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} \mathrm{D}(\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 \|\rho\|_1=\Tr\sqrt{\rho^\dagger \rho}. \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. Important Observations
John Watrous's The Theory of Quantum Information, Theorem 4.24 in Section 4.2.2 gives a more general existence result. On a finite-dimensional Hilbert space $\mathcal{X}$, every positive, trace-preserving linear map $\Phi:\mathrm{L}(\mathcal{X})\to\mathrm{L}(\mathcal{X})$ leaves at least one density operator unchanged:
\begin{equation} \exists\,\rho^\star \in \mathrm{D}(\mathcal{X}): \qquad \Phi(\rho^\star)=\rho^\star. \end{equation}The result here addresses a different problem. Rather than starting with a channel and asking whether a fixed state exists, we start with an arbitrary target qubit state $\rho^\star$ and explicitly construct a channel $\Phi$ that fixes the target. Moreover, $\rho^\star$ is the unique fixed density operator such that
\begin{equation} \lim_{n\to\infty}\Phi^n(\rho)=\rho^\star \qquad \text{for every } \rho\in\mathrm{D}(\mathbb{C}^2). \end{equation}Note that we could have used the reset channel $\mathcal{R}_{\rho^\star}(\rho)=\Tr(\rho)\rho^\star$ as it sends every density operator to $\rho^\star$ in one step. In our case, we wanted to construct a channel using physically meaningful maps, such as the depolarizing and amplitude damping channels plus a unitary rotation, and show the path that $\rho_0$ follows as it converges to $\rho^\star$.