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I have a completely positive map $T$ and a sequence of $d\times d$ states $S_1,S_2,\ldots$ obtained by applying $T$ repeatedly to the identity matrix.

I'm interested in quantifying what happens to purity of $S_i$ as $i$ grows. In particular, is it possible to quantify how fast the purity grows with $i$ and what it converges to, from the spectrum of the Choi matrix?

Toy example, taking channel $T$ with the following Kraus operators $$ \left( \begin{array}{cc} 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 \\ \end{array} \right),\left( \begin{array}{cc} 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} \\ \end{array} \right)$$

I'm seeing Choi matrix have eigenvalues $1,\frac{1}{2}$

And the purity over time below:

Notebook

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    $\begingroup$ Interesting question. Note that you shouldn't expect the purity to increase (or decrease) in general; any unital channel will leave the purity exactly where it is in this case. $\endgroup$
    – forky40
    Dec 22, 2023 at 0:38
  • $\begingroup$ Isn't purity the reciprocal of Schatten-2 norm? This one says unital positive maps are contractive wrt to Schatten-2 norm, so purity would grow -- arxiv.org/pdf/math-ph/0601063.pdf $\endgroup$ Dec 22, 2023 at 6:15
  • $\begingroup$ Why do you expect it to have some connection with the spectrum of the Choi matrix? Also, an unital map preserves the identity operator. $\endgroup$
    – narip
    Dec 22, 2023 at 6:18
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    $\begingroup$ @narip (ChatGPT told me there is connection) background on question here $\endgroup$ Dec 22, 2023 at 18:01
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    $\begingroup$ Another possibly useful toy example is the replacement channel $T(X) = \text{Tr}(X) \sigma$ for any state $\sigma$. Then the spectrum of the Choi matrix is proportional to the spectrum of $\sigma$. This a channel for which $T \circ T = T$ so $S_i = S_{i+1}$, but the spectrum can take on a lot of different behaviors... $\endgroup$
    – forky40
    Dec 22, 2023 at 18:56

1 Answer 1

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TL;DR: For many channels, as the number $i$ of iterated applications of the channel grows, the state $T^i(\rho_0)$ converges to a fixed point of $T$. Then, the purity of $T^i(\rho_0)$ converges to the purity of a fixed point of $T$, not to an eigenvalue of its Choi matrix.

Example where purity is outside Choi spectrum

Let $\rho$ denote a quantum state with eigenvalues $p_1,\dots,p_d$ and define $\mathcal{C}_\rho(X):=\rho\,\mathrm{tr}(X)$, the state preparation channel for $\rho$. The purity of $\rho$ is $\gamma=\sum_ip_i^2$, but the eigenvalues of the Choi matrix $J(\mathcal{C}_\rho)$ coincide with those of $\rho$. But generally $\gamma\notin\{p_1,\dots,p_d\}$.

Contractive channels

By Banach fixed-point theorem quantum channels that are contractive maps, such as non-trivial depolarizing and amplitude damping channels, have exactly one fixed point $\rho_{\text{fix}}$. In this case, the purity of $T^i(\rho_0)$ tends to $\mathrm{tr}(\rho_{\text{fix}}^2)$ independently of the initial state.

Non-contractive channels

The situation is more complicated for non-contractive channels. If $T$ is non-contractive, then $T^i(\rho_0)$ may fail to converge, as happens in the case of the bit-flip channel acting on $|0\rangle$.

Nevertheless, by Schauder fixed-point theorem, every channel, contractive or otherwise, has at least one fixed-point. Moreover, when $T^i(\rho_0)$ does converge, then it necessarily converges to a fixed point$^1$. However, Banach fixed point theorem no longer applies and the fixed point may not be unique. For example, every point on the line connecting the poles of the Bloch sphere is a fixed point of the phase damping channel. In general, which fixed-point $T^i(\rho_0)$ converges to depends on $\rho_0$.

Non-contractive qubit channels

The situation is a little simpler if $T$ acts on a qubit and when $\rho_0$ is the maximally mixed state. In this case, if $T$ is non-contractive then the maximally mixed state $\frac{I}{2}$ is necessarily$^2$ among the fixed points of $T$. Therefore, the purity of $T^i\left(\frac{I}{2}\right)$ is $\frac12$.


$^1$ This can be deduced from the continuity argument similar to that in the proof of Banach fixed-point theorem: $$ \rho_*=\lim_{i\to\infty}T^i(\rho_0)=T\left(\lim_{i\to\infty}T^{i-1}(\rho_0)\right)=T(\rho_*). $$
$^2$ We can make a simple geometric argument. Suppose $T$ is non-contractive. For contradiction, it is sufficient to show that $T$ shrinks the trace distance between any pair of antipodal points. If $T$ displaces the center of the Bloch sphere $B(O,1)$ from $O$ to $A\ne O$, then more than half of the surface of the ball $B(A,1)$ lies outside of $B(O,1)$. However, the image $T(B(O,1))$ must lie within $B(A,1)$. Therefore, distance between any two antipodal points is necessarily reduced by the action of $T$.

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  • $\begingroup$ Thanks for the in-depth reply. I'm wondering if there is a way to link asymptotic purity to properties of channel $T$. For instance, if $T$ is a diagonal matrix, then asymptotic purity is 1. This means that spectrum of T does not determine asymptotic purity either, so what determines it? $\endgroup$ Jan 9 at 21:20
  • $\begingroup$ Asymptotic purity is related to fixed-points of the channel, but has a connection to the eigendecomposition of a certain matrix, too. However, it is not the Choi matrix $J(T)$ of $T$, but simply the matrix $K(T)$ representing linear (super)operator $T$ in some basis. Indeed, $x$ is a fixed-point of linear operator $A$ iff $x$ is the eigenvector of $A$ corresponding to eigenvalue $1$. Every channel has eigenvalue $1$, but they differ in the corresponding eigenstate. If the maximally mixed state $I/d$ is an eigenstate of $T$ corresponding to eigenvalue $1$, then asymptotic purity is $1/d$. $\endgroup$ Jan 9 at 23:50
  • $\begingroup$ We can detect this by inspecting $K(T)$ in the Pauli basis (ordered with identity first): $I/d$ is a fixed-point if and only if the first column is $[1,0,\dots,0]$. Also, if the eigenspace $V_1$ corresponding to eigenvalue $1$ is one-dimensional, then the quantum state $\rho_*$ that spans $V_1$ is the sole fixed-point of $T$, so the asymptotic purity is the purity of $\rho_*$ (regardless of the initial state). In fact, since $1$ is the largest eigenvalue of $T$, this is a special case of the power method. $\endgroup$ Jan 9 at 23:53

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