Search Results

It suffices to prove that if $P$ and $Q$ are positive semidefinite operators, then $$ \operatorname{im}(P) \subseteq \operatorname{im}(P+Q). $$ Once you have this, the statement follows by taking $P = …

First let me mention a minor point concerning terminology. The type of channel you are suggesting is often called a Pauli channel; the term depolarizing channel usually refers to the case where $p_x = …

For the specific linear function you are interested in, the solution turns out to be trivial: you can take the channel to be $N_{X\rightarrow Y}(\rho) = \operatorname{Tr}(\rho) |\psi\rangle\langle \ps …

Here's the simplest argument I could come up with for the statements Rammus made about unital channels. We actually don't need complete positivity, just positivity is enough. Suppose that $T$ is a pos …

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 th …

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 …

No, not necessarily. For example, take $\rho$ to be a GHZ state and let $\sigma$ be the completely mixed state of one qubit. We then have $\lambda=4$ and $\mu=2$.

Let us first suppose more generally that we have a map defined as $$ \Psi(X) = \sum_k \langle B_k, X \rangle A_k $$ for all $X$. The adjoint mapping $\Psi^{\ast}$ must satisfy \begin{multline} \langle …

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: $$ \l …

Yes, the channel $\tilde{N}$ necessarily exists. Notice first that the state $\rho_B$ is the completely mixed state $\mathbb{1}/d$. So, in order for $\tilde{\rho}_{A'B}$ to be contained in $S$, three …

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 natu …

The inequality can be proved as follows. First, let $P$ and $Q$ be any two positive semidefinite operators, and consider the operator $\Psi(P - Q)$. Because $\Psi$ is positive, this is a Hermitian ope …

It appears that the statement is not true in general. Suppose $X = Y = \{0,1\}$, $\mathcal{H}$ is the Hilbert space corresponding to a single qubit, and $W$ is defined as \begin{align} W(0,0) & = | 0 …

No, such a dilation might not exist. For a counter-example that can easily be generalized, let $\Phi$ be any non-isometric channel from $A$ to $B$, let $n=2$, and let $N = \Phi\otimes \Phi$. By a non- …

No, not necessarily. For example, the channel $\Phi(\rho) = \operatorname{Tr}(\rho) \vert 0 \rangle \langle 0 \vert$ makes $\{\Phi(\sigma_0), \ldots, \Phi(\sigma_3)\}$ linearly dependent. (In fact, th …