6

I want to ask what is the difference between the three types of bosonic reservoirs that we use in the theory of quantum decoherence: sub-ohmic, ohmic and super-ohmic. I know that there is a parameter "s" that changes in the mathematical relation of the spectral density; for $s = 1$, the reservoir is of ohmic type, for $0 < s < 1$, the reservoir is sub-ohmic and for $s > 1$, the reservoir is called super-ohmic.

What is the physical meaning of each type?

glS
  • 15,488
  • 5
  • 42
  • 114
Abdellah Salaoui
  • 69

1 Answers1

-2

This is the quantum dissipation. Quantum dissipation is the branch of physics, that studies the QM processes of irreversible loss of energy viewed at the classical level. Its main purpose is to derive the laws of classical dissipation from QM view. The main problem is to show the irreversible loss of energy, but QM usually uses Hamiltonian, where the total energy is conserved. The solution is to split the system into two:

  1. the QM, where the dissipation works
  2. and the environment or bath, where the energy will flow. The simplest way to model the bath is with infinite number of harmonic oscillators, in QM a set of bosonic particles. In the harmonic bath model, the good description of the dissipation is the bath spectral function.

$$ J(\omega) = \frac{\pi}{2}\sum_{i}\frac{C^{2}_{i}}{m_{i}\omega_{i}}\delta(\omega-\omega_{i}) $$

The spectral function shows constraints in the $C_{i}$, and when it is in the form of $J(\omega)=\eta\omega$ then the classical view of dissipation can be shown to be Ohmic. A more generic form is $J(\omega)\propto\omega^{s}$ In this case, when s>1 the dissipation is super-Ohmic. If s<1, it is called sub-Ohmic.The EM field is under some circumstances super-Ohmic.

stafusa
  • 13,064
Árpád Szendrei
  • 30,008