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We know that POVMs are applied in the more general cases where the system is not necessarily closed. So mathematically, how does going from open to closed system change the scenario in case of POVM so that it becomes a projective measurement?

glS
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user27286
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1 Answers1

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An $m$-outcome positive-operator value measure or POVM consists of $m$ positive operators $A_1,\ldots,A_m$ on $\mathbb{C}^n$ such that $$\sum_{i=1}^m A_i=I.$$ Each $A_i$ is positive iff $A_i=B_i^*B_i$ for some operator $B_i$ on $\mathbb{C}^n$, thus each POVM is a projection-valued measure or PVM iff each $B_i$ is an orthogonal projection, if so then $B_i=B_i^*=B_i^2$ and $B_1,\ldots,B_m$ is a PVM since $$\sum_{i=1}^n B_i=\sum_{i=1}^n B_i^2=\sum_{i=1}^n B_i^*B_i=\sum_{i=1}^n A_i=I.$$ As mentioned in the comments, Naimarks dilation theorem shows that for any POVM on $\mathbb{C}^n$ there is an isometry $V:\mathbb{C}^n \rightarrow \mathbb{C}^{N}$ where $N\geq n$ and each positive operator $A_i$ gets mapped to an orthogonal projection $\widetilde{A_i}$ in the larger space where the "completeness relation" of summing to the identity still holds, thus turning any POVM into a PVM on the larger space.

Condo
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