量子から古典的なチャネルのクラウス表現とは何ですか?


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Watrous' bookで説明したように、量子からクラシックチャネルは、出力が常に完全に偏光解消されるCPTPマップです。これらは常に次のように書くことができます $$ \ Phi_ \ mu(X)= \ sum_a \ langle X、\ mu(a)\ rangle E_ {a、a} $$ $ \ sum_a \ mu(a)= Iを満たすいくつかの正の演算子 $ \ mu(a)\ ge0 $ の場合$

これらのマップの(a)Kraus表現は何ですか?

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

(Natural representations) To derive the natural representation of the map, note that $$\Phi_\mu(E_{k,\ell})=\sum_a\mu(a)_{\ell,k} E_{a,a}.$$ It follows that $$K(\Phi_\mu)_{ij,k\ell} = \langle i\rvert \Phi_\mu(E_{k,\ell})\lvert j\rangle = \sum_a \mu(a)_{\ell,k} \langle i\rvert E_{a,a}\lvert j\rangle=\delta_{ij} \mu(i)_{\ell,k},$$ where $E_{a,b}\equiv\lvert a\rangle\!\langle b\rvert$ and $K(\Phi)$ denotes the natural representation of $\Phi$. As an operator, this reads $$K(\Phi_\mu) %= \sum_a \lvert a,a\rangle \langle \mu(a)^T\rvert \equiv \sum_a \ket{a,a}\!\operatorname{vec}(\mu(a)^*)^T.$$

(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 $\langle i,j\rvert J(\Phi)\lvert k,\ell\rangle = \langle i,k\rvert K(\Phi)\lvert j,\ell\rangle$.

(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 $\ket{v_{a,j}}\equiv \ket a\otimes \ket{p_{a,j}^*}$ 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: $$ A_{a,j} = \sqrt{p_{a,j}} \lvert a\rangle\!\langle p_{a,j}\rvert \Longleftrightarrow (A_{a,j})_{ik} = \sqrt{p_{a,j}}\langle i,k\ket{v_{a,j}} = \sqrt{p_{a,j}} \delta_{a,i}\langle k\rvert p_{a,j}^*\rangle. \tag1 $$ 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 $$\Phi_\mu(X) = \sum_{a,jk\ell} E_{a,a}(M_a^*)_{j,k} X_{\ell,k} (M_a)_{j,\ell} E_{a,a}.$$ The corresponding Kraus operators thus have the form $$A_{a,j}= \lvert a\rangle\!\langle j\rvert M_a.\tag2$$ 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 $\lvert u_{a,\ell}\rangle=\lvert \ell\rangle$ to retrieve (1).