How can I find a quantum channel connecting two arbitrary quantum states?

Question

Given two arbitrary density matrices $\rho, \sigma\in \mathcal{H}$ (they have unit trace and are positive), how do I go about finding a possible quantum channel $\mathcal{E}$ such that $\mathcal{E}(\rho)=\sigma$? $\mathcal{E}$ is a general CPTP map, as such it is 1) trace preserving, 2) convex-linear, and 3) completely positive. It admits a Kraus operator representation or can be expressed as a unitary operation (gate) on an extended Hilbert space via Stinespring dilation. Can one say something about the remaining degree of freedom in the choice of $\mathcal{E}$?

I am simply wondering how one goes about constructing a valid quantum channel (representing the most general form of evolution of a quantum system) which connects two fixed states. This is a very general problem: One can think of a situation where a quantum system is initialized in some fixed state $\rho$ and one would like to manipulate it ( $\leftrightarrow$ subject it to a given quantum channel) such that it ends up in a target state $\sigma$. As such, this question must be addressed in a plethora of quantum experiments... (Maybe someone can also simply point me to some relevant literature?)

Answer 1

Given a state $\sigma$, the replacement channel is defined by the action $$ \mathcal{E}_{\sigma}(\rho) = \mathrm{Tr}(\rho) \sigma. $$ This channel trivially connects any state to $\sigma$. As Norbert pointed out this can be thought of operationally as first throwing away the system you have and then preparing a new system in the state $\sigma$. Indeed, we can view this channel as the composition of the trace channel $\mathrm{Tr}: \mathcal{H}_1 \rightarrow \mathbb{C}$ and a preparation channel $\mathcal{E}_{\mathrm{prep}} : \mathbb{C} \rightarrow \mathcal{H}_2$ where the action of the second channel is defined as $\mathcal{E}_{\mathrm{prep}}(\alpha) \rightarrow \alpha \sigma$.

To show the replacement channel is indeed a channel, by the spectral theorem we can define an orthonormal basis $\{|\psi_i\rangle \}_i$ of $\mathcal{H}_2$ such that $\sigma = \sum_i \lambda_i |\psi_i \rangle \langle \psi_i |$. Then take an orthonormal basis $\{|i\rangle\}_i$ of $\mathcal{H}_1$ and define the Kraus operators $$ K_{i,j} = \sqrt{\lambda_i} |\psi_i \rangle \langle j |. $$ Then we have $$ \begin{aligned} \mathcal{E}(\rho) &= \sum_{i,j} K_{i,j} \rho K_{i,j}^* \\ &= \sum_{i,j} \lambda_i |\psi_i \rangle \langle j | \rho |j \rangle \langle\psi_i | \\ &= \sum_i \lambda_i |\psi_i \rangle \langle \psi_i | \sum_j \langle j | \rho | j \rangle \\ &= \mathrm{Tr}[\rho] \sigma \end{aligned} $$ and also $$ \begin{aligned} \sum_{i,j} K_{i,j}^* K_{i,j} &= \sum_{i,j} \lambda_i |j \rangle \langle \psi_i | \psi_i \rangle \langle j | \\ &= \sum_{i,j} \lambda_i |j \rangle \langle j | \\ &= \sum_j |j \rangle \langle j | \\ &= I \end{aligned} $$

Answer 2

Given two states $\rho, \sigma$, consider their spectral decomposition, $$\rho = \sum\limits_{j=1}^{d} p_{j} | \psi_{j} \rangle \langle \psi_{j} | , \sigma = \sum\limits_{j=1}^{d} q_{j} | \phi_{j} \rangle \langle \phi_{j} |. $$ I'm assuming, for simplicity, that $\rho, \sigma$ are have non-degenerate eigenvalues -- this is not a strict requirement for the argument that follows but does simplify the analysis. Then, the problem of $\rho \mapsto \sigma$ breaks down into two steps: (i) transforming their eigenvectors and (ii) transforming their eigenvalues.

To transform their eigenvectors, consider the following unitary, $U = \sum\limits_{j=1}^{d} | \phi_{j} \rangle \langle \psi_{j} | $. It is easy to check that the action of the unitary channel is to transform the eigenvectors, $$\mathcal{U}( | \psi_{j} \rangle \langle \psi_{j} | ) := U ( | \psi_{j} \rangle \langle \psi_{j} | ) U^{\dagger} = | \phi_{j} \rangle \langle \phi_{j} | ~~\forall j.$$ Therefore, $\mathcal{U}(\rho) = \sum\limits_{j=1}^{d} p_{j} | \phi_{j} \rangle \langle \phi_{j} | $, that is, the eigenvectors have been transformed. More generally, any time one wants to transform an orthonormal set of states $\{ |\psi_{j} \rangle \} \mapsto \{ |\phi_{j} \rangle \}$, we construct a unitary of the form above.

To transform the eigenvalues, first note that unitary operators cannot change the spectrum of a state, therefore, we need a non-unitary channel. Also, with the action of $\mathcal{U}$ above, both $\mathcal{U}(\rho)$ and $\sigma$ are in the same eigenbasis, so transforming the eigenvalues has a "classical" flavor to it. I can't think of an answer for the most general case (off the top of my head), but if $\{ p_{j} \}$ is ``less disordered'' than $\{ q_{j} \}$ (in the sense of vector majorization), then, one can show that $$ \operatorname{spec}(\rho) \succ \operatorname{spec}(\sigma) \Longleftrightarrow \exists \mathcal{E}(\rho)=\sigma, $$ where, $\vec{v} \succ \vec{w}$ is vector majorization, $\mathcal{E}$ is a unital CPTP map, and $\text{spec}(\rho)$ the spectrum of $\rho$. A proof of this can be found in Nielsen's (other!) book (warning: the book is in a .ps format).

Therefore, given two states, $\rho, \sigma$, if $\operatorname{spec}(\rho) \succ \operatorname{spec}(\sigma)$ then this transformation can be achieved by using a unitary channel $\mathcal{U}$ to transform the eigenvectors and a non-unitary channel $\mathcal{E}$ to transform the eigenvalues; composing these two, we have, $\mathcal{E} \circ \mathcal{U}$ is the channel that does the transformation.

Edit: For $\rho,\sigma$ pure, the above construction tells us that we only need a unitary transformation to connect them, as expected.