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Unformatted text preview: Ph219a/CS219a Solutions to Hw 7 May 5, 2009 Problem 1 (a) We can recall the action of the Hadamard ( H = H 1 ) and the phase gates ( P ) on the Pauli operators from an earlier Hw problem (HW4, Problem 3a), where we had shown HXH 1 = Z , HYH 1 = Y , HZH 1 = X (1) PXP 1 = Y , PYP 1 = X , PZP 1 = Z (2) From the definition of the normalizer group, this immediately shows that H , P ∈ C 1 . To show Λ( X ) ∈ C 2 , it suffices to see how the CNOT gate acts by conjugation on the generators of P 2 : { XI , IX , ZI , IZ } . Using Λ( X ) =  ih ⊗ I +  1 ih 1 ⊗ X and noting that (Λ( X )) 1 = Λ( X ), we have, Λ( X )( XI )Λ( X ) = XX ∈ P 2 , Λ( X )( IX )Λ( X ) = IX ∈ P 2 Λ( X )( ZI )Λ( X ) = ZI ∈ P 2 , Λ( X )( IZ )Λ( X ) = ZZ ∈ P 2 (3) Since all the generators of P 2 are mapped to elements of the group itself under the action of Λ( X ), we see that Λ( X ) ∈ C 2 . (b) We know that under the action of H , X → Z , Z → X , Y →  Y ⇒ XYZ →  ZYX = YXZ = XYZ (4) Similarly under the action of the phase gate P , X → Y , Y →  X , Z → Z ⇒ XYZ →  YXZ = XYZ (5) Thus, H and P are permutations of X , Y , Z that preserve the inner product XYZ = i I . Fur thermore, since H is a permutation of ( X , Z ), and P permutes ( X , Y ), using these we can generate all other inner product preserving permutations. Since all elements of C 1 are inner 1 product preserving permutations of X , Y and Z , H and P generate C 1 . (c) Using eqns.(1) and (2), it follows that Λ( Z ) = H 2 Λ( X ) H 2 (where H 2 denotes a Hadamard gate acting on the target qubit), and Λ( Y ) = P 2 Λ( X ) P 2 . Now, consider Λ( σ ) Z 1 Λ( σ ). When the control qubit is  i , the target is left unchanged by both the Λ( σ ) gates. When the control qubit is  1 i , Λ( σ ) Z 1 Λ( σ )  1 i . i = Λ( σ ) Z 1 σ 2  1 i . i = ( σ 2 ) 2  1 i . i = Z 1  1 i . i , since σ 2 = I . Similarly, we can analyze the action of Λ( σ ) X 1 Λ( σ ) as follows: Λ( σ ) X 1 Λ( σ )  i . i = Λ( σ )  1 i . i = X 1 σ 2  i . i Λ( σ ) X 1 Λ( σ )  1 i . i = Λ( σ ) X 1 σ 2  1 i . i = Λ( σ ) σ 2  i . i = X 1 σ 2  1 i . i (6) Thus we have shown that Λ( σ ) Z 1 Λ( σ ) = Z 1 Λ( σ ) X 1 Λ( σ ) = X 1 σ 2 (7) (d) First, note that UX 1 U 1 anticommutes with UZ 1 U 1 ( UX 1 U 1 )( UZ 1 U 1 ) = UX 1 Z 1 U 1 = UZ 1 X 1 U 1 = ( UZ 1 U 1 )( UX 1 U 1 ) (8) implying that UX 1 U 1 and UZ 1 U 1 must differ in their action on atleast one of the n + 1 qubits. Suppose they differ in their action on the first qubit, we can write U : X 1 → M 1 M , Z 1 → N 1 N (9) where M 1 6 = N 1 ∈ P 1 . In this case, we can simply choose W to be an element of C 1 that acts such that W : M 1 → X 1 , N 1 → Z 1 ....
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 Spring '09
 Vector Space, Trigraph, NZ, NX, nX Xtype generators

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