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Quantum Information Stephen M. Barnett University of Strathclyde [email protected] The Wolfson Foundation 1. Probability and Information 2. Elements of Quantum Theory 3. Quantum Cryptography 4. Generalized Measurements 5. Entanglement 4.1 4.2 4.3 4.4 4.5 6. Quantum Information Processing 7. Quantum Computation 8. Quantum Information Theory Ideal von Neumann measurements Non-ideal measurements Probability operator measures Optimised measurements Operations 4.1 Ideal von Neumann measurements Measurement model What are the probabilities for the measurement outcomes? How does the measurement change the quantum state? Red ‘Black box’ von Neumann measurements Mathematical Foundations of Quantum Mechanics (i) An observable A is represented by an Hermitian operator: Aˆ n n n (ii) A measurement of A will give one of its eigenvalues as a result. The probabilities are P(n ) n or 2 n n P(n ) n ˆ n Tr n n ˆ (iii) Immediately following the measurement, the system is left in the associated eigenstate: ˆ n n More generally for observables with degenerate spectra: (i’) Let P̂n be the projector onto eigenstates with eigenvalue n : Aˆ Pˆn n Pˆn (ii’) The probability that the measurement will give the result n is P(n ) Tr Pˆn ˆ (iii’) The state following the measurement is Pˆn ˆPˆn ˆ Tr ( Pˆn ˆ ) b PˆDetect x x dx a PˆNo detect Î - PˆDetect a b Detector Properties of projectors † ˆ Pn Pˆn I. They are Hemitian II. They are positive Pˆn 0 III. They are complete ˆ P n Î Observable Probabilities Probabilities n IV. They are orthonormal Pˆi Pˆj ij Pˆi ?? 4.2 Non-ideal measurements Real measurements are ‘noisy’ and this leads to errors P(0) 1 p 0 ‘Black box’ P (1) p P(0) p 1 ‘Black box’ P(1) 1 p For a more general state: P(0) (1 p) 0 ˆ 0 p 1 ˆ 1 P(1) (1 p) 1 ˆ 1 p 0 ˆ 0 We can write these in the form P(0) Tr (ˆ 0 ˆ ) P(1) Tr (ˆ1 ˆ ) where we have introduced the probability operators ˆ 0 (1 p) 0 0 p 1 1 ˆ1 (1 p) 1 1 p 0 0 These probability operators are † ˆ 0,1 ˆ0,1 I. Hermitian II. positive ˆ 0 (1 p) 0 ˆ1 (1 p) 1 III. complete 2 2 p 1 2 0 p 0 2 0 ˆ 0 ˆ1 0 0 1 1 Î But IV. they are not orthonormal ˆ0ˆ1 p(1 p)Î 0 We seem to need a generalised description of measurements 4.3 Probability operator measures Our generalised formula for measurement probabilities is P(i) Tr ˆi ˆ The set probability operators describing a measurement is called a probability operator measure (POM) or a positive operator-valued measure (POVM). The probability operators can be defined by the properties that they satisfy: Properties of probability operators I. They are Hermitian II. They are positive III. They are complete ˆ ˆ n † n ˆ n 0 ˆ n Î Observable Probabilities Probabilities n IV. Orthonormal ˆ iˆ j ijˆ i ?? Generalised measurements as comparisons Prepare an ancillary system in a known state: A S Perform a selected unitary transformation to couple the system and ancilla: S+A S S A Uˆ S A A Perform a von Neumann measurement on both the system and ancilla: i Si Ai The probability for outcome i is P(i) i Uˆ A S S A Uˆ † i S A Uˆ † i i Uˆ A S ˆi The probability operators ˆ i act only on the system state-space. POM rules: I. Hermiticity: A Uˆ † i i Uˆ A † ˆ AU i † i Uˆ A II. Positivity: ˆi i Uˆ S A 2 III. Completeness follows from: i i i Î A,S 0 Generalised measurements as comparisons We can rewrite the detection probability as P(i) A S Pˆi S A † ˆ ˆ Pi U i i Uˆ is a projector onto correlated (entangled) states of the system and ancilla. The generalised measurement is a von Neumann measurement in which the system and ancilla are compared. ˆi A Pˆi A ˆ nˆ m A Pˆn A A Pˆm A 0 Simultaneous