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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 ( 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 ( X ) =  ih  I +  1 ih 1  X and noting that (( X )) 1 = ( X ), we have, ( X )( XI )( X ) = XX P 2 , ( X )( IX )( X ) = IX P 2 ( X )( ZI )( X ) = ZI P 2 , ( X )( IZ )( X ) = ZZ P 2 (3) Since all the generators of P 2 are mapped to elements of the group itself under the action of ( X ), we see that ( 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 ( Z ) = H 2 ( X ) H 2 (where H 2 denotes a Hadamard gate acting on the target qubit), and ( Y ) = P 2 ( X ) P 2 . Now, consider ( ) Z 1 ( ). When the control qubit is  i , the target is left unchanged by both the ( ) gates. When the control qubit is  1 i , ( ) Z 1 ( )  1 i . i = ( ) Z 1 2  1 i . i = ( 2 ) 2  1 i . i = Z 1  1 i . i , since 2 = I . Similarly, we can analyze the action of ( ) X 1 ( ) as follows: ( ) X 1 ( )  i . i = ( )  1 i . i = X 1 2  i . i ( ) X 1 ( )  1 i . i = ( ) X 1 2  1 i . i = ( ) 2  i . i = X 1 2  1 i . i (6) Thus we have shown that ( ) Z 1 ( ) = Z 1 ( ) X 1 ( ) = X 1 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

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