# Tag Info

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 $\mathcal{... 8 This question is posed, and answered positively, in Nielsen & Chuang in a subsection of chapter 8 entitled "System-environment models for and operator-sum representation". In my version, it can be found on page 365. Imagine$|\psi\rangle$is an arbitrary pure state on the space upon which you wish to enact the operators. Let$|e_0\rangle$be some fixed ... 8 You cannot always find such a Kraus decomposition. Notice that any CPTP map$\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: $$\mathcal E(\tfrac{1}{d} \mathbf 1) = \tfrac{1}{... 7 Definitions Denoting the Haar measure of some function f\left(x\right) over d-dimensional unitaries as \int_{\mathrm U\left(d\right)}f\left(x\right)d\mu\left(x\right), twirling some arbitrary channel \varepsilon can be defined as the operation \varepsilon \mapsto\int_{\mathrm U\left(d\right)}U^\dagger\varepsilon U dU, which, when \varepsilon is ... 6 Basically, it means that the correlations could be used to send a message. Or simply that Bob’s measurement outcomes can reveal some details of Alice’s actions. This is impossible when Alice and Bob each hold one qubit of a Bell pair. Despite the entanglement present, as well as contextuality, signaling in this case would result faster than light ... 5 A channel \Phi is said to be degradable if there exists another channel \Xi such that \Xi\Phi is complementary to \Phi. The idea here is as follows. Suppose \Phi is a channel and \Psi is complementary to \Phi. If \Phi is applied to a state \rho, then the output of the channel is \Phi(\rho) (of course), while \Psi(\rho) represents ... 5 One way to understand the relationship between the Choi representation of a channel and its possible Kraus representations is to use the vectorization map. Suppose that we have two finite-dimensional Hilbert spaces \mathcal{X} and \mathcal{Y}, and that we have fixed a standard basis \{|1\rangle,\ldots,|n\rangle\} of \mathcal{X} and a standard basis ... 5 If P is (ortho)projector, that is P^2=P=P^\dagger, then we can define unitary U = I - 2P. You can verify$$ 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 There are several ways that you could realise the depolarising map$ \mathcal N_p(\rho) = (1\!-\!p)\:\!\rho + p \!\!\:\cdot\!\tfrac{1}{2}\mathbf 1$map on a quantum computer — including an idealised quantum computer, in which waiting around for the noise to do the work for you would not be an available method.$\def\ket#1{\lvert#1\rangle}$We start ... 5 "Church of the higher hilbert space" is a term coined by John Smolin. According to quantiki it is: for the dilation constructions of channels and states, which [...] provide a neat characterization of the set of permissible quantum operations and to quote wikipedia, it: describe[s] the habit of regarding every mixed state of a quantum system as a pure ... 5 Any map which is not Completely Positive, Trace Preserving (CPTP), is not possible as an "allowed operation" (a more-or-less complete account of how some system transforms) in quantum mechanics, regardless of what states it is meant to act upon. The constraint of maps being CPTP comes from the physics itself. Physical transformations on closed systems are ... 5 Mathematically it is a relationship between a bipartite linear operator vector space$L(X\otimes Y)$and a superoperator vector space$C(X): L(X)\to L(Y)$(maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum ... 5 For the specific linear function you are interested in, the solution turns out to be trivial: you can take the channel to be$N_{X\rightarrow Y}(\rho) = \operatorname{Tr}(\rho) |\psi\rangle\langle \psi|$for$|\psi\rangle$being an eigenvector of$\sigma_Y$having the largest possible eigenvalue. More generally, however, you can optimize any real-valued ... 4 First let me mention a minor point concerning terminology. The type of channel you are suggesting is often called a Pauli channel; the term depolarizing channel usually refers to the case where$p_x = p_y = p_z$. Anyway, it is not really correct to say that Pauli channels are the channel model considered for quantum error correction. Standard quantum error ... 4 Let's recap a bit: In classical information theory, the analogous formula is the Shannon noisy channel coding theorem. It's charming, because it is basically just a very simple optimization of the mutual information. The quantum channel capacity is that it is given by $$\lim\limits_{n\to\infty} \frac{1}{n}Q(T^{\otimes n})$$ where$T$is the quantum ... 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 Let$\mathcal{N}be the channels which subscripts for which conventions. $$\mathcal{N}_{N.C.} (\rho) = \begin{pmatrix} \rho_{00} & \rho_{01} \sqrt{1-\lambda}\\ \rho_{10} \sqrt{1-\lambda} & \rho_{11} \end{pmatrix}$$ As compared to $$\mathcal{N}_{P} (\rho) = \begin{pmatrix} \rho_{00} & \rho_{01} (1-\lambda)\\ \rho_{10} (1-\lambda) & \... 4 If you're told about an operation on a single qubit, then to convert it into an operation on both qubits, you just include the identity matrix on the other qubit. So, contrast the single qubit state |\psi\rangle going through the bit-flip channel$$ |\psi\rangle\rightarrow(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X with what happens on two ... 4 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 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_\... 3 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\... 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 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. 2 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(... 2 The channel\mathcal{E}$is explicitly defined in the preceding paragraph as being the depolarising channel. Thus, all you need to calculate is $$F=\sqrt{\langle 0|\mathcal{E}(|0\rangle\langle 0|)|0\rangle}.$$ 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|+|\...