1. Introduction
I recently learned about the Banach fixed-point theorem in my applied functional analysis course. Later that day at home, me and my roommates Daniel and Marc were discussing about possible applications of the Banach fixed-point theorem for quantum information. A trivial example we found was the depolarizing channel
\begin{equation} \mathcal{D}_p(\rho) = (1-p)\rho + p\,\frac{\mathds{1}}{d}\Tr(\rho) \end{equation}Throughout this post, we take $0 \lt p \leq 1$ and $0 \lt \gamma \leq 1$, where $\gamma$ is the amplitude damping parameter introduced below. These assumptions exclude the identity-channel cases $p=0$ and $\gamma=0$.
It is easy to see that
\begin{equation} \lim_{n\to \infty} \mathcal{D}^n_p(\rho_0) = \frac{\mathds{1}}{d}, \qquad \mathcal{D}_p\left(\frac{\mathds{1}}{d}\right) = \frac{\mathds{1}}{d} \end{equation}for any arbitrary initial state $\rho_0$. So we asked ourselves: For a single qubit, given any $\rho_0$ and target state $\rho^\star$, can we find a channel $\Phi$ such that
\begin{equation} \lim_{n\to \infty} \Phi^n(\rho_0) = \rho^\star, \qquad \Phi\left(\rho^\star\right) = \rho^\star \end{equation}The answer is yes!
2. Construction of the Channel
The starting point is to consider the convex combination of a depolarizing channel $\mathcal{D}_p$ and the amplitude damping channel
\begin{equation} \mathcal{A}_{\gamma}(\rho) = K_0 \rho K_0^\dagger + K_1 \rho K_1^\dagger, \qquad K_0 = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{pmatrix}, \quad K_1 = \begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix} \end{equation}Taking into account that
\begin{equation} \lim_{n\to \infty} \mathcal{A}^n_\gamma(\rho_0) = \ketbra{0}, \qquad \mathcal{A}_\gamma(\ketbra{0}) = \ketbra{0} \end{equation}for any $\rho_0$, we figured that a convex combination
\begin{equation} \lambda \mathcal{D}_p(\rho) + (1-\lambda)\mathcal{A}_\gamma(\rho), \qquad 0 \leq \lambda \leq 1 \end{equation}composed infinite times should converge to a density matrix lying on the positive side of $Z$ axis of the Bloch sphere, depending on the value of $\lambda$. We proved that this is true. Consider the Pauli transfer matrix (PTM) picture of the depolarizing and amplitude damping channel
\begin{equation} [\mathcal D_p] = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1-p & 0 & 0 \\ 0 & 0 & 1-p & 0 \\ 0 & 0 & 0 & 1-p \end{pmatrix}, \qquad [\mathcal A_\gamma] = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \sqrt{1-\gamma} & 0 & 0 \\ 0 & 0 & \sqrt{1-\gamma} & 0 \\ \gamma & 0 & 0 & 1-\gamma \end{pmatrix} \end{equation}Then,
\begin{equation} \lim_{n\to\infty} \left( \lambda[\mathcal D_p]+(1-\lambda)[\mathcal A_\gamma] \right)^n = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \frac{(1-\lambda)\gamma}{\lambda p+(1-\lambda)\gamma} & 0 & 0 & 0 \end{pmatrix} \end{equation}Therefore, for a density matrix $\rho$ written in the Pauli basis
\begin{equation} \rho = \frac{1}{2}\left( \mathds{1} + r_xX+r_yY+r_zZ \right), \qquad \operatorname{vec}(\rho) = \frac{1}{2} \begin{pmatrix} 1 \\ r_x \\ r_y \\ r_z \end{pmatrix} \end{equation}Here, $\operatorname{vec}(\rho)$ denotes the coefficient vector in the ordered, unnormalized Pauli basis $(\mathds{1},X,Y,Z)$, rather than column-stacking the density matrix.
