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?

  • 3
    $\begingroup$ 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... $\endgroup$ – Roger Vadim Mar 18 at 17:29
  • $\begingroup$ Could you tell me the book $\endgroup$ – amilton moreira Mar 18 at 17:29
  • $\begingroup$ 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. $\endgroup$ – Roger Vadim Mar 18 at 17:31
  • $\begingroup$ @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. $\endgroup$ – user132849 Mar 18 at 17:39

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