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I have been reading the book Decoherence and the quantum-to-classical transition by Maximilian Schlosshauer, section 5.3.2

A classical example is the spin-boson model which has the following Hamiltonian (g is the coupling strength) $$H=-\frac{1}{2}\Delta_o \sigma_x+\sum_i \omega_i (a_i^{\dagger}a_i)+\sum_i\sigma_z(g_ia_i^{\dagger}+g_i^* a_i)$$

We solve the master equation under the Born-Markov approximation. The decoherence rate D, as discussed in the book, is given by the following expression $$\nu(\tau)=\int_{0}^{\infty} d\omega J(\omega)coth(\frac{\omega}{2k_bT}) cos(\omega \tau)$$ $$D=\int_{0}^{\infty} d\tau \nu(\tau)cos(\Delta \tau)$$

My question is that it seems that for the calculation of D, if we do the integration on $\tau$ first, the $cos(\omega\tau)cos(\Delta\tau)$ term will produce something like $\delta(\omega-\Delta)$ (assuming $\Delta$ is positive)

This seems that only this single mode with $\omega=\Delta$ is contributing to the decoherence, and it is quite counter-intuitive, because naively I would expect that all the modes should contribute to the decoherence.

How to interpret this result? Does anyone know any reference that discuss this problem?

Tan Tixuan
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1 Answers1

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It is true that the dissipation rate will depend only on the spectral density at the system's frequency, but that does not imply that only the mode at the system's frequency is causing the dissipation. The other modes are essential in order to guarantee that the bath's correlations decay.

Note that this feature that only the spectral density at the resonant frequency is directly relevant happens because of these approximations made, and one of those are that the bath correlation decay faster then the typical time scales of the system, and for that to happen, you need a continuum of modes.

peep
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