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15

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 $...


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 ...


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

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

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)...


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|...


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

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

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

Since it hasn't been mentioned so far, and I think it's an interesting aspect: A weighted ensemble $(p_i,U_i)$ of unitaries in $U(d)$ such that $$ \sum_i p_i U_i X U_i^\dagger = \operatorname{tr}(X) \mathbb{I}/d, $$ is called a weighted unitary 1-design. If the weights can be chosen uniformly, i.e. $p_i \equiv 1/N$ where $N$ is the size of the ensemble, this ...


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 $$(...


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

As an alternative to the other answer (which is perfectly valid), we can use the formalism specified in the question. We're given a $U=P_0\otimes I+P_1\otimes X$. We know that the environment starts in state $|e_0\rangle=|0\rangle$. We need to establish an orthonormal basis for the environment, $|e_k\rangle$. In this case, it's straightforward as the ...


3

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 ...


3

QETLAB usually deals with channels as Choi operators. You can convert your Kraus operators to the Choi matrix by providing the Kraus operators as a cell array. Example with the amplitude damping channel below. >> damp = 0.3; >> K = { diag([1,sqrt(1-damp)]); [0,sqrt(damp);0,0] }; >> ChoiMatrix(K) ans = 1.0000 0 0 0....


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

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 \...


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