Deduce the Kraus operators of the dephasing channel using the Choi

Question

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.

Answer 1

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.

Answer 2

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 $$ A_1 = \frac{\sqrt{3-2p}}{\sqrt3} I, \quad A_2 = \frac{\sqrt{p}}{\sqrt2} (E_{11} - E_{22}), \quad A_3 = \frac{\sqrt{p}}{\sqrt6} (E_{11} + E_{22} - 2 E_{33}). $$