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I'm trying to deduce the Kraus representation of the dephasing channel using the Choi operator (I know the Kraus operators can be guessed in this case, I want to understand the general case).

The dephasing channel maps a density operator $\rho$ as $$\rho\rightarrow D(\rho)=(1-p)\rho+ p\textrm{diag}(\rho_{00},\rho_{11}) $$

The Choi operator acts on a channel as

$$C(D)=(I \otimes D)\sum_{k,j=0}^1 \vert k\rangle \langle j \vert \otimes \vert k\rangle \langle j \vert=\sum_{k,j=0}^1\vert k\rangle \langle j \vert \otimes D(\vert k\rangle \langle j \vert)=\\=|0\rangle\langle 0|\otimes|0\rangle\langle 0|+p|0\rangle\langle 1|\otimes|0\rangle\langle 1|+p|1\rangle\langle 0|\otimes|1\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|=\\=|00\rangle\langle00|+p|01\rangle\langle01|+p|10\rangle\langle10|+|11\rangle\langle11|= \sum_{j=0}^3 |\psi_j\rangle\langle\psi_j|$$

Now, to find the Kraus operators, I should just find some $K_j$ such that $|\psi_j\rangle =(I\otimes K_j) \sum_{k=0}^1 \vert k\rangle \otimes \vert k\rangle$. These operators are simply

$$ K_0=|0\rangle\langle 0|\quad K_1=\sqrt{p}|1\rangle\langle 0| \quad K_2=\sqrt{p}|0\rangle\langle 1|\quad K_3=|1\rangle\langle 1|$$

And I should have $$D(\rho)=\sum_{j=1}^3 K_j\rho K_j^\dagger$$

But $$ \sum_{j=1}^3 K_j\rho K_j^\dagger=(\rho_{00}+p\rho_{11})|0\rangle\langle0| + (\rho_{11}+p\rho_{00})|1\rangle\langle1|$$

Which is most certainly not what I should get. I'm sure I'm either doing a massive calculation error, or I have massively misunderstand everything. Moreover, doing this I should only be able to find 4 Kraus operator, while I know that the representation is not unique and in particular this channel can be represented by only two Kraus operators. Any help is appreciated.

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Acting with the dephasing channel on the possible states of a single qubit:

\begin{align}D\left(\left|0\rangle\langle0\right|\right) &= \left|0\rangle\langle0\right| \\ D\left(\left|0\rangle\langle1\right|\right) &= \left(1-p\right)\left|0\rangle\langle1\right|\\ D\left(\left|1\rangle\langle0\right|\right) &= \left(1-p\right)\left|1\rangle\langle0\right|\\ D\left(\left|1\rangle\langle1\right|\right) &= \left|1\rangle\langle1\right|.\end{align}

This gives that \begin{align}C\left(D\right) &= \sum_{k,j=0}^1\vert k\rangle \langle j \vert \otimes D(\vert k\rangle \langle j \vert) \\ &= |0\rangle\langle 0|\otimes|0\rangle\langle 0|+\left(1-p\right)|0\rangle\langle 1|\otimes|0\rangle\langle 1|+\left(1-p\right)|1\rangle\langle 0|\otimes|1\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|\\ &= |00\rangle\langle00|+|00\rangle\langle11|+|11\rangle\langle00|+|11\rangle\langle11|- p|00\rangle\langle 11|-p|11\rangle\langle 00|\\ &=\sum_{k, j=0}^1\left(1-p\right)\vert k\rangle \langle j \vert \otimes \vert k\rangle \langle j \vert + p\left(|0\rangle\langle 0|\otimes|0\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|\right) \\ &= \sum_{j=0}^N |\psi_j\rangle\langle\psi_j|.\end{align}

Now using $|\psi_j\rangle =(I\otimes K_j) \sum_{k=0}^1 \vert k\rangle \otimes \vert k\rangle$ to get $$\sum_{j=0}^N |\psi_j\rangle\langle\psi_j| = \sum_{j=0}^N\sum_{k,l=0}^1\vert k\rangle\langle l\vert \otimes K_j\vert k\rangle \langle l\vert K_j^{\dagger},$$ which equals $C\left(D\right)$ when the Kraus operators $K_0 = \sqrt{1-p}I,\, K_1 = \sqrt p |0\rangle\langle 0|$ and $K_2 = \sqrt p |1\rangle\langle 1|$.

Taking an arbitrary (single qubit) density matrix $$\rho = \rho_{00}|0\rangle\langle 0| + \rho_{01}|0\rangle\langle 1| + \rho_{10}|1\rangle\langle 0| + \rho_{11}|1\rangle\langle 1|$$ and acting on this using the above Kraus operators gives \begin{align}D(\rho)&=\sum_{j=1}^3 K_j\rho K_j^\dagger \\ &=\left(1-p\right)\rho + p\rho_{00}|0\rangle\langle 0| + p\rho_{11}|1\rangle\langle 1|,\end{align} as expected for the dephasing channel.

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Write the dephasing channel, acting on arbitrary $d$-dimensional states, as $$D(\rho)\equiv D_p(\rho) = (1-p)\rho + p \sum_k\operatorname{Tr}(E_{kk}\rho)E_{kk}.$$ Using the notation $E_{ij}\equiv|i\rangle\!\langle j|$.

