DENSITY MATRICES, traces, Operators and Measurements Lectures 10 ,11 and 12 Richard Cleve Sources: Michael A. Nielsen Michele Mosca 1 Density matrices of pure states Review: We have represented quantum states as vectors (e.g. , 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 is the matrix = Example: the density matrix of 0 + 1 is

2 2 2 Reminder: Trace of a matrix The trace of a matrix is the sum of its diagonal elements a00 a01 a02 e.g. Tr a10 a20 a11 a12 a00 a11 a22 a21 a22

Some properties: Tr xA yB xTr A yTr B Tr AB Tr BA Tr [ ABC ] Tr [CAB ] Tr UAU t Tr A Orthonormal basis {i Tr A i A i } 3 Example: Notation of Density Matrices and traces 0 0 1 1 Notice that 0=0|, and 1=1|. So the probability of getting 0 when measuring | is: 2 p ( 0) 0 0 0 0 2

0 0 0 0 Tr 0 0 Tr 0 0 Tr 0 0 where = || is called the density matrix for the 4 state | 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): 1 , p1 , 2 , p2 , ... 5 Density matrices of mixed states A probability distribution on pure states is called a mixed state:

( (1, p1), (2, p2), , (d, pd)) The density matrix associated with such a mixed state is: d pk k k k 1 Example: the density matrix for ((0, ), (1, )) is: 1 1 0 1 0 0 1 1 0 2 0 0 2 0 1 2 0 1 Question: what is the density matrix of ((0 + 1, ), (0 1, )) ? 6 Density matrix of a mixed state (use of trace) then the probability of measuring 0 is given by conditional probability: p(0) pi prob. of measuring 0 given pure state i i

pi Tr 0 0 i i i Tr pi 0 0 i i i Tr 0 0 wher pi i i is the density matrix for the mixed i e state Density matrices contain all the useful information about 7 an arbitrary quantum state. Recap: operationally indistinguishable 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 8 Applying Unitary Operator to a Density Matrix of a pure state If we apply the unitary operation U to the resulting stateUis with density matrix U U t U U t 9 Applying Unitary Operator to a Density Density Matrix ofMatrix a mixed state How do quantum operations work for these mixed states? qk , k If we apply the unitary operation U to

the resulting state is qk , U k with density matrix t q U U k k k k t U qk k k U k t UU 10 Operators on Density matrices of mixed states. Effect of a unitary operation on a density matrix: applying U to still yields U U Thus this

is true always Effect of a measurement on a density matrix: measuring state with respect to the basis 1, 2,..., d, th still yields the k outcome with probability k k Why? 11 How do quantum operations work using density matrices? 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 (this is because k k = k k = k2 ) and the state collapses to k k 12

More examples of density matrices The density matrix of the mixed state d ((1, p1), (2, p2), ,(d, pd)) is: pk k k Examples (from previous lecture): k 1 1 1 1 1. & 2. 0 + 1 and 0 1 both have 1 1 2 3. 0 with prob. 1 with prob. 4. 0 + 1 with prob. 0 1 with prob. 6. 0 1 0 + 1 0 1 with prob. with prob. with prob.

with prob. 1 1 0 2 0 1 13 More examples of density matrices Examples (continued): 5. 0 with prob. 0 + 1 with prob. has: 1 1 0 1 1 / 2 1 / 2 3 / 4 1 / 2 0 0 1 / 2 1 / 2

1 / 2 1 / 4 2 2 7. The first qubit of 01 10 ...? (later) 14 To Remember: Three Properties of Density Matrices d pk k k k 1 Three properties of : Tr = 1 (Tr M = M11 + M22 + ... + Mdd ) = (i.e. is Hermitian) 0, for all states Moreover, for any matrix satisfying the above properties,

there exists a probabilistic mixture whose density matrix is Exercise: show this 15 Use of Density Matrix and Trace to Calculate the probability of obtaining state in measurement If we perform a Von Neumann of the state measurement wrt a basis containing , the probability of obtaining is 2 Tr This is for a pure

