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While Hadamard gate is defined as $$H= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},$$ $y$-rotation by $\pi/2$ leads to gate $$Ry(\pi/2)= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.$$ So, there is a difference in position of -1 in the second column. Application of the $X$ gate returns the -1 in $... 6 If you are trying to implement a fault-tolerant quantum computation, you need to implement unitary gates that act on logical qubits. You typically have a finite set of these gates available, and what you really care about is making your operations in such a way as to keep the fault-tolerant threshold as small as possible. If you calculate a fault-tolerant ... 5 The matrix $$M = \frac{1}{\sqrt{2}}\begin{bmatrix}-i & 1\\-1 & i\end{bmatrix}$$ resembles $$X/2 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & -i\\-i & 1\end{bmatrix}\tag1$$ where we follow the notation$\pm X/2$for the$\pm\frac{\pi}{2}$rotation around the$X$axis as used in the table B.6 on page 101 in Julian Kelly's PhD thesis. We can make ... 3 The decomposition they give is the following: $$R_z(\theta) = R_x\left(\pi / 2\right) R_y(\theta) R_x\left(-\pi / 2\right)$$ Therefore, the Qiskit code would look like: from qiskit import QuantumCircuit from qiskit.quantum_info import Operator import numpy as np theta = np.pi / 4 qc = QuantumCircuit(1) qc.rx(-np.pi / 2, 0) qc.ry(theta, 0) qc.rx(np.pi / 2,... 3 You can decompose the T gates themselves to create a Toffoli Gate. Here is one way of doing this:- You can refer to this Qiskit chapter if you are interested and want to understand gate decomposition: https://qiskit.org/textbook/ch-gates/more-circuit-identities.html 1 There currently isn't a controlled Pauli product gate in Stim. You have to decompose it into a series of CX, CY, and CZ gates. # Apply X1*Y2*Z3 controlled by qubit 0 CX 0 1 CY 0 2 CZ 0 3 # Apply X1*Y2*Z3 if latest measurement result was True CX rec[-1] 1 CY rec[-1] 2 CZ rec[-1] 3 1 Qiskit defines the$RX$gate as follows: $$RX(\theta) = \exp\left(-i \frac{\theta}{2} X\right) = \begin{pmatrix} \cos{\frac{\theta}{2}} & -i\sin{\frac{\theta}{2}} \\ -i\sin{\frac{\theta}{2}} & \cos{\frac{\theta}{2}} \end{pmatrix}$$ Thus, setting$\theta = \pi$, would give us: $$RX(\pi) = \... 1 Retrieving an arbitrary qubit, even using an ancilla qubit, would violate the no-cloning theorem. This can be seen in the simplest case where the array stores a single qubit and is indexed by a single qubit. So the array is of the form$$ \left|\psi\right> = a \left|00\right> + b \left|01\right> + c \left|10\right> + d\left|11\right>$$where$|...