8

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 ...


6

Stinespring dilation can be thought of as a way of representing an arbitrary completely positive trace preserving map $\Lambda$ on a system $A$ as a composition of two simpler maps: a unitary evolution $U_{AE}$ in a Hilbert space obtained by adjoining an auxiliary system $E$ followed the partial trace over $E$ $$ \Lambda(\rho_A) = \mathrm{tr}_E\left(U_{AE} (\...


5

Suppose that $$ \mathrm{tr}\left(\sum_k E_k\rho E_k^\dagger\right) = \mathrm{tr}(\rho) $$ for all $\rho$. Then $$ \mathrm{tr}\left(\sum_k E_k^\dagger E_k\rho\right) = \mathrm{tr}(I\rho) $$ for all $\rho$. The last equation can be rewritten in terms of Hilbert-Schmidt inner product as $$ \left\langle \sum_k E_k^\dagger E_k,\rho\right\rangle_{HS} = \left\...


5

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|...


5

You have $$\newcommand{\ket}[1]{\lvert #1\rangle}U(\ket\psi\otimes\ket0) = \bigg(I\otimes \underbrace{\sum_\ell \ket\ell\!\langle\ell|}_{\equiv I}\bigg) U (\ket\psi\otimes\ket0) \\ = \sum_\ell (I\otimes \ket\ell\!\langle\ell|)U(\ket\psi\otimes\ket0) = \sum_\ell (U_{(\ell,0)}\ket\psi)\otimes\ket\ell $$ where $$\Pi_\ell\equiv U_{(\ell,0)} \equiv (I\otimes \...


5

Quoting from the linked source: "thus SWAP has negative eigenvalues, which means that $T\otimes I$ is not positive and therefore $T$ is not completely positive", where $T$ is the transpose. So they are not saying that the SWAP is not a realisable operation; they are saying that $T$ isn't. As you note, the SWAP is a perfectly fine unitary gate. That ...


4

Another way is to observe that Choi $J(\Phi)\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ and Kraus operators $\{A_a\}_a\subset\mathrm{Lin}(\mathcal X,\mathcal Y)$ of a map $\Phi:\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal Y)$ are directly related via $$J(\Phi) = \sum_a \operatorname{vec}(A_a)\operatorname{vec}(A_a)^\dagger,$$ where $\operatorname{vec}(...


4

More generally, given any two states, you can always find some channel sending one into the other. Consider for example replacement maps, which have the form $$\Phi_Y(X) = \operatorname{Tr}(X) Y.$$ Given any pair of states $\rho$ and $\sigma$, the channel $\Phi_\sigma$ will send $\rho$ (as well as any other state) into $\sigma$. The (or a) set of Kraus ...


4

As Adam Zalcman has stated in his answer, channels whose Kraus operators are proportional to unitary operators are called mixed-unitary channels (or, alternatively, random unitary channels). Every mixed-unitary channel is unital (meaning that it maps the identity operator to itself), so if you want a channel that is not mixed unitary, just pick any non-...


4

TL;DR Quantum capacity of $\mathcal{N}_2\circ\mathcal{N}_1$ can be anywhere between zero and the minimum of the quantum capacities of $\mathcal{N}_1$ and $\mathcal{N}_2$. Background Quantum capacity of a quantum channel $\mathcal{N}$ is defined as the greatest real number $Q(\mathcal{N})$ such that for any $R < Q(\mathcal{N})$ (representing a transmission ...


4

I think this question is generally difficult because there is no standard metric for non-Markovianity, for example this paper would suggest you try to express the evolution in a time-local canonical (Lindblad) form and then look at the negativity of the rates, but other metrics may not agree for certain channels. Perhaps it is easier to answer the simpler ...


4

Assuming w.l.o.g. that $p\in\mathbb{R}$, the linear map in the question may be rewritten as $$ \mathcal{E}(\rho) = p^2\rho+p^2X\rho X = 2p^2\left(\frac12\rho + \frac12 X\rho X\right) $$ where $X$ is the Pauli matrix. Thus, the action of $\mathcal{E}$ can be understood as the composition $\mathcal{E}=\mathcal{S}_{2p^2}\circ\mathcal{X}_{\frac12}$ of a scaling ...


3

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=\...


3

No. The key thing about what Preskill is saying is that all the Kraus operators must be proportional to the same unitary. Your two Kraus operators are proportional to different Kraus operators.


3

I guess from the operator sum representation $\rho(t) = \sum_k K_k \rho(t_0) K_k^\dagger$ alone you won't be able to make any statement about non-Markovianity, since you are missing interesting features: (Domain) Is this dynamical map valid for all initial reduced density matrices? If so, the initial state might be a product state $\rho(t_0) \otimes \rho_E$ ...


3

Why not just concatenate the Kraus operators from each channel, i.e., take $\{Z_{k}\}_{1 \leq k \leq d_1 + d_1}$ where $$ Z_k = \begin{cases} \sqrt{p} M_k \qquad \quad &\text{if } 1 \leq k \leq d_1 \\ \sqrt{1-p}N_{k - d_1} &\text{if } d_1 < k \leq d_1 + d_2 \end{cases} $$ You can check that these operators satisfy completeness property for Kraus ...


