soln7_09

# soln7_09 - Ph219a/CS219a Solutions to Hw 7 May 5 2009...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

soln7_09 - Ph219a/CS219a Solutions to Hw 7 May 5 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online