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A serious answer: you pretty much can't. It's not that you in particular can't, it's that no one can. Huge companies pour in huge amounts of money to try and make a proof-of-concept quantum computer (there is actually no 'proper' quantum computer yet). A slightly less serious answer: some odd $10-100$ Million would get you started I would say. It all ...

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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 With current technology, there's not much of a chance to build a true quantum computer, but you may be able to build some interesting quantum circuits with a fairly sizable (but still on the scale of "self-funded" for the ordinary person) budget, using the optical photon model. For instance, one could use the linear optical quantum computing model. Using ... 6 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 ... 6 Note that while you probably can't build a quantum computer at home, you can simulate one with a classical computer, at the cost of merely an exponential slowdown. There's a rather long list of available software at https://www.quantiki.org/wiki/list-qc-simulators. 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 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 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 \... 4 Let me quote my answer from over at physics.SE: The intuition Let us consider a channel \mathcal E, which we want to apply to a state \rho. (This could equally well be part of a larger system.) Now consider the following protocol for applying \mathcal E to \rho: Denote the system of \rho by A. Add a maximally entangled state |\omega\rangle=\... 3 I think it's just notational inconsistency. If you look at the page code, the symbols are generated in two different ways: in the text, someone has just inserted the greek letter symbol (presumably a unicode character) whereas in the equation, they've used LaTeX. They're clearly both supposed to be capital phi. 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 Right, they are quite similar. The Holevo bound is a bound on the amount of accessible information between your quantum system and your classical system. The I(X;B) object written in the HSW theorem wikipedia page is actually this bound, while the \chi there is the Holevo rate, or product state capacity. What HSW showed was that if you took many copies of ... 3 It's actually an interesting question. And the previous answers ("strictly speaking, you can't," "you can simulate quantum computers," and "photonics-based processing holds some promise") are all true. According to a pioneer in photonics computing we won't see real quantum computing until around 2035. I've not yet seen anything to justify thinking to the ... 2 I am by no means an expert in this sort of calculation, but I think I (mostly) agree with you. I divided the calculation up slightly differently, which simplified things notationally. Firstly, I considered the input |0\rangle_A|0\rangle_B. Clearly, H acting on this is just 0, so this state doesn't evolve. So, the lop-left element of E_0 is 1, and that ... 2 While the procedure in the existing answer, based on channel-state duality, applies to general channels, there's a more direct way to obtain Kraus operators for this particular case of the depolarising channel, \mathcal{E}(\rho) = (1-p)\rho + p \frac{I}{d}, where d is the Hilbert space dimension. Since \mathrm{tr}(\rho) = 1 for any density operator, we ... 2 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 ... 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 ... 2 It is exactly what you say: A "Choi state" which is only positive semi-definite corresponds to a completely positive map which is not necessarily trace preserving. Fixing the trace of the Choi state rescales the "success probability" when implementing the CP map via the isomorphism - if you wish, it rescales the trace of all output states \mathcal E(\rho) ... 2 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 ... 2 For arbitrary one qubit density matrix we have:$$\rho = \frac{I}{2} + \frac{r_x \sigma_x + r_y \sigma_y + r_z \sigma_z}{2}$$where |r| \le 1. Here we should take into account that \sigma_i \sigma_j \sigma_i = -\sigma_j where i \ne j and i, j \in \{x, y, z\}, and, also \sigma_i\sigma_i=I. With this we will obtain the equality presented in the ... 2 This method is largely similar to Davit's (this covers a slightly more general case where \rho is any arbitrary matrix with trace 1, and you easily see how to adjust it without the trace 1 condition). Any 2\times 2 matrix can be decomposed as aI+\vec{n}\cdot\vec{\sigma} if we allow a and \vec{n} to take on arbitrary complex values. Moreover, two 2\... 2 From my understanding of what you are asking, you may take the product of two depolarization operations, using the reduced density matrix of each qubit in the Bell state in the expression. Let's denote our two qubits as \mathrm{A} and \mathrm{B}. The Bell state of these two qubits is then:$$ |\beta_{00} \rangle =\frac{|0 \rangle_\mathrm{A} \otimes |0 \...

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Adding my comment as an answer: I think this is a typo. The equation contains a scalar, namely $S(\cdot)$ and an operator, the output of the channel $\mathcal{E}(\cdot)$ in linear combination (their dimensions do not match). Therefore, this has to be a typo (there are no hidden Identity operators in the scalar terms as is easy to check).

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Ok, so starting with Trace Preserving, since it's easier: $$Tr(I/2) = 1$$ $$Tr(\rho) = 1$$ $$Tr((1-\lambda)I/2 + \lambda\rho) = (1-\lambda)Tr(I/2) + \lambda Tr(\rho) = 1$$ Now for a map to be completely positive, it must take positive elements to positive elements. So since depolarizing noise is essentially just adding in a bit of the identity operator (who'...

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Here's an alternative proof: first note that any quantum map, $\Phi(\rho) \mapsto \sigma$ that can be written in the Kraus form, that is, as $\Phi(\cdot) = \sum_{j} K_j (\cdot) K_j^\dagger$, with, $K_j^\dagger K_j \geq 0, \sum_j K_j^\dagger K_j = \mathbb{I}$ is a CP map (see for example, Nielsen and Chuang, or Page 26 of https://arxiv.org/abs/1902.00967). ...

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This is a special instance of a general linear algebra result. Note that the identity matrix $I$ can be decomposed as $I=\sum_k v_k\otimes v_k^*$ for any orthonormal basis $\{v_k\}_k$, and vice versa any such decomposition identifies the identity matrix. Now notice that the Pauli matrices are an orthonormal basis in an enlarged Hilbert space, meaning that $... 1 Chapter VII. E. in Daniel Lidar's notes. Use$\rho = \frac{1}{2}(I + \vec{v}\cdot\vec{\sigma})$and products of Pauli matrices: Check for each pair that:$\sigma_i \sigma_j = \delta_{ij} I + i \epsilon_{ijk}\sigma_k$Use it to show:$ \sigma_i \sigma_j \sigma_k = \delta_{ij} \sigma_k - \delta_{ik} \sigma_j + \delta_{jk} \sigma_i + i \epsilon_{ijk} I \$ One ...

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