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Unformatted text preview: C/CS/Phy191 Problem Set 5 Out: October 21, 2007 1. The uncertainty principle bounds how well a quantum state can be localized simultaneously in the standard basis and the Fourier basis. In this question, we will derive an uncertainty principle for a discrete system of nqubits. Let  i = x { , 1 } n x  x i be the state of an nqubit system. A measure of the spread of  i is S (  i ) x  x  . For example, for a completely localized state  i =  y i ( y { , 1 } n ), the spread is S (  i ) = 1. For a maximally spread state  i = 1 2 n x  x i , S (  i ) = 2 n 1 2 n = 2 n . a) Prove that for any quantum state  i on n qubits, S (  i ) 2 n / 2 . (Hint: use the CauchySchwarz inequality, v w k v kk w k .) Answer : We need to show that if x { , 1 } n  x  2 = 1, then x  x  2 n / 2 . Using the CauchySchwarz inequality v w k v kk w k , we get x  x  = x (  x  1 ) x  x  2 1 / 2 x 1 2 1 / 2 = 1 2 n / 2 = 2 n / 2 , with equality iff  x  = 1 / 2 n / 2 for all x . b) Suppose that  x  a for all x . Prove that S (  i ) 1 a . (Hint: think about normalization....) Answer : Using the normalization condition, 1 = x  x  2 x a  x  = aS ( ) . (Notice that this inequality is an equality iff all x are zero or exactly a that is, to minimize the spread, concentrate the probability mass as much as possible while still satisfying the constraint  x  a .) Now let H n  i = x x  x i , where (by homework 2) x = 1 2 n / 2 y ( 1 ) x y y . ( x y n i = 1 x i y i .) c) In Problem Set 3 you showed that H n x = y ( 1 ) x y y ( x y n i = 1 x i y i ). Hence we can obtain the action of H n on as H n  i = x x  x i , where x = 1 2 n / 2 y ( 1 ) x y y . Use this to prove that for all y ,  y  1 2 n / 2 S (  i ) . (Hint: use the triangle inequality.) Answer : Using the triangle inequality  a + b   a  +  b  ,  y  = 1 2 n / 2  y ( 1 ) x y y  1 2 n / 2 y  ( 1 ) x y y  = 1 2 n / 2 y  y  = 1 2 n / 2 S ( ) ....
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 Fall '07
 BIRGITTAWHALEY
 mechanics

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