Then,
\begin{equation} \lim_{n\to \infty} \operatorname{vec}\left[\left(\lambda \mathcal{D}_p + (1-\lambda)\mathcal{A}_\gamma\right)^n(\rho)\right] = \lim_{n\to\infty} \left( \lambda[\mathcal D_p]+(1-\lambda)[\mathcal A_\gamma] \right)^n \operatorname{vec}(\rho) = \frac{1}{2} \begin{pmatrix} 1 \\ 0 \\ 0 \\ \frac{(1-\lambda)\gamma}{\lambda p+(1-\lambda)\gamma} \end{pmatrix} \end{equation}Therefore, the final state $\rho'$ is
\begin{equation} \rho' = \frac{1}{2}\left(\mathds{1} + \frac{(1-\lambda)\gamma}{\lambda p+(1-\lambda)\gamma}Z\right) \end{equation}Let $\mathcal{U}(\rho) = U\rho U^\dagger$ be a unitary channel. Let $\Vec{n}$ be the unit vector pointing in the axis that intersects the target state $\rho^\star$. Then, we can find a unitary $U$ such that $UZU^\dagger = n_xX+n_yY+n_zZ$. For $\rho^\star = \mathds{1}/2$, the direction is arbitrary and we can take $\Vec{n}=\Vec{z}$ and $U=\mathds{1}$. Consider the following channel:
\begin{equation} [\Phi] := [\mathcal{U}]\left(\lambda[\mathcal D_p]+(1-\lambda)[\mathcal A_\gamma] \right)[\mathcal{U}^\dagger] \end{equation}which can also be written as
\begin{equation} \Phi(\rho) = U\left(\lambda\mathcal{D}_p(U^\dagger \rho U) + (1-\lambda)\mathcal{A}_\gamma(U^\dagger \rho U)\right)U^\dagger \end{equation}Since $[\mathcal{U^\dagger}] = [\mathcal{U}]^\dagger$, and the composition of superoperators in the PTM picture is by left multiplication, it is easy to see that
\begin{equation} [\Phi]^n = [\mathcal{U}]\left(\lambda[\mathcal D_p]+(1-\lambda)[\mathcal A_\gamma] \right)^n[\mathcal{U}^\dagger] \end{equation}We claim that, for non-trivial $p,\, \gamma$, any arbitrary state $\rho_0$, and target state $\rho^\star$, we can always find a $U$ and $\lambda$ such that
\begin{equation} \lim_{n \to \infty} [\Phi]^n\operatorname{vec}(\rho_0) = \operatorname{vec}(\rho^\star), \qquad [\Phi]\operatorname{vec}(\rho^\star) = \operatorname{vec}(\rho^\star) \end{equation}or equivalently,
\begin{equation} \lim_{n \to \infty} \Phi^n(\rho_0) = \rho^\star, \qquad \Phi(\rho^\star) = \rho^\star \end{equation}We have mentioned already how to find $U$. Let $\textbf{r}^\star = (r_x^\star,r_y^\star,r_z^\star)$ be the target vector of $\rho^\star$. Then, the value of $\lambda$ is independent of $\rho_0$, and is given by
\begin{equation} \lambda = \frac{\gamma\left(1 - \|\textbf{r}^\star\|_2\right)} {\gamma\left(1 - \|\textbf{r}^\star\|_2\right) + p\|\textbf{r}^\star\|_2}. \end{equation}The denominator is positive under our parameter assumptions, and $0\leq\lambda\leq1$. A maximally mixed target gives $\lambda=1$, while a pure target gives $\lambda=0$. Both endpoints are included in the construction.
Here is a visual animation of the trajectory from $\rho_0$ to $\rho^\star$ as we compose the channel $\Phi$ defined above. You can change the coordinates of the two density operators and see how the convergence path changes!
3. Connection to the Banach Fixed-Point Theorem
The PTM calculation above gives the limit explicitly. To connect it to the Banach fixed-point theorem, we need a complete metric space and a contraction on that space. Consider the set of single-qubit density operators
\begin{equation} \mathcal{S}(\mathbb{C}^2) = \left\{\rho\in M_2(\mathbb{C}) : \rho\geq0,\ \Tr(\rho)=1\right\} \end{equation}with trace distance
\begin{equation} D(\rho,\sigma)=\frac12\|\rho-\sigma\|_1, \qquad \|A\|_1=\Tr\sqrt{A^\dagger A}. \end{equation}This set is nonempty and closed in the finite-dimensional normed space $M_2(\mathbb{C})$, so it is complete in this metric. Also, $\Phi$ maps this set into itself: convex combinations and compositions of quantum channels are quantum channels.
