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Symmetry and Exact Solutions of General Spin-Boson Models

Yifan Sun, Lian-Ao Wu

Abstract

Spin-boson models are the canonical benchmark for quantum dissipation. We show the symmetry structure of general spin-boson Hamiltonians and obtain their spectra explicitly by exploiting the symmetry. As an illustration of the general case, we numerically demonstrate the exact solution for the two-mode case.

Symmetry and Exact Solutions of General Spin-Boson Models

Abstract

Spin-boson models are the canonical benchmark for quantum dissipation. We show the symmetry structure of general spin-boson Hamiltonians and obtain their spectra explicitly by exploiting the symmetry. As an illustration of the general case, we numerically demonstrate the exact solution for the two-mode case.
Paper Structure (4 sections, 62 equations, 3 figures)

This paper contains 4 sections, 62 equations, 3 figures.

Figures (3)

  • Figure 1: A schematic illustration of how would the state evolves under Hamiltonian \ref{['eq:diago']}. The energy of the system depends on the parities of both the spin and the bosonic modes. If the number of bosons changes, for example, when the system is excited from an odd-boson-number (lower-energy) state to an even-boson-number (higher-energy) state, it could later relax by flipping the spin parity from positive to negative.
  • Figure 2: The behaviour of $G_2^{+}$ (dotted blue curve) and $G_2^{-}$ (dotted Green curve) when $X\in[-0.1,3.1]$. The parameters are $\omega_1=1,~w_2=0.92$, $g_1=0.7,~g_2=0.78$, and $\Delta=0.68$. The vertical dashed rel lines marks the poles. They are given by $X=n_1\omega_1+n_2\omega_2$, and marked by $(n_1,n_2)$ in the upper place. The line $y=0$ is also marked out.
  • Figure 3: Landscape of the first energy level when $g_1,g_2\in[0,0.4]$. (a) The surfaces of the positive and the negative sectors. (b) The plot of the $g_2=0$ edge in (a). (c) The plot of diagonal line ($g_1=g_2$) in (a). The dashed lines in (a) mark out the position for obtains the curves in (b) and (c), colored accordingly. $\omega_1=1$; $\omega_2=0.92$.