3

(Stinespring) Given a basis $\{|u_k\rangle\}$ for the input space, you want an isometry $V$ such that $V |u_k\rangle=|0\rangle\otimes |u_k\rangle$ for some orthonormal set $\{|u_k\rangle\}$. Then you have $$\mathrm{Tr}_2\Big(V|u_k\rangle\!\langle u_k|V^\dagger\Big) = |0\rangle\!\langle0|,$$ and thus the same hold for any pure state $|\psi\rangle$ (because ...


3

I don't know the fully general answer, but have found a solution for channels acting on a single qubit. Mixed-unitary channels Quantum channels that admit a Kraus representation consisting solely of multiples of unitary operators are known as mixed-unitary channels, i.e. $\mathcal{E}$ is a mixed-unitary channel if there exists unitary operators $U_i$ and ...


3

Control errors The term control error is generally used to refer to errors due to imperfections of the qubit control system. Hardware devices that control qubit evolution have a number of knobs that the control system sets to various values. In the process known as calibration we learn the settings of the knobs that correspond to each of the supported gates. ...


3

Let $\newcommand{\rmD}{\mathrm{D}}\newcommand{\rmU}{\mathrm{U}}\newcommand{\calU}{\mathcal{U}}\newcommand{\CC}{\mathbb{C}}\rho\in\rmD(\CC^n),\sigma\in\rmD(\CC^m)$ be two arbitrary finite-dimensional quantum states, and let $\calU\in \rmU(\CC^n\otimes\CC^m)$ be a unitary in the total space. We have $$[U(\rho\otimes\sigma)U^\dagger]_{ij,k\ell} = \sum_{mnpq} U_{...


3

Just plug in all of the relevant stuff you state in the question, i.e. $$ U = |0\rangle \langle 0 | \otimes I + |1 \rangle \langle 1 | \otimes X $$ and $$ \rho_{\mathrm{env}} = |0\rangle \langle 0 |. $$ Then expand and simplify $$ \begin{aligned} \mathcal{E}(\rho) &= \mathrm{Tr}_{\mathrm{env}}[(P_0 \otimes I + P_1 \otimes X)(\rho \otimes |0\rangle \...


3

The first condition is satisfied for example by unitaries of the form $U = e^{i\theta}I_A \otimes U_B$ where $I_A$ is identity on subsystem $A$, $U_B$ is any unitary on subsystem $B$ and the phase factor $e^{i\theta}$ is irrelevant. Let us consider the second condition. It turns out that the condition cannot be guaranteed for all states $\sigma_{AB}$. More ...


3

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is ...


3

You are asking whether any unitary operator $U:\mathcal H\otimes\mathcal K\to \mathcal H\otimes\mathcal K$ can be written, given $|\psi\rangle\in\mathcal H$ and $|e_0\rangle\in\mathcal K$, as $$U |\psi\rangle|e_0\rangle = \sum_k E_k |\psi\rangle|e_k\rangle,$$ for some $E_k$ and $|e_k\rangle$. The answer is yes, because this essentially just amounts to a ...


3

Examples The condition $[A_a, B_b]=0$ is sufficient, but not necessary for $\Phi$ and $\Psi$ to commute. Indeed, the standard Kraus representations of many commuting pairs of channels have non-commuting Kraus operators, e.g. bit- and phase-flip channel, depolarizing and unitary channel, amplitude damping and phase-flip channel etc. However, Kraus operators ...


2

Assuming $|\phi^+\rangle = |00\rangle+|11\rangle$ and $|\phi^-\rangle = |00\rangle-|11\rangle$ we compute $$ c = \begin{pmatrix} 1 & 0 & 0 & 2\lambda-1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2\lambda-1 & 0 & 0 & 1 \end{pmatrix}\tag1. $$ A useful property of the $d^2\times d^2$ Choi-Jamiołkowski matrix is that ...


2

This is because: $$ \sum_n \langle n | M_m | \psi\rangle \langle \psi | M_m^{\dagger} | n\rangle = \sum_n \langle \psi | M_m^{\dagger} | n\rangle \langle n | M_m | \psi\rangle = \langle \psi | M_m^{\dagger} I M_m | \psi\rangle = \langle \psi | M_m^{\dagger} M_m | \psi\rangle $$ note that $\sum_n|n\rangle\langle n| = I $


2

Let's start by considering specific density matrices $\rho=|i\rangle\langle i|$. This immediately tells you that $$ \langle i|\sum_kE_k^\dagger E_k|i\rangle=1, $$ and hence all diagonal elements of $\sum_kE_k^\dagger E_k$ are 1. Next, consider a more general $\rho$, which we divide into diagonal and off-diagonal components, $$ \rho=\rho_d+\rho_o. $$ We ...


2

For every matrix $A=\begin{pmatrix}a & x-iy\\x+iy & b\end{pmatrix}$(hermitian here) with real number $a,b,x,y$. And $A$ satisfy $Tr(A\rho)$ and $Tr(\rho)=1$, let's consider two by two matrix for example, we can choose $\rho=|0\rangle\langle 0|$ to make sure $Tr(A\rho)=I$. Then $A_{11}$ must be 1. For the same reason we can get $A_{22}=1$. Now $A=\...


2

The extra transformation $U_{\mathrm{env}}$ is equivalent to choosing a new basis for the environment states $$|l\rangle\to U_{\mathrm{env}}^\dagger |l\rangle.$$ This is the same freedom as the freedom to choose the basis for $|l\rangle$; equivalently, your question is whether there is additional freedom between the choices $$\Pi_l=(I\otimes \langle l|)U(I\...


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