soln1 - EE 520 Quantum Information Processing Solution to...

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EE 520: Quantum Information Processing Solution to Homework # 1 Exercise 2.2 In terms of the basis {| 0 , | 1 } ˆ A = 0 1 1 0 = | 1 0 | + | 0 1 | . We can transform ˆ A by choosing a different basis. For instance, in the basis = ( | 0 ± | 1 ) / 2 ˆ A = 1 0 0 - 1 = | + + | - |- -| . Exercise 2.9 ˆ X = | 1 0 | + | 0 1 | , ˆ Y = i | 1 0 | - i | 0 1 | , ˆ Z = | 0 0 | - | 1 1 | , Exercise 2.10 In the | v i basis, the dyad | v j v k | is the matrix with the jk element equal to 1 and all other elements 0. Exercise 2.11 All three of the Pauli matrices have eigenvalues ± 1, which means they all have the same diagonal representation 1 0 0 - 1 The ± 1 eigenvectors for ˆ X, ˆ Y , ˆ Z (respectively) are { ( | 0 ± | 1 ) / 2 } , { ( | 0 ± i | 1 ) / 2 } , and {| 0 , | 1 } (up to a phase). 1
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Exercise 2.17 If a normal matrix ˆ H is Hermitian ˆ H = ˆ H , then it is Hermitian in any basis: ( ˆ U ˆ H ˆ U ) = ˆ U ˆ H ˆ U . In particular, it must be Hermitian in the diagonal representation, which implies that all of its eigenvalues λ i (along the diagonal) must satisfy λ i = λ * i . The converse is even simpler. If ˆ H is normal and has real eigenvalues, then in the diagonal representation it is clearly Hermitian (since λ i = λ * i , which means that it is Hermitian in any basis (as shown above). Exercise 2.19 We can simply check Hermiticity for each matrix: ˆ X = 0 1 1 0 = 0 1 * 1 * 0 = 0 1 1 0 = ˆ X ˆ Y = 0 - i i 0 = 0 ( i ) * ( - i ) * 0 = 0 - i i 0 = ˆ Y ˆ Z = 1 0 0 - 1 = 1 * 0 0 ( - 1) * = 1 0 0 - 1 = ˆ Z As for unitarity, all three matrices are idempotent: ˆ X ˆ X = ˆ X 2 = ˆ Y ˆ Y = ˆ Y 2 = ˆ Z ˆ Z = ˆ Z 2 = ˆ I .
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