<p>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. </p>

<p>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.</p>

CC BY-SA 3.0
2018-03-12T11:49:34.487Z
2018-03-12T11:49:34.487Z
Stephen Diadamo
<p>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.</p>

<p>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 
$$  \left(\bigcup_{i\in J} A_i\times B_i\right)^c.$$ </p>

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

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

<p>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.</p>

<p>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</p>

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

<p>My first thought was to have </p>

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

<p>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 $A_i\times B_i^c$.</p>

<p>Then I thought, in a De Morgan sense</p>

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

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

<p>Thanks for your time and sorry if the question isn't clear.</p>

CC BY-SA 3.0
2018-03-08T19:56:32.747Z
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?
biryani
CC BY-SA 3.0
2018-03-08T20:17:22.947Z
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.
Stephen Diadamo
CC BY-SA 3.0
2018-03-08T20:22:19.590Z
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.
Stephen Diadamo
CC BY-SA 4.0
2018-05-18T12:57:01.293Z
quantumcomputing.stackexchange may be a better place for such questions.
Jalex Stark
CC BY-SA 4.0
2018-05-18T15:52:47.060Z
Thanks. I recently learned about it and will post future questions there.
Stephen Diadamo
<quantum-computing><quantum-information>
Complement for joint POVMs?
Stephen Diadamo
# Complement for joint POVMs?

(Asked by: username_0 on 2018-03-08)

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.

**Tags:** quantum-computing, quantum-information

> **Comments:**
> **username_1**: 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?  
> **username_0**: 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.  
> **username_0**: 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.  
> **username_2**: quantumcomputing.stackexchange may be a better place for such questions.  
> **username_0**: Thanks. I recently learned about it and will post future questions there.  

---

## 1 answers

### Answer by username_0 on 2018-03-12. Score: 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.


---

