Physics Asked by qc2014 on May 20, 2021
It is claimed in many papers that the two-time Green’s function in time periodic Hamiltonian case is periodic in the average time, i.e.
begin{equation}
G(t+T,t’+T)=G(t,t’)
end{equation}
when $H(t+T)=H(t)$. I wonder if there is any rigorous proof of this property starting from the definition of Green’s function?
I am going to assume the reader is familiar with the (contour-ordered) definition of the Green's function:
$$begin{align} iG(t_1,t_2) &equiv langle T_c psi_H(x_1,t_1) psi^dagger_H(x_2,t_2) rangle &equiv mathop{Tr}left[rho(H) T_c psi_H(x_1,t_1) psi^dagger_H(x_2,t_2) right] / mathop{Tr}left[rho(H) right] end{align} $$
where $psi_H$ is the annihilation operator in the Heisenberg picture, whose evolution is dictated by the time evolution operator $U$, starting from some time $t_0$, where the Heisenberg and Schrodinger pictures coincide.
$$ psi_H(x,t) = U^dagger(t,t_0) psi(x) U(t,t_0) $$
Now here's the proof:
$$begin{align} frac{d}{dt}U_{H}left(t,t_{0}right)&=-iHleft(tright)U_{H}left(t,t_{0}right) Sleft(tright)&equiv U_{H}left(t+T,t_{0}right)U_{H}^{dagger}left(T,t_{0}right)quadtext{define helper quantity} frac{d}{dt}Sleft(tright)&=frac{d}{dt}U_{H}left(t+T,t_{0}right)U_{H}^{dagger}left(T,t_{0}right)&=-iHleft(t+Tright)U_{H}left(t+T,t_{0}right)U_{H}^{dagger}left(T,t_{0}right)&=-iHleft(tright)Sleft(tright)quadtext{using periodicity of }H Rightarrow Sleft(tright)&=U_{H}left(t,t_{0}right)U_{H}^{dagger}left(0,t_{0}right)quadtext{because }Sleft(0right)=1Rightarrow U_{H}left(t+T,t_{0}right)&=U_{H}left(t,t_{0}right)U_{H}^{dagger}left(0,t_{0}right)U_{H}left(T,t_{0}right)&=U_{H}left(t,t_{0}right)Xtext{where }X&equiv U_{H}^{dagger}left(0,t_{0}right)U_{H}left(T,t_{0}right) text{now } psi_{H}left(x,t+Tright)&=U_{H}^{dagger}left(t+T,t_{0}right)psileft(x,t_{0}right)U_{H}left(t+T,t_{0}right) &=X^{dagger}U_{H}^{dagger}left(t,t_{0}right)psileft(x,t_{0}right)U_{H}left(t,t_{0}right)X &=X^{dagger}psi_{H}left(x,tright)X iGleft(t_{1}+T,t_{2}+Tright)&=leftlangle T_{c}psi_{H}left(x_{1},t_{1}+Tright)psi_{H}^{dagger}left(x_{2},t_{2}+Tright)rightrangle &=leftlangle T_{c}X^{dagger}psi_{H}left(x_{1},t_{1}right)XX^{dagger}psi_{H}^{dagger}left(x_{2},t_{2}right)Xrightrangle &=leftlangle T_{c}X^{dagger}psi_{H}left(x_{1},t_{1}right)psi_{H}^{dagger}left(x_{2},t_{2}right)Xrightrangle &=leftlangle T_{c}psi_{H}left(x_{1},t_{1}right)psi_{H}^{dagger}left(x_{2},t_{2}right)XX^{dagger}rightrangle quadtext{using cyclic property of trace}&=leftlangle T_{c}psi_{H}left(x_{1},t_{1}right)psi_{H}^{dagger}left(x_{2},t_{2}right)rightrangle &=iGleft(t_{1},t_{2}right) end{align} $$
q.e.d.
Answered by Pooya on May 20, 2021
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