I'm trying to relate some notions of set theory to POVMs. I firstly explain the scenario with set theory and then in the POVM setting.

For some finite $N \in \mathbb{N}$, let $A_i$ and $B_i$ for $i=1,...,N$ be sets and assume there is some larger set $C$ such that $A_i\subset C, B_i\subset C$ for all $i$. Let's say for $J=\{1,...,N/2\}$ (just take $N$ even), I'd like to compute $$ \left(\bigcup_{i\in J} A_i\times B_i\right)^c.$$

By De Morgan's laws we can write, $$\left(\bigcup_{i\in J} A_i\times B_i \right)^c = \bigcap_{i\in J} (A_i\times B_i)^c = \bigcap_{i\in J} A_i^c\times B_i \cup A_i\times B_i^c \cup A_i^c \times B_i^c$$

Now I'd like to do something similar for POVMs. Let's say I have two POVMs, $A = \{A_1,...,A_N \}$ and $B=\{B_1,...,B_N\}$.

For either POVM, I think it's clear how to define a complement for a subset. For example, if I want the complement of $\{A_2, A_3\}$, then I can do $\mathbb{1} - A_2 - A_3$. My interpretation is, $\mathbb{1} - A_2 - A_3$ encompasses all events except $A_2$ or $A_3$. Now I want to define a (or be given a known) complement for a joint measurement.

Let's again say I have some $J = \{1, ..., N/2\}$ and I want to determine the complement of the sum joint measurements indexed over $J$. That is, I want somehow to define

$$\left(\sum_{i\in J}A_i \otimes B_i\right)^c$$

My first thought was to have

$$\left(\sum_{i\in J}A_i \otimes B_i\right)^c = \mathbb{1} - \sum_{i\in J}A_i \otimes B_i$$

But this only encompasses the analog of $\bigcup_{i \in J} A_i^c \times B_i^c$ and misses the analogs of $A_i^c \times B_i$ and $A_i\times B_i^c$.

Then I thought, in a De Morgan sense

$$\left(\sum_{i\in J}A_i \otimes B_i\right)^c = \prod_{i\in J} (A_i \otimes B_i)^c = \prod_{i\in J} (\mathbb{1} - A_i)\otimes B_i + A_i \otimes (\mathbb{1}-B_i) + (\mathbb{1} - A_i)\otimes (\mathbb{1}-B_i)$$

This I think is logical, but I've not yet convinced myself that it's true. Has anyone had a similar scenario?

Thanks for your time and sorry if the question isn't clear.

  • $\begingroup$ Both cases are the same if the operators are projections. Why would you think that a De Morgan like law holds for general a general POVM? $\endgroup$
    – biryani
    Mar 8, 2018 at 19:56
  • $\begingroup$ In my case they are general POVMs. I think that this a reasonable complement for a joint measurement because it encompasses every "way" to make a complement measurement, if we agree that a complement measurement as I stated it for a single POVM is a valid complement. Further, it seems valid because a POVM is indexed the same way a set is, so a complement of a tensor operator seems like it should have a similar form of the complement of set. This is why I think it should hold, but I still have some work to do. $\endgroup$ Mar 8, 2018 at 20:17
  • $\begingroup$ I think I should clarify my last point. Since a POVM has a set structure, taking a complement of some subset of "elements"/measurements is pretty straight forward, simply sum the elements outside of the subset. When taking a joint measurement, still we can consider subsets, so the complement is essentially just like the normal set complement. At least that's what I think. $\endgroup$ Mar 8, 2018 at 20:22
  • $\begingroup$ quantumcomputing.stackexchange may be a better place for such questions. $\endgroup$ May 18, 2018 at 12:57
  • $\begingroup$ Thanks. I recently learned about it and will post future questions there. $\endgroup$ May 18, 2018 at 15:52

1 Answer 1


I've come to the conclusion that the definition of the complement as I've written doesn't make sense for POVMs. It assumes that the product of two elements is a simultaneous event which is false.

What I found is, because I'm dealing with probabilities and for me this was a "fail" event, I just switch my notion to 1 - "success" and I don't have to use any complements.


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