Given finitely many pairs of pure states $|x_1\rangle,|y_1\rangle,\ldots,|x_k\rangle,|y_k\rangle\in\mathcal{H}_n$, we can decide if there exists a unitary operator $U$ such that $U|x_i\rangle=|y_i\rangle$ for all $i$ by verifying the following:

  • Linearity: whenever $\alpha_1|x_1\rangle+\cdots+\alpha_k|x_k\rangle=0$, it holds $\alpha_1|y_1\rangle+\cdots+\alpha_k|y_k\rangle=0$.
  • Inner Product Preserving: For all $i,j$, it holds $\langle x_i|x_j\rangle = \langle y_i|y_j\rangle$.

It is not hard to show that there exists such $U$ if and only if both linearity and inner product preserving properties hold.

Now given finitely many pairs of density operators $\rho_1,\sigma_1,\ldots,\rho_k,\sigma_k\in L(\mathcal{H}_n)$, is there a similar process that could decide whether there exists a quantum channel $\Phi$ such that $\Phi(\rho_i)=\sigma_i$ for all $i$? Linearity is still a requirement and is easy to check, but how to check complete positivity? Choi's theorem seems hard to use when we don't have full information about $\Phi$.


2 Answers 2


The problem can clearly be solved with an SDP. Let $F :=\mathfrak{C}(\Phi)$ be $\Phi$'s Choi matrix. Then the channel exists iff \begin{gather*} \operatorname{tr}_\text{in}[ (\rho_i^T \otimes I)F] = \sigma_i\quad \forall i \\ \operatorname{tr}_\text{out}[F] = I \\ F \ge 0, \end{gather*} which are linear constraints and PSD conditions, as we need.

Now, perhaps you wanted some analytical condition? I don't think such a thing exists. An easy necessary (but not sufficient) condition is that the channel is contractive, i.e., $$\|\sigma_i-\sigma_j\|_1 \le \|\rho_i-\rho_j\|_1\quad \forall i,j$$


This is not a full solution, but rather a way to see why the problem is harder than it might seem on a first look.

As you yourself note, the map has to be linear and that's a first necessary requirement that is relatively easy to check. However, it's far from sufficient. Consider for example what happens in the single qubit case. Suppose I tell you that a channel $\Phi$ is such that $$\Phi(I)=I, \quad \Phi(X)=X, \quad \Phi(Z)=Z. $$ Or equivalently, I tell you that it preserves the eigenvectors of $Z$ and $X$. Observe that, as far as linearity is concerned, this says absolutely nothing about how the channel is supposed to act on $Y$, as $Y$ is outside of the span of $\{I,X,Z\}$ (where I'm talking about span in the operator space). The requirements of positivity and trace-preservingness impose some further easy constraints. But most importantly, if you want complete positivity, than the above three conditions imply that you must also have $\Phi(Y)=Y$.

One way to see it is looking at the Choi operator, defined as $J(\Phi)=(\Phi\otimes I)\mathbb{P}_+$ where $$\mathbb{P}_+\equiv \sum_{ij}|i,i\rangle\!\langle j,j| = \begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix} = \frac14(I\otimes I+X\otimes X-Y\otimes Y+Z\otimes Z).$$ The last identity is in particular useful to understand when the Choi is positive semidefinite (which is an iff condition for complete positivity of $\Phi$ itself). The Choi of our $\Phi$ equals $$J(\Phi) = \frac14[I+X\otimes X-\Phi(Y)\otimes I+Z\otimes Z].$$ It is then straightforward to see by direct analysis that this object is only positive-semidefinite if $\Phi(Y)=Y$, for example writing $\Phi(Y)=\alpha I+\beta X + \gamma Y+\delta Z$ and looking at the constraints on the coefficients $\alpha,\beta,\gamma,\delta$ that don't give negative eigenvalues. Note in particular how $\Phi(Y)=-Y$ corresponds to the transpose channel $\Phi(\rho)=\rho^T$, which is a standard example of a positive non-CP map.

Another approach to see why defining the action on a subset of the operators, for proper channels, imposes constraints on how the channel acts on other linearly independent operators, we can use the dilation representation of the channel. Namely, any channel can be represented via its Stinespring dilation, which amounts to representing it as an isometry $V$. Let's see what this isometry looks like if the constraints on $\Phi$. Saying $\Phi(I)=I$ and $\Phi(Z)=Z$ means that computational basis vectors are preserved, and thus $$V|0\rangle = |0\rangle|v_0\rangle, \qquad V|1\rangle = |1\rangle|v_1\rangle$$ for arbitrary states $|v_0\rangle,|v_1\rangle$. But if we now use $\Phi(X)=X$, which means $|\pm\rangle$ are also preserved, then we get $$V|\pm\rangle = \frac{1}{\sqrt2}(|0\rangle|v_0\rangle\pm |1\rangle|v_1\rangle)= |\pm\rangle|v_\pm\rangle.$$ The only way this is verified for some $|v_\pm\rangle$ is to have $|v_0\rangle=|v_1\rangle$, and then $|v_\pm\rangle=|v_0\rangle$ as well. But this, in turn, implies that we must also have $V|L\rangle=|L\rangle|v_0\rangle$ and $V|R\rangle=|R\rangle|v_0\rangle$, which means $\Phi(Y)=Y$.

You can easily make a lot of other (more or less general) statements from this. For example, if $\Phi$ preserves computational basis elements, but sends $|\pm\rangle$ into maximally mixed states, then it must necessarily also send $|L\rangle,|R\rangle$ into maximally mixed states (as the first two conditions imply $|v_0\rangle$ and $|v_1\rangle$ are orthogonal).

More generally, you can characterise a generic channel by specifying the corresponding isometry. This can be convenient because it's easy to characterise the set of possible isometries, and you also don't have to consider isometries with output spaces of arbitrary dimension. For example for a single qubit, you can characterise the possible channels as the set of $4\times2$ isometries. Though one must take care of the fact that multiple isometries can correspond to the same channel. For example $$V=\begin{pmatrix}1&0\\0&1\\0&0\\0&0\end{pmatrix}, \quad V=\begin{pmatrix}0&0\\0&0\\1&0\\0&1\end{pmatrix} $$ both correspond to the identity channel, while $$ V=\begin{pmatrix}1&0\\0&0\\0&0\\0&1\end{pmatrix}, \quad V=\begin{pmatrix}0&0\\0&1\\1&0\\0&0\end{pmatrix}, $$ both correspond to the channel such that $\Phi(I)=I$, $\Phi(Z)=Z$, $\Phi(X)=\Phi(Y)=0$.


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