# Complete positivity: why is the condition sufficient for quantum maps?

I know that when we define quantum maps, we need the map to be completly positive, to ensure that if our system $$A$$ is entangled with some extra system $$B$$, the evolution on $$H_A \otimes H_B$$ will also be positive (not only the evolution on $$H_A$$).

For this purpose, we say that the map must be completly positive.

But why is this condition enough. Like, why complete positivity will ensure me that I will never find a non positive global transformation on $$H_A \otimes H_B$$?

My definition of complete positivity is :

$$\forall |\phi^{AB}\rangle \in H_A \otimes H_B : \langle \phi^{AB} | \mathcal{L}_A \otimes 1 (\rho_{AB}) | \phi^{AB} \rangle \geq 0$$ where $$\mathcal{L}_A$$ is the operator that I want completly positive.

So it assumes that the operator acting on $$\rho_{AB}$$ has the form $$\mathcal{L}_A \otimes 1$$ which is not obvious for me.

[edit 2] actually my question is very closely related to another one I asked here Quantum map and preservation of trace

But this question is physically more general. Also, it would allow first, to check that my assumptions were correct in this other post, and also the answer given is from my perspective a little complicated.

Apparently the motivation behind the proof comes from classical probability theory that I don't master enough to really understand how the ideas came in the proof proposed.

Thus, I would like a different way of answering the problem, if it exists.

• Complete positivity is defined as the property that you will never find a non positive global transformation on $H_A \otimes H_B$. Did you have some other definition in mind? If so, what? – Emilio Pisanty Oct 22 '18 at 8:19
• @Emilio Pisanty I edited – StarBucK Oct 22 '18 at 8:35
• What is $\phi^{AB}$? You never define it, so I don't know if you're wondering why you can assume $\phi^{AB}$ is a maximally entangled state, or whether you have a different question that I don't completely understand. – Peter Shor Oct 22 '18 at 18:43
• @PeterShor I edited : it is any vector belonging to $H_A \otimes H_B$. – StarBucK Oct 22 '18 at 20:01

Let me start by asking the question: What do you think the definition of complete positivity should be instead? You want it to "ensure that I will never find a non positive global transformation", but that can not be possible in that generality. What I am trying to say here is: if complete positivity is supposed to be a condition on $$\mathcal L_A$$, then $$\mathcal L_A$$ must appear in the definition.

What we really want is that maps of the form $$\mathcal L_A \otimes \mathcal L_B$$ is positive, for arbitrary $$H_B$$ and $$\mathcal L_B$$. We only consider global transformations that are a product of two maps $$\mathcal L_A$$ and $$\mathcal L_B$$ that act on their individual Hilbert spaces and leave the other space alone.

Now note that $$\mathcal L_A \otimes \mathcal L_B = (\mathcal L_A \otimes 1_B) (1_A \otimes \mathcal L_B)$$ and it should become clear why complete positivity is defined the way it is.

The definition guarantees that, if $$\mathcal L_A$$ and $$\mathcal L_B$$ are completely positive, also $$\mathcal L_A \otimes \mathcal L_B$$ is completely positive. It is initially surprising that the same property does not hold if we replace "completely positive" with "positive" and I recommend you to come up with a counterexample to that.

Edit in response to the comments.

1. Yes, we could make other definitions, but that is not what is meant by "completely positive".
2. Complete positivity usually crops up in the context of CPTP maps. CPTP maps are the answer to the question:
If I have access to a subsystem $$H_A$$ only, and I study its time evolution: what is the absolute minimum of properties I can be certain that the time evolution will have?
The answer is that the evolution operator must be (convex) linear and map a density matrix to a density matrix, and the positivity of the resulting density matrix must not depend on what happens in unrelated other experiments. And that corresponds to the definition of complete positivity.
3. So far we didn't talk about two systems being entangled or anything at all. That comes in via the Stinespring theorem which tells us: a map is CPTP if and only if it can be written in the form $$\rho \mapsto \operatorname{tr}_B \{ U (\rho \otimes \rho_B) U^\dagger \} .$$ Here, $$U$$ is a unitary time evolution on the full system.
4. So what exactly are the conditions you want for a map $$\mathcal L_A$$ to be, let's say, "StarBucK-positive"? Some options:

$$\mathcal L_A$$ is StarBucK-positive iff: for all Hilbert spaces $$H_B$$ and all (all positive?) (all StarBuck-positive?) (all unitary?) time evolutions $$\mathcal L$$ on $$H_A \otimes H_B$$ so that $$\operatorname{tr}_B \circ \mathcal L = \mathcal L_A$$ (and so that $$\operatorname{tr}_A \circ \mathcal L$$ is also StarBucK-positive?) ... what condition holds?

