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Introduction to Quantum Information Processing CS 467 / CS 667 Phys 467 / Phys 767 C&O 481 / C&O 681 Lecture 12 (2005) Richard Cleve DC 3524 [email protected] Course web site at: http://www.cs.uwaterloo.ca/~cleve/courses/cs467 1 Contents • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 2 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 3 Correction to Lecture 11: 2/3 instead of 2/3 General quantum operations (III) Example 3 (trine state “measurent”): Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20 3/21 Define A0 =2/300 2 1 0 3 0 0 1 2 3 2 A1=2/311 4 2 6 Then A0 t t 1 2 3 2 A2=2/322 4 2 6 t A0 A1 A1 A2 A2 I The probability that state k results in “outcome” Ak is 2/3, and this can be adapted to actually yield the value of k with this success probability 4 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 5 Distinguishing mixed states (I) What’s the best distinguishing strategy between these two mixed states? 0 with prob. ½ 0 + 1 with prob. ½ 3 / 4 1 / 2 ρ1 1 / 2 1 / 4 0 with prob. ½ 1 with prob. ½ 1 1 also arises from this orthogonal mixture: 1 1 0 ρ2 2 0 1 + 0 … as does 2 from: 0 0 with prob. cos2(/8) 1 with prob. sin2(/8) 0 with prob. ½ 1 with prob. ½ 6 Distinguishing mixed states (II) We’ve effectively found an orthonormal basis 0, 1 in which both density matrices are diagonal: cos 2 π / 8 0 ρ2 2 0 sin π / 8 1 1 0 ρ1 2 0 1 1 1 Rotating 0, 1 to 0, 1 the scenario can now be examined using classical probability theory: + 0 0 Distinguish between two classical coins, whose probabilities of “heads” are cos2(/8) and ½ respectively (details: exercise) Question: what do we do if we aren’t so lucky to get two density matrices that are simultaneously diagonalizable? 7 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 8 Basic properties of the trace The trace of a square matrix is defined as Tr M d M k 1 It is easy to check that k ,k Tr M N Tr M Tr N and Tr M N Tr N M λ The second property implies Tr M Tr U 1MU d k 1 k 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 9 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 10 Recap: general quantum ops General quantum operations (a.k.a. “completely positive trace preserving maps” ): m Let A1, A2 , …, Am be matrices satisfying t A j Aj I j 1 m t ρ A ρ A Then the mapping j j is a general quantum op j 1 Example: applying U to yields U U † 11 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 12 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 13 Partial trace (II) We’ve already seen this defined in the case of 2-qubit systems: discarding the second of two qubits 1 0 0 0 0 1 0 0 Let A0 = I0 and A = I 1 1 0 0 1 0 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 14 Partial trace (III) For 2-qubit systems, the partial trace is explicitly ρ00,00 ρ Tr2 01,00 ρ10,00 ρ11,00 ρ00,01 ρ01,01 ρ10,01 ρ11,01 ρ00,10 ρ01,10 ρ10,10 ρ11,10 ρ00,11 ρ01,11 ρ00,00 ρ01,01 ρ10,11 ρ10,00 ρ11,01 ρ11,11 ρ00,10 ρ01,11 ρ10,10 ρ11,11 ρ00,01 ρ01,01 ρ10,01 ρ11,01 ρ00,10 ρ01,10 ρ10,10 ρ11,10 ρ00,11 ρ01,11 ρ00,00 ρ10,10 ρ10,11 ρ01,00 ρ11,10 ρ11,11 ρ00,01 ρ10,11 ρ01,01 ρ11,11 and ρ00,00 ρ Tr1 01,00 ρ10,00 ρ11,00 15 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 16 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 Aj ρ Atj produces the outcome: j (classical information) A j ρ Atj Tr A j ρ Atj (the collapsed quantum state) Example 1: Aj = jj (orthogonal projectors) This reduces to our previously defined measurements … 17 POVMs (II) When Aj = jj are orthogonal projectors and = , Tr Aj ρ Atj = Trjjjj = jjjj = j2 Moreover, A j ρ Atj Tr A j ρ A t j φj φj ψ ψ φj φj φj ψ 2 φj φj 18 POVMs (III) Example 3 (trine state “measurent”): Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20 3/21 Define A0 =2/300 2 1 0 3 0 0 1 2 3 2 A1=2/311 4 2 6 Then t t 1 2 3 2 A2=2/322 4 2 6 t A0 A0 A1 A1 A2 A2 I If the input itself is an unknown trine state, kk, then the probability that classical outcome is k is 2/3 = 0.6666… 19 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 20 Simulations among operations (I) Fact 1: any general quantum operation can be simulated by applying a unitary operation on a larger quantum system: input 0 0 0 Example: decoherence 0 + 1 0 output U discard α 2 ρ 0 0 2 β 21 Simulations among operations (II) Fact 2: any POVM can also be simulated by applying a unitary operation on a larger quantum system and then measuring: input 0 0 0 U quantum output j classical output 22 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 23 Separable states A bipartite (i.e. two register) state is a: • product state if = • separable state if m p j 1 j j j ( p1 ,…, pm 0) (i.e. a probabilistic mixture of product states) Question: which of the following states are separable? ρ1 12 00 11 00 11 ρ2 12 00 11 00 11 12 00 11 00 11 24 • Correction to Lecture 11: 2/3 2/3 • Distinguishing mixed states • Basic properties of the trace • Recap: general quantum operations • Partial trace • POVMs • Simulations among operations • Separable states • Continuous-time evolution 25 Continuous-time evolution Although we’ve expressed quantum operations in discrete terms, in real physical systems, the evolution is continuous Let H be any Hermitian matrix and t R 1 Then eiHt is unitary – why? λ1 † D H = U DU, where λd eiλ1t t † iDt iHt Therefore e = U e U = U U iλd t e 0 (unitary) 26 27