Iii23 on a state xi by checking the cases x 0 and x

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Unformatted text preview: first calculate the L(Hnof the Hadamard transform in the form of a it helps transform Uf 2 effect +1 , H) (Fig. III.23). on a state |xi. By checking the cases x = 0 and x = 1 separately we see that for a single p P qubit H |xi = z ( 1)xz |z i/ 2. Thus ¶3. We are told only that fP either constant or balanced, which means is x1 z1 +·· +xn zn ( |z1 , . , zn i that it is 0non , half xn i = z1 ,...,zand1)1 on the other. .half. .Our task is to p (1.49) H |x1 . . . , its domain n n 2 determine into which class a given f falls. This can be summarized more succinctly in the very useful equation P ¶4. Classical: Consider first the classical·zsituation. We can try di↵erent ( 1)x |z i n , (1.50) H |xi = z p input bit strings x. 2n We might (if we’re lucky) discover after the second query of f that it where x · z constant. is not is the bitwise inner product of x and z , modulo 2. Using this equation and (1.48) we can now evaluate | 3 i, But we might require as many as 2n /2+1 queries to answer the question. X ) function (x) |z i |0i So we’re facing O(2n 1 X ( 1)x·z+fevaluations.|1i p | 3i = . (1.51) 2n 2 z x ¶5. Initial state:...
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