Solving for unitary operation using perturbation theory

Question

Let the time-dependent Hamiltonian be \begin{equation} H(t) = H_0(t) + \lambda H_1(t), \end{equation} where $\lambda$ is a small parameter. In the interaction picture (i.e. rotating frame w.r.t unitary generated by $H_0(t)$), the unitary operation obeys the following Schrodinger equation: \begin{equation} i\dot{U_I}(t) = \lambda H_I(t) U_I(t), \quad H_I(t) = U_0^\dagger(t) H_1(t) U_0(t), \quad U_0(t) = \mathcal{T}e^{-i\int_0^t H_0(t') dt'}. \end{equation} My objective is to solve $U_I(t)$ at final time $t = t_f$ using perturbation theory.

Following some references such as this one, since $\lambda$ is small, we can first express the unitary $U_I(t)$ as the power series of $\lambda$: \begin{equation} U_I(t) = \sum_{k=0}^\infty \lambda^k U_I^{(k)}(t). \end{equation} Substituting this into the Schrodinger equation above, we have the following equation: \begin{equation} i\partial_t(U_I^{(0)}(t) + \lambda U_I^{(1)}(t) + \lambda^2 U_I^{(2)}(t) + \dotsc) = \lambda H_I(t) (U_I^{(0)}(t) + \lambda U_I^{(1)}(t) + \lambda^2 U_I^{(2)}(t) + \dotsc). \end{equation} Since the coefficients of each power of $\lambda$ must vanish, we have \begin{align} i\partial_tU_I^{(0)}(t) = 0, \quad i\partial_tU_I^{(k)}(t) = H_I(t) U_I^{(k-1)}(t) \quad \forall k \geq 1, \end{align} where we can assume $U_I^{(0)}(t) = I$ and $U_I^{(k)}(0) = 0$ for all $k$.

Hence, we can iteratively solve the equation from $k=1, \dotsc$, to obtain $U_I^{(k)}(t_f)$. For instance, to solve for $k=1$, we have to solve: \begin{equation} i\partial_t U_I^{(1)}(t) = H_I(t) \quad \leftrightarrow \quad U_I^{(1)}(t_f) = -i\int_0^{t_f} H_I(t) dt. \end{equation} To find $U_I^{(2)}(t_f)$, we can solve: \begin{equation} i\partial_t U_I^{(2)}(t) = H_I(t) U_I^{(1)}(t), \quad U_I^{(1)}(t) = -i\int_0^t H_I(t')dt', \end{equation} and so on. I hope what I explained is mathematically correct, but I am happy to be corrected and learn from my mistakes. Given the set-ups, here are my two inter-connected questions:

$\textbf{Question 1.}$ What is the benefit of using perturbation theory to find the unitary operation $U_I(t_f)$? The reason I am asking this is because it seems to me that to calculate $U_I^{(k)}(t_f)$, one needs to perform $k$-fold integrals, which is essentially the same cost when one tries to solve the unitary operation via Dyson or Magnus series.

$\textbf{Question 2.}$ Is there any special cases (e.g. $H_I(t)$ being a periodic Hamiltonian with the fact that I'm only interested in unitary at final time $t_f$ instead of arbitrary time $t$) where one can solve for $U_I^{(k)}(t_f)$ efficiently without invoking multi-integrals, or efficient iterative algorithm to solve the equation up to the desired order of $k$?

I know that this might be a research-level question, so I don't expect detailed answers. Instead, I would be very thankful if anyone could point me towards related research papers.

Answer 1

Let us for clarity review the derivation.

1. We start with the Hamiltonian in the Schrödinger (S) picture: $$ H_S(t)~=~H_{0,S}(t)+V_S(t) .\tag{A}$$ Define evolution operators $U_0(t)$ and $U(t)$ satisfying $$\begin{align} i\hbar d_tU_0(t)~=~&H_{0,S}(t)U_0(t),\qquad U_0(t\!=\!0)~=~{\bf 1},\cr i\hbar d_tU(t)~=~&H_S(t)U(t),\qquad U(t\!=\!0)~=~{\bf 1},\cr U_0(t_2,t_1)~:=~&U_0(t_2)U_0(t_1)^{-1}, \end{align}\tag{B}$$ cf. e.g. this Phys.SE post.

2. Define the interaction (I) picture $$\begin{align} A_I(t)~:=~&U_0(t)^{-1}A_S(t)U_0(t), \cr U_I(t)~:=~&U_0(t)^{-1}U(t),\qquad U_I(t\!=\!0)~=~{\bf 1},\cr i\hbar d_tU_I(t)~=~&U_0(t)^{-1}V_S(t)U(t)~=~V_I(t)U_I(t). \end{align}\tag{C}$$ Here $A(t)$ denotes any operator that depends on only 1 time. One useful feature of the interaction picture is that if the perturbation $V_I(t)$ is small, then $U_I(t)$ is almost conserved in time.

3. The Schrödinger eq. (C) can be rewriten on integral form: $$\begin{align} U_I(t)~=~&{\bf 1} -\frac{i}{\hbar}\int_0^t \!ds~U_0(s)^{-1}V_S(s)U(s), \cr U(t)~=~&U_0(t)-\frac{i}{\hbar}\int_0^t \!ds~U_0(t,s)V_S(s)U(s). \end{align}\tag{D}$$

4. The last eq. (D) is a fixed-point equation that can be used repeatedly to generate a formula for $U(t)$ to higher and higher order in the perturbation $V_S$.

   This in turn yields a perturbative formula for OP's sought-for quantity $U_I(t)=U_0(t)^{-1}U(t)$.

   Alternatively, such formula follows by expanding the (anti)time-ordered exponential $$\begin{align} U_I(t)~=~&(A)T\exp\left[-\frac{i}{\hbar}\int_0^t\! ds~V_I(s)\right]\cr ~=~&{\bf 1}-\frac{i}{\hbar}\int_0^t \!ds~V_I(s)+\ldots\cr ~=~&{\bf 1}-\frac{i}{\hbar}\int_0^t \!ds~U_0(0,s)V_S(s)U_0(s,0)+\ldots.\end{align}\tag{E}$$

5. The formula (E) has a nice graphical interpretation: At $n$th order we have to insert $n$ perturbations at all possible $n$-tuple of intermediate times. So OP is right: We cannot avoid multi-integrals.

6. The interaction picture is in practice only useful if $H_{0,S}(t)$ does not depend on time, and if the perturbation is of the form $$ V_S(t)~=~f(t, z_S(t),\dot{z}_S(t),\ldots),\tag{F}$$ where $z_S(t)$ denote the fundamental dynamical variables of the theory (e.g. positions and momenta). Then $$ V_I(t)~=~f(t, z_I(t),\dot{z}_I(t),\ldots),\tag{G}$$ i.e. we no longer have to refer to the Schrödinger picture.