10

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{...


8

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}{...


7

Definitions Denoting the Haar measure of some function $f\left(x\right)$ over $d$-dimensional unitaries as $\int_{\mathrm U\left(d\right)}f\left(x\right)d\mu\left(x\right)$, twirling some arbitrary channel $\varepsilon$ can be defined as the operation $\varepsilon \mapsto\int_{\mathrm U\left(d\right)}U^\dagger\varepsilon U dU$, which, when $\varepsilon$ is ...


5

A channel $\Phi$ is said to be degradable if there exists another channel $\Xi$ such that $\Xi\Phi$ is complementary to $\Phi$. The idea here is as follows. Suppose $\Phi$ is a channel and $\Psi$ is complementary to $\Phi$. If $\Phi$ is applied to a state $\rho$, then the output of the channel is $\Phi(\rho)$ (of course), while $\Psi(\rho)$ represents ...


5

Mathematically it is a relationship between a bipartite linear operator vector space $L(X\otimes Y)$ and a superoperator vector space $C(X): L(X)\to L(Y)$ (maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum ...


5

If $P$ is (ortho)projector, that is $P^2=P=P^\dagger$, then we can define unitary $U = I - 2P$. You can verify $$ UU^\dagger = U^2 = (I-2P)(I-2P) = I-4P+4P = I $$ Now we can express $P=\frac{1}{2}(I-U), Q=\frac{1}{2}(I+U)$ and calculate $$ P\rho P + Q\rho Q = \frac{1}{4}(I-U)\rho(I-U) + \frac{1}{4}(I+U)\rho(I+U)= $$ $$ = \frac{1}{4}(\rho - U\rho - \rho U +...


5

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 $...


4

If you're told about an operation on a single qubit, then to convert it into an operation on both qubits, you just include the identity matrix on the other qubit. So, contrast the single qubit state $|\psi\rangle$ going through the bit-flip channel $$ |\psi\rangle\rightarrow(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X $$ with what happens on two ...


4

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 ...


4

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 ...


3

The first and foremost thing to realize is that the partial trace over a density matrix is indeed a linear CPTP map $\Lambda$, but it is not a map from any $\mathcal{C}^{n\times n} \rightarrow \mathcal{C}^{n\times n}$ (i.e. to `itself' - the same dimension), but rather to a density operator space with a lower dimension: $\mathcal{C}^{n\times n} \rightarrow \...


3

Josu, You might be mis-understanding something. Your equation for the Pauli channel says that when $t\rightarrow \infty$, all operators ($X,Y,Z,I$) have an equal probability (1/4) of transforming the density matrix $\rho$. You seem to be suggesting that $t\rightarrow \infty$, the probability of the $I$ operator acting on $\rho$ whould go to 0. Keep in mind ...


3

Quantum measurement (without results recording) is just a special case of quantum operation (quantum channel). So, yes, measurement operators (as in general measurement formalism) are indeed Kraus operators. But Kraus operators are more general. For example, they can be "rectangular", while measurement operators can't.


3

These are not really the definitions of classical and quantum capacity, as I will explain. Before doing that, let me adjust the notation being used slightly: let $\Phi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Y})$ be the channel whose capacities we are interested in and let $\Psi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Z})$ be a ...


3

I'll cover a slightly more general case. Let $P_k$, $k=1,...,N$ a complete set of orthogonal projectors: $\sum_k P_k=I$ and $P_j P_k=\delta_{jk}P_j$. Consider the map $\mathcal E(\rho)=\sum_k P_k \rho P_k$. We want to find a set of unitaries $\mathcal U_k$ and probabilities $p_k$ such that $\mathcal E(\rho)=\sum_\ell p_\ell\mathcal U_\ell\rho\,\mathcal U_\...


2

The general expression for the fidelity is $$ F(\rho,\sigma)=\left(\text{Tr}\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}\right)^2=(\text{Tr}|\sqrt{\rho\sigma}|)^2. $$ Assume $\rho$ and $\sigma$ are $2\times 2$ matrices. Then $\sqrt{\rho\sigma}$ is also a $2\times 2$ matrix which we assume to have eigenvalues $\lambda_1$ and $\lambda_2$. Thus, $$ F=(|\lambda_1|+|\...


