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In this post I read that "quantum measurements are special cases of quantum channels (CPTP maps). Stinespring's dilation states that any quantum channel is realized by partial tracing a unitary operator acting on a possibly bigger Hilbert space."
Is it possible to get a table of examples of such dilation-channel correspondence?
Obviously, there's infinitely many channels, so you cannot have an exhaustive table. I'll just give the Stinespring isometries corresponding to some notable ones. I'll use the examples from this post on physics.SE.
Let $\Phi:\operatorname{Lin}(\mathcal{X})\to \operatorname{Lin}(\mathcal{Y})$ be a channel with Kraus representation $\Phi(X)=\sum_a A_a X A_a^\dagger$, with Kraus operators $A_a\in\operatorname{Lin}(\mathcal{X},\mathcal{Y})$. Then its Stinespring representatoin is $$\Phi(X)=\operatorname{Tr}_{\cal Z}(VXV^\dagger)$$ where the isometry $V:\mathcal{X}\to \mathcal{Y}\otimes\mathcal{Z}$ is related to the Kraus operators by $$V = \sum_a A_a \otimes e_a, \qquad V\psi \equiv\sum_a (A_a \psi)\otimes e_a,\,\,\forall \psi\in\mathcal X.$$ Here I'm denoting with $e_a\in\mathcal Z$ the elements of an orthonormal basis for $\mathcal Z$ (you can write these as $|a\rangle$ if you prefer).
Often the channel $\Phi$ is not given via its Kraus representation. In such cases, one can first find the Kraus representation via the Choi $J(\Phi)\equiv\sum_{ij}\Phi(E_{ij})\otimes E_{ij}$. See also How does the spectral decomposition of the Choi operator relate to Kraus operators? and How does the Kraus decomposition imply the Stinespring representation? for more details about this process.
It is worth stressing that the Stinespring isometry is not unique. Different choices of Kraus operators will lead to different dilations. In the following, I will present a somewhat "special" representation, which is the one obtained from Kraus operators which are orthogonal (that is, obtained from the eigendecomposition of the Choi). Even so, there's additional sources of "freedom" in the choice of $V$:
Furthermore, just like one can describe a channel using a number of Kraus operators larger than the rank of the Choi, it is always possible to find Stinespring dilations that work on ancillary spaces that are "larger than they need to be".
If $\Phi(\rho)=\rho$, then $V=I$ is the identity matrix, and we don't need an auxiliary space $\mathcal Z$. Or if we want to stick with the general definition, we can use $V=I \otimes e_1$ for any normalised vector $e_1\in\mathcal Z$ in an auxiliary one-dimensional space $\mathcal Z$.
These are the channels $\Phi:\operatorname{Lin}(\mathbb{C}^d)\to\operatorname{Lin}(\mathbb{C}^d)$ with $$\Phi_d(\rho) = p\rho + (1-p)\operatorname{Tr}(\rho) \frac{I}{d}, \qquad p\in[0,1].$$ The corresponding Chois are $$J(\Phi_d) = p \mathbb{P}(|m\rangle) + (1-p) \frac{I\otimes I}{d}, \qquad \mathbb{P}(|m\rangle)\equiv \sum_{ij} |ii\rangle\!\langle jj|.$$ For example, for $d=2$, this has matrix representation $$J(\Phi_2)= \begin{pmatrix}\frac{1+p}{2} & 0 & 0 & p \\ 0 & \frac{1-p}{2} & 0 & 0 \\ 0 & 0 & \frac{1-p}{2} & 0 \\ p & 0 & 0 & \frac{1+p}{2}\end{pmatrix},$$ with associated Kraus operators $$A_1 = \sqrt{\frac{1+3p}{4}} \,I, \qquad A_2 = \sqrt{\frac{1-p}{4}} \, Z, \qquad A_3 = \sqrt{\frac{1-p}{2}} E_{01}, \qquad A_4 = \sqrt{\frac{1-p}{2}} E_{10},$$ where $E_{ij}\equiv |i\rangle\!\langle j|$, and $Z\equiv E_{00}-E_{11}$. We conclude that the (a choice of) Stinespring isometry is (can be represented as): $$V_2 = \begin{pmatrix} \sqrt{\frac{1+3p}{4}} \,I \\ \sqrt{\frac{1-p}{4}} \,Z \\ \sqrt{\frac{1-p}{2}} E_{01} \\ \sqrt{\frac{1-p}{2}} E_{10} \end{pmatrix}.$$ Note how this collapses to the identity channel case for $p=1$.