Another edit, I want to make one thing clear: If $$\mathcal L_A$$ is CP, that does not guarantee that every $$\mathcal L$$ with $$\operatorname{tr}_B \circ \mathcal L = \mathcal L_A$$ is positive. Such a guarantee is impossible as $$\mathcal L_A$$ does not contain all information about $$\mathcal L$$.

• The thing I don't understand is that you assume the global transformation is $\mathcal{L}_A \otimes \mathcal{L}_B$ which is a big assumption. We could imagine that the density matrix follow $\rho_A \rightarrow \mathcal{L}_A(\rho_A)$ and $\rho_B \rightarrow \mathcal{L}_B(\rho_B)$ but the global transformation is not $\mathcal{L}_A \otimes \mathcal{L}_B(\rho_{AB})$. So in general the transformation would be some $\mathcal{L}$ acting on $\rho_{AB}$ but verifying $Tr_A(\mathcal{L}(\rho_{AB}))=\mathcal{L}_B(\rho_B)$ and $Tr_B(\mathcal{L}(\rho_{AB}))=\mathcal{L}_A(\rho_A)$. – StarBucK Oct 22 '18 at 11:09
• In this way we have $\mathcal{L}_A$ and $\mathcal{L}_B$ that appear in the definition (so this would be the most general evolution for $\rho_{AB}$, thus $\mathcal{L}$ is the quantity we have to constraint with positivity.) But maybe I didn't understood enough what you wanted to say – StarBucK Oct 22 '18 at 11:09
• @StarBucK I edited – Noiralef Oct 22 '18 at 13:11
• Actually with your very last edit you are exactly in the point that is confusing me. What I didn't understand was that the CPTP condition is not enough to ensure that $\mathcal{L}$ will be positive, even if $B$ doesn't evolve, it justs deals with some very particular evolutions $\mathcal{L}$. So the CPTP condition is necessary but not sufficient to ensure that $\mathcal{\rho_{AB}}$ will still be positive, right ? – StarBucK Oct 22 '18 at 13:26
• Pity that I wrote so many other things before realizing what your problem was^^ You are right: Just knowing $\mathcal L_A$ does not tell you how $\rho_{AB}$ evolves, and it could lose positivity. In physical situations, we usually assume that $\rho_{AB}$ evolves unitarily -- then it will always remain positive, of course. – Noiralef Oct 22 '18 at 13:39

My definition of complete positivity is : $$\langle \phi^{AB} | \mathcal{L}_A \otimes 1 (\rho_{AB}) | \phi^{AB} \rangle \geq 0$$ where $$\mathcal{L}_A$$ is the operator that I want completely positive.

This is indeed the definition of a completely positive map, and it is defined in that way to guarantee that the evolution caused by $$\mathcal{L}_A$$ is physical even if the system turns out to be part of a larger system in an entangled state.

The reason why we restrict the evolution operator in question to the form $$\mathcal{L}_A \otimes 1$$ instead of $$\mathcal{L}_A \otimes \mathcal{L}_B$$ or even $$\mathcal{L}_{AB}$$ is because we want complete positivity to be a statement that is exclusively about $$\mathcal{L}_A$$ and not something else. For this criterion, $$\mathcal{L}_A$$ is fixed: it is what it is and it has the action it has on $$\mathscr H_A$$, and we don't care where it came from or what else is acting in any other parts of the system.

In particular, that means that if the way that you generate $$\mathcal{L}_A$$ is that you have some ancilla system $$\mathscr H_{A'}$$ and you have some bigger channel $$\mathcal L_{AA'}$$ acting on $$\mathscr H_A\otimes \mathscr H_{A'}$$, which you then partial-trace down to $$\mathcal L_A = \mathrm{tr}_{A'}(\mathcal L_{AA'})$$, then as far as the complete-positivity of $$\mathcal L_A$$ as a quantum channel goes, we don't care that that's how it was produced. We care that it's a functional quantum channel which represents a physical evolution on $$\mathscr H_A$$ even if there are other systems (not the ancilla) which are entangled with the system of interest.