2

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 ...


2

Suppose that $\mathsf{X}$ is a register that can store each possible choice for $x$, as a classical state, while $\mathsf{Y}$ is a register that can store each possible state $\rho_x$. It is then natural to associate the classical-quantum state $$ \rho = \sum_x p_x |x\rangle \langle x| \otimes \rho_x $$ with the ensemble $\{(p_x,\rho_x)\}$. Now try taking ...


2

In my understanding, the key part for entanglement to increase capacity is to have a suboptimal channel. Suppose the input of you channel can take value in the set $X$, and note $G(X)$ the graph where nodes are possible inputs and there is an edge between to inputs if the range of their corresponding output is overlapping, i.e. When going through the channel ...


2

I will try to succinctly answer your first question given that I possess little knowledge regarding entanglement assistance. Shannon's capacity theorem (the noisy channel coding theorem) states that for any code $\mathbf{C}$ of code rate $R \leq C$, where $C$ represents the channel capacity, an encoding and decoding scheme of rate $R$ with a probability of ...


2

The form (A) above is known as a separable superoperator. The effect of any LOCC protocol can be described as a separable superoperator, or as a separable POVM with POVM elements $N_i = A_i\otimes B_i$. This can be seen as follows (adapted from this answer - I will focus on the POVM case, the superoperator is obtained by ignoring the final outcome of the ...


2

When I was doing some work on quantum cloning (so, slightly different applications to the channels you were asking about), I basically ended up setting up the Choi matrix as a description of the action that I'd like to achieve (perfect cloning), averaged over all possible inputs. In that case, the maximum eigenvalue tells you quite a lot - up to a scale ...


1

Please be careful with your notation; don't get confused with the number of qubits that are in $|0_{Eve}\rangle$ (I'm not necessarily saying that you are - just mentioning this as a precaution). Since the state $|\psi\rangle = \frac{1}{\sqrt{N}}\sum_{j \in 0,1,2....N-1}|j\rangle$ that Alice wants to send is a $n$-qubit state with $N=2^{n}$, Alice needs to ...


1

It does exist. It's basically the controlled-not gate generalised to higher dimensional systems. The important thing to realise is that means that the bit that Bob ends up with will be highly entangled with what Eve has, and that will have significant observable consequences. As part of a cryptographic protocol, for exampe, Bob could detect that Eve is ...


1

The basic answer is to just write, for a given bipartite state $X$, $\Phi(X)\equiv\operatorname{Tr}_B(X)$. This defines a quantum channel which acts on the input as a partial trace. You can check that this does indeed qualify as a quantum channel etc. A misconception that might have led you to ask this question can be found in your saying that "I have no ...


1

Let's say $|a\rangle$ is the pure state qubit and $|b\rangle$ is the bell state. Putting $|a\rangle$ through the Pauli channel depends on which of the Pauli gates (which of the $\sigma_j$) you want acting on it. Once you choose this (call it $\sigma_a$), apply $2\times2$ Identity operators to it on either side until it's the same size as $\rho$, then just ...


1

This probably only addresses half of your question... Most of the time, estimates of the error probability are not so important. If you're using a non-degenerate error correcting code for a single logical qubit, all you need to know (effectively) is enough to resolve a choice between two different error corrections (i.e. you want to know that the number of ...


1

The answer is yes, you need $d_1 d_2$ operators, as already pointed out in the other answer. Here I'll show explicitly why this is the case. Let $\Phi\in\mathrm{T}(\mathcal X,\mathcal Y)$ be a CPTP operator sending maps in $\mathcal X$ into maps in $\mathcal Y$. The spaces $\mathcal X$ and $\mathcal Y$ have dimensions $d_1$ and $d_2$, respectively. Short ...


1

I will give here another way to prove explicitly the equivalence of the two expressions. From the Kraus representation, $\Phi(\rho)=\sum_a A^a \rho A^{a\dagger}$, making the indices explicit, we get $$\Phi(\rho)_{ij}=\sum_{a,k,\ell}A^a_{ik}A^{a*}_{j\ell}\rho_{k\ell}.\tag1$$ On the other hand, unravelling the second expression we have for a generic $\sigma$ ...


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