# Asymptotic purity from the spectrum of the Choi matrix?

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

• 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. Dec 22, 2023 at 0:38
• 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 Dec 22, 2023 at 6:15
• Why do you expect it to have some connection with the spectrum of the Choi matrix? Also, an unital map preserves the identity operator. Dec 22, 2023 at 6:18
• @narip (ChatGPT told me there is connection) background on question here Dec 22, 2023 at 18:01
• 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... Dec 22, 2023 at 18:56

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 $$T$$.

• 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? Jan 9 at 21:20
• 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 $1/d$. Jan 9 at 23:50
• 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. Jan 9 at 23:53