16

One way to understand the relationship between the Choi representation of a channel and its possible Kraus representations is to use the vectorization map. Suppose that we have two finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{Y}$, and that we have fixed a standard basis $\{|1\rangle,\ldots,|n\rangle\}$ of $\mathcal{X}$ and a standard basis $...


12

Let's start by finding a complementary channel for any channel given by a Kraus representation $$ \Phi(X) = \sum_{k=1}^n A_k X A_k^{\dagger}. $$ To make the necessary equations clear, let us assume that the channel has the form $\Phi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Y})$ for finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{...


10

Matrix inequalities of the form $A\ge B$ should be read as $$ A-B\ge 0\ , $$ which in turn means that all eigenvalues of $A-B$ are larger or equal than zero. In the given case, $M\le I$ means that all eigenvalues of $M$ are smaller or equal than one. (Note that this convention for $\ge$ used on matrices depends on the field. In other fields, "$\ge0$" ...


10

This question is posed, and answered positively, in Nielsen & Chuang in a subsection of chapter 8 entitled "System-environment models for and operator-sum representation". In my version, it can be found on page 365. Imagine $|\psi\rangle$ is an arbitrary pure state on the space upon which you wish to enact the operators. Let $|e_0\rangle$ be some fixed ...


9

You cannot always find such a Kraus decomposition. Notice that any CPTP map $\mathcal E$ which does have a decomposition as a probabilistic mixture unitaries is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state: $$ \mathcal E(\tfrac{1}{d} \mathbf 1) = \tfrac{1}{...


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 ...


7

This really depends where you want to start from. For instance, you can construct the Choi state of $\mathcal E$, i.e., $$ \sigma = (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle\Omega|)\ , $$ with $\Omega = \tfrac{1}{\sqrt{D}}\sum_{i=1}^D |i,i\rangle$, and then extract the Kraus operators of $\mathcal E(\rho)=\sum M_i\rho M_i^\dagger$ by taking any ...


7

There is an ambiguity in the choice of Kraus operators: If $\{E_a\}$ is a set of Kraus operators for a channel $\mathcal E$, so is $\{F_b\}$ with $F_b=\sum_a v_{ab} E_a$, with $(v_{ab})$ an isometry. In particular, you can choose a $(v)$ which diagonalizes the matrix $X_{ac}=\mathrm{tr}[E_a^\dagger E_b]$, in which case $\{F_b\}$ satisfies your ...


7

No, this is not always possible. A counterexample is given by $\sigma=I/d'$ and $\Phi(X)=\mathrm{tr}(X)|0\rangle\langle0|$. To see this, note that for $X=I/d$, \begin{align} 2(1-1/d) & = \|\,|0\rangle\langle0|-I/d\|_1 \\ &= \|\Phi(X)-I/d\|_1 \\ &\le \left\|U\left(X\otimes \frac{I}{d'}\right)U^\dagger-U\left(\frac{I}{d}\otimes\frac{I}{d'}\right)...


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} (\...


6

Summary Below, we prove that $$ \|\hat\Phi\|_{\rm op}=\sup_{\rho\in D(\mathcal{X})}\sqrt\frac{\gamma(\Phi(\rho))}{\gamma(\rho)} $$ where $D(\mathcal{X})$ denotes the set of density matrices on the Hilbert space $\mathcal{X}$ associated with the input of the quantum channel $\Phi$ and $\gamma(\rho)=\mathrm{tr}(\rho^2)$ is the purity of $\rho$. This justifies ...


5

So, let the system be $\rho$, and the environment $|0\rangle \langle 0|$. The given operation (which you can check is unitary, and incidentally happens to be the CNOT operation), is applied on $\rho \otimes |0\rangle \langle 0|$. So, you have: $$(P_0 \otimes I + P_1 \otimes X)(\rho \otimes |0\rangle \langle 0|)(P_0 \otimes I + P_1 \otimes X)$$ Now, you ...


5

From N&C: Assuming the environment is in some pure state we recall that Kraus representations comes from the unitary evolution $$\sum_{k}E_k\rho E_k^*=\sum_k \langle e_k |U\left(\rho\otimes|e_0\rangle \langle e_0|\right)U^*|e_k\rangle$$ for some unitary $U$. So we need a unitary such that $E_k=\langle e_k|U|e_0\rangle$. We can pick $$U=\begin{pmatrix} ...


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

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

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

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 ...


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

I think it helps here to write things explicitly. Suppose $\mathcal E(\rho)=\operatorname{Tr}_E[U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]$. Pick a basis for the environment in which $|\mathbf e_0\rangle$ is the first element. Note that here $U$ is a unitary matrix in a bipartite system. The operator before taking the partial trace ...


4

As matrices, the natural representation and Choi representation of a map $\Phi$ have exactly the same entries, but arranged into matrices in different ways. One way to express this is like this: $$ \langle k,\ell | K(\Phi) | i,j \rangle = \langle k | \Phi(|i\rangle\langle j|) | \ell \rangle = \langle k,i | J(\Phi) | \ell,j \rangle, $$ where $K(\Phi)$ denotes ...


4

Let's start with a general state $$ \rho\otimes\rho_0=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |0\rangle\langle 0|. $$ If we apply the controlled-not, we have $$ \rightarrow\rho_{\text{final}}=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |x\rangle\langle y|. $$ Now we want to take the partial trace over ...


4

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\...


4

You can obtain the Kraus operators of the combined channel by taking products of the Kraus operators of the individual channels (using the notation from the paper you linked): Amplitude damping: $E^{AD}_1 = \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-p_{AD}} \end{bmatrix}$, $E^{AD}_2 = \begin{bmatrix} 0 & \sqrt{p_{AD}} \\ 0 & 0 \end{bmatrix}$ Phase ...


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

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

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

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 ...


3

It's true for any matrix $A$ that $A^\dagger A\ge 0$. It's because $(A^\dagger A v,v)=(Av, Av)$, where $(,)$ is the inner product and $v$ is any vector.


3

As pointed out in a comment, what you wrote as $\rho$ should more precisely be written as $\rho\otimes\mathbb 1$ (although the Kraus operator decomposition can be obtained similarly with any initial ancilla state, in which case you have $\rho\otimes|\phi\rangle\!\langle\phi|$). The standard algebraic properties of tensor product spaces then tell you that $$(...


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