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Suppose you have a quantum state $|w\rangle$ consisting of $m + n$ qubits, and you set up a measurement that measures the first $m$ qubits in the standard basis. What are the matrices in the corresponding POVM?

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Well, since these are projective measurements on the subspace of the first $m$ qubits, we can just list all projectors on the computational basis of this first subspace and 'pad' them with $I$'s on the second subspace:

$$ P_{j} = |j\rangle\langle j|_{m} \otimes I_{|n|},\,\,\, \forall j \in \{0,1\}^{m}, $$ which gives exactly $|\{0,1\}^{m}| = 2^{m}$ different operators for the POVM. If you identify distinct measurement outcomes with every operator, say $\lambda_{j} = j_{d}$ (e.g. $j$ in decimal form), you can easily write down a measurement operator as well:

$$ M = \sum_{j} \lambda_{j}P_{j} = \sum_{j} j_{d}|j\rangle \langle j \otimes I_{n}| $$

See also, for instance, this nice answer by Daftwullie for a different measurement operator. Note that that answer omits the extra subspace of $n$, but you can just treat that by padding with $I$'s again.

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    $\begingroup$ Thank you so much. $\endgroup$ – Anurag Singla Jan 4 at 11:17

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