Consider a quantum system with the following Hamiltonian: $$H(t)=H_0+H_1(t),\tag{1}$$ where $H_0$ is a noninteracting Hamiltonian and $H_1(t)$ a time-dependent perturbation.

To formulate the linear response theory, one first need figure out the time evolution operator $U(t,t_0)$ by solving the corresponding Schrodinger equation: $$U(t,t_0)=T\left[e^{-\dfrac{i}{\hbar}\int_{t_0}^t d\bar{t}H(\bar{t})}\right].\tag{2}$$

Furthermore, one can argue that $H_0$ and $H_1(t)$ commutes under time ordering and then obtain the following relation: $$T\left[e^{-\dfrac{i}{\hbar}\int_{t_0}^t d\bar{t}H(\bar{t})}\right]=T\left[e^{-\dfrac{i}{\hbar}\int_{t_0}^t d\bar{t}H_0}e^{-\dfrac{i}{\hbar}\int_{t_0}^t d\bar{t}H_1(\bar{t})}\right].\tag{3}$$

How can I convince me to believe this relation (3)? Can someone help me to prove this?

  1. One proof of eq. (3): Everything commute$^1$ by definition under the time (normal, radial, etc) ordering symbol: $$T\left([A,B]\right)~=~0,\tag{A}$$ so BCH formula simplifies $$ T\left(e^Ae^B\right)~=~T\left(e^{A+B+\frac{1}{2}[A,B]+[\text{nested commutator terms}]}\right)~=~ T\left(e^{A+B}\right) ,\tag{B}$$ which proves OP's sought-for eq. (3).

  2. Another proof of eq. (3): Use the Trotter formula.


$^1$ The main point is that the time ordering procedure $T$ does not take operators to operators, but symbols/functions to operators, cf. this & this related Phys.SE post and links therein.

  • $\begingroup$ If this would be true, interaction picture would not make sense. $\endgroup$ – Jon Dec 13 '17 at 14:26
  • $\begingroup$ @Jon: Actually, even in the interaction picture you will meet the same problem. $\endgroup$ – Jack Dec 14 '17 at 1:17
  • $\begingroup$ @Qmechanic: Can you provide more details? Because I come from the community of condensed matter physics. And I know the introduction of the time-ordering operator is just the product of a series of Heaviside step function with different permutation. $\endgroup$ – Jack Dec 14 '17 at 1:43
  • $\begingroup$ Very nice but $T([q,p])\stackrel{?}{=}0$ or you have to fix a picture? $\endgroup$ – Jon Dec 24 '17 at 14:34
  • $\begingroup$ The caveat is that one should not apply the CCR relations inside the $T$ argument, cf. e.g. this Phys,SE post. $\endgroup$ – Qmechanic Dec 24 '17 at 20:41

In practically all the interesting cases it is $[H_0,H_1(t)]\ne 0$ and so, you cannot write the time evolution operator as the product of two exponentials. In this particular case, the approach to use is the following. Let us consider the Schroedinger equation for the time-evolution operator $U=U(t,t_0)$ $$ i\hbar\frac{\partial U}{\partial t}=(H_0+H_1(t))U. $$ Let us put $$ U=e^{-\frac{i}{\hbar}H_0(t-t_0)}U_I(t,t_0) $$ and let us compute the evolution equation for $U_I$. By substitution, we get $$ i\hbar\frac{\partial U_I}{\partial t}=e^{\frac{i}{\hbar}H_0(t-t_0)}H_1(t)e^{-\frac{i}{\hbar}H_0(t-t_0)}U_I. $$ This can be solved by writing $$ U_I(t,t_0)=Te^{-\frac{i}{\hbar}\int_{t_0}^tH_I(t')dt'} $$ where we have set $H_I(t)=e^{\frac{i}{\hbar}H_0(t-t_0)}H_1(t)e^{-\frac{i}{\hbar}H_0(t-t_0)}$. Then, finally $$ U(t,t_0)=e^{-\frac{i}{\hbar}H_0(t-t_0)}Te^{-\frac{i}{\hbar}\int_{t_0}^tH_I(t')dt'}. $$ but you are entitled now to put the first exponential inside the $T$ operator and this is what you were asking for. Please, note that all this is known in literature as interaction picture and is the starting point for doing time-dependent perturbation theory.


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