: Decoherence Operations Decoherence Operations

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

33

General quantum operations: measurement General quantum operations: measurement operationsoperations

Example 3 (trine state “measurement”):

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

Then IAAAAAA 221100ttt

•The probability that state k results in “outcome” state Ak is 2/3.

•This can be adapted to actually yield the value of k with this success probability

00

01

3

2Define A0 = 2/300

A1= 2/311 A2= 2/322

62

232

4

1

62

232

4

1

We apply the general quantum mapping operator

m

jjj AA

1

t

Condition Condition satisfiedsatisfied

34

General quantum operations: General quantum operations: Partial tracePartial trace discards discards the second of two qubitsthe second of two qubits

Example 4 (discarding the second of two qubits):

Let A0 = I0 and A1 = I1

0100

0001

1000

0010

State becomes

State becomes 110011002

12

12

12

1

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

10

01

2

1

Note 2: the operation is the partial trace Tr22

We apply the general quantum mapping operator

m

jjj AA

1

t

35

Distinguishing Distinguishing mixed statesmixed states

Several mixed states can have the same density matrix – we cannot distinguish between them.

How to distinguish by two different density matrices?

Try to find an orthonormal basis 0, 1 in which both density matrices are diagonal:

36

Distinguishing mixed states (I)Distinguishing mixed states (I)

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:

/8=180/8=22.5

37

Distinguishing mixed states (II)Distinguishing mixed states (II)

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

Density matrices 1 and 2 are simultaneously diagonalizable

38

Reminder:Reminder: Basic properties of Basic properties of the tracethe trace

d

kk,kMM

1

Tr

NMNM TrTrTr

NMNM TrTrTr

MNNM TrTr

adcbdcba Tr

d

kkMUUM

1

1 λTrTr

The trace of a square matrix is defined as

It is easy to check that

The second property implies

and

Calculation maneuvers worth remembering are:

aMMa bb Tr and

Also, keep in mind that, in general,

39

Partial Partial TraceTrace How can we compute probabilities

for a partial system? E.g.

yxp

p

yx

yx

y x y

xyy

y xxy

yxxy

,

Partial measurement

40

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

From previous slide

41

Partial Trace: we derived an important formula to use 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

Derived in previous slide

42

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

43

Partial Trace can be calculated in arbitrary basis

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

44

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

(cont) Partial Trace can be calculated in arbitrary basis

Which is the same as in Which is the same as in previous slide for other previous slide for other basebase

45

Methods to calculate the Partial Trace

Partial Trace 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

46

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

Tr Tr 22

47

ExamplesExamples: Partial trace (I): Partial trace (I)

In such circumstances, if the second register (say) is discarded then the state of the first register remains

Two quantum registers (e.g. two qubits) in states and (respectively) are independent if then 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)

We can define the partial trace, Tr2 , as the unique linear

operator satisfying the identity Tr2( ) = For example, it turns out that

110011002

12

12

12

1

10

01

2

1Tr2( ) =

index means 2nd system traced out

48

Examples:Examples: Partial trace (II) Partial trace (II)We’ve already seen this defined in the case of 2-qubit systems: discarding the second of two qubits

Let A0 = I0 and A1 = I1

0100

0001

1000

0010

For the resulting quantum operation, state becomes

For d-dimensional registers, the operators are Ak = Ik ,

where 0, 1, …, d1 are an orthonormal basis

As we see in last slide, partial trace is a matrix.

How to calculate this matrix of partial trace?

49

Examples:Examples: Partial trace (III): Partial trace (III): calculating calculating matrices of partial tracesmatrices of partial traces

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

50

Unitary transformations don’t change the local density matrix

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

51

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

52

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.

53

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

54

POVMs (I)POVMs (I)Positive operator valued measurementPositive operator valued measurement (POVM):

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

m

jj

1

t

Then the corresponding POVM is a stochastic operation on

that, with probability produces the outcome:

j (classical information)

tjj AρATr

t

t

jj

jj

AρA

AρA

Tr(the collapsed quantum state)

Example 1Example 1: Aj = jj (orthogonal projectors)

This reduces to our previously defined measurements …

55

POVMs (II):POVMs (II): calculating the measurement calculating the measurement outcome and the collapsed quantum stateoutcome and the collapsed quantum state

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

(the collapsed quantum state)

probability of the outcome:

56

The measurement postulate The measurement postulate formulatedformulated

in terms of “observables”in terms of “observables”

A measurement is described by a complete set of projectors onto orthogonal subspaces. Outcome occurs with probability Pr( ) .The corresponding post-measure

Our

ment state is

f orm

:j

j

P j

j P

P

.j

jP

This is a projector matrix

57

The measurement postulate formulatedThe measurement postulate formulatedin terms of “observables”in terms of “observables”

