# Tag Info

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As long as your final state is a product state, everything is a product, and the probabilities for the individual qubits will just multiply. So compute the probabilities for each qubit to be in the respective state and just multiply them. But even if this is not the case, you can compute $\langle \psi\vert D_p^{\otimes k}\vert\psi\rangle$ rather than the ...

3

Noise effects introduce classical uncertainty in what the underlying state is. A mixed state is a statistical ensemble of several quantum states $|\psi_i\rangle$ (not necessarily orthogonal), with respective probabilities $p_i$. With the state vector you can represent pure states, not mixed ones. Instead, with the density operator you can represent both pure ...

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To be more specific and as an addendum to Drito's answer which provides a good starting point for your search, I would like to narrow it down for you, by recommending Quantum Information Processing. Since this journal is good, and moreover it has papers regarding Image Processing related topics like its Representation, security.

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Norbert's answer is correct, but just for the sake of being explicit: $$\langle 0|^{\otimes N} D_P^{\otimes N}|0\rangle^{\otimes N}=\left(\langle 0|D_p|0\rangle\right)^{\otimes N}=\langle 0|D_p|0\rangle^{N}=\frac{1}{2^N}.$$

2

You're missing a bit of algebraic trickery. Remember that $\frac{1}{\sqrt{2}}=\sin(\pi/4)=\cos(\pi/4)$. Thus, $$\cos(\pi/8)/\sqrt{2}+\sin(\pi/8)/\sqrt{2}=\cos(\pi/8)\cos(\pi/4)+\sin(\pi/8)\sin(\pi/4)=\cos(\pi/4-\pi/8)=\cos(\pi/8)$$ by the double angle formula. Also, be careful of signs. It might be an amplitude is $\pm\sin(\pi/8)$, but when you take the ...

2

No, there's not a lot you can say. Consider these two cases, both with $\epsilon=0$. First, the obvious one, $\rho=\sigma=|\psi\rangle\langle\psi|\otimes |\psi\rangle\langle\psi|$. Clearly $\rho_r-\sigma_r=0$. Second, let $|\psi^\perp\rangle$ be orthogonal to $|\psi\rangle$. You can have $$\rho=(|\psi\rangle\langle\psi|\otimes |\psi^\perp\rangle\langle\psi^\... 2 Apparently, the specific question posed here has been answered in the affirmative--at least (first, we point out) through numerical means--by Arsen Khvedelidze and Ilya Rogojin in Table 2 of their 2018 paper, "On the Generation of Random Ensembles of Qubits and Qutrits: Computing Separability Probabilities for Fixed Rank States" ArsenIlya They ... 2 Remember that the partial transpose condition is generally good for detecting entanglement, i.e. a bipartite state \rho is certainly entangled if the partial transpose is not non-negative. In other words, if there exists a state |\psi\rangle such that$$ \langle\psi|I\otimes\text{T}(\rho)|\psi\rangle<0, $$then the state is certainly entangled. If you ... 2 This is a good list made by Prof. Rod Van Meter. Maybe you could look at the papers you based your work off of and see where they were published? That should give you a sense of which journals are interested in the sort of work you've done. 2 As @rnva points out these are not the same quantities. To give some clarity as to why they are both referred to as D_{\min} it is best to look at the as limiting cases of \alpha-R'enyi divergences. First, we have the sandwiched divergences which for \alpha \in (0, 1) \cup (1, \infty) are defined as$$ \widetilde{D}_{\alpha}(\rho\|\sigma) = \frac{1}{\...

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If it holds for hermitian matrices, it holds for all matrices due to linearity: Over $\mathbb C$, the hermitian matrices span the full matrix space.

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The reason for this viewpoint on measurement is primarily historical. Physicists often think of measurements in terms of observables (aka hermitian operators). The way this is related to the more mathematical notion of projective measurements (a type of POVM measurement) is by thinking about the spectral decomposition of a hermitian operator M=\sum_i \...

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You do not need to use the density matrix approach. However, as the most general representation of a quantum state, doing so has several advantages. You can simulate noise using just statevectors using probabilistic approaches, eg wavefunction monte-carlo, that converge to the density matrix results in the limit of many repetitions. Along this same thread of ...

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This problem can be approached without regards to Kraus representations (even if the motivation is to prove the convexity of entropy) or whether A is a normal matrix or not. Rather, this is a feature of the choice of $\{ U_{j} \}$. In particular, there exists a choice such that their action is to coarse-grain'' all the information in a state. Here's a ...

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This is a very interesting question. Indeed, CP maps - and this includes the operations used in the error correction (measurement and subsequent unitaries) - will always decrease the trace norm. The answer is that if you take a (strictly) contractive map on, say, a qubit, and consider how it acts if you apply it to many qubits, there will always be some ...

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Like many ideas in quantum information theory, I think this is best understood using a $2$-party communication scenario. Suppose Alice has a classical random variable, $X$ which can take values $1,2, \cdots, k$ with probabilities $p_{X}(1), p_{X}(2), \cdots, p_{X}(k)$. Alice then encodes this information by encoding the classical index $j$ in the state $\rho^... 1 Some thoughts: A theoretical perspective From a theoretical perspective, the depolarizing channel is the 'standard' (if there is such a thing) or by some means the most applicable. Because the Paulis (together with the identity operator) form a basis for$SU(2)$, if a code can correct the$X, Y$and$Z$flips on a certain qubit (and it it is able to ... 1 Indeed, the PLOB bound is an ultimate upper bound for repeaterless quantum communications, and is thus derived by averaging over$n\rightarrow \infty$uses of the communications channel - Hence why the capacity is given as bits per channel use. If you wish to more closely compare the PLOB bound to other well known protocols, look at Supplementary Note 6 in ... 1 Classical information is stored the same way as it is for classical computers. You cannot use quantum states to store (in a retrievable way) more information that you would with classical bits. 1 You can represent the "A" as a series of qubits in the$\vert 01000001\rangle$-state. Any other representation would work as well, as long as there is consensus on the notation. Regarding usefulness, most envisioned applications of quantum computers are as subpart of classical algorithms, where computationally hard problems are solved. Hence, '... 1 The first answer discusses what the pseudothreshold is and how to find it, but I will try to give a few details on the difference between thresholds and pseudothresholds, since your first question does ask for both definitions. In quantum error correction (QEC), a logical qubit is encoded in many physical qubits. Given some underlying physical error rate$p$... 1 So the general definition of pseudothreshold is when the logical qubit outperforms a physical qubit. If your error model is idling error, for example, you want to find the physical value of T1 and T2 for which the physical qubits lifetime is lower than the logical qubit's lifetime. The easiest way to calculate this would be to write a simulation of the ... 1 (General result) The main thing to keep in mind is that this is a result about a type of channel, not about specific states. Suppose$\operatorname{tr}(U_i U_j^\dagger)=\delta_{ij}$for some set of matrices$U_i$. This is equivalent to$\sum_{k\ell}(U_i)_{k\ell} (U_j^*)_{k\ell}=\delta_{ij}$. If$U_i$form a basis (i.e. there are$n^2\$ of them), then we must ...

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