Introduction to quantum information processing

Transcription

1 Introduction to quantum information processing Measurements and quantum probability Brad Lackey 25 October 2016 MEASUREMENTS AND QUANTUM PROBABILITY 1 of 22

2 OUTLINE 1 Probability 2 Density Operators 3 Review: the Spectral Theorem MEASUREMENTS AND QUANTUM PROBABILITY 2 of 22

3 PROBABILITY AND MEASUREMENT AS TAUGHT IN QUANTUM MECHANICS In traditional quantum mechanics courses: 1 A system is associated with a Hilbert space H. In quantum information we typically work with finite dimensional spaces. 2 Probability is determined by the state of the system ψ H. This is a unit vector, but only determined up to phase. 3 Observables are given by Hermitian operators A, and A = ψ A ψ. In particular, if φ is an eigenvector of A with eigenvalue λ then Pr{Getting outcome λ upon measuring A} = Pr{Seeing the state φ upon observing A} = ψ φ 2. This produces the idea of a measurement basis: the eigenbasis of A. MEASUREMENTS AND QUANTUM PROBABILITY 3 of 22

4 OUTLINE 1 Probability 2 Density Operators 3 Review: the Spectral Theorem MEASUREMENTS AND QUANTUM PROBABILITY 4 of 22

5 EIN GEDANKENEXPERIMENT Alice flips a (fair) coin, and without telling us the result does the following: on heads, she prepares the state 0 ; and on tails, she prepares the state 1. Then sends the resulting state. We make a measurement. What do we get? Say our measurement basis is Then φ = α 0 + β 1, φ = β 0 α 1. Pr{Seeing φ } = φ φ 2 = α 2 + β 2 Heads Tails 2 = 1 2 Pr{Seeing φ } = φ φ 2 = β 2 + α 2 2 = 1 2. Heads Tails MEASUREMENTS AND QUANTUM PROBABILITY 5 of 22

6 QUANTUM PROBABILITY Hmmm... Alice managed to prepare a state where any measurement basis { φ, φ } we use gives us the statistics: So what is this state? Pr{Seeing φ } = Pr{Seeing φ } = 1 2. Suppose this state is ψ. But then measuring in the basis { ψ, ψ } gives Pr{Seeing ψ } = 1 and Pr{Seeing ψ } = 0, not as we ve already shown it must be. So this state is never of the form ψ! What is it? Answer: states of the form ψ are special, called pure states. There s many more quantum states than just these. Alice prepared something called a mixed state. MEASUREMENTS AND QUANTUM PROBABILITY 6 of 22

7 MIXED STATES VERSUS PURE STATES Let figure it out what it is by pushing symbols around: Pr{Seeing φ } = φ φ 2 = 1 2 ( 0 φ ) 0 φ ( 1 φ ) 1 φ = 1 2 φ 0 0 φ φ 1 1 φ ( ) ( ) 1 1 = φ φ + φ φ ( ) 1 = φ ( ) φ 2 = φ ρ φ where ρ = 1 ( ). 2 MEASUREMENTS AND QUANTUM PROBABILITY 7 of 22

8 MIXED STATES VERSUS PURE STATES What s with this notation? Let s really compute this mixed state: 1 2 φ 0 0 φ = 1 [ ( α β ) ( )] [ 1 ( ) ( )] α β = ( α β ) [ ( ) 1 1 ( ) ] ( ) α β So is the matrix (or operator) = 1 2 ( 1 0 ) ( 1 0 ) = ( ) And similarly for , and so ρ = 1 2 ( ) = ( ) = MEASUREMENTS AND QUANTUM PROBABILITY 8 of 22

9 MIXED STATES VERSUS PURE STATES Now things are falling into place. The state Alice prepared is an operator ρ = We ve shown the general probability rule So we simply compute: for any φ, Pr{Seeing φ } = φ ρ φ. 1 2 φ 1 φ = 1 2 φ φ = 1 2. MEASUREMENTS AND QUANTUM PROBABILITY 9 of 22

10 OUTLINE 1 Probability 2 Density Operators 3 Review: the Spectral Theorem MEASUREMENTS AND QUANTUM PROBABILITY 10 of 22

