What is the Kraus representation of quantum-to-classical channels?

We start from the defining form of the channel as $Phi_mu(X)=sum_a operatorname{tr}(mu(a)X)E_{a,a}$.$newcommand{PP}{mathbb{P}}newcommand{tr}{operatorname{tr}}newcommand{calX}{mathcal X}newcommand{calY}{mathcal Y}newcommand{calZ}{mathcal Z}newcommand{ket}[1]{lvert #1rangle}newcommand{bs}[1]{boldsymbol{#1}}$

(Natural representations) To derive the natural representation of the map, note that $$Phi_mu(E_{k,ell})=sum_amu(a)_{ell,k} E_{a,a}.$$ It follows that $$K(Phi_mu)_{ij,kell} = langle irvert Phi_mu(E_{k,ell})lvert jrangle = sum_a mu(a)_{ell,k} langle irvert E_{a,a}lvert jrangle=delta_{ij} mu(i)_{ell,k},$$ where $E_{a,b}equivlvert arangle!langle brvert$ and $K(Phi)$ denotes the natural representation of $Phi$. As an operator, this reads $$K(Phi_mu) %= sum_a lvert a,arangle langle mu(a)^Trvert equiv sum_a ket{a,a}!operatorname{vec}(mu(a)^*)^T.$$

(Choi representation) Consider now the Choi operator, defined as $J(Phi)equiv sum_{i,j}Phi(E_{i,j})otimes E_{i,j}$. From this we get $$J(Phi_mu) = sum_{a,i,j} mu(a)_{j,i} E_{a,a}otimes E_{i,j} = sum_a E_{a,a}otimes mu(a)^T.$$ We can also get this from $K(Phi)$, using the relation $langle i,jrvert J(Phi)lvert k,ellrangle = langle i,krvert K(Phi)lvert j,ellrangle$.

(Kraus representation from Choi) One way to get the Kraus representation is via the spectral decomposition of the Choi. From the relations above, we see that the spectral decomposition of the Choi is in this case quite easy: define $ket{v_{a,j}}equiv ket aotimes ket{p_{a,j}^*}$ with $ket{p_{a,j}}$ the eigenvector of $mu(a)$ with eigenvalue $p_{a,j}$, and using $ket{p_{a,j}^*}$ to denote the complex conjugate of $ket{p_{a,j}}$.

From this we get the Kraus operators as the maps $A_{a,j}$ of the form: $$ A_{a,j} = sqrt{p_{a,j}} lvert arangle!langle p_{a,j}rvert Longleftrightarrow (A_{a,j})_{ik} = sqrt{p_{a,j}}langle i,kket{v_{a,j}} = sqrt{p_{a,j}} delta_{a,i}langle krvert p_{a,j}^*rangle. tag1 $$ With these operators, we can write $$Phi_mu(X) = sum_{a,j} A_{a,j} X A_{a,j}^dagger.$$

(Direct derivation) For a direct route that doesn't require passing through the Choi representation, let us write down the explicit form of $Phi_mu(X)$: $$Phi_mu(X) = sum_{a,ell k} mu(a)_{k,ell}X_{ell,k} E_{a,a}.$$ Because, by hypothesis, $mu(a)ge0$, we can find some operator $M_a$ such that $mu(a)=M_a^dagger M_a$. Componentwise, this reads $mu(a)_{k,ell} = sum_j(M_a^*)_{j,k}(M_a)_{j,ell}.$ Using this in the expression above we get $$Phi_mu(X) = sum_{a,jkell} E_{a,a}(M_a^*)_{j,k} X_{ell,k} (M_a)_{j,ell} E_{a,a}.$$ The corresponding Kraus operators thus have the form $$A_{a,j}= lvert arangle!langle jrvert M_a.tag2$$ Of course, this now begs the question: are the Kraus operators in (2) compatible with those previously derived in (1)? The answer is: not necessarily. Equation (2) is more general, due to the freedom in the choice of $M_a$, and in particular doesn't necessarily lead to orthogonal Kraus operators, like (1) does. To see this, notice that we can generally express $M_a$ in terms of the eigendecomposition of $mu(a)$ as $$M_a = sum_ell sqrt{p_{a,ell}} lvert u_{a,ell}rangle!langle p_{a,ell}rvert,$$ for any choice of orthonormal vectors $lvert u_{a,ell}rangle$. In particular, we can choose $lvert u_{a,ell}rangle=lvert ellrangle$ to retrieve (1).