state. How it would be for a 16 mixed state? Use of Density Matrix and Trace to Calculate the probability of obtaining state in measurement (now DensityaMatrix for measuring mixed state) If we perform a Von Neumann qof k , the k state measurement wrt a basis containing the probability ofobtaining is q k k k 2

qk Tr k k k Tr qk k k k Tr The same state 17 Conclusion: Density Matrix Has Complete Information In other words, the density matrix contains all the information necessary to compute the probability of any outcome in any future measurement. 18 Spectral decomposition can be used to represent a useful 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 19 orthonormal basis Continue - Spectral decomposition used to diagonalize the density matrix In other words, we can always diagonalize a density matrix so that it is written as pk k k k where k is an eigenvector with eigenvalue pk and k forms an orthonormal basis

20 Taxonomy of various normal matrices 21 Normal matrices Definition: A matrix M is normal if MM = MM Theorem: M is normal iff there exists a unitary U such that M = UDU, where D is diagonal (i.e. unitarily diagonalizable) 1 0 0 2 D 0 0 0 0 d

Examples of abnormal matrices: 1 0 1 is not even 1 diagonalizable 1 1 is diagonalizable, 0 2 but not unitarily eigenvectors: 22 Unitary and Hermitian matrices Normal: 1 0 0 2 M 0 0

0 0 d with respect to some orthonormal basis Unitary: MM = 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) 23 Positive 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) 24 Projectors 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

d k 1 k 1 Question: which matrices are both projectors and density matrices? Answer: rank-one projectors (k = 1 if k = k0 and k = 0 if k k0 ) 25 Taxonomy of normal matrices normal unitary If Hermitian then normal Hermitian reflection positive semidefinite density matrix

projector rank one projector 26 Review: Bloch sphere for qubits Consider the set of all 2x2 density matrices They have a nice representation in terms of the Pauli matrices: 0 1 x X 1 0 1 0 z Z 0 1 0 i y Y

i 0 Note that these matricescombined with Iform a basis for the vector space of all 2x2 matrices We will express density matrices in this basis 27 Note that the coefficient of I is , since X, Y, Z have trace zero Bloch sphere for qubits: polar coordinates We will express I c x X c yY c z Z 2 First consider the case of pure states , where, without loss of generality, = cos()0 + e2isin()1 (, R) cos 2

e i 2 cossin 1 1 cos 2 e i 2sin 2 i 2 i 2 2 2 sin e cossin e sin 2 1 cos 2 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 28 Bloch sphere for qubits: location of pure and mixed states 0 + = 0 +1 i = 0 1 + +i +i = 0 + i1 i = 0 i1

1 Note that orthogonal corresponds to antipodal here Pure states are on the surface, and mixed states are inside (being weighted averages of pure states) 29 General quantum operations Decoherence, partial traces, measurements. 30 General quantum operations (I) General quantum operations are also called completely positive trace preserving maps, or admissible operations Let A1, A2 , , Am be matrices satisfying m condition t A j A j I j 1

Then the mapping m t A A j j is a general quantum j 1 operator Example 1 (unitary op): applying U to yields U U 31 General quantum operations: Decoherence Operations Example 2 (decoherence): let A0 = 00 and A1 = 11 This quantum op maps to 0000 + 1111 For = 0 + 1, 2 2

2 0 0 2 Corresponds to measuring without looking at the outcome After looking at the outcome, becomes 00 with prob. || 2 11 with prob. | | 2 32 General quantum operations: measurement operations Example 3 (trine state measurement): Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20 3/21 2 1 0 Define A0 = 2/300 3 0 0

1 2 3 2 A1= 2/311 4 2 6 1 2 3 2 A2= 2/322 4 2 6 Then A t A A t A A t A I 0 0 1 1 2 2 Condition satisfied We apply the general quantum mapping operator m

A A j t j j 1 The probability that state k results in outcome state Ak is 2/3. This can be adapted to actually yield the value of probability k with this success 33 General quantum operations: Partial trace discards the second of two qubits Example 4 (discarding the second of two qubits): 1 0 0 0 Let A0 = I0 0 0 1 0 0 1 0 0 and A1 = I1 0 0 0 1

