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}{...


5

Not exactly sure what you find confusing, but the ultimate need for Stinespring dilation theorem is that in quantum mechanics the dynamics is in general defined by a completely positive trace preserving map (CPTP) $\rho \mapsto \Lambda(\rho)$. Now, we have a belief (rightly or wrongly) that all there is is a unitary evolution governed by Schrodinger's ...


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 $...


5

This is not the unitary that you have to implement: you need a two-qubit unitary $$ \frac{1}{\sqrt{3}}\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 & \omega^2 & \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \end{array}\right), $$ where $\omega=e^{2i\pi/3}$, the point being that if you introduce ...


4

In the paper that you refer to, they are essentially asking "when can we implement the partial transpose map $\Theta=I_2\otimes\Lambda$?". So, that means the SPA of this map must be positive. What you have calculated, by comparison, is to ask when the SPA of the transpose map $\Lambda$ can be made positive. It might sound like these ought to be the same ...


4

The partial transpose is not the only positive but not completely positive operation that is possible on 2x2 and 2x3 systems. Trivially, any completely positive operation (such as a local unitary) combined with the partial transpose is a different positive operation. The point is that, as wikipedia puts it every such map $\Lambda$ can be written as $...


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 ...


4

The existing answers have quite an elegant idea behind them. However, my concern is that they don't seem to allow for the introduction of an ancilla. If we introduced an ancilla in a fixed state, and applied a unitary across the two systems, then it could be that all the eigenvectors are entangled across the two subsystems, and the basic argument wouldn't ...


4

In general, "operating on a state with an observable" does not have direct physical meaning (i.e. you cannot think of it as evolving the state doing something to it). What does have physical meaning, is applying a unitary operation to a state. Every unitary operator corresponds to a physical operation that you can (in principle) implement, transforming (...


3

Unitary operations are a subset of CPTP operations. You can think of a CPTP operation as the description of a unitary over a larger system. The advantage of using CPTP maps is that you increase the generality of your statement. Think about, for example, a proof of the no-cloning theorem. People usually start talking about the input state, the target state, ...


3

For amplitude damping, $\gamma$ is something like $e^{-\Delta t/T_1}$ where $\Delta t$ is how long the Kraus operator is supposed to act. But be very careful, Kraus evolution assumes your system has no initial correlations, that every qubit interacts with identical baths and that every qubit is identical. All the assumptions are most likely violated and so ...


3

The state $\mid \psi \rangle$ is fixed. You can write it as $a |0 \rangle + b |1 \rangle$. If you write that in the other basis and get $ (a,b) = (1,0) \implies (c,d) = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) $ They are the same state so it is still true that if you do a measurement in the $|0 \rangle$, $|1 \rangle$ basis, you will surely get $\mid 0 \...


3

Those maps are linear, so if it preserves the trace of trace 1 matrices then it preserves the trace of any other matrix. We just don't need to restrict ourselves by considering only trace 1 matrices (from mathematical point of view).


3

I agree with the main points that Niel makes: not all operators are observables, and the purpose of the ones you list is typically to transform states, not to be measured. However, the operators you list happen to be hermitian (allowing them also to represent observables) as well as unitary (allowing them to represent transformations), so in this case we ...


3

In quantum mechanics, not all operators are observables. Many operators are observables, and in the first year or two of treatments in some physics courses you will only see operators which are observables; but not all operators of interest are observables. The operators you have mentioned all happen to be Hermitian, and could therefore be interpreted as ...


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

This gate, which I'll denote as $U$ is essentially the square root of not gate, whose decomposition has already been discussed elsewhere. Hence, we only need the conversion: Specifically, if we perform the corresponding matrix multiplications, we have $$ \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & ...


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

This is easiest to show by contradiction, so lets suppose that there exists an operation TransformToOrthogonal that maps the state of its input qubits to an orthogonal state. Reversible quantum operations can be simulated using unitary operators, so we can simulate TransformToOrthogonal by a matrix $U$ such that $UU^\dagger = 𝟙$. Since $U$ is unitary, $U$ ...


3

Every unitary matrix can be diagonalized as $U=S^\dagger \Lambda S$, where $S$ is unitary and $\Lambda$ is diagonal and contains the eigenvalues ($\lambda_i$) of $U$. So you can consider instead $$\langle x \vert U \vert x \rangle=\langle x \vert S^\dagger \Lambda S \vert x \rangle = 0.$$ Defining $\vert \tilde x \rangle \equiv S \vert x \rangle$, you only ...


3

This is a neat idea. However, having individual overlaps being small isn't sufficient in a quantum system. For example, imagine the set of overlaps $$ \langle V_1|V_N\rangle=0,\qquad \langle V_1|V_n\rangle=\langle V_N|V_n\rangle=\epsilon \quad \forall n=2,\ldots,N-1, \langle V_n|V_m\rangle=0 \quad \forall n,m=2,\ldots,N-2 $$ Remember that these are supposed ...


2

First of all, $\|A|\psi\rangle\|^2 = (A|\psi\rangle)^\dagger(A|\psi\rangle) = \langle \psi | A^\dagger A|\psi\rangle = 1$. To see why your last expression equals to $1$ note that $$ \langle\psi\rvert \mu(a)\rvert\psi\rangle = \sum_{k} \lambda^a_k \lvert\langle\lambda^a_k\rvert\psi\rangle\rvert^2 ~~\text{and}~~ \sum_a \mu(a) = \mathbb 1 $$ So, it's ...


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

This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as. Just as an example of what I mean, let $$ W=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & ...


2

Check this $$\frac{1}{\sqrt{3}}\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \\ \end{array}\right) = \left(\begin{array}{cc} H & 0 \\ 0 & 1 \\ \end{array}\right) \cdot \frac{1}{\sqrt{3}}\left(\begin{array}{ccc} \sqrt{2} & 0 & 1 \\ 0 & \sqrt{3} & 0 \\ 1 & 0 & -\sqrt{...


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 ...


2

An isometry $U:S\rightarrow S\otimes E$, where $S$ is your system and $E$ is an environment, such that $$\mathrm{Tr}_E(U\rho U^\dagger)=\Phi(\rho)$$ is called a Stinespring dilation of $\Phi$. An easy way to construct a Stinespring dilation from the Kraus operators is to consider $\mathcal{H}_E=\mathrm{span}\{|j\rangle\}_{j=1}^r$ and $$U=\sum_j F_j^S\...


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