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Based on my former questions and the kind help from the answers (especially Norbert Schuch), it seems that the maximally mixed state is not that special as I thought. But still I have to admit that it is a special state.

We know that there exist such operators $M_A$ on a system A, that for some initial state of A, $\rho_A$, the iterative implementation of $M_A$ on $\rho_A$ will not lead to a convergence. For example, implementing a simple operator given as $M_A=|1><0|+|0><1|$ on an input $\rho_A=p|0><0|+(1-p)|1><1|, p\neq 1/2$ will not lead to a converged state, instead the iteration will result in a bouncing between two states.

So I'm curious if there exist such kind of operators that will not converge on maximally mixed state of A. In my imagination, it maybe behave like jumping among different fixed states periodically or just evolve non-periodically. But I have no solid example of it.

I tried to construct such an operator but without success.

Can anybody tell me if there is such kind of operators or give me an example of it? Thanks.

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  • $\begingroup$ I assume that by "such operators" you in fact refer to trace-preserving quantum channels, as in your previous questions. $\endgroup$
    – Norbert Schuch
    Commented Feb 28, 2016 at 17:18

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The following channel on a $3$-level system will do: $$ \mathcal E(\rho) = |0\rangle\langle0|\,\big[\langle1|\rho|1\rangle +\langle2|\rho|2\rangle\big] + |2\rangle\langle2|\,\langle0|\rho|0\rangle\ . $$ With the initial state being the maximally mixed state, it will oscillate between $$ \tfrac23|0\rangle\langle0| + \tfrac13|2\rangle\langle2| $$ and $$ \tfrac13|0\rangle\langle0| + \tfrac23|2\rangle\langle2|\ . $$

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  • $\begingroup$ Thanks so much for showing me the example. A quick question, as you mentioned in your previous response, most of quantum channels will converge to a fixed state, may I guess that such unconverged case is rare (maybe with probability 0 w.r.t. Haar measure)? $\endgroup$
    – XXDD
    Commented Feb 29, 2016 at 0:36
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    $\begingroup$ I'm not sure what you mean by the Haar measure for quantum channels, but for any reasonable (i.e., non-delta) measure, I would indeed expect that all channels except a set of measure 0 have a unique fixed point. (A second fixed point corresponds to a second eigenvalue of $\mathcal E$ on the unit circle, which should have probability 0 w.r.t. any smooth measure.) $\endgroup$
    – Norbert Schuch
    Commented Feb 29, 2016 at 11:41
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As long as the operation is unitary, it won't increase the "mixed-ness" of the state. Basically every operation in quantum computing, except noise and measurement, is unitary. Including the one you gave as an example (it's an X gate; see common quantum gates).

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