# Hamiltonian commutators and the time evolution operator

I am reading through a quantum optics book at the moment in which the authors state that given the interaction Hamiltonian:

$$V(t) = \int \vec{J}(\vec{r},t) \cdot \hat{A}(\vec{r},t) \hspace{1mm} d^3r$$

(Where $$\hat{A}$$ is the normal vector potential operator and $$\vec{J}$$ is a classical current density), then the commutators of the hamiltonian with itself at different times (which are non-zero) should allow us to write the time evolution operator as:

$$U(t) = \exp{\frac{-i}{\hbar} \int_0^t V(t') dt'}$$

In addition to some constant phase factor. They then state that therefore we can just leave it as what is written above. I began to consider how to obtain this form, and I got a little lost. I know that for non-commuting hamiltonians, we get the Dyson series for the time evolution operator, and I got that the commutator at different times is a purely imaginary factor (although I don't think it's constant in time). However, I don't see how this leads to the above form for $$U(t)$$. Can anyone help with this?

• Your expression for $U(t)$ lacks a time ordering operator - it is just a formal sum of the series. Apart from that, the evolution operator in the interaction picture is derived in many textbooks... – Roger Vadim Mar 18 at 17:29
• Could you tell me the book – amilton moreira Mar 18 at 17:29
• Depends on what you have accessible - e.g., any introduction to QFT for condensed matter: AGD, Fetter&Walecka, Mahan, Doniach and Sondheimer, or any newer one. – Roger Vadim Mar 18 at 17:31
• @Vadim I agree that the expression lacks the time ordering operator. I want to know how, given that the Hamiltonians don't commute and therefore the TO operator should be included, but the book I have says that when included, the series will give the operator written above with an additional phase. – user132849 Mar 18 at 17:39