11 MIXED STATES AND ENSEMBLES Instead, imagine Alice prepared a huge number of quantum states: half of them are prepared in 0, and the other half in 1 ; we take one at random and measure it. In general: an ensemble is a collection {( ψ j, p j )} N j=1 ψ j H is any pure state, and p j is the proportion of the ensemble prepared in ψ j. Then our statistics are given by Pr{Seeing φ } = = N p j ψ j φ 2 j=1 N N p j φ ψ j ψ j φ = φ p j ψ j ψ j φ. j=1 j=1 So ensembles are (generally) types of mixed states. MEASUREMENTS AND QUANTUM PROBABILITY 11 of 22

12 POSITIVE OPERATORS The state of an ensemble has the form ρ = N j=1 p j ψ j ψ j. This operator is Hermitian since p j is real. Definition A Hermitian operator ρ is positive (denoted ρ 0) if all its eigenvalues are nonnegative. Why isn t this called nonnegative? I don t know. Proposition An operator ρ 0 if and only if for every φ we have φ ρ φ 0. We ll prove this soon, but we can apply it now. The state of any ensemble is positive. In fact, φ ρ φ = N p j ψ j φ 2 = Pr{Seeing φ } 0. j=1 MEASUREMENTS AND QUANTUM PROBABILITY 12 of 22

13 Recall from the linear algebra: TRACE the trace of a matrix is the sum of its diagonal entries. Important fact 1: tr(ab) = tr(ba). Important fact 2: tr(c 1 A + c 2 B) = c 1 tr(a) + c 2 tr(b). Fact 1 implies that the trace can be computed in any basis: if P is the change-of-basis matrix, then tr(p 1 AP) = tr((p 1 A)P) = tr(p(p 1 A)) = tr(pp 1 A) = tr(a). In Dirac notation the trace is expressed as follows. Let { φ j } be any basis of H (typically we use an orthonormal one). The trace of an operator is given by tr(ρ) = j φ j ρ φ j. (Warning: in general φ j refers to the dual basis; this can be a tricky to compute if the basis is not orthonormal!) MEASUREMENTS AND QUANTUM PROBABILITY 13 of 22

14 DENSITY OPERATORS The state of an ensemble is ρ = N j=1 p j ψ j ψ j. Proposition The state of any ensemble has trace one. Now we use an orthonormal basis, { φ k }. We first compute tr( ψ j ψ j ) = k φ k ψ j ψ j φ k = k ψ j φ k 2 = φ j 2 = 1 (Pythagoras Theorem). But then N tr(ρ) = tr p j ψ j ψ j = j=1 N p j tr( ψ j ψ j ) = j=1 N p j = 1. j=1 Definition A density operator is an operator ρ on H with (i) ρ 0 and (ii) tr(ρ) = 1. MEASUREMENTS AND QUANTUM PROBABILITY 14 of 22

15 EXAMPLE: DENSITIES ON THE BLOCH SPHERE Recall the Bloch sphere: a qubit can be written as ψ = cos(θ/2) 0 + e iϕ sin(θ/2) 1, on the Bloch sphere it is (r x, r y, r z ) = (cos ϕ sin θ, sin ϕ sin θ, cos θ). Consider ψ as an ensemble by itself ( ψ, 1), so ρ = ψ ψ. We compute: ( cos(θ/2) ρ = e iϕ sin(θ/2) ) ( cos(θ/2) e iϕ sin(θ/2) ) ( = cos 2 (θ/2) e iϕ sin(θ/2) cos(θ/2) e iϕ sin(θ/2) cos(θ/2) sin 2 (θ/2) = 1 2 ( 1 + cos θ cos ϕ sin θ i sin ϕ sin θ cos ϕ sin θ + i sin ϕ sin θ 1 cos θ ) ) = 1 (1 + cos θz + cos ϕ sin θx + sin ϕ sin θy) 2 = 1 (1 + rxx + ryy + rzz). 2 So the Bloch sphere coordinates of ψ are just the coefficients of ρ = ψ ψ with respect to the Pauli matrices. MEASUREMENTS AND QUANTUM PROBABILITY 15 of 22

16 OUTLINE 1 Probability 2 Density Operators 3 Review: the Spectral Theorem MEASUREMENTS AND QUANTUM PROBABILITY 16 of 22