We apply the general quantum mapping operator m t A A j j j 1 State becomes State 1 2 00 1 2

11 1 2 00 1 2 11 1 1 0 becomes 2 0 1 Note 1: its the same density matrix as for ((0, ), (1, )) Note 2: the operation is the partial trace Tr2 34 Distinguishing mixed 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: 35 Distinguishing mixed states (I) Whats the best distinguishing strategy between these two mixed states? 0 with prob. 0 + 1 with prob. 3 / 4 1 / 2 1 1 / 2 1 / 4 1 1 also arises from this orthogonal mixture: 0 with prob.

1 with prob. 1 1 0 2 2 0 1 + 0 as does 2 from: 0 0 with prob. cos2(/8) 1 with prob. sin2(/8)) 0 with prob. 1 with prob. 36 /8=180/8=22.5 Distinguishing mixed states (II) Density matrices 1 and 2 are simultaneously diagonalizable Weve effectively found an orthonormal basis 0, 1 in which both density matrices are diagonal: 1 1 2 1 0 cos / 8)

0 2 2 0 sin / 8) 1 1 2 0 1 + 0 Rotating 0, 1 to 0, 1 the scenario can 0 now be examined using classical probability theory: Distinguish between two classical coins, whose probabilities of heads are cos2(/8)) and respectively (details: exercise)

Question: what do we do if we arent so lucky to get two density matrices that are simultaneously diagonalizable? 37 Reminder: Basic properties of the trace The trace of a square matrix is defined as Tr M d M k ,k k 1 It is easy to check that Tr M N Tr M Tr N and Tr M N Tr N M d

The second property implies Tr M Tr U 1MU k k 1 Calculation maneuvers worth remembering are: Tr a b M b M a and Tr a b c d b c d a Also, keep in mind that, in general, Tr M N Tr M Tr N 38 Partial Trace How can we compute probabilities for a partial system? E.g. xy x y x ,y

xy x y y x y xy py x y x py Partial measurement 39 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

From previous slide E.g. xy y p x y y x p y Trace 2 xy p y , x 2 Tr2 p x y

40 Partial Trace: we derived an important formula to use partial trace y xy py x y x py Derived in previous slide

Trace 2 xy p y , x 2 Tr2 p x y Tr2 p y y y y y x xy py x 41 Why?

w the probability of measuring e.g. in the first register depends onlyTr2 on 2 y 2 wy p y y wy py p yTr w w y y y

Tr w w p y y y y Tr w w Tr2 42 Partial Trace can be calculated in arbitrary basis Notice that it doesnt matter in which orthonormal basis we trace out the 2nd system, e.g. Tr2 2 2 00 11 0 0 1 1 In a different basis

1 1 1 0 1 0 1 00 11 2 2 2 1 1 1 0 1 0 1 2 2 2 43 (cont) Partial Trace can be calculated in Partial Trace arbitrary basis 1 0 1 2

1 0 1 2 1 Tr2 0 2 1 0 2 2 1 0 2 1 0 2 1 1 2 1 1 2

1 * 0 * 1 1 * 0 * 1 2 0 0 1 1 Which is the same as in previous slide for other base 44 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 Tr 2 i k j l i k Tr j l i k l j l j i k 45 Partial Trace using matrices a00 a 10 a20 a30 Tracing out the 2nd system a01 a02

a11 a12 a21 a22 a31 a32 Tr 2 a03 a00 Tr a13 Tr2 a10 a20 a23 Tr a33 a30 a01 a11 a21 a31 a02 Tr a12 a22

Tr a32 a03 a13 a23 a33 a00 a11 a02 a13 a20 a31 a22 a33 46 Examples: Partial trace (I) Two quantum registers (e.g. two qubits) in states and (respectively) are independent if then the combined system is in state = In such circumstances, if the second register (say) is discarded then the state of the first register remains 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( ) = index means

