Search results
Results from the WOW.Com Content Network
[2] [3] [4] This set of gates is minimal in the sense that discarding any one gate results in the inability to implement some Clifford operations; removing the Hadamard gate disallows powers of / in the unitary matrix representation, removing the phase gate S disallows in the unitary matrix, and removing the CNOT gate reduces the set of ...
The Hadamard transform H m is a 2 m × 2 m matrix, the Hadamard matrix (scaled by a normalization factor), that transforms 2 m real numbers x n into 2 m real numbers X k.The Hadamard transform can be defined in two ways: recursively, or by using the binary (base-2) representation of the indices n and k.
The Hadamard test produces a random variable whose image is in {} and whose expected value is exactly | | . It is possible to modify the circuit to produce a random variable whose expected value is I m ψ | U | ψ {\displaystyle \mathrm {Im} \langle \psi |U|\psi \rangle } by applying an S † {\displaystyle S^{\dagger }} gate after the first ...
Download QR code; Print/export Download as PDF; Printable version; In other projects ... We apply the Hadamard gate again only to the input qubits, causing the ...
Arbitrary Clifford group element can be generated as a circuit with no more than (/ ()) gates. [6] [7] Here, reference [6] reports an 11-stage decomposition -H-C-P-C-P-C-H-P-C-P-C-, where H, C, and P stand for computational stages using Hadamard, CNOT, and Phase gates, respectively, and reference [7] shows that the CNOT stage can be implemented using (/ ()) gates (stages -H- and -P ...
Download as PDF; Printable version; In other projects ... state using two and single qubit quantum gates, i.e. a Ry-rotation gate, a controlled Hadamard gate, ...
The figures below are examples of implementing a Hadamard gate and a Pauli-X-gate (NOT gate) by using beam splitters (illustrated as rectangles connecting two sets of crossing lines with parameters and ) and mirrors (illustrated as rectangles connecting two sets of crossing lines with parameter ()).
Let H be a Hadamard matrix of order n.The transpose of H is closely related to its inverse.In fact: = where I n is the n × n identity matrix and H T is the transpose of H.To see that this is true, notice that the rows of H are all orthogonal vectors over the field of real numbers and each have length .