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 $H_A \otimes H_B$?
[Edit]
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.
 A: 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.


*

*Yes, we could make other definitions, but that is not what is meant by "completely positive".

*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.

*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.

*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 $\mathcal L$.
A: 
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.
A: 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.  
