# Complete positivity and quantum dynamics

1. Aug 7, 2014

### JorisL

Hi again,

Another, possibly trivial, question.
In quantum dynamics we consider maps containing the evolution of a system.
Suppose we have a completely positive (CP henceforth) map
(1) $\Gamma:\mathcal{M}_k\rightarrow \mathcal{M}_n$

This map has following properties:
• Trace-preserving
• Complex linear
• Continuous in time (not important here)

This map is CP if it is d-positive for all d. I gather from this it should be possible to trivially extend the system we are interested in to the system+an environment 'of dimension d'.
More usable is the fact that the extended map $id_d\otimes \Gamma$ is positive.

This map is defined for the Schrödinger picture, acting on density matrices. There exists a dual map $\Gamma^*$ in the Heisenberg picture, i.e. acting on observables. (this dual is unity/unital preserving)

In various texts I've found that it should be easy to show that
$\Gamma$ is CP iff $\Gamma^*$ is CP

Now some class mates used some really fishy manipulation (perhaps bad mathematical notation/practice) which seems to have gotten me completely lost.

Here's what they did.
in the schrödinger picture the map is trace-preserving thus
$$tr \rho X = tr \Gamma(\rho)X = tr \rho\Gamma^*(X)$$
So far it seems ok to agree except for the first equality. In general the dimensions k and n in (1) aren't the same so the matrix product in one of the sides will not be defined.

Then they extend the map to use d-positivity of $\Gamma$. From this they infer that the dual map is also d-positive. (symbolic, no explanation)

I'm not sure how they justified all this hand-wavy(at best) stuff. Another problem is that they don't appreciate attaining maximal rigour in this kind of things so they might have overlooked these problems.

Looking forward to some suggestions,

Joris

2. Aug 7, 2014

### kith

What are Mk and Mn? Some kind of state space? If the CP map acts on the density matrix of a certain open system, the dimensions of these spaces must be the same.

This identity not only uses the trace-preserving property of Γ but implies that ρ and Γ(ρ) are the same state. This is certainly wrong in general. If Γ is the time evolution of the open system your equation states that the expectation value of all observables X is constant.

3. Aug 7, 2014

### JorisL

You are right, the spaces Mk are the observable on a k-dimensional system.
The identity should in my opinion hold if we don't have the observable X in there since the trace of a d.m. should be normalised (in our convention, one can use a normalisation constant).

I'll try to figure some more of this out and give an expanded explanation of the context. Although it's only 2 pages long in my notes (a little bit short if you ask me). Some reference texts would be nice.

4. Aug 7, 2014

### Staff: Mentor

Two things that may interest you in this context is Gleason's Theorem:
http://kof.physto.se/cond_mat_page/theses/helena-master.pdf [Broken]

And especially Wigner's theorem:
http://arxiv.org/abs/0808.0779

Thanks
Bill

Last edited by a moderator: May 6, 2017