Solutions to Assignment 3

# Solutions to Assignment 3 - March 2 2006 Physics 681-481 CS...

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March 2, 2006 Physics 681-481; CS 483: Discussion of #3 I. (a) For each of the four possibilities for the unknown function f , the corresponding forms for the state | ψ i = | 0 i| f (0) i + | 1 i| f (1) i (1) are | ψ i 00 = 1 2 ( | 0 i + | 1 i ) | 0 i , f (0) = 0 , f (1) = 0; (2) | ψ i 11 = 1 2 ( | 0 i + | 1 i ) | 1 i , f (0) = 1 , f (1) = 1; (3) | ψ i 01 = 1 2 ( | 0 i| 0 i + | 1 i| 1 i ) , f (0) = 0 , f (1) = 1; (4) | ψ i 10 = 1 2 ( | 0 i| 1 i + | 1 i| 0 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 ( | 0 i| 0 i + | 0 i| 1 i ) , f (0) = 0 , f (1) = 0 , (6) ( H H ) | ψ i 11 = 1 2 ( | 0 i| 0 i - | 0 i| 1 i ) , f (0) = 1 , f (1) = 1 , (7) ( H H ) | ψ i 01 = 1 2 ( | 0 i| 0 i + | 1 i| 1 i ) , f (0) = 0 , f (1) = 1 , (8) ( H H ) | ψ i 10 = 1 2 ( | 0 i| 0 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 Case-1 states, (6) or (7), we get 00 half the time and 01 half the time, while if we have one of the Case-2 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 two-Qbit 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 computational-basis states that appear in the Case-1 expansions cannot appear in the Case-2 expansions, for otherwise there would be a non-zero probability of a measurement outcome that did not enable us to discriminate between the two cases. Consequently 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 1

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impossible, because unitary transformations preserve inner products, while (2)-(5) show that the inner product of any Case-1 state with any Case-2 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 ) Qbits. These might be used to further process the input and output registers through some elaborate quantum subroutine, producing an arbitrary unitary transformation W that acts on all n + 2 Qbits before a final measurement of all n + 2 Qbits is made. This reduces to the simpler case if
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