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

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?)

• Note $\mathcal E(x) = \mathrm{tr} (x) \sigma$ will always work. Nov 3, 2020 at 8:13
• @Rammus I don't quite understand your comment. For a valid density matrix we have ${\rm tr}(\rho)=1$ so you may as well write $\mathcal{E}(\rho)=\sigma$ Nov 3, 2020 at 12:48
• That's a bit of a subtle issue. Note that with your definition, $\mathcal{E}$ is not a linear map. Albeit, it doesn't need to be since it's anyway only defined on unit-trace matrices which form an affine not a linear subspace. Now you can easily check that the map is affine: $\mathcal{E}(\sum_i c_i \rho_i) = \sigma = (\sum_i c_i)\sigma = \sum_i c_i \mathcal{E}(\rho_i)$ for any affine combination $c_i\in\mathbb{R}$, $\sum_i c_i =1$. However, it is often simpler to work with linear maps defined on all matrices instead, and then Rammus' definition is the right one. Nov 3, 2020 at 13:05
• @LenusStueli the replacement channel which Rammus suggests is indeed linear, CPTP, has a Kraus decomposition, a Stinespring representation etc. Nov 3, 2020 at 19:27
• " One can think of a situation where a quantum system is initialized in some fixed state 𝜌 and one would like to manipulate it ( ↔ subject it to a given quantum channel) such that it ends up in a target state 𝜎. " --- Throw the system away and prepare a new one in $\sigma$. Nov 3, 2020 at 20:36

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.

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}