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I am reading the paper Duality of Quantum and Classical Error Correction Codes: Design Principles & Examples and at the beginning of it, the authors describe how the general Pauli channel is related to more realistic channels such as amplitude/phase damping channels. The Pauli channel is defined as:

$\mathcal{N}_P(\rho)=(1-p_x-p_y-p_z)I\rho I + p_xX\rho X+p_y Y\rho Y + p_z Z\rho Z$

In order to describe how such probabilities of $X,Y,Z$ events happen depending on the relaxation ($T_1$) and dephasing ($T_2$) times of a qubit, they give the next set of equations:

$p_x=p_y=\frac{1}{4}(1-e^{-\frac{t}{T_1}})$

$p_z=\frac{1}{4}(1+e^{-\frac{t}{T_1}}+ 2e^{-\frac{t}{T_2}})$

However, my doubt comes from the fact that intuitively, when the time goes to infinity, the probability of no error in the qubits should go to $0$, that is, complete decoherence of the system. However, making an analysis of this, this is not what happens:

$\lim_{t\rightarrow\infty} 1 - p_x-p_y-p_z = \lim_{t\rightarrow\infty}\frac{1}{4}(1+e^{-\frac{t}{T_1}}+2e^{-\frac{t}{T_2}})=\frac{1}{4}$.

So my question here is to clarify why such probability should go to $1/4$ instead of going to $0$, or in the case that someone knows more about this kind of relationships, to give some useful reference about it.

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Josu, You might be mis-understanding something. Your equation for the Pauli channel says that when $t\rightarrow \infty$, all operators ($X,Y,Z,I$) have an equal probability (1/4) of transforming the density matrix $\rho$. You seem to be suggesting that $t\rightarrow \infty$, the probability of the $I$ operator acting on $\rho$ whould go to 0. Keep in mind that $I$ is an operator just like $X,Y,Z$ and it makes sense that one might model it to be just as likely as $X,Y$ and $Z$.

Another way to think of it, is that as time goes to infinity, all sorts of things may have happened, such as the $X$ operator having acted twice, which would mean that the overall transformation on $\rho$ would be described by $I$.

There is no reason why there should be a 100% chance that $\rho$ would be transformed either by $X$, $Y$, or $Z$. You are saying that there should be a 100% chance that $\rho$ changes in some way, whereas I don't see any reason why there should not be an equal chance of it staying the same as there is chance for it to be transformed by $X$ or by $Y$ or by $Z$.

Finally, all of this is just some extremely simplified model of how decoherence really works. Therefore it is up to the authors how they want to model the decoherence, and up to the referees and the readers of the paper, whether they like the paper or not. In this case they chose a model where all four SU(2) operators have an equal probability (1/4) of acting on $\rho$ as $t\rightarrow \infty$. I personally would have modeled the qubit's bath with a Hamiltonian, and simulated the $\rho_{\textrm{qubit}}(t)$ using a quantum master equation or a numerically exact method rather than this extremely simplified model, but I do not blame them. The freedom of choice goes to the author, and we are given the freedom of whether or not we like their paper :)

There is a further point: I believe that in the real story (if everything follows the Schroedinger equation exactly, including system and bath), the limit is "un-defined". This is because decoherence will happen, then when you wait long enough, there will be a re-vival (Poincaré recurrence), then a decoherence again, then a re-vival, then a decoherence again, then a re-vival, and so on. Therefore what state $\rho$ will be in, will depend on what time you make the measurement. There is NO defined limit for sin(x) as x -> $\infty$. Likewise, there is no defined limit for $\rho$.

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  • $\begingroup$ @Josu, there is a further point: I believe that in the REAL story (if everything follows the Schroedinger equation exactly, including system and bath), the limit is "un-defined". This is because decoherence will happen, then when you wait long enough, there will be a re-vival (Poincaré recurrence), then a decoherence again, then a re-vival, then a decoherence again, then a re-vival, and so on. Therefore what state $\rho$ will be in, will depend on what time you make the measurement. There is NO defined limit for sin(x) as x -> $\infty$. Likewise, there is no defined limit for $\rho$. $\endgroup$ – CARNEGIE Mar 20 at 20:22
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    $\begingroup$ It's probably worth adding that the model where "four SU(2) operators have an equal probability (1/4) of acting on $ρ$ as $t→∞$" is an extremely common one when it comes to modelling open systems - modelling the bath with a Hamiltonian is considerably harder and markovianity is a pretty good assumption for real systems, over realistic time-scales, although non-markovian effects do need to be taken into account on top of this. I'd also argue that this more like the definition of decoherence and that what you're saying is that the decoherence channel isn't a perfect description of real noise $\endgroup$ – Mithrandir24601 Mar 20 at 23:24
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    $\begingroup$ @Mithrandir24601: I didn't realize it was so common, and I guess it is less and less common when you go beyond 1 qubit. I agree that modeling the bath with a Hamiltonian adds a lot more complexity. I also note that in the system-bath model, either a Markovian or non-Markovian method can be used to simulate the dynamics (e.g. Markovian master equation). In any case, Josu probably agrees now that it's not unreasonable to have "no overall change" as one of he possible outcomes after a long period of time (this doesn't mean it never changed at all, e.g. two X's can happen and give a net $I$). $\endgroup$ – CARNEGIE Mar 20 at 23:32
  • $\begingroup$ Guy's is this really a "resource" request? I couldn't change the tag. $\endgroup$ – CARNEGIE Mar 20 at 23:56
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    $\begingroup$ But changes in some way can just be absorbed into X, Y and Z. So X acting 3 times and Y acting once will just turn into Z acting with some coefficent. So that third paragraph is already taken care of. There are only $2^2$ needed anyway for Choi's theorem as already used. So already have the most general form. The question about limits or lack thereof about the p's remains especially asking for an intuitive explanation thereof. $\endgroup$ – AHusain Mar 21 at 0:22

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