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2

I'll answer the first question; these might be better served as two distinct questions. Let's start with the mathematical definition quoted in the paper: for a quantum channel $\Phi(\rho)$ acting on density operators and a set of unitary operators $U$ belonging to some group, the equality $$\Phi(U \rho U^\dagger)=U \Phi(\rho)U^\dagger$$ implies that the ...


4

The given form of the Kraus operators, though not unique, tells us what is physically happening. In the case of pure-state outputs, each Kraus operator takes one of the possible basis states and swaps it with the desired output state. All basis states thus become $|0\rangle$ through some sort of swapping mechanism. In the case of mixed-state outputs, each ...


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The set $\mathrm L(\mathcal X)\equiv\mathrm L(\mathcal X,\mathcal X)$ is the set of linear maps from $\mathcal X$ to $\mathcal X$. In other words, $A\in\mathrm L(\mathcal X)$ iff $A$ is a linear function of the form $A:\mathcal X\to\mathcal X$. Note that $\mathrm L(\mathcal X,\mathcal Y)$ is itself a vector space. That means you can have, for example, $\...


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Let $M\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ be some linear operator whose input and output spaces are both $\mathcal Y\otimes\mathcal X$, for some pair of finite-dimensional Hilbert spaces $\mathcal X,\mathcal Y$. Moreover, suppose $M$ is positive semidefinite: $M\ge0$. It being positive semidefinite implies it admits a decomposition of the form $M=\...


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Choi operator of a linear map $\mathcal{E}$ is defined as $$ J(\mathcal{E}) = \sum_{ij} \mathcal{E}(|i\rangle\langle j|)\otimes |i\rangle\langle j|.\tag1 $$ Substituting $\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$ into $(1)$, we have $$ \begin{align} J(\mathcal{E}) &= \sum_{ijk} \left(E_k|i\rangle\langle j| E_k^\dagger\right)\otimes |i\rangle\langle j|...


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Every quantum channel has many Kraus representations that may differ in the number of Kraus operators. For example, for any positive integer $n$ and numbers $p_i$ with $i=1,\dots,n$ and $\sum_{i=1}^np_i=1$ the matrices $E_i=\sqrt{p_i}I$ form a valid, if impractical, Kraus representation of the identity channel with $n$ Kraus operators. This example also ...


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There's no need to increase the dimension of $\newcommand{\calH}{{\mathcal H}}\calH_E$ to $D+1$. All you care about is that $U$ must send orthonormal vectors to orthonormal vectors. What the state $|0\rangle_E$ actually represents is purely conventional. You could put an arbitrary (pure) state there, and get the same channel (as long as you also change the ...


7

More precisely, they are described by affine maps. This is also not special to quantum mechanics, but rather a more general fact in physical theories (keyword: generalized probabilistic theory). Let me elaborate. Suppose you can prepare two states $s_1$ and $s_2$ of a physical system, then it is always possible to prepare a probabilistic mixture of those. ...


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Source of the problem The purported contradiction arises due to the use of incorrect assumptions for Klein equality $$ S(\rho||\sigma) \ge 0. $$ The inequality does not require any particular relationship$^1$ between the support of $\rho$ and the support of $\sigma$. However, it does require that $\rho$ and $\sigma$ be states, i.e. unit trace positive ...


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Finding Kraus operators The method of determining the Kraus operators of a quantum channel $\Phi: L(\mathcal{X})\to L(\mathcal{Y})$ from the knowledge of its action on a set of inputs is called quantum process tomography. See section 8.4.2 on page 389 in Nielsen & Chuang for details. In particular, see equation $(8.168)$ on page 392 for how Kraus ...


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