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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?

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  • $\begingroup$ you are asking for a table showing the correspondence between every possible channel and its dilation? or are you asking how to derive the Stinespring isometry corresponding to an otherwise specified channel? $\endgroup$
    – glS
    Mar 14, 2022 at 11:23
  • $\begingroup$ I think you want to understand the formula/method for deriving the Stinespring representation of a quantum channel. Asking for a table is a bit like asking for a multiplication table instead of learning how to multiply—it probably won’t be nearly as useful as just learning how to do it yourself! $\endgroup$ Mar 15, 2022 at 4:14
  • $\begingroup$ i would like to see something like the Fourier Transform tables. with on the left the main used logical gates (Pauli, hadamard...) and on the right their dialation map. and on the left several cases of non unitary gates (random measurement) $\endgroup$
    – Naima
    Mar 15, 2022 at 12:32
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    $\begingroup$ Given a CPTP map, the Stinespring dilation is not unique, so you cannot generally have such a table. Also if you are only interested in Stinespring dilations of unitary channels, this is trivial since the unitaries are themselves the Stinespring dilations. That being said, you can construct a Stinespring dilation from a "measurement channel" if you know the Kraus operators. If this is what you want, I can provide the construction. $\endgroup$
    – Condo
    Mar 15, 2022 at 18:30

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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.

General correspondence

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 $V:\mathcal{X}\to \mathcal{Z}\otimes\mathcal{Y}$ is related to the Kraus operators by $$V = \sum_a e_a\otimes A_a, \qquad V\psi \equiv\sum_a e_a\otimes (A_a \psi),\,\,\forall \psi\in\mathcal X.$$ 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). Note that it's also common to define the isometry with $\mathcal Z$ and $\mathcal Y$ inverted in the definition above. The only reason I define it this way is that it makes $V$ visualisable as the Kraus operators stacked in a column.

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. In short, if you can eigendecompose the Choi as $J(\Phi)=\sum_a \lambda_a \mathbb{P}(v_a)$, where $\mathbb{P}(v_a)\equiv v_a v_a^\dagger$, $\lambda_a>0$ are the eigenvalues, and $\{v_a\}$ is an(y) orthonormal set of eigenvectors, then the operators $A_a=\sqrt{\lambda_a} \operatorname{unvec}(v_a)$ give you a (minimal) Kraus decomposition for the channel. More generally, given any rank-1 decomposition for the Choi, $J(\Phi) = \sum_a \mathbb{P}(w_a)$ with the vectors $w_a$ not necessarily orthogonal nor normalised, the operators $A_a\equiv \operatorname{unvec}(w_a)$ give you a Kraus decomposition of the channel. All Kraus decompositions can be obtained this way.

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 $V$:

  1. Whenever the Choi has degenerate eigenvalues, there is unitary freedom in picking its eigenvectors, which translates into different equivalent choices of Kraus operators, and thus of dilations. See also Prove that different Kraus decompositions are related through a unitary, using the Choi isomorphism for more details.
  2. For any isometry $V$, the isometry $(U\otimes I)V$ will produce the same channel, for any isometry $U$. If $V$ is the isometric dilation corresponding to a set of orthogonal Kraus operators, that is, those obtained from the eigendecomposition of the Choi, then all isometric dilations can be obtained this way. More generally, if $V,V'$ are two isometric dilations of a given channel, then there's some partial isometry $U$ such that $(U\otimes I)V=V'$, with either $U$ or $U^\dagger$ being an isometry.

Note that, 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" (that is, larger than the Choi rank of the channel).

Examples

Identity channel

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 $\mathcal Z$.

Fully depolarising channel

Consider the channel $\Phi:\operatorname{Lin}(\mathbb{C}^d)\to\operatorname{Lin}(\mathbb{C}^d)$ defined as $$\Phi_d(\rho) = \operatorname{tr}(\rho) \frac{I}{d}.$$ Its Choi representation is then the $d^2$-dimensional matrix $$J(\Phi_d) = \frac{I_d\otimes I_d}{d}.$$ Modulo a $d$ factor, this is just the maximally mixed state in dimension $d^2$. Its eigendecomposition is trivial: it has a single eigenvalue $\lambda=1/d$, corresponding to a $d^2$-dimensional eigenspace. For example, for $d=2$ this is $$J(\Phi_2) = \frac12\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$ A standard choice of eigenvectors is then the computational basis in this space, which corresponds after unvectorisation to the Kraus operators $A_{ij} = \frac{1}{\sqrt2} E_{ij}$, $E_{ij} \equiv |i\rangle\!\langle j|,$ for $i,j=1,2$. The dilation isometry corresponding to this decomposition is then $$V = \frac1{\sqrt2}\begin{pmatrix}1&0\\0&0\\0&1\\0&0\\0&0\\1&0\\0&0\\0&1\end{pmatrix}.$$ What if you use a different eigendecomposition, i.e. different Kraus decomposition? For example, say we take the Bell states as eigenvectors of $J(\Phi_2)$: $$v_1 = \frac{|00\rangle+|11\rangle}{\sqrt2}, \quad v_2 = \frac{|01\rangle+|10\rangle}{\sqrt2}, \quad v_3 = \frac{|01\rangle-|10\rangle}{\sqrt2}, \quad v_4 = \frac{|00\rangle-|11\rangle}{\sqrt2}.$$ Then the corresponding Kraus operators read $$A_1 = \frac{1}2 I, \quad A_2 = \frac{1}2 X, \quad A_3 = \frac{1}2 iY, \quad A_4 = \frac{1}2 Z,$$ and the corresponding isometry is $$V' = \frac{1}{2}\begin{pmatrix}I\\X\\iY\\Z\end{pmatrix}.$$ Both $V$ and $V'$ generate the same identical channel, and they are related as $V'=(U\otimes I)V$ with the unitary $U$ given by $$U = \frac{1}{\sqrt2}\begin{pmatrix}1&0&0&1\\0&1&1&0\\0&1&-1&0\\1&0&0&-1\end{pmatrix}.$$ The matrix $U$ can be obtained as the unitary connecting the two choices of eigenbases of $J(\Phi_2)$.

Another convenient way to see that $V,V'$ correspond to the same channel is to write their explicit action on pure states, which reads: $$V|0\rangle = \frac{|1,0\rangle+|3,1\rangle}{\sqrt2}, \qquad V|1\rangle = \frac{|2,0\rangle+|4,1\rangle}{\sqrt2},\\ V'|0\rangle = \frac{1}{\sqrt2}\left[\left(\frac{|1\rangle+|4\rangle}{\sqrt2}\right)|0\rangle + \left(\frac{|2\rangle-|3\rangle}{\sqrt2}\right)|1\rangle \right], \\ V'|1\rangle = \frac{1}{\sqrt2}\left[\left(\frac{|2\rangle+|3\rangle}{\sqrt2}\right)|0\rangle + \left(\frac{|1\rangle-|4\rangle}{\sqrt2}\right)|1\rangle \right]. $$ The reason these both work equally is that they're both isometries of the form $$|0\rangle\to \frac1{\sqrt2}(|\psi_1\rangle|0\rangle+|\psi_2\rangle|1\rangle), \quad |1\rangle\to \frac{1}{\sqrt2}(|\psi_3\rangle|0\rangle+|\psi_4\rangle|1\rangle),$$ for some set of orthonormal states $\{|\psi_i\rangle\}_{i=1}^4$.

Depolarising channel

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.

Dephasing channel

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.

"Wirf weg und mach neu" channels

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.

Entanglement breaking channels

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.

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