This is why the criterion is written in the way you've set down: there might be other tensor factors around, but the physical transformation in question is $$\mathcal L_A$$ and $$\mathcal L_A$$ only (which is why you have an $$\otimes 1_B$$ trailing it). Everything else is just there to ensure that the transformation is physical in the most general possible setting that $$\mathcal L_A$$, as a unit, might encounter itself.

• If I understand you : we don't care how $\mathcal{L}_A$ has been produced : it can has been done through something very different than $\mathcal{L}_A \otimes 1$. But it also could have been done through this transformation. So it must at least check the fact that $\mathcal{L}_A$ preserve positivity : it is a necessary condition. – StarBucK Oct 22 '18 at 17:34
• And here, your second point is that $\mathcal{L}_A \otimes 1$ is a particular case of evolution that only depends on $\mathcal{L}_A$, and I totally agree. But the thing I don't get is : how do we know it is the only one (the only particular evolution that only depends on $\mathcal{L}_A$) ? Maybe we don't care in the end because we just need necessary conditions but I would like to check this. (I rewrote my comments to try to make my question more clear) – StarBucK Oct 22 '18 at 17:35

There seem to be a bunch of questions here. I'm answering some questions that I think you are asking, but I'm not sure whether I really am addressing your questions — I really only understand one of your questions.

(1) Why does the operator acting on $$\rho_{AB}$$ have the form $${\cal L}\otimes I$$?

Generally, we assume that we are given a map that acts only on system $$A$$.

Now, if we apply $${\cal L}$$ to $$A$$, and we do nothing to $$B$$, then the combined operator is $${\cal L}\otimes I$$.

A brief intuition for this: if applying some operator just to system $$A$$ has some non-identity effect on a second system $$B$$ that was uncorrelated to $$A$$, that would be very weird. And linearity of quantum mechanics then implies that the combined operator is $${\cal L}\otimes I$$.

(2) Maybe your real question is: suppose you have a quantum map $${\cal L}$$ that acts on multiple systems. How do you know that when you just look at $$A$$, that quantum map is completely positive?

Suppose there some system $$C$$, somewhere in the universe (maybe on Alpha Centauri), which $${\cal L}$$ acts on as the identity. Then $${\cal L} \otimes I$$ is the map that acts on $$A \otimes C$$, and the argument above show that $${\cal L}$$ is completely positive. I assume that this addresses your question for practical real-world quantum maps.

(3) If you have a very, very small universe — say your Hilbert space $$A$$ is a subspace of the universe $$U$$ with $$\dim U < (\dim A)^2$$ — then I believe it's quite possible that $${\cal L}$$ isn't completely positive.

• Your answer is actually related to this one physics.stackexchange.com/questions/434702/… the answer uses a lot of "tricks" (extending the action of the operator, writing "smart" decompositions etc), and I wonder if there is a more straightforward way to see it because even if I can follow line by line the thing I can't really say I understand the proof as a whole. And as I am not sure to totally understand all the details of this proof I am not sure I can accept it for now. – StarBucK Oct 22 '18 at 20:10
• But even, actually the point of view you propose is different than the two other answers. In these ones the authors considers that we don't care what happens in $H_B$ (it could evolve via $\mathcal{L}_B$) but we just want to constraint $\mathcal{L}_A$ by only looking at the particular $\mathcal{L}_A \otimes 1$ and constraining $\mathcal{L}_A$ with those. Which would give a necessary condition on $\mathcal{L}_A$. And the questions I asked in the comment is if it is a arbitrary choice to only constraint with those or if there is a deeper reason. – StarBucK Oct 22 '18 at 20:18
• I think my point of view is fairly consistent with Emilio Pisanty's. I assume that the map only acts on $A$, and does nothing to any quantum system that is unentangled with $A$. He says let's consider a map, and only worry about its effects on $A$. Then let's extend it to ${\cal L}\otimes I$. – Peter Shor Oct 22 '18 at 20:32