(1) Compute the Choi representation

It follows that $D(E_{ij})=(1-p)E_{ij}$ for $i\neq j$, and $D(E_{ii})= E_{ii}$. Denote the Choi representation of $D$ as $J(D)$. One way to define the Choi of a channel is as $$J(D)\equiv \sum_{ij} D(E_{ij})\otimes E_{ij},$$ which using the above rules immediately becomes $$J(D) = (1-p)\sum_{i\neq j} E_{ij}\otimes E_{ij} + \sum_i E_{ii}\otimes E_{ii} \\ = \sum_i E_{ii}\otimes E_{ii} + (1-p)\sum_{i\neq j}E_{ij}\otimes E_{ij} \\ \equiv \sum_i |ii\rangle\!\langle ii| + (1-p)\sum_{i\neq j}|ii\rangle\!\langle jj|.$$ The same result could also equivalently be obtained observing that $$D_p=(1-p)\operatorname{Id} + p \operatorname{FullyDephasingCh},$$ that the map $D\mapsto J(D)$ is linear, and that $J(\operatorname{Id})=|m\rangle\!\langle m|$ with $|m\rangle\equiv\sum_i |ii\rangle$, and $$ J(\operatorname{FullyDephasingCh}) = \sum_i |ii\rangle\!\langle ii|. $$

(2) Eigendecomposition of the Choi

A straightforward way to get a Kraus decomposition is via the eigendecomposition of the Choi. To this end, start by observing that support and image of $J(D)$ are spanned by the vectors $\{|ii\rangle : i=1,...,d\}$. We can thus make a formal simplification replacing $|ii\rangle\to|i\rangle$. Upon this replacement, $J(D_p)$ gets a nice matrix representation: $$J(D_p)=I + (1-p)(\boldsymbol1-I),$$ where $I$ is the identity matrix, and $\boldsymbol1$ the constant matrix with $(\boldsymbol1)_{ij}=1$ for all $i,j$. For concreteness, in the $d=2$ case the above amounts to $$J(D_p) = \begin{pmatrix}1 & 1-p \\ 1-p & 1\end{pmatrix}.$$ It is not too hard to verify that the eigenvalues of $J(D_p)$ are a nondegenerate $\lambda_1=d(1-p)+p$, and a $(d-1)$-fold degenerate $\lambda_2=p$. A corresponding set of eigenvectors is $$|\lambda_1\rangle = \frac{1}{\sqrt d}\sum_i |i\rangle, \\ |\lambda_2^j\rangle = \frac{1}{\sqrt2} (|j\rangle - |1\rangle), \qquad j=2,...,d.$$ The $|\lambda_2^j\rangle$ here are eigenvectors corresponding to $\lambda_2$, but note that they are not orthogonal. An orthonormal system of eigenvectors can be written in the form $$|\lambda_2^j\rangle = \frac{1}{\sqrt{j(j+1)}}\left(\sum_{i=1}^j|i\rangle - j|j+1\rangle\right), \qquad j=1,...,d-1. $$ These correspond to the following spectral decomposition for the Choi: $$J(D_p) = ((1-p)d+p)\frac{\boldsymbol1}{d} + p\left(I-\frac{\boldsymbol1}{d}\right),$$ where we observe that $\frac{\boldsymbol1}{d}$ is a unit-trace orthoprojection, and $\left(I-\frac{\boldsymbol1}{d}\right)$ an orthogonal projection with trace $d-1$, as expected.

(3) Kraus operators from Choi's eigendecomposition

Given a map $\Phi\in\mathrm T(\mathcal X,\mathcal Y)$ we have $$\Phi(X)=\sum_{a}A_a X B_a^\dagger \iff J(\Phi) = \sum_a \operatorname{vec}(A_a)\operatorname{vec}(B_b)^\dagger, $$ for some linear operators $A_a,B_b:\mathcal X\to\mathcal Y$. Also, one can always find such a representation, with operators such that $\langle A_a,A_b\rangle=\langle B_a,B_b\rangle=0$ for any $a\neq b$. In particular, if the map $\Phi$ is CP, and thus its Choi is positive semidefinite, we get the "standard" Kraus decomposition, and $$\Phi(X)=\sum_a A_a X A_a^\dagger \iff J(\Phi) = \sum_a \operatorname{vec}(A_a)\operatorname{vec}(A_a)^\dagger. $$ This means that the eigenvectors of the Choi are tightly related to (one choice of) Kraus operators. More specifically, if we have an eigendecomposition of the form $$J(\Phi) = \sum_a \lambda_a \mathbb P(v_a),$$ with $v_a\in\mathcal Y\otimes\mathcal X$ such that $\|v_a\|=1$, and $\mathbb P(v_a)$ denoting the corresponding rank-1 projection, then a valid choice of Kraus operators is $$A_a = \sqrt{\lambda_a} \operatorname{unvec}(v_a) \in\mathrm{Lin}(\mathcal X,\mathcal Y).$$ Going back to the case at hand, using the eigenvalues and orthonormal eigenvectors previously derived, we get the Kraus operators $$ A_1 = \frac{\sqrt{d+p(1-d)}}{\sqrt d} \sum_i E_{ii} \equiv \frac{1}{\sqrt d}I, \qquad A_2^k = \frac{\sqrt{p}}{\sqrt{k(k+1)}} \left( \sum_{j=1}^k E_{jj} - k E_{k+1} \right), $$ for $k=1,...,d-1$.

(4) Examples

For $d=2$, the above amount to $$A_1 = \frac{\sqrt{2-p}}{\sqrt2} I, \qquad A_2 = \frac{\sqrt{p}}{\sqrt2} Z. $$ For $d=3$, we get instead $$ A_1 = \frac{\sqrt{3-2p}}{\sqrt3} I, \quad A_2 = \frac{\sqrt{p}}{\sqrt2} (E_{11} - E_{22}), \quad A_3 = \frac{\sqrt{p}}{\sqrt6} (E_{11} + E_{22} - 2 E_{33}). $$

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