Correlation function of damped harmonic oscillator

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
rbwang1225
Messages
112
Reaction score
0
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.
 
Physics news on Phys.org
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!