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In the case of projective measurements, we have a set of projectors $\{P_i\}$ satisfying the completeness relation $$ \sum_iP_i=I. $$ Note that this also means they satisfy $\sum_iP_i^\dagger P_i$, which I would argue is more relevant. If you have a state $|\phi\rangle$, then with probability $p_i=\langle\phi|P_i^\dagger P_i|\phi\rangle=\langle\phi|P_i|\phi\...


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Your first set of projectors is correct. There's a general rule - if you're not supposed to do anything to a particular qubit, that's described by the identity matrix. How you perform the projection is exactly what you would normally do. If $|\psi\rangle$ is a 3-qubit state, you calculate, for example, the probability of getting the 00 result: $$ p_{00}=\...


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POVMs are more general than projective measurements. Thus, every projective measurement is also a POVM, by choosing $E_i=P_i$.


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This can actually be easily done using the Qiskit Terra qiskit.quantum.info.analysis.average.average_data function that takes the counts data returned by a backend and the desired observable defined by a dict, list, or ndarray. The doc-string for that function actually has the ZZ your looking for as an example.


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There is no single set that is "the" complete set of measurement operators. You can have any set that you want so long as they add up to the identity. So, one such set is $P_0\otimes I, P_1\otimes I$, which describes measuring only qubit 1 in the standard basis, and there's $I\otimes P_0, I\otimes P_1$ which describes only measuring qubit 2 in the standard ...


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A valid measurement consists of operators $\{M_k\}$ such that $\sum_k M^\dagger_k M_k = I$. You need appropriate normalization factors in front for both cases but since $P_i^\dagger P_i = P_i$ and $P_0 + P_1 = I$, they both do correspond to valid measurements.


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These measreuements describe a non-projective measurement. We typically convert these into projective measurements by introducing ancilla qubits. In this case, define a unitary $U$ such that $$ U|0\rangle=\alpha|0\rangle+\sqrt{1-\alpha^2}|1\rangle. $$ Take the qubit that we want to measure, and introduce an ancilla in the state $|0\rangle$. Apply controlled-...


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Qiskit currently supports measurements in the computational basis from Qiskit Terra and Aer, that is, returning 1 if the qubit is in state $|1\rangle$, and 0 if the qubit is measured to be in state $|0\rangle$. However, it is relatively easy to perform a change of basis unitary to our quantum circuit just prior to measurement, in order to instead measure in ...


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