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I wrote an ipython script to perform a polynomial Pauli decomposition for a general matrix. It can be found here: https://github.com/trahancj/quantum_algorithms.git There are examples as well.


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I wrote an ipython script to perform this decomposition for a general matrix. It can be found here: https://github.com/trahancj/quantum_algorithms.git There are examples as well.


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There are two reasons this works and you have identified the first one. Namely, the fact that a Pauli operator anywhere in a Clifford circuit is equivalent to a Pauli operator immediately preceding a terminal measurement. The second reason is the simple predictable effect that a Pauli operator immediately preceding a computational basis measurement has on ...


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$Ry$ gate is defined as $$ Ry(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}. $$ Hence $$ Ry(\pi/2) = \begin{pmatrix} \cos(\pi/4) & -\sin(\pi/4) \\ \sin(\pi/4) & \cos(\pi/4) \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}. $$ Similarly for $...


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You are using rz in your code, while the identity you are asking about uses $R_y$ Here is a qiskit code to check the identity: from qiskit import QuantumCircuit from qiskit.quantum_info.operators import Operator from qiskit.visualization import array_to_latex import numpy as np circ = QuantumCircuit(1) circ.ry(-np.pi / 2, 0) circ.z(0) circ.ry(np.pi / 2, 0) ...


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Summary: There is a solution for expressing a tridiagonal matrix of the form you've provided for arbitrary $n$ in terms of Pauli operators using recursion. This procedure is given at the bottom of this answer. Expressing a tridiagonal matrix recursively Writing an $n$-qubit tridiagonal matrix in terms of Pauli operators can be done recursively. Ignoring the ...


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