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Constructing a CP map with some decaying property

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Given some observable $\mathcal O \in \mathcal H$ it is simple to construct a CP (completely positive) map $\Phi:\mathcal{H}\mapsto \mathcal{H}$ that conserves this quantity. All one has to observe is that $$ \text{Tr}(\mathcal O \, \Phi[\rho]) = \text{Tr}(\Phi^*[\mathcal O] \rho).$$ Therefore, if we impose $\Phi^*[\mathcal O] = \mathcal O$, then $\text{Tr}(\mathcal O \, \Phi[\rho])=\text{Tr}(\mathcal O \rho), \; \forall \rho\in \mathcal H$. That amounts to impose that the Kraus operators of $\Phi^*$ should commute with $\mathcal O$.

I'd like, however, to construct a trace-preserving CP map for which the expectation value of $\mathcal O$ does not increase for any $\rho \in \mathcal H$. More explicitly, given $\mathcal O\in \mathcal H$, I want to construct $\Gamma:\mathcal H \mapsto \mathcal H$ such that $$ \text{Tr}(\mathcal O\, \Gamma[\rho]) \le \text{Tr}(\mathcal O \rho), \; \forall \rho \in \mathcal H .$$

How would you go about that? Any ideas?

This post has been migrated from (A51.SE)
asked Feb 22, 2012 in Theoretical Physics by Fernando (20 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
Do you want to construct $\Gamma$, or to characterize all possible $\Gamma$ ?

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3 Answers

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I'll restrict myself to trace-preserving CP-maps.

One can rewrite $\mathcal O=\sum_{k,l}o_k|k,l\rangle\langle k,l|$, where the $o_k$ are in decreasing order. The non-increasing condition $\langle\mathcal O\rangle$ corresponds then to an non-increasing condition on $k$. Writing $\Gamma$ in terms of Kraus operators, one has $\Gamma(\rho)=\sum_i B_i\rho B_i^*$ with $\sum_iB_iB_i^*=\mathbb1$. The condition on $k$ given above is then translated into the following writing of the Kraus operators: $$B_i=\sum_{\substack{k,l,k',l'\\k\le k'}}B_i^{klk'l'}|k,l\rangle\langle k',l'|.$$ Another way to say the same thing is the condition $B_i^{klk'l'}=0$ if $k>k'$.

Then, of course, the normalization condition imposes $$ \sum_{\substack{i,k',l'//k\le k'}}\left|B_i^{klk'l'}\right|^2=1, \forall k,l.$$

If you apply the same reasoning with a non-increasing and non-decreasing condition, you find that $k$ has to be conserved, and this is then equivalent to the commutativity condition you give in your question. In the same way, this answer is not general: you have operations which preserve $\langle\mathcal O\rangle$ without commuting with $\mathcal O$.

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answered Feb 23, 2012 by Frédéric Grosshans (210 points) [ no revision ]
I got it. The idea is to shift "population" of $rho$, in the eigenbasis of $\mathcal O$, towards eigenvectors that will have smaller contributions to $\langle \mathcal O \rangle_\rho$. Thanks a lot. Just one last comment, I believe the trace-preserving condition should be $\sum_l B_l^* B_l = \idty$, which then imposes a couple of other constraints in $B_l$ (maybe already implied by your conditions, but not explicitly). Let me know if you want to change something in your answer, and I'll accept it right afterwards. Thanks.

This post has been migrated from (A51.SE)
Exactly. The trace-preserving condition is the last condition I give $ \sum_{\substack{i,k',l'//k\le k'}}\left|B_i^{klk'l'}\right|^2=1, \forall k,l.$

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The negativity of the $o_k$ is not a problem. You probably missed the $k\le k'$ condition in my sum, which, with the ordering of the $o_k$ is crucial. I'll try to give a more detailed answer tomorrow, but I can't now. Sorry.

This post has been migrated from (A51.SE)
Thanks for the quick reply! Indeed the r.h.s. is $\text{Tr}(\rho \mathcal O)$, and the l.h.s is $\text{Tr}(\Gamma[\rho]\mathcal O)$. The inequality is what I want to obtain by putting constraints on the $B_i$'s. (Note that the l.h.s is $\text{Tr}(\Gamma[\rho]\mathcal O)$ even without imposing $s=t$). Of course I could ask $B_i$ to be diagonal in the basis of $\mathcal O$, but still there is the problem that some $o_k$ can be negative. Am I loosing something?

This post has been migrated from (A51.SE)
Thanks for the answer Frédéric, but I don't totally follow it. To start with, I believe you need just one index for the eigenbasis of $\mathcal O$, right? Assuming that, and decomposing the Kraus operators in the eigenbasis of $\mathcal O$, as you suggested, I get the condition: $\sum_i \sum_{k,s,t} o_k B_i^{ks} {B^*}_i^{tk} \langle s|\rho |t\rangle \le \sum_k o_k \langle k|\rho |k\rangle $. How do I conclude something from this?

This post has been migrated from (A51.SE)
@Fernando : if the observable is non-degenerate, the index $l$ is indeed is useless. But if the observable is degenerate, *i.e.* if two or more state correspond to the same value $o_k$ of the observable, this second index helps to have a more general solution.

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@Fernando : the right hand term is $\mathrm{Tr}(\rho\mathcal O)$, and, if $s=t$ in the left-hand term, the left hand term becomes $\mathrm{Tr}(\Gamma(\rho)\mathcal O)$.

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+ 1 like - 0 dislike

This is probably not exactly what you had in mind, but how about the channel that discards its input and always outputs the state corresponding the the minimum eigenvalue of $\mathcal{O}$?

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answered Feb 23, 2012 by Dan Stahlke (70 points) [ no revision ]
Thanks Dan, but indeed that's not what I had in mind. I'd like something smoother than that. More a characterization of the channels than a single construction.

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The formal condition---the correspondent of $\Phi^*[\mathcal{O}]=\mathcal{O}$ in the other case---is $\Gamma^*[\mathcal{O}]\leq \mathcal{O}$. For $\mathcal{O}>0$ (all eigenvalues strictly positive), if one multiplies on the right and on the left by $\mathcal{O}^{-1/2}$, one can see that this condition is equivalent to the map with Kraus operators $\mathcal{O}^{1/2}K_i\mathcal{O}^{-1/2}$ being trace non-increasing.

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answered Feb 24, 2012 by Marco (260 points) [ no revision ]

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