hw5sol_fa07

# hw5sol_fa07 - C/CS/Phy191 Problem Set 5 Out 1 The...

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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 n-qubits. Let | ψ i = ∑ x ∈{ , 1 } n α x | x i be the state of an n-qubit system. A measure of the spread of | ψ i is S ( | ψ i ) ≡ ∑ x | α x | . For example, for a completely localized state | ψ i = | y i ( y ∈ { , 1 } n ), the spread is S ( | ψ i ) = 1. For a maximally spread state | ψ i = 1 √ 2 n ∑ x | x i , S ( | ψ i ) = 2 n · 1 √ 2 n = √ 2 n . a) Prove that for any quantum state | ψ i on n qubits, S ( | ψ i ) ≤ 2 n / 2 . (Hint: use the Cauchy-Schwarz inequality, v w ≤ k v k·k w k .) Answer : We need to show that if ∑ x ∈{ , 1 } n | α x | 2 = 1, then ∑ x | α x | ≤ 2 n / 2 . Using the Cauchy-Schwarz inequality v w ≤ k v k·k w k , we get ∑ x | α x | = ∑ x ( | α x |· 1 ) ≤ ∑ x | α x | 2 1 / 2 ∑ x 1 2 1 / 2 = 1 · 2 n / 2 = 2 n / 2 , with equality iff | α x | = 1 / 2 n / 2 for all x . b) Suppose that | α x | ≤ a for all x . Prove that S ( | ψ i ) ≥ 1 a . (Hint: think about normalization....) Answer : Using the normalization condition, 1 = ∑ x | α x | 2 ≤ ∑ x a | α x | = aS ( ψ ) . (Notice that this inequality is an equality iff all α 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 | α x | ≤ a .) Now let H ⊗ n | ψ i = ∑ x β x | x i , where (by homework 2) β x = 1 2 n / 2 ∑ y ( − 1 ) x · y α 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 ψ as H ⊗ n | ψ i = ∑ x β x | x i , where β x = 1 2 n / 2 ∑ y ( − 1 ) x · y α y . Use this to prove that for all y , | β y | ≤ 1 2 n / 2 S ( | ψ i ) . (Hint: use the triangle inequality.) Answer : Using the triangle inequality | a + b | ≤ | a | + | b | , | β y | = 1 2 n / 2 | ∑ y ( − 1 ) x · y α y | ≤ 1 2 n / 2 ∑ y | ( − 1 ) x · y α y | = 1 2 n / 2 ∑ y | α y | = 1 2 n / 2 S ( ψ ) ....
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hw5sol_fa07 - C/CS/Phy191 Problem Set 5 Out 1 The...

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