The correct set of measurement operators on a mutiple qubit system

Question

I was wondering if the complete set of measurement operators for a state :

$|\phi \rangle=c_{00}|00\rangle+c_{01}|01\rangle+c_{10}|10\rangle+c_{11}|11\rangle$

Would be given by :

$P_0\otimes I=|00\rangle\langle00|+|01\rangle\langle01|$

$P_1\otimes I=|10\rangle\langle10|+|11\rangle\langle11|$

$I\otimes P_0=|00\rangle\langle00|+|10\rangle\langle10|$

$I\otimes P_1=|01\rangle\langle01|+|11\rangle\langle11|$

Or would it be given by :

$P_0\otimes I=|00\rangle\langle00|+|01\rangle\langle01|$

$P_1\otimes I=|10\rangle\langle10|+|11\rangle\langle11|$

$I\otimes P_0=|00\rangle\langle00|+|10\rangle\langle10|$

$I\otimes P_1=|01\rangle\langle01|+|11\rangle\langle11|$

$P_0\otimes P_0=|00\rangle\langle00|$

$P_1\otimes P_1=|11\rangle\langle11|$

$P_0\otimes P_1=|01\rangle\langle01|$

$P_1\otimes P_0=|10\rangle\langle10|$.

I feel like it may be the second one by intuition as it seems to allow for more distinction between measurements but intuition is often misleading in mathematical contexts and I feel like I recall seeing the first one written somewhere. Could anyone clear this up for me please?

Answer 1

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 basis, and there's also $P_0\otimes P_0, P_0\otimes P_1, P_1\otimes P_0, P_1\otimes P_1$ which describes measuring both qubits in the standard basis.

Answer 2

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.