For example, it turns out that Tr2( 1 2 00 1 2 11 1 2 00 1 2 11 ) = 1 1 0 2 0 1

2nd system traced out 47 Examples: Partial trace (II) Weve already seen this defined in the case of 2-qubit systems: discarding the second of two qubits 1 0 0 0 Let A0 = I0 0 0 1 0 0 1 0 0 and A1 = I1 0 0 0 1 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? 48 Examples: Partial trace (III): calculating matrices of partial traces For 2-qubit systems, the partial trace is explicitly 00 ,00 Tr2 01,00 10 ,00 11,00 00 ,01 00 ,10 01,01 10 ,01 01,10 10 ,10

11,01 11,10 00 ,01 00 ,10 01,01 01,10 10 ,01 11,01 10 ,10 11,10 00 ,11 01,11 00 ,00 01,01 10 ,11 10 ,00 11,01 11,11 00 ,10 01,11 10 ,10 11,11 00 ,11 01,11 00 ,00 10 ,10

10 ,11 01,00 11,10 11,11 00 ,01 10 ,11 01,01 11,11 and 00 ,00 Tr1 01,00 10 ,00 11,00 49 Unitary transformations dont 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. p y y U y I U

y Trace2 p y , y 2 Tr2 50 Distant transformations dont 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 py , y y I.e.

Trace2 p y , y 2 Tr2 51 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. 52

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 53 POVMs (I) Positive operator valued measurement (POVM): m t A Let A1, A2 , , Am be matrices satisfying j A j I j 1 Then the corresponding POVM is a stochastic operation on that, with probability Tr A j Atj

produces the outcome: j (classical information) A j Atj Tr A j Atj (the collapsed quantum state) Example 1: 1 Aj = jj (orthogonal projectors) This reduces to our previously defined measurements 54 POVMs (II): calculating the measurement outcome and the collapsed quantum state When Aj = jj are orthogonal projectors and = , Tr A j Atj = Trjjjj = jjjj 2 = j Moreover,

A j Atj Tr A j A t j probability of the outcome: j j j j j 2 j j (the collapsed quantum state) 55 The measurement postulate formulated in terms of observables

Our f orm: A measurement is described by a complete set of proj ectors Pj onto orthogonal subspaces. Outcome j occurs with probability Pr(j ) Pj . The corresponding post-measurement state is Pj . Pj This is a projector matrix 56 The measurement postulate formulated in terms of observables Our f orm: A measurement is described by a complete set of proj ectors Pj onto orthogonal subspaces. Outcome j occurs with probability Pr(j ) Pj . The corresponding post-measurement state is Pj . Pj Old f orm: A measurement is described by an observable, a Hermitian operator M , with spectral decomposition M j j Pj . The possible measurement outcomes correspond to the eigenvalues j , and the outcome j occurs with probability

Pr(j ) Pj . The corresponding post-measurement state is Pj . The same Pj 57 An example of observables in action Example: Suppose we "measure Z". Z has spectral decomposition Z 0 0 - 1 1, so this is j ust like measuring in the computational basis, and calling the outcomes "1" and "-1", respectively, f or 0 and 1. Exercise: Find the spectral decomposition of Z Z . Show that measuring Z Z corresponds to measuring the parity of two qubits, with the result +1 corresponding to even parity, and the result -1 corresponding to odd parity. Hint: Z Z 00 00 11 11 10 10 01 01 58 An example of observables in action Exercise: Suppose we measure the observable M f or a state which is an eigenstate of that observable. Show that, with certainty, the outcome of the measurement is

the corresponding eigenvalue of the observable. 59 What can be measured in 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? 60 What can be measured in quantum mechanics? Traditional approach to quantum measurements: A quantum measurement is described by an observable M M is a Hermitian operator acting on the state space of the system. Measuring a system prepared in an eigenstate of M gives the corresponding eigenvalue of M as the measurement outcome. The question now presents itself Can every observable be measured? The answer theoretically is yes. In practice

it may be very awkward, or perhaps even beyond the ingenuity of 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 61 Von Neumann measurement in the computational basis Suppose we have a universal set of quantum gates, and the ability to measure each {0qubit , 1 } in the basis If we measure ( 0 1 ) we get 0 1 2 with probability b b 62

