Integral of time dependent Hamiltonian

Question

Computing the time evolution of a quantum system described by a time-dependent Hamiltonian, $H(t)$, amounts to constructing the time evolution operator $$U = \mathcal{T} \exp \Biggl( -i \int_{0}^{t} \mathrm{d} \tau \ H(\tau) \Biggr) \ . $$ What if the time-dependence in $H(t)$ can be integrated analytically, e.g., if the Hamiltonian is of the form $$ H(t) = H_1 + t H_2$$ with $H_1, H_2$ time-independent? In that case, can I write the following? $$U = \exp \Bigl( -i (H_1 t + \frac{t^2}{2} H_2) \Bigr)$$

This seems naïve, but take for example the Hamiltonian in this question. Could one not simply compute the $t$-integral over $\sin(\omega_0 t)$?

Answer 1

You have to be careful about the expression of $U$. The expression you put down mathematically means \begin{align} U & = \lim_{n \rightarrow +\infty}{\big(e^{-{i \over \hbar}{t \over n}H(t)}\big)\big(e^{-{i \over \hbar}{t \over n}H(t(1-{1 \over n}))}\big)\big(e^{-{i \over \hbar}{t \over n}H(t(1-{2 \over n}))}\big)\cdots \big(e^{-{i \over \hbar}{t \over n}H({t \over n})}\big)} \\ & = \lim_{n \rightarrow +\infty}{\big(e^{-{i \over \hbar}{t \over n}(H_1+t H_2)}\big)\big(e^{-{i \over \hbar}{t \over n}(H_1+t(1-{1 \over n}) H_2)}\big)\big(e^{-{i \over \hbar}{t \over n}(H_1+t(1-{2 \over n}) H_2)}\big)\cdots \big(e^{-{i \over \hbar}{t \over n}(H_1+{1 \over n} H_2)}\big)} \\ & = \lim_{n \rightarrow +\infty}{\Pi_{j=1}^{n}{e^{-{i \over \hbar}{t \over n}(H_1+j{t \over n}H_2)}}} \end{align} If you want to integrate over $t$ to get the result you wrote, you will need $[H_1,H_2]=0$.