measurement of position and momentum The simultaneous perfect measurement of x and p would violate complementarity. p Position measurement gives no momentum information and depends on the position probability distribution. x Simultaneous measurement of position and momentum The simultaneous perfect measurement of x and p would violate complementarity. Momentum measurement gives no position information and p depends on the momentum probability distribution. x Simultaneous measurement of position and momentum The simultaneous perfect measurement of x and p would violate complementarity. p Joint position and measurement gives partial information on both the position and the momentum. x Position-momentum minimum uncertainty state. POM description of joint measurements Probability density: ( xm , pm ) Trˆˆ ( xm , pm ) Minimum uncertainty states: xm , pm 2 2 1 / 4 ( x xm ) 2 pm x dx exp 4 2 i x 1 dxm dpm xm , pm xm , pm Î 2 This leads us to the POM elements: 1 ˆ ( xm , pm ) xm , p m xm , p m 2 The associated position probability distribution is ( x xm ) 2 ( xm ) dx x ˆ x exp 2 2 Var ( xm ) x 2 2 2 & Var ( pm ) p 2 4 2 Increased uncertainty is the price we pay for measuring x and p. The communications problem ‘Alice’ prepares a quantum system in one of a set of N possible signal states and sends it to ‘Bob’ i selected. prob. pi Preparation device ̂ i Measurement device P( j | i) Tr ˆ j ˆ i Bob is more interested in Tr ˆ j ˆ i pi P (i | j ) Tr (ˆ j ˆ ) Measurement result j In general, signal states will be non-orthogonal. No measurement can distinguish perfectly between such states. Were it possible then there would exist a POM with 1 ˆ1 1 1 2 ˆ 2 2 2 ˆ1 2 0 1 ˆ 2 1 Completeness, positivity and 1 ˆ1 1 1 ˆ1 1 1 Aˆ 2 ˆ1 2 1 2 Aˆ positive and Aˆ 1 0 2 2 Aˆ 2 1 2 2 0 What is the best we can do? Depends on what we mean by ‘best’. Minimum-error discrimination We can associate each measurement operator ˆ i with a signal state ̂ i . This leads to an error probability Pe 1 p jTr ˆ j ˆ j N j 1 Any POM that satisfies the conditions ˆ j ( p j ˆ j pk ˆ k )ˆ k 0 N p ˆ ˆ k 1 k k k p j ˆ j 0 will minimise the probability of error. j , k j For just two states, we require a von Neumann measurement with projectors onto the eigenstates of p1 ˆ1 p2 ˆ 2 with positive (1) and negative (2) eigenvalues: Pemin 12 1 Tr p1 ˆ1 p2 ˆ 2 Consider for example the two pure qubit-states 1 cos 0 sin 1 1 2 cos(2 ) 2 cos 0 sin 1 The minimum error is achieved by measuring in the orthonormal basis spanned by the states 1 and 2 . 1 1 2 2 We associate 1 with 1 and 2 with 2 : Pe p1 1 2 2 p2 2 1 2 The minimum error is the Helstrom bound Pemin 1 2 1 1 4 p1 p2 1 2 1/ 2 2 A single photon only gives one “click” P = |a|2 a +b P = | b| 2 But this is all we need to discriminate between our two states with minimum error. A more challenging example is the ‘trine ensemble’ of three equiprobable states: 0 31 1 12 0 3 1 p1 2 12 p2 3 0 1 3 p3 1 3 1 3 It is straightforward to confirm that the minimum-error conditions are satisfied by the three probability operators ˆi 23 i i Simple example - the trine states Three symmetric states of photon polarisation 3 2 3 1 2 3 2 Minimum error probability is 1/3. This corresponds to a POM with elements 2 ˆ j j j 3 How can we do a polarisation measurement with these three possible results? Polarisation interferometer - Sasaki et al, Clarke et al. 1 / 6 2 / 3 1 / 6 0 0 0 0 0 0 1 / 6 1 / 6 2 / 3 PBS 2 0 0 0 0 3 / 2 3 / 2 PBS 1 1 / 2 1 / 2 0 3 / 2 3 / 2 2 11/ 2 1/ 2 0 0 0 PBS 0 0 0 2 / 3 1 / 6 1 / 6 1 / 3 1 / 12 1 / 12 2 / 3 1 / 6 1 / 6 Unambiguous discrimination The existence of a minimum error does not mean that