This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: March 3, 2005 Physics 681481; CS 483: Discussion of #3 ( with postscripts on the Schmidt decomposition theorem) I. (a) For each of the four possibilities for the unknown function f , the corresponding forms for the state  ψ i =  i f (0) i +  1 i f (1) i (1) are  ψ i 00 = 1 √ 2 (  i +  1 i )  i , f (0) = 0 , f (1) = 0; (2)  ψ i 11 = 1 √ 2 (  i +  1 i )  1 i , f (0) = 1 , f (1) = 1; (3)  ψ i 01 = 1 √ 2 (  i i +  1 i 1 i ) , f (0) = 0 , f (1) = 1; (4)  ψ i 10 = 1 √ 2 (  i 1 i +  1 i i ) , f (0) = 1 , f (1) = 0 . (5) We know that  ψ i has one of these four forms, and wish to distinguish between two cases: Case 1:  ψ i =  ψ i 00 or  ψ i 11 ; Case 2:  ψ i =  ψ i 01 or  ψ i 10 . By applying Hadamard transformations we can change the four possible states to ( H ⊗ H )  ψ i 00 = 1 √ 2 (  i i +  i 1 i ) , f (0) = 0 , f (1) = 0 , (6) ( H ⊗ H )  ψ i 11 = 1 √ 2 (  i i   i 1 i ) , f (0) = 1 , f (1) = 1 , (7) ( H ⊗ H )  ψ i 01 = 1 √ 2 (  i i +  1 i 1 i ) , f (0) = 0 , f (1) = 1 , (8) ( H ⊗ H )  ψ i 10 = 1 √ 2 (  i i   1 i 1 i ) , f (0) = 1 , f (1) = 0 . (9) If we now make a measurement then if we have one of the Case1 states, (6) or (7), we get 00 half the time and 01 half the time, while if we have one of the Case2 states, (8) or (9), we get 00 half the time and 11 half the time. So regardless of what the state is, half the time we get 00 and learn nothing whatever, and half the time we get 01 (Case 1) or 11 (Case 2) and learn which case we are dealing with. (b) We know that the state (1) is one of the four states (2)(5). We are allowed to apply an arbitrary twoqubit unitary transformation U to  ψ i before we make the measurement. If every possible measurement outcome must rule out one or the other of the two cases, then if U  ψ i 00 , U  ψ i 11 , U  ψ i 01 , and U  ψ i 10 are all expanded in the computational basis, then those computationalbasis states that appear in the Case1 expansions cannot appear in the Case2 expansions, for otherwise there would be a nonzero probability of a measurement outcome that did not enable us to discriminate between the two cases. Consequently 1 U  ψ i 00 and U  ψ i 11 must each be orthogonal to each of U  ψ i 01 and U  ψ i 10 . But this is impossible, because unitary transformations preserve inner products, while (2)(5) show that the inner product of any Case1 state with any Case2 state is 1 2 . (c) We’re asked to show that no matter what unitary transformation is applied to the state (1) prior to a measurement, the probability of being able to learn from the measurement whether or not f (0) = f (1) cannot exceed 1 2 . I show this below in an even more general situation, in which we bring in n additional ( ancillary ) qubits and let an arbitrary unitary transformation W act on all n + 2 qubits before making a final...
View
Full
Document
This note was uploaded on 09/28/2008 for the course PHYS 481 taught by Professor Anon during the Spring '05 term at Cornell.
 Spring '05
 ANON
 Physics

Click to edit the document details