17 EIGENVECTORS AND EIGENVALUES We re familiar with eigenvectors and eigenvalues of matrix M v = λ v. Dirac notation: M ψ = λ ψ. Important fact: let H be Hermitian on H, 1 then H has a orthonormal basis of eigenvectors for H, { ψ j } dim H j=1, 2 and all the eigenvalues of H are real. Proposition The operator Π = ψ ψ is the orthogonal projection onto span{ ψ }. In other words: (i) Π ψ = ψ and (ii) if ψ φ = 0 then Π φ = 0. Using the notation above Π j = ψ j ψ j is an eigenprojector of H. The eigenvalue equation reads HΠ j = λ j Π j. Orthogonality implies Π j Π k = 0 whenever j k. MEASUREMENTS AND QUANTUM PROBABILITY 17 of 22

18 SPECTRAL MEASURES What if H has degenerate eigenvalues? E.g. H = Z 1: eigenvalue +1 has eigenvectors 00 and 01, eigenvalue -1 has eigenvectors 10 and 11. This would come up when we measure the first qubit of a register. The eigenprojection Π +1 maps onto span{ 00, 01 }. The eigenprojection Π 1 maps onto span{ 10, 11 }. If { ψ k } d k=1 is a orthonormal basis of eigenvectors for eigenvalue λ: the eigenprojection is Π λ = ψ 1 ψ ψ d ψ d. Still, if λ µ are eigenvalues then Π λ Π µ = 0. Definition A spectral (or projection-valued) measure is a collection of orthogonal projections {Π j } such that (i) Π j Π k = 0 whenever j k, and (ii) j Π j = 1. MEASUREMENTS AND QUANTUM PROBABILITY 18 of 22

19 THE SPECTRAL THEOREM Theorem (The Spectral Theorem) Let H be Hermitian, σ(h) = {λ 1,..., λ r } be its eigenvalues, and {Π 1,..., Π r } be the associated eigenprojections. Then for any function f defined on σ(h), we have f (H) = r k=1 f (λ k)π k. This is not just notation: consider H 2 = r k=1 λ2 k Π k. I know how to calculate each side independently. The spectral theorem shows they always give the same answer. More important examples: (spectral resolution) for f (x) = x we have H = r k=1 λ kπ k ; (square-root) if ρ 0 and f (x) = x we have ρ 1/2 = r k=1 λk Π k ; (modulus) for f (x) = x we have A = r k=1 λ k Π k ; (propagation) for f (x) = e ixt we have e iht = r k=1 eiλkt Π k ; (resolution-of-the-identity) for f (x) = 1 we have 1 = r k=1 Π k. MEASUREMENTS AND QUANTUM PROBABILITY 19 of 22

20 EXAMPLE: DENSITIES VERSUS ENSEMBLES We claim every density operator is just an ensemble. Use the spectral resolution to write ρ = dim H j=1 λ j ψ j ψ j. Here, if an eigenvalue is degenerate (say λ k ) we split up its projection Π k = ψ k1 ψ k1 + + ψ kd ψ kd. Always λ j 0 since ρ 0. Trace can be computed in any basis, so use the basis { ψ j }: tr(ρ) = jk λ j ψ j ψ k 2 = j λ j = 1. Therefore {( ψ j, λ j )} is an ensemble. So we see (i) mixed states, (ii) ensembles, and (iii) density operators are all the same. MEASUREMENTS AND QUANTUM PROBABILITY 20 of 22

21 Some unfinished business: Proposition POSITIVE OPERATORS An operator A 0 if and only if for every ψ, we have ψ A ψ 0. Proof. Suppose ψ A ψ 0 for all ψ. Let λ be an eigenvalue of A, so that A φ = λ φ for some φ. Then 0 φ A φ = φ λ φ = λ. Since λ was arbitrary, all eigenvalues of A are nonnegative. Now suppose A 0, and write A = n j=1 λ j φ j φ j using the spectral theorem. Then since each λ j 0, every ψ has ψ A ψ = n λ j ψ φ j 2 0. j=1 MEASUREMENTS AND QUANTUM PROBABILITY 21 of 22

22 NEXT TIME... Schmidt decomposition. Partial trace. State purification. Superoperators. MEASUREMENTS AND QUANTUM PROBABILITY 22 of 22