Quantum Transport in Presence of Bound States – Noise Power

Abstract

International audienceThe impact of bound states in Landauer-Buttiker scattering approach tonon-equilibrium quantum transport is investigated. We show that the noise powerat frequency $\nu$ is sensitive to all bound states with energies $\omega_b$satisfying $|\omega_b| < \nu$. We derive the exact expression of the boundstate contribution and compare it to the one produced by the scattering statesalone. It turns out that the bound states lead to specific modifications ofboth space and frequency dependence of the total noise power. The theoreticaland experimental consequences of this result are discussed

Microscopic Features of Bosonic Quantum Transport and Entropy Production

“…If bound states are present, the solution involves an additional term established in ref. []. As explained there, this term contributes to the correlation functions and , but not to their zero frequency limits and , we are focusing on in this paper.…”

“…The LB framework has been further generalized in refs. [] and finds nowadays various applications, ranging from the computation of the noise power to the full counting statistics . Most of the quoted studies have been performed for fermionic systems.…”

Entropy Production in Systems with Spontaneously Broken Time-Reversal

“…One can easily verify that this scattering matrix is unitary S(k)S * (k) = I and satisfies Hermitian analyticity S * (k) = S(−k). The dynamics fixed by (18)(19)(20) is invariant under time-reversal if and only if U and consequently S(k) are symmetric.…”

“…We will show below that the above system, called in what follows Schrödinger junction, nicely illustrates the program from the previous section and works simultaneously for both Fermi (+) and Bose (−) statistics. For simplicity we assume in what follows that S(k) has no bound states, referring for the general case to [19]. Then, the general solution of (18)(19)(20) is given by…”

“…For simplicity we assume in what follows that S(k) has no bound states, referring for the general case to [19]. Then, the general solution of (18)(19)(20) is given by…”