Correlation function of damped harmonic oscillator

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The discussion focuses on the correlation functions of a damped harmonic oscillator, specifically examining the operators in the interaction picture and their expected values. The initial reservoir density is defined, and calculations show that certain correlation functions, such as <tildeΓ_j(t)>_R, are zero, raising questions about the underlying reasons. The thread explores the implications of these results, particularly in relation to the partial trace over reservoir states and how it affects the final equalities in the equations presented. Participants seek clarification on the calculations and the significance of the zero values in the context of the damped harmonic oscillator model.
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The model of damped harmonic oscillator is given by the composite system with the hamiltonians ##H_S\equiv\hbar \omega_0 a^\dagger a##, ##H_R\equiv\sum_j\hbar\omega_jr_j^\dagger r_j##, and ##H_{SR}\equiv\sum_j\hbar(\kappa_j^*ar_j^\dagger+\kappa_ja^\dagger r_j)=\hbar(a\Gamma^\dagger+a^\dagger\Gamma)##.
Now, the initial reservior density is ##R_0=\prod_je^{-\hbar\omega_jr_j^\dagger r_j/k_BT}(1-e^{-\hbar\omega_j/k_BT})##
We have the operators in the interaction picture,
##\tilde\Gamma_1(t)=\tilde\Gamma^\dagger(t)=\sum_j\kappa_j^*r_j^\dagger e^{i\omega_jt}## and ##\tilde\Gamma_2(t)=\tilde\Gamma(t)=\sum_j\kappa_jr_j e^{-i\omega_jt}##
I want to calculate ##<\tilde\Gamma_j(t)>_R=Tr_R(R_0\tilde\Gamma_j(t))## which is identical to zero.
I have no idea why ##<\tilde\Gamma_j(t)>_R##'s are zero.
 
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few more reservoir correlation functions

##<\tilde\Gamma^\dagger(t)\tilde\Gamma^\dagger(t')>_R\equiv Tr_R(R_0\tilde\Gamma^\dagger(t)\tilde\Gamma^\dagger(t'))\\ =\sum_{k,l} \kappa _k^* \kappa _l^* e^{i\omega_kt}e^{i\omega_lt'}\prod_j(1-e^{-\hbar\omega_j/k_BT})tr_R(e^{-\hbar\omega_jr_j^\dagger r_j/k_BT}r_k^\dagger r_l^\dagger)=0##

##<\tilde\Gamma(t)\tilde\Gamma(t')>_R\equiv Tr_R(R_0\tilde\Gamma(t)\tilde\Gamma(t'))\\ =\sum_{k,l}\kappa_k \kappa_le^{-i\omega_kt}e^{-i\omega_lt'}\prod_j(1-e^{-\hbar\omega_j/k_BT})tr_R(e^{-\hbar\omega_jr_j^\dagger r_j/k_BT}r_k r_l)=0##

##<\tilde\Gamma^\dagger(t)\tilde\Gamma(t')>_R\equiv Tr_R(R_0\tilde\Gamma^\dagger(t)\tilde\Gamma(t'))\\ =\sum_{k,l} \kappa _k^* \kappa _l e^{i\omega_kt}e^{-i\omega_lt'}\prod_j(1-e^{-\hbar\omega_j/k_BT})tr_R(e^{-\hbar\omega_jr_j^\dagger r_j/k_BT}r_k^\dagger r_l)\\ = \sum_j|\kappa_j|^2e^{i\omega_j(t-t')}\bar n(\omega_j,T)##

##\bar n(\omega_j,T)=tr_R(R_0r_j^\dagger r_j)=\frac{e^{-\hbar\omega_j/k_BT}}{1-e^{-\hbar\omega_j/k_B}}##

##<\tilde\Gamma(t)\tilde\Gamma^\dagger(t')>_R\equiv Tr_R(R_0\tilde\Gamma(t)\tilde\Gamma^\dagger(t'))\\ =\sum_{k,l} \kappa _k \kappa _l^* e^{-i\omega_kt}e^{i\omega_lt'}\prod_j(1-e^{-\hbar\omega_j/k_BT})tr_R(e^{-\hbar\omega_jr_j^\dagger r_j/k_BT}r_k r_l^\dagger)\\ = \sum_j|\kappa_j|^2e^{-i\omega_j(t-t')}[\bar n(\omega_j,T)+1]##

##tr_R## is the partial trace over the reservoir states.

I can't figure out last equalities in the above equations, mostly because of the partial traces.
Any advice would be very appreciated!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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