For qubits, the eigenvalues of $\rho-\sigma$ are $\pm\|\mathbf{r}_\rho-\mathbf{r}_\sigma\|_2/2$, so
\begin{equation} D(\rho,\sigma)=\frac12\|\mathbf{r}_\rho-\mathbf{r}_\sigma\|_2. \end{equation}Let
\begin{equation} \Psi=\lambda\mathcal{D}_p+(1-\lambda)\mathcal{A}_\gamma. \end{equation}The PTMs above show that its action on a Bloch vector is
\begin{equation} \mathbf{r}\longmapsto M\mathbf{r}+\mathbf{t},\qquad M=\begin{pmatrix}a&0&0\\0&a&0\\0&0&b\end{pmatrix},\qquad \mathbf{t}=\begin{pmatrix}0\\0\\(1-\lambda)\gamma\end{pmatrix}, \end{equation}where
\begin{equation} \begin{aligned} a&=\lambda(1-p)+(1-\lambda)\sqrt{1-\gamma},\\ b&=\lambda(1-p)+(1-\lambda)(1-\gamma). \end{aligned} \end{equation}When we subtract the output Bloch vectors of two states, the translation $\mathbf{t}$ cancels. Therefore,
\begin{equation} \begin{aligned} D(\Psi(\rho),\Psi(\sigma)) &=\frac12\|M(\mathbf{r}_\rho-\mathbf{r}_\sigma)\|_2\\ &\leq q\,D(\rho,\sigma), \end{aligned} \qquad q=\max\{|a|,|b|\}=a\lt1. \end{equation}Here $0\leq b\leq a\lt1$ follows from $0\lt p\leq1$, $0\lt\gamma\leq1$, and $0\leq\lambda\leq1$. Thus the same constant $q\lt1$ works for every pair of states, including when $\lambda=0$ or $\lambda=1$. The contraction is on differences of normalized states, not a claim that the full PTM has norm less than one.
Unitary conjugation preserves trace distance, so $\Phi=\mathcal{U}\circ\Psi\circ\mathcal{U}^\dagger$ has the same contraction constant:
\begin{equation} D(\Phi(\rho),\Phi(\sigma))\leq q\,D(\rho,\sigma),\qquad 0\leq q\lt1. \end{equation}We can now apply the Banach fixed-point theorem: $\Phi$ has exactly one fixed state, and its iterates converge to that state from every initial density operator. Our choice of $U$ and $\lambda$ identifies that state as $\rho^\star$. We also get a quantitative convergence bound:
\begin{equation} D(\Phi^n(\rho_0),\rho^\star) \leq q^nD(\rho_0,\rho^\star),\qquad n\geq1. \end{equation}4. Relation to Known Fixed-Point Results
John Watrous's The Theory of Quantum Information, Theorem 4.24 in Section 4.2.2 (pp. 228–229), gives a more general existence result. On a finite-dimensional Hilbert space $\mathcal{X}$, every positive, trace-preserving linear map $T:\mathrm{L}(\mathcal{X})\to\mathrm{L}(\mathcal{X})$ leaves at least one density operator unchanged:
\begin{equation} \exists\,\omega\geq0:\qquad \Tr(\omega)=1,\qquad T(\omega)=\omega. \end{equation}In particular, this applies to every quantum channel with the same input and output space. Despite the section's focus on unital channels, this theorem does not require unitality or strict contractivity. Watrous notes that existence follows from Brouwer's fixed-point theorem, then gives a direct proof using averages of powers of the map and nested compact sets.
This guarantees existence, but not uniqueness or convergence of the iterates. For example, every density operator is fixed by the identity channel. Also, the channel $\Theta(\rho)=X\rho X$ fixes $\mathds{1}/2$, but starting from $\ketbra{0}$ gives
\begin{equation} \Theta^{2n}(\ketbra{0})=\ketbra{0},\qquad \Theta^{2n+1}(\ketbra{0})=\ketbra{1},\qquad n\geq0, \end{equation}so the sequence does not converge.
Our example adds more conclusions for a more restricted family of channels. We choose the target state first, then choose $U$ and $\lambda$ so that the target is the unique fixed state and every initial state converges to it at the rate bounded above. This does not strengthen Watrous's theorem for arbitrary channels; it uses the extra strict-contraction property of our construction.
The connection between Banach's theorem and quantum channels is already known. Raginsky studies strictly contractive quantum channels, their unique fixed states, convergence under iteration, and their Bloch-ball geometry. His paper also shows that convex combinations of strictly contractive channels remain strictly contractive.
Prescribing a target state is also possible more directly: the reset channel $\mathcal{R}_{\rho^\star}(A)=\Tr(A)\rho^\star$ sends every density operator to $\rho^\star$ in one step. Constant-output channels appear in Raginsky's construction as well. The point of this post is the explicit construction using depolarizing and amplitude damping channels plus a unitary rotation, and its connection to a theorem from functional analysis, rather than a claim of a new general fixed-point theorem.