A measurement is described by a complete set of projectors onto orthogonal subspaces. Outcome occurs with probability Pr( ) .The corresponding post-measure

Our

ment state is

f orm

:j

j

P j

j P

P

.j

jP

A measurement is described by an ,

a Hermitian operator , with spectral decomposiOld f orm: o

tion

bservable

.j jj

MM P

The possible measurement outcomes correspond to theeigenvalues , and the outcome occurs with probability Pr( ) .

j j

j jP

The corresponding post-measurement state is

.j

j

P

P

The same

58

An example of observables in action

Suppose we "measE uxample: re ".Z has spectral decomposition 0 0 - 1 1, so

this is just like measuring in the computational basis,and calling the outcomes "1" and "-1", respectively, f or0 and 1.

Z Z

Find the spectral decomposition of .Show that measuring corresponds to measuringthe parity of two qubits, with the result +1 correspondingto even parity, and the result

Exercis

-1 correspon

:

i

e

d

Z ZZ Z

ng to oddparity.

00 00 11Hint: 11 10 10 01 01Z Z

59

An example of observables in action

Suppose we measure the observable f or astate which is an eigenstate of that observable. Showthat, with certainty, the outcome of the measurement isthe corresponding eigenvalue

Exerci

of the ob

se: M

servable.

60

What can be measured in What can be measured in quantum mechanics?quantum mechanics?

Computer science can inspire fundamental questions about physics.

We may take an “informatic” approach to physics.(Compare the physical approach to information.)

Problem: What measurements can be performed in quantum mechanics?

61

•“Traditional” approach to quantum measurements: A quantum measurement is described by an observable M•M is a Hermitian operator acting on the state spaceof the system.Measuring a system prepared in an eigenstate of

M gives the corresponding eigenvalue of M as themeasurement outcome.

“The question now presents itself – Can every observablebe measured? The answer theoretically is yes. In practiceit may be very awkward, or perhaps even beyond the ingenuityof the experimenter, to devise an apparatus which could measure some particular observable, but the theory always allows one to imagine that the measurement could be made.” - Paul A. M. Dirac

What can be measured in quantum mechanics?

62

““Von Neumann measurement Von Neumann measurement in the computational basis”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

63

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

64

What is an “Expected value” What is an “Expected value” of an observableof 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

65

““Von Neumann measurement in Von Neumann measurement in the computational basis”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

66

Partial measurementsPartial 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

67

Most general measurementMost general measurement

kk

000U

68

In section 2.2.5

This partial measurement corresponds to measuring the observable

1n1n I111I000M

69

Von Neumann MeasurementsVon 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

70

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

71

Von Neumann Measurements

Note that22

10

22

10 **

22

10

2

10Tr

2

10

2

10

2

72

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

73

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

74

Another approach

kk U 1U

kk

000

kkk

000k000

kkk

kkk

2

kprob

These two approaches will These two approaches will be illustrated in next slidesbe illustrated in next slides

75

Example:Example: Bell basis changeBell basis change

100101

Consider the orthonormal basis consisting of the “Bell” states

110000

110010 100111 Note that

xyx

y

H

We discussed Bell basis in lecture about superdense coding and teleportation.

76

Bell measurements: destructivedestructive and non-destructive

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

77

Most general measurement Most general measurement

000U

0000002 Tr

78

SimulationsSimulations among operations: among operations: general quantumgeneral quantum operations operations

Fact 1:Fact 1: any general quantum operation can be simulated by applying a unitary operation on a larger quantum system:

U000

Example: decoherence

0

0 + 1

2

2

0

0

β

αρ

output

discard

input

zeros discard

79

Simulations among operations: Simulations among operations: simulations of POVMsimulations of POVM

Fact 2:Fact 2: any POVM can also be simulated by applying a unitary operation on a larger quantum system and then measuring:

U000

quantum outputinput

classical outputj

80

Separable statesSeparable states

m

jjjjp

1

• product state if =

• separable state if

A bipartite (i.e. two register) state is a:

Question: which of the following states are separable?

1100110011001100 21

21

2 ρ

(i.e. a probabilistic mixture of product states)

( p1 ,…, pm 0)

1100110021

1 ρ

81

Continuous-time evolutionContinuous-time evolutionAlthough we’ve expressed quantum operations in discrete terms, in real physical systems, the evolution is continuous

0

1Let H be any Hermitian matrix and t R

Then eiHt is unitary – why?

H = U†DU, where

dλ

λ

D 1

Therefore eiHt = U† eiDt U = U

e

e

Utλi

tλi

d

1

t (unitary)

82

Partially covered in 2007:

• Density matrices and indistinguishable states• Taxonomy of normal operators• General Quantum Operations• Distinguishing states• Partial trace• POVM• Simulations of operators• Separable states• Continuous time evolution