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Unformatted text preview: 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 twolevel 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 twolevel unitary transformations: The matrix representation of a twolevel unitary transformation must have at most two nonzero 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  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 α 1 α 2 α 3 = xα + yα 1 zα + wα 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 α 1 α 2 α 3 = xα + yα 1 zα + wα 1 uα + α 2 + vα 3 α 3 We notice that in the second case, when there are four nonzero offdiagonal 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 x y z w ¶ x * z * y * w * ¶ = 1 0 0 1 ¶ We now examine a twoqubit 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 x y z w 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 y 0 1 0 0 z w 0 0 0 1 , U 5 = x 0 0 y 0 1 0 0...
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This note was uploaded on 08/08/2011 for the course COT 6600 taught by Professor Staff during the Fall '08 term at University of Central Florida.
 Fall '08
 Staff

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