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Unformatted text preview: Ph219a/CS219a Solutions to Hw 8 May 5, 2009 Problem 1 (a) We can write ¯ X a as a product of X α ’s and Z β ’s: ¯ X a = (sgn) n O α =1 X u α α ! n O β =1 Z v β β , u α ,v β ∈ { , 1 } (1) Then, ¯ X A,a ⊗ ¯ X B,a can be written as ¯ X A,a ⊗ ¯ X B,a = n O α =1 ( X A,α ⊗ X B,α ) u α ! n O β =1 ( Z A,β ⊗ Z B,β ) v β (2) Therefore, if we can do bitwise Bell measurements, ¯ X A,a ⊗ ¯ X B,a is given by the product (parity) of all those X A,α ⊗ X B,α and Z A,β ⊗ Z B,β outcomes for which u α ,v β = 1 in Eqn.(2). In a similar way, we can get the values of ¯ Z A,a ⊗ ¯ Z B,a . (b) Consider the stabilizer generators for the two code blocks of the form M i ⊗ M i . These generators are symmetric between the two blocks and hence can be written as a product of X A,α ⊗ X B,α and Z A,α ⊗ Z B,α . By taking products of the appropriate bitwise Bell measure ment outcomes (similar to part (a)), we can figure out the outcome of measuring M i ⊗ M i . When an error that anticommutes with M i occurs, we get 1 on measuring M i . For M i ⊗ M i we get 1 when that error occurs in either block A or block B, but not when it occurs in both simultaneously (i.e. same error acting on the same qubits in each block). Therefore, the set of stabilizer generators { M i ⊗ M i } allows us to correct all the errors (up to t of them) that the original singleblock code can correct, except that the error can now occur in either block, but not both simultaneously 1 . Note that you cannot identify which block the error occurred in. The simultaneous error does not matter, because it does not change the value of the bitwise Bell measurements. Being unable to identify which block the error occurred in is also of no 1 For example, for a CSS code with t = 1, the symmetrized generator set can correct X A,α ⊗ I B,α and X A,α ⊗ Z B,α , but not X A,α ⊗ X B,α 1 consequence because, whether it was in block A or block B, the error flips the same bitwise Bell measurement outcomes. To correct for it, we use the symmetrized generators to identify which qubits had errors, and then flip the bitwise Bell measurement results affected by those errors. (c) We want to show that the given circuit satisfies the following two properties: We are given that the state preparation circuit satisfies the property shown in Figure 1. Further more, from class, we know that we can perform Pauli gates faulttolerantly using any stabilizer code, i.e. the encoded Pauli operation circuit satisfies the property shown in Figure 2. Given properties A and B, and assuming that given any set of bitwise Bell measurement results, the classical processing always tells us a particular Pauli gate to apply (can be chosen by convention for uncorrectable outcomes), we can easily show Property 1 (see Figure 3)....
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 Spring '09
 Konrad Zuse, Wolfgang Pauli, Pauli matrices, Quantum gate

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