1. 1 Introduction to Quantum Information Processing CS 667 / PH 767 / CO 681 / AM 871 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 14 (2009)

2. 2 Bloch sphere for qubits

3. 3 Bloch sphere for qubits (1) Consider the set of all 2x2 density matrices  Note that the coefficient of I is ½, since X, Y, Y are traceless They have a nice representation in terms of the Pauli matrices: 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

4. 4 Bloch sphere for qubits (2) We will express First consider the case of pure states  , where, without loss of generality,  = cos (  )  0  + e 2i  sin (  )  1  ( ,   R ) Therefore c z = cos (2  ), c x = cos (2  ) sin (2  ), c y = sin (2  ) sin (2  ) These are polar coordinates of a unit vector ( c x, c y, c z )  R 3

5. 5 Bloch sphere for qubits (3) ++ 00 11 –– +i+i –i–i +i = 0 + i1+i = 0 + i1 –i = 0 – i1–i = 0 – i1 – = 0 – 1– = 0 – 1 + = 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

6. 6 Distinguishing mixed states

7. 7 Distinguishing mixed states (1)  0  with prob. ½  0  +  1  with prob. ½  0  with prob. ½  1  with prob. ½  0  with prob. cos 2 (  /8)  1  with prob. sin 2 (  /8) 00 ++ 00 1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. 8 Distinguishing mixed states (2) 00 ++ 00 11 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 cos 2 (  /8) and ½ respectively (details: exercise) 11

9. 9 General quantum operations

10. 10 General quantum operations (1) Example 1 (unitary op): applying U to  yields U  U † Also known as: “quantum channels” “completely positive trace preserving maps”, “admissible operations” Let A 1, A 2, …, A m be matrices satisfying Then the mappingis a general quantum op Note: A 1, A 2, …, A m do not have to be square matrices

11. 11 General quantum operations (2) Example 2 (decoherence): let A 0 =  0  0  and A 1 =  1  1  This quantum op maps  to  0  0  0  0  +  1  1  1  1  Corresponds to measuring  “ without looking at the outcome” For  ψ  =  0  +  1 , After looking at the outcome,  becomes  0  0  with prob. |  | 2  1  1  with prob. |  | 2

12. 12 General quantum operations (3) Example 3 (discarding the second of two qubits): Let A 0 = I   0  and A 1 = I   1  States of the form    (product states) become  State becomes Note 1: it’s the same density matrix as for ( (½,  0  ), (½,  1  ) ) Note 2: the operation is called the partial trace Tr 2 

13. More about the partial trace If the 2 nd register is discarded, state of the 1 st register remains  Two quantum registers in states  and  (resp.) are independent when 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) The partial trace Tr 2 , can also be characterized as the unique linear operator satisfying the identity Tr 2 (    ) =  For d -dimensional registers, Tr 2 is defined with respect to the operators A k = I   k , where  0 ,  1 , …,  d  1  can be any orthonormal basis The partial trace Tr 2 gives the effective state of the first register

14. 14 Partial trace continued For 2-qubit systems, the partial trace is explicitly and

15. 15 General quantum operations (4) Example 4 (adding an extra qubit): Just one operator A 0 = I   0  States of the form  become   0  0  More generally, to add a register in state , use the operator A 0 = I  

16. 16 POVM measurements (POVM = Positive Operator Valued Measre)

17. 17 POVM measurements (1) Let A 1, A 2, …, A m be matrices satisfying Corresponding POVM measurement is a stochastic operation on  that, with probability, produces outcome: j (classical information) (the collapsed quantum state) Example 1: A j =  j  j  (orthogonal projectors) This reduces to our previously defined measurements …

18. 18 POVM measurements (2) Moreover, When A j =  j  j  are orthogonal projectors and  = , = Tr  j  j  j  j  =  j  j  j  j  =  j  2

19. 19 POVM measurements (3) Example 3 (trine state “measurent”): Let  0  =  0 ,  1  =  1/2  0  +  3/2  1 ,  2  =  1/2  0    3/2  1  Then If the input itself is an unknown trine state,  k  k , then the probability that classical outcome is k is 2/3 = 0.6666… Define A 0 =  2/3  0  0  A 1 =  2/3  1  1  A 2 =  2/3  2  2 

20. 20 POVM measurements (4) Simplified definition for POVM measurements: Let E 1, E 2, …, E m be positive definite and such that The probability of outcome j is Often POVMs arise in contexts where we only care about the classical part of the outcome (not the residual quantum state) The probability of outcome j is This is usually the way POVM measurements are defined

21. 21 “Mother of all operations” Let A 1,1, A 1,2, …, A 1, m 1 satisfy A 2,1, A 2,2, …, A 2, m 2 A k,1, A k,2, …, A k, m k Then there is a quantum operation that, on input , produces with probability the state: j (classical information) (the collapsed quantum state)