error-free or unambiguous state discrimination is impossible. A von Neumann measurement with Pˆ1 1 1 ˆ P1 1 1 will give unambiguous identification of 2 : result 1 2 error-free result 1 ? inconclusive There is a more symmetrical approach with ˆ1 ˆ 2 1 1 1 2 1 1 1 2 2 2 1 1 ˆ ? Î ˆ1 ˆ 2 Result 1 Result 2 Result ? State 1 1 1 2 0 1 2 State 2 0 1 1 2 1 2 How can we understand the IDP measurement? Consider an extension into a 3D state-space b a a b Unambiguous state discrimination - Huttner et al, Clarke et al. a ? b a b A similar device allows minimum error discrimination for the trine states. Measurement model What are the probabilities for the measurement outcomes? How does the measurement change the quantum state? Red ‘Black box’ 4.5 Operations How does the state of the system change after a measurement is performed? Problems with von Neumann’s description: 1) Most measurements are more destructive than von Neumann’s ideal. 2) How should we describe the state of the system after a generalised measurement? Physical and mathematical constraints What is the most general way in which we can change a density operator? Quantum theory is linear so … ˆ Aˆ ˆBˆ or more generally ˆ Aˆi ˆBˆi i BUT this must also be a density operator. Properties of density operators I. They are Hermitian ˆ † ˆ II. They are positive ˆ 0 III. They have unit trace Tr ˆ 1 The first of these tells us that the second is satisfied. † ˆ ˆ Bi Ai and this ensures that The final condition tells us that i † ˆ ˆ Ai Ai Î The operator Aˆi† Aˆi is positive and this leads us to associate ˆi Aˆi† Aˆi (Knowing ˆ i does not give us Âi .) If the measurement result is i then the density operator changes as Aˆi ˆAˆi† Aˆi ˆAˆi† ˆ † Tr ( Aˆi Aˆi ˆ ) Tr (ˆi ˆ ) This replaces the von Neumann transformation Pˆn ˆPˆn ˆ Tr ( Pˆn ˆ ) If the measurement result is not known then the transformed density operator is the probability-weighted sum † ˆ ˆ ˆ A A i i ˆ ˆAˆ † ˆ Tr ( Aˆi† Aˆi ˆ ) A i i † ˆ ˆ Tr ( Ai Ai ˆ ) i i which has the form required by linearity. We refer to the operators or an an ‘effect’. Âi and ˆA† i as Krauss operators Repeated measurements Suppose we perform a first measurement with results i and effect operators Âi and then a second with outcomes j and effects B̂ j . The probability that the second result is j given that the first was i is Tr ( Bˆ †j Bˆ j Aˆi ˆAˆi† ) P( j | i ) Tr ( Aˆ † Aˆ ˆ ) i i P(i, j ) P( j | i) P(i) Tr ( Aˆi† Bˆ †j Bˆ j Aˆi ˆ ) Hence the combined probability operator for the two measurements is ˆij Aˆi† Bˆ †j Bˆ j Aˆi If the results i and j are recorded then ˆ † ˆ† ˆ ˆ ˆ ˆ B j Ai Ai B j P (i, j ) If they are not known then † ˆ† ˆ ˆ ˆ ˆ ˆ B j Ai Ai B j i, j Unitary and non-unitary evolution The effects formalism is not restricted to describing measurements, e.g. Schroedinger evolution ˆ (0) ˆ (t ) exp iHˆ t / ˆ (0) exp iHˆ t / which has a single Krauss operator Aˆ exp iHˆ t / . We can also use it to describe dissipative dynamics: e Spontaneous decay rate 2g g We can write the evolved density operator in terms of two effects: ˆ (t ) Aˆ N (t ) ˆ (0) Aˆ N† AˆY (t ) ˆ (0) AˆY† gt ˆ AN (t ) e e e g g 2gt ˆ AY (t ) 1 e g e Measurement interpretation? ˆ N Aˆ N† Aˆ N e 2gt e e g g ˆY AˆY† AˆY (1 e 2gt ) e e Has the atom decayed? e g e g No detection ee ˆ ge eg ee e 2gt gg gee gt eg e gt ( ee e 2gt gg ) 1 gg e g Detection ̂ g g Optimal operations: an example What processes are allowed? Those that can be described by effects. State separation Suppose that we have a system known to have been prepared 1 1 in one of two non-orthgonal states and . Our task is to separate the states, i.e. to transform them into 2 and 2 