I'm trying to solve a problem with a time dependent Hamiltonian $$H = H_0 + V(t)$$ where $H_0$ is a Hamiltonian with a non explicit time-dependence and $V(t)$ has an explicit time dependence. My system consists of three parts. Each is an harmonic oscillator and there's interaction with them, that is $$H_0 = H_L+ H_R + H_C +H_{coup}$$

where $L$ stands for left, $R$ for right, $C$ for center and $coup$ for the coupling and

$$H_\alpha = \sum_n \frac{p_{\alpha,n}^2}{2m_\alpha}+ \frac{k_\alpha}{2}(x_{\alpha,n}-x_{\alpha,n+1})^2$$

for $\alpha = L,R,C$ and

$$H_{coup}= \frac{k_L}{2}(x_{L,1}-x_{C,1})^2+\frac{k_R}{2}(x_{R,1}-x_{C,N})^2$$

that is, I have a chain of spring and masses where I can recognise three parts and a coupling.

so I want to solve $(\imath \partial_t -H)G=\delta(t-t')\delta_{l,l'}$ or in matricial form $$\begin{pmatrix} \omega-H_L &-H_{LC} &0 \\ -H_{LC} &\omega-H_C &-H_{RC} \\ 0& -H_{LC} & \omega-H_R \end{pmatrix}. \begin{pmatrix} G_L &G_{L,C} &0 \\ G_{L,C} & G_C &G_{R,C} \\ 0&G_{R,C} &G_R \end{pmatrix}= \begin{pmatrix} 1 & & \\ &1 & \\ & &1 \end{pmatrix}$$

The answer is $$G(t,t')=G_0+\int_{t'}^{t}dt_1G(t,t_1)V(t_1)G_0(t_1-t)$$.

where $G_0$ is the green function of $H_0$.

So to get $G$ I saw that is needed to make a gradient expansion and only can be solve to an order (for example at first order)

$$G(t,\omega) = G_0 + GVG_0+ \imath \partial_\omega G\partial_t V G_0$$ and so on.

My question are why I need to do a gradient expansion? Why isn't $$ G = [G_0^{-1}-V(t)]^{-1}$$ the solution? How can I compute the solution (fix point for example)? I saw that the solution is get by replacing $\tilde{G}=[G_0^{-1}-V(t)]^{-1}$ in the formula above so $$G = \tilde{G}+ \imath \partial_\omega \tilde{G} \partial_t V G_0$$ but I don't realise how to get it.

Edit : here is a Pic of math I'm trying to reproduce and understand


The expression below is close to, but not exactly the answer you have. But perhaps it can help.

Start with the Green's function eq. in the form $$ i \frac{\partial G(t, t')}{\partial t} = [ H_0 + V(t) ] G(t, t') + \delta(t - t') $$ and look for a solution of the form $$ G(t, t') = G_0(t - t') W(t, t') $$ where $G_0(t-t')$ is the Green's function for $H_0$, satisfying $$ i \frac{\partial G_0(t-t')}{\partial t} = H_0 G_0(t-t') + \delta(t - t') $$ and $W$ satisfies the equal-time condition $W(t, t) = I$. This gives $$ i \frac{\partial G_0}{\partial t}W + i G_0\frac{\partial W}{\partial t} = H_0 G_0 W + V(t) G_0 W + \delta(t - t') $$ $$ H_0 G_0 W + \delta(t - t') I + i G_0\frac{\partial W}{\partial t} = H_0 G_0 W + V(t) G_0 W + \delta(t - t') $$ $$ i \frac{\partial W}{\partial t} = G_0^{-1} V(t) G_0 W $$ and eventually $$ W = I - i \int_{t'}^t{dt_1 G_0^{-1}(t_1-t')V(t_1) G(t_1, t') } $$ Take this back into the expression for $G$, $$ G(t, t') = G_0(t-t') - i \int_{t'}^t{dt_1 G_0(t-t')G_0^{-1}(t_1-t')V(t_1) G(t_1, t') } $$ use the fact that $G_0(t-t')G_0^{-1}(t_1-t') = G_0(t-t_1)$ and obtain $$ G(t, t') = G_0(t-t') - i \int_{t'}^t{dt_1 G_0(t-t_1)V(t_1) G(t_1, t') } $$ As I already mentioned, this is formally similar to your answer, but not exactly identical. Maybe you can use a hermitian conjugate to rearrange, but there still remains a factor of (-i). Also, if you want to check that $G(t, t')$ is indeed a solution of the Green's function eq. keep the factors $G_0(t-t')G_0^{-1}(t_1-t')$ under the integral separate when you take the derivative on t.

  • $\begingroup$ I get your point but I read that to solve $G(t,'t)$ is needed a gradient expansion in the way I mentioned above. I don't understand why is needed that expansion given that you found another solution defining $W$ $\endgroup$ – Daniel Aug 17 '16 at 12:55
  • $\begingroup$ Do you have a reference for the solution you have? Or can you post an image? $\endgroup$ – udrv Aug 17 '16 at 14:52
  • $\begingroup$ I edited the post with a pic of the math I'm trying to reproduce and understand :) $\endgroup$ – Daniel Aug 17 '16 at 16:31
  • 1
    $\begingroup$ I see. Then your answer is in the first phrase of Sec. C: the authors are looking at "the limit of low driving frequency $\Omega_0$". This is not so much about an exact solution to the Dyson eq., they already have that in Eq.(37), but about examining some physically relevant limits. Basically, they look at the truncation of Eq.(37) to 1st and 2nd order in $\Omega_0$, and express it using the time derivative of $M_1(t)$. Note their use of $dM_1(t)/dt$, $d^2M_1(t)/dt^2$ instead of $dM_1(0)/dt$, $d^2M_1(0)/dt^2$, which is ok on time scales where approx. to $O(\Omega_0)$, $O(\Omega_0^2)$ hold. $\endgroup$ – udrv Aug 17 '16 at 17:02

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