In Cohen-Tannoudji's Atom-Photon Interactions, he gives the integral form of the time evolution operator in the Schrodinger representation as \begin{equation} U(t_{f},t_{I}) = U_{0}(t_{f},t_{i}) + \frac{1}{i\hbar}\int^{t_{f}}_{t_{i}}U_{0}(t_{f},t)VU(t,t_{i})\, dt\tag{1} \end{equation} where \begin{equation} U_{0}(t_{f},t_{i}) = \exp{\left(\frac{-iH_{0}(t_{f}-t_{i})}{\hbar}\right)}.\tag{2} \end{equation} He then says that in order to prove that this is equivalent to the equation of motion of the operator, it is satisfactory to prove that it satisfies the equation of motion \begin{equation} i\hbar\frac{d}{dt_{f}}U(t_{f},t_{i}) = (H_{0}+V)U(t_{f},t_{i})\tag{3} \end{equation} as well as having the property \begin{equation} U(t_{i},t_{i}) = \mathbb{1}\tag{4} \end{equation} What's confusing me here is I don't really see how he has differentiated the integral with respect to $t_{f}$ as well as proving that this integral must be 0 when the two time indexes coincide. Likewise, he doesn't really say why he takes this approach when every other textbook I have read on the matter take the time evolution to have the form \begin{equation} U(t,t_{0}) = \mathbb{1} - \frac{1}{i\hbar} \int^{t}_{t_{0}}V(t')U(t',t_{0})\, dt'.\tag{5} \end{equation} I guess my questions are is there some trick to the integral that I'm not seeing, as well as his method providing a better approach?
2 Answers
Hints for eq. (4) using eq. (1):
- $U_0(t_i,t_i)=\mathbb{1}$ and $\int_{t_i}^{t_i} \!dt =0$.
Hints for eq. (3):
Notice that $t_f$ appears in 3 places on the RHS of eq. (1):
Inside $U_0$ in 2 places and
as the upper limit of the integral.
Apply $i\hbar\frac{d}{dt_{f}}$ to the RHS of eq. (1):
$U_0$ gets replaced with $H_0U_0$ in 2 places, cf. eq. (2).
the integral gets replaced with the integrand with the integration variable $t=t_f$.
Recognize that the result is precisely the RHS of eq. (3).
I will provide another approach to this problem, imagine you have a state ket at $t_{0}$ and you evolved this state with an unknown time evolution operator $\mathcal{U}$ as
$$|\alpha, t_{0};t\rangle = \mathcal{U(t,t_0)}|\alpha, t_{0}\rangle$$
if our initial state ket is normalized like
$$\langle\alpha, t_{0}|\alpha, t_{0}\rangle = 1$$
it will imply it is normalized as time evolves and this will produce the unitarity of time evolution operator.
$$|\alpha, t_{0};t_0+dt\rangle = \mathcal{U(t_0+dt,t_0)}|\alpha, t_{0}\rangle$$
we expect time evolution operator to be unity as $dt\rightarrow 0$ thus we expect $\mathcal{U}(t_0+dt,t_0)$ to be first order in dt
based on these, you assume a solution and check if it satisfies the unitarity property of time evolution equation. Since from classical mechanics we know Hamiltonian is the generator of time evolution like momentum is the generator of translation in space.
$$\mathcal{U}(t_0+dt,t_0)=1-\frac{iHdt}{\hbar}$$
As we know this is the infinitesimal time evolution operator but one can calculate the time evolution operator with N-tiple time evolution operator acting on a ket as N goes to infinity which will result with an exponential, yet it's easier to see the pattern here: it looks like a taylor expansion of an exponential. You can prove that it indeed is, finally evolution operator becomes
$$\mathcal{U}(t, t_0)=e^{-\frac{iH(t-t_0)}{\hbar}}$$
