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12 measurement therefore we can determine whether f 0

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Unformatted text preview: the two cases by measuring the first qubit in the sign basis. ¶12. Measurement: Therefore we can determine whether f (0) = f (1) or not by measuring the first bit of | 2 i in the sign basis, which we can do with the Hadamard gate (recall H |+i = |0i and H | i = |1i): | 3i = ( H ⌦ I )| 2 i ⇢ ±|0i| i, if f (0) = f (1) = ±|1i| i, if f (0) 6= f (1) = ±|f (0) f (1)i| i. ¶13. Therefore we can determine whether or not f (0) = f (1) with a single evaluation of f . (This is very strange!) ¶14. In e↵ect, we are evaluating f on a superposition of |0i and |1i and determining how the results interfere with each other. As a result we get a definite (not probabilistic) determination of a global property with a single evaluation. ¶15. This is a clear example where a quantum computer can do something faster than a classical computer. ¶16. However, note that Uf has to uncompute f , which takes as much time as computing it, but we will see other cases (Deutsch-Jozsa) where the speedup is much more than 2⇥. D. QUANTUM ALGORITHMS Quantum algori...
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