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?)

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    $\begingroup$ Note $\mathcal E(x) = \mathrm{tr} (x) \sigma$ will always work. $\endgroup$
    – Rammus
    Commented Nov 3, 2020 at 8:13
  • $\begingroup$ @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 $\mathcal{E}(\rho)=\sigma$ $\endgroup$ Commented Nov 3, 2020 at 12:48
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    $\begingroup$ 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. $\endgroup$ Commented Nov 3, 2020 at 13:05
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    $\begingroup$ @LenusStueli the replacement channel which Rammus suggests is indeed linear, CPTP, has a Kraus decomposition, a Stinespring representation etc. $\endgroup$ Commented Nov 3, 2020 at 19:27
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    $\begingroup$ " 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 $\sigma$. $\endgroup$ Commented Nov 3, 2020 at 20:36

2 Answers 2


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 $$ \begin{aligned} \sum_{i,j} K_{i,j}^* K_{i,j} &= \sum_{i,j} \lambda_i |j \rangle \langle \psi_i | \psi_i \rangle \langle j | \\ &= \sum_{i,j} \lambda_i |j \rangle \langle j | \\ &= \sum_j |j \rangle \langle j | \\ &= I \end{aligned} $$


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.

  • $\begingroup$ As far as I know, in Nielsen's theorem, the Majorization condition is for the spectrum of the reduced density matrices, and not the complete states. Also, the link you have provided is not working anymore. It would be great if you could clarify, and/or provide the reference again. $\endgroup$
    – Cain
    Commented Jun 23, 2023 at 12:41

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