In section 2.2.5, this is described as follows We have the projection operators P0 0 0 and P1 1 1 satisfying P0 P1 I We consider the projection operator or observable M 0P0 1P1 P1 Note that 0 and 1 are the eigenvalues When we measure this observable M, the b probability of getting the eigenvalue is 2 and we are Pr(b) Pb b Pb b in that case left with the state b 63 b p(b) b What is an Expected value of an observable If we associate with outcomeb the eigenvalueb

then the expected outcome is b Pr(b) b b Pb bPb b b Tr bPb Tr M b 64 Von Neumann measurement in the computational basis Suppose we have a universal set of quantum gates, and the ability to measure each {0qubit , 1 } in the basis x x Say we have the state x{0,1}n

If we measure all n qubits, then we 2 x obtain with probability x Notice that this means that probability of measuring a 0 in the first qubit 2 x equals 65 x0 {0,1}n 1 Partial measurements If we only measure the first qubit and 0 get leave the rest alone, 2then we still p 0

with probabilityx0{0,1} x The remaining n-1 qubits are then in the renormalized state x n 1 x0 {0,1}n 1 p0 x (This is similar to Bayes Theorem) 66 Most general measurement k k 000

U 67 In section 2.2.5 This partial measurement corresponds to measuring the observable M 0 0 0 I n 1 11 1 I n 1 68 Von Neumann Measurements A Von Neumann measurement is a type of projective measurement. Given an orthonormal basis { k } , if we perform a Von Neumann measurement with { k } to state k k respect

of the then wemeasure with probability k 2 k k 2 k k Tr k k Tr k k 69 Von Neumann Measurements E.x. Consider Von Neumann (of 0the state 1) measurement 0 1 0 1 basis

with respect to the orthonormal , 2 Note that 2 0 1 0 1 2 2 2 2 0 1

2 We therefore get with probability 2 2 70 Von Neumann Measurements Note that 0 1 2 2 0 1 * *

2 2 0 1 0 1 2 2 0 1 0 1 2 Tr 2 2 2 71 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 { k } basis as follows. 72 How do we implement Von Neumann measurements? Construct a quantum network that implements the unitary transformation U k k Then conjugate the measurement operation with the operationU k

k U k prob k U 2 1 k 73 Another approach k U k U 1 k k

prob k 000 k k k 000 k k k k k k 2 000 k These two approaches74 will be illustrated in next slides

Example: Bell basis change Consider the orthonormal basis consisting of the Bell states 00 00 11 01 01 10 10 00 11 11 01 10 Note that x y H xy We discussed Bell basis in lecture about superdense 75 coding and teleportation.

Bell measurements: destructive and non-destructive We can destructively measure x H x,y xy y x, y prob xy 2 Or non-destructively project x, y xy

x, y 00 H H x,y xy prob 76xy 2 Most general measurement 000 U Tr2 000 000 77 Simulations among operations: general quantum operations Fact 1: any general quantum operation can be simulated by applying a unitary operation on a larger quantum system: input zeros

output 0 0 0 Example: decoherence 0 + 1 0 U discard 2 0 discard 0 2 78 Simulations among operations: simulations of POVM Fact 2: any POVM can also be simulated by applying a

unitary operation on a larger quantum system and then measuring: quantum output input 0 0 0 U j classical output 79 Separable states A bipartite (i.e. two register) state product state if is a: = m p separable state if j

j j j 1 ( p1 ,, pm 0) (i.e. a probabilistic mixture of product states) Question: which of the following states are separable? 1 12 00 11 00 11 2 12 00 11 00 11 00 1 2 11 00 11

80 Continuous-time evolution Although weve expressed quantum operations in discrete terms, in real physical systems, the evolution is continuous Let H be any Hermitian matrix and t R Then eiHt is unitary why? 1 D H = U DU, where 1 d e i1t t iDt iHt Therefore e = U e U = U

0 U eid t (unitary) 81 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 82