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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?

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  • $\begingroup$ Do you want to construct $\Gamma$, or to characterize all possible $\Gamma$ ? $\endgroup$
    – Frédéric Grosshans
    Commented Feb 23, 2012 at 11:22

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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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  • $\begingroup$ 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? $\endgroup$
    – Fernando
    Commented Feb 23, 2012 at 18:26
  • $\begingroup$ @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. $\endgroup$
    – Frédéric Grosshans
    Commented Feb 23, 2012 at 18:31
  • $\begingroup$ @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 $\mathrm{Tr}(\Gamma(\rho)\mathcal O)$. $\endgroup$
    – Frédéric Grosshans
    Commented Feb 23, 2012 at 18:34
  • $\begingroup$ 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? $\endgroup$
    – Fernando
    Commented Feb 23, 2012 at 18:48
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    $\begingroup$ 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. $\endgroup$
    – Fernando
    Commented Feb 24, 2012 at 10:31
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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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  • $\begingroup$ 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. $\endgroup$
    – Fernando
    Commented Feb 23, 2012 at 18:15
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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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