Someone can help me to prove this identity, it comes from a normalization in MQ. From perturbation theory time dependent we have $$H(t)=H_0+H’(t)$$ $$|Ψ>=c_a(t)e^{-iE_at/\hbar}|Ψ_a>+c_b(t)e^{-iE_bt/\hbar}|Ψ_b>$$ $$\dot{c_a(t)}=-i/\hbar H_{ab}’(t)c_b(t)e^{-iwt}$$ $$\dot{c_b(t)}=-i/\hbar H_{ba}’(t)c_a(t)e^{iwt}$$ Where $H_{ij}’=<Ψ_i|H’|Ψ_j>=H_{ji}’^{*} $, $H_{ii}’=0$ , $w=(E_b-E_a)/\hbar$,
Solving iteratively, at zero order: $$c_a^0(t)=1$$$$c_b^0(t)=0$$ At first order $$c_a^1(t)=1$$$$c_b^1(t)=-i/\hbar \int_{0}^{t}H_{ba}’(t’)e^{iwt’}dt’ $$ At second order
$$c_a^2(t)=1-1/\hbar^2\int_{0}^{t}H_{ab}’(t’)e^{-iwt’}[\int_{0}^{t’}H_{ba}’(t’’)e^{iwt’’}dt’’]dt’$$$$c_b^2(t)=-i/\hbar \int_{0}^{t}H_{ba}’(t’)e^{iwt’}dt’ $$
But $|c_a^n(t)|^2+|c_b^n(t)|^2=1+o(t^n)$ So for first order is obviously true But for second order comes this relation:
$$2Re{[\int_{0}^{t}f(t’)[\int_{0}^{t’} f(t”)^{*}dt”]dt’]}=\int_{0}^{t} f(t’)^{*}dt’\int_{0}^{t}f(t'')dt''$$
Could someone prove this?
