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?

  • $\begingroup$ I don't understand the question. What do you mean exactly with "how does going from open to closed change the scenario"? $\endgroup$
    – glS
    Commented Feb 16, 2021 at 9:25
  • $\begingroup$ @glS.: We can't apply projective measurement when the system is not closed. In an open system the evolution of a state can't be described by unitary operation. We often hear that a general measurement will be projective measurement if it follows the additional condition that $$M_{m}^\dagger M_{m'}= \delta_{m,m'}M_{m}$$. I want to know if there is some relation holds between these two. $\endgroup$
    – user27286
    Commented Feb 16, 2021 at 11:51
  • $\begingroup$ Are you aware of the concept of a Naimark dilation? $\endgroup$
    – Rammus
    Commented Feb 16, 2021 at 11:59
  • $\begingroup$ @Rammus.: Heard first time in my life a min ago from you. $\endgroup$
    – user27286
    Commented Feb 16, 2021 at 12:01
  • $\begingroup$ sure you can do projective measurements when the system is not closed. The system being "open" or "closed" refers to its dynamics, i.e. how it evolves. That doesn't affect the types of measurements you can perform on the state itself. About the relation between POVMs and projective measurements, have you seen quantumcomputing.stackexchange.com/q/7004/55? See also quantumcomputing.stackexchange.com/q/12275/55 $\endgroup$
    – glS
    Commented Feb 16, 2021 at 12:26

1 Answer 1


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.


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