QCV2Appendix

# QCV2Appendix - 1 Simulation of Quantum Circuits The set of...

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1 Simulation of Quantum Circuits The set of unitary operations performed by quantum gates is continuous; thus, the problem of defining a finite, and hopefully small, set of universal quantum gates is slightly more convoluted than the universality of classical gates. In quantum computing we say that a set of quantum gates is universal if any unitary operation can be approximated with ar- bitrary accuracy by a circuit including only gates from this discrete set. We notice that in the case of quantum circuits we have relaxed our requirement of exact simulation. The Solovay- Kitaev theorem tells us that sufficiently good approximations for quantum gates exist. We now discuss the three stages of the protocol to simulate a quantum circuit operating on n qubits using only CNOT, H, S and T gates. Stage 1 . We define a two-level unitary matrix as a transformation involving two or fewer computational basis states. Then we show that any unitary transformation A ∈ H 2 n can be carried out by unitary transformations U 1 , U 2 , . . . U k , . . . U m , with m 2 n - 1, which act only upon two or fewer computational basis states. First, we present an important property of two-level unitary transformations: The matrix representation of a two-level unitary transformation must have at most two non-zero off- diagonal elements. Indeed, the transformation carried out by such a matrix should affect only two projections of a vector. For example, consider two matrices in H 4 , U 1 and U 2 , acting upon an ensemble of two qubits in state | ψ i = α 0 | 00 i + α 1 | 01 i + α 2 | 10 i + α 3 | 11 i : U 1 | ψ i = x y 0 0 z w 0 0 0 0 1 0 0 0 0 1 α 0 α 1 α 2 α 3 = 0 + 1 0 + 1 α 2 α 3 and U 2 | ψ i = x y 0 0 z w 0 0 u 0 1 v 0 0 0 1 α 0 α 1 α 2 α 3 = 0 + 1 0 + 1 0 + α 2 + 3 α 3 We notice that in the second case, when there are four non-zero off-diagonal elements three components of the vector are affected by the transformation U 2 . If U is a unitary transformation of one qubit then the elements of the matrix: U = x y z w , x, y, z, w C must satisfy the following relations: | x | 2 + | y | 2 = 1 | z | 2 + | w | 2 = 1 xz * + yw * = 0 x * z + y * w = 0 . Indeed, U is unitary thus UU = U U = I : 1

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x y z w x * z * y * w * = 1 0 0 1 We now examine a two-qubit transformation U ∈ H 4 and consider a decomposition: U = U 1 U 2 U 3 U 4 U 5 U 6 = U = U 6 U 5 U 4 U 3 U 2 U 1 . with U k , 1 k 6, given by: U 1 = x y 0 0 z w 0 0 0 0 1 0 0 0 0 1 , U 2 = 1 0 0 0 0 x y 0 0 z w 0 0 0 0 1 , U 3 = 1 0 0 0 0 1 0 0 0 0 x y 0 0 z w , U 4 = x 0 y 0 0 1 0 0 z 0 w 0 0 0 0 1 , U 5 = x 0 0 y 0 1 0 0 0 0 1 0 z 0 0 w , U 6 = 1 0 0 0 0 x 0 y 0 0 1 0 0 z 0 w .
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