2017
DOI: 10.1007/s11128-017-1672-1
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Asymptotic entanglement in quantum walks from delocalized initial states

Abstract: We study the entanglement between the internal (spin) and external (position) degrees of freedom of the one-dimensional discrete time quantum walk starting from local and delocalized initial states whose time evolution is driven by Hadamard and Fourier coins. We obtain the dependence of the asymptotic entanglement with the initial dispersion of the state and establish a way to connect the asymptotic entanglement between local and delocalized states. We find out that the delocalization of the state increases th… Show more

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Cited by 21 publications
(25 citation statements)
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“…The initial qubits which lead to the maximal entanglement necessarily also lead to a symmetrical probability distribution regardless of the initial position state [18]. At the same time, while there are only two particular initial qubits over one position (local state) which lead the quantum walk to asymptotically time-evolve to the maximal entanglement condition, when a quantum walk starts from a delocalized state, there are infinite available initial qubits to reach this condition [19]. Figure 1 shows the probability distribution for Hadamard quantum walks starting from local and two Gaussian states after 3000 time steps.…”
Section: Resultsmentioning
confidence: 99%
“…The initial qubits which lead to the maximal entanglement necessarily also lead to a symmetrical probability distribution regardless of the initial position state [18]. At the same time, while there are only two particular initial qubits over one position (local state) which lead the quantum walk to asymptotically time-evolve to the maximal entanglement condition, when a quantum walk starts from a delocalized state, there are infinite available initial qubits to reach this condition [19]. Figure 1 shows the probability distribution for Hadamard quantum walks starting from local and two Gaussian states after 3000 time steps.…”
Section: Resultsmentioning
confidence: 99%
“…Obviously C(k) is symmetric under the interchanging part 1 and 2, so T r 1 (P 0 (k) ⊗ I C(k)) can replaced by T r 2 (I ⊗ P 0 (k) C(k)) in (20).…”
Section: Formalismmentioning
confidence: 99%
“…when we use projector of initial state as P 0 ⊗ I and it equals to summation of all diagonal blocks of C, similarly T r 2 (.) will be applied when we use I ⊗ P 0 in (20) in this case each element of result matrix is the trace of corresponding block. Therefore the action of T r 1 (.)…”
Section: Formalismmentioning
confidence: 99%
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