See also Can a Kraus representation act as the identity on any operator? for more detail.
This is $\Phi:\operatorname{Lin}(\mathbb{C}^2)\to\operatorname{Lin}(\mathbb{C}^2)$ with $$\Phi(\rho) \equiv p \rho + (1-p) Z\rho Z.$$ Its Choi is $$J(\Phi) = 2p\, \mathbb{P}(|\Phi^+\rangle) + 2(1-p) \mathbb{P}(|\Phi^-\rangle) \doteq \begin{pmatrix} 1 & 0 & 0 & 2p-1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 2p-1 & 0 & 0 & 1 \end{pmatrix},$$ where the Bell states are $|\Phi^+\rangle\equiv\frac1{\sqrt2}(|00\rangle+|11\rangle)$ and $|\Phi^-\rangle\equiv\frac1{\sqrt2}(|00\rangle-|11\rangle)$.
the Kraus operators are thus $$A_1 = \sqrt{p} \, I, \qquad A_2 = \sqrt{1-p} \, Z,$$ and therefore the Stinespring isometry reads $$V = \begin{pmatrix}\sqrt p & 0 \\ 0 & \sqrt p \\ \sqrt{1-p} & 0 \\ 0 & -\sqrt{1-p}\end{pmatrix}.$$
More details about Choi and Kraus of dephasing channels can be found here.
Channels $\Phi:\operatorname{Lin}(\mathbb{C}^n)\to\operatorname{Lin}(\mathbb{C}^n)$ of the form $\Phi(\rho)\equiv \operatorname{Tr}(\rho) \sigma$ for some pair of states $\rho,\sigma$. The Choi is then $$J(\Phi) = \sigma\otimes I_n,$$ and a choice of Kraus operators is $$A_{a,b} = \sqrt{p_a} |v_a\rangle\!\langle b|, \qquad a,b=1,...,n,$$ where $|v_a\rangle$ are the (normalised) eigenvectors of $\sigma$, and $p_a$ the corresponding eigenvalues, and $|b\rangle$ is an arbitrary choice of orthonormal basis for $\mathbb{C}^n$.
The Stinespring isometry will obviously depend on the choice of $\sigma$, but you can again obtain it by just stacking the Kraus matrices one on top of the other.
Those of the form $\Phi(\rho) = \sum_a \langle \mu_a,\rho\rangle \,\sigma_a$ for some collection of states $\sigma_i$ and POVM $\{\mu_a\}$. Then $$J(\Phi) = \sum_a \sigma_a\otimes \mu_a^T,$$ and the Kraus will strongly depend on the eigendecompositions of $\sigma_a$ and $\mu_a$, as will the Stinespring isometry.
One thing you can notice to significantly simplify the calculation is however that everything is linear, and therefore if $\mu_a=\sum_b p_{ab} \mathbb{P}(|\mu_{ab}\rangle)$ and $\sigma_a=\sum_c q_{ac}\mathbb{P}(|\sigma_{ac}\rangle)$, then $\Phi(\rho)$ is a linear combination of the channels $\Phi_{abc}(\rho)=\langle \mu_{ab}|\rho|\mu_{ab}\rangle \mathbb{P}(|\sigma_{ac}\rangle)$, with Chois $$J(\Phi_{abc}) = \mathbb{P}(|\sigma_{ab}\rangle)\otimes\mathbb{P}(|\bar\mu_{ac}\rangle) \equiv |\sigma_{ab}\rangle\!\langle\sigma_{ab}|\otimes|\bar\mu_{ac}\rangle\!\langle\bar\mu_{ac}|.$$ Now, with this you can easily write the Choi of $\Phi$ itself, but the catch is that this doesn't translate into an easy expression for the Kraus operators, and thus the dilation. The fundamental reason being that the Kraus operators are related to the eigendecomposition of the Choi, and the eigendecomposition of a sum of operators is not straightforwardly related to the eigendecompositions of the elements of the sum.