with 2 2 1 1 so that the states are more orthogonal and hence more distinguishable. This process cannot be guaranteed but can succeed with some probability PS . How large can this be? We introduce an effect associated with successful state separation Aˆ S 1 2 PS | |2 We can bound the success probability by considering the action 1 of ÂS on a superposition of the states and noting that the result must be an allowed state: PS | |2 1 1 1 1 2 2 There is a natural interpretation of this in terms of unambiguous state discrimination: 1( 2 ) Conc P 1 P 1( 2 ) ? 1 1( 2 ) 1( 2 ) Because this is the maximum allowed, state separation cannot better it so 2 1 PS PConc PConc PS 1 1 1 1 2 2 No cloning theorem - Wootters & Zurek, Dieks Can we copy an unknown state s ? Suppose it is possible: s A s s A A But the superposition principle then gives: a b A a A b A clone a b a b a b Exact cloning? - Duan & Guo, Chefles Can we clone exactly a quantum system known to be in the state |a or |b? Pclone a A b b b A 1 Pclone Error-free discrimination probability PIDP 1 a b For the cloned states a a ? Cloning cannot increase the discrimination probability cloned PIDP PclonePIDP cloned PIDP 1 a a b b 1 Pclone 1 a b 1 a b separation bound Summary • The projective von Neumann measurements do not provide the most general description of a measurement. • The use of probability operators (POMs) allows us to seek optimal measurements for any given detection problem. • The language of operations and effects allows us to describe, in great generality, the post-measurement state. Naimark’s theorem All POMs correspond to measurements. Consider a POM for a qubit with N probability operators: ˆ j j j , j j 0 0 j1 1 N POM conditions: j 1 N j 1 2 j0 N 1 j1 2 j 1 N j 0 0 j1 * j1 j 1 * j0 Can we treat this as a von Neumann measurement in a enlarged state-space? Our task is to represent the vectors j as the components in the qubit space of a set of orthonormal states j in an enlarged space. Enlarged space spanned by N orthornormal states j . The extra states can be other states of the system or by introducing an ancilla: 0 A0 , 1 A0 , 0 A1 , 0 A2 ,, 0 AN 2 An alternative orthonormal basis is N i *ji j Uˆ j j 1 N m n jm *jn mn j 1 Because Û is unitary, we can also form the orthonormal basis: N 1 j Uˆ † j ji i i 0 These are the required orthonormal states for our von Neumann measurement: P( j ) j ˆ j j ˆ j Tr ˆ j ˆ All measurements can be described by a POM and all POMs describe possible measurements. Summary The probabilities for the possible outcomes of any given measurement can be written in the form: P(i) Tr ˆˆ i The probability operators satisfy the three conditions ˆ i† ˆ i , ˆ i 0, ˆ i Î i All measurements can be described in this way and all sets of operators with these properties represent possible measurements. Spin and polarisation Qubits Poincaré and Bloch Spheres Two state quantum system Bloch Sphere Electron spin Poincaré Sphere Optical polarization States of photon polarisation Horizontal 0 Vertical 1 Diagonal up 1 2 Diagonal down 1 2 Left circular 1 2 Right circular 1 2 0 1 0 1 0 i 1 0 i 1 Example from quantum optics: How can we best discriminate (without error) between the coherent states a and a ? Coherent states: a n 0 an n! n Interfere like classical amplitudes. ra tb rb t a a b Symmetric beamsplitter Unambiguous discrimination between coherent states - Hutner et al 2 1/ 2 (i i )a 2 ia or 0 1/ 2 a 2 1/ 2 (1 1)a 0 or 21/ 2a ia This interference experiment puts all the light into a single output. Inconclusive results occur when no photons are detected. This happens with probability P? 2 a 0 1/ 2 2 exp( 2 | a |2 ) a a