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How to show that a sum of Pauli operators is non-zero?

The set of $n$-qubit Pauli matrices $\{P_i\}_{i=1,\dots,4^n}$ forms a (orthogonal) basis for the vector space of complex matrices. In particular, the set of Pauli matrices is linearly independent. ...
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How to show that a sum of Pauli operators is non-zero?

I realized that this can be checked rather easily. We compute $$\left( \sum_{(i,j) \in E} Y_i Z_j \right) |0\rangle^{\otimes n}$$ and see that it is the sum of a subset of standard basis vectors: $$\...
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What are the Pauli-Y eigenvectors?

The eigenvectors that you have written above (both from Wikipedia and those plotted with Numpy) are valid eigenstates of the $pauli-Y$ operator. I have done the same calculations that you have ...
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What are the Pauli-Y eigenvectors?

Your first equation is wrong. If $|i\rangle=\frac1{\sqrt2} (|0\rangle + i|1\rangle)$, then its dual is the tranposed-conjugated version $\langle i|=\frac1{\sqrt2} (\langle 0| - i\langle 1|)$, which ...
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Why doesn't Z-gate change phase of |0⟩

A rotation by $\pi\over2$ around the Z-axis is done by applying $iZ$. However, this is equivalent to $Z$ up to unobservable global phase.
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Why doesn't Z-gate change phase of |0⟩

You can consider $Z$ gate to be a "negation" in Hadamard basis. If you apply the gate on state $|+\rangle$, you get $|-\rangle$ and vice versa. In computational basis, state $|0\rangle$ ...
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How to decompose a multi qubit Clifford unitary into a sequence of clifford gates

I don't know if there is a formal way to decompose a n-qubits Clifford operation with gates from {H,X,Y,Z,CNOT}, but I know a paper that uses Monte Carlo Three Search for this purpose: Automated ...
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