# Transforming simple Hamiltonian to interaction picture

I am trying to follow the math in a paper, and in it they do a lot of transforming into the interaction frame. It has been awhile since I have done these kind of calculations explicitly by hand and I don't remember everything, so I apologize if this question is trivial.

Just so I can get back into the swing of things, let us say that we have the following Hamiltonian:

$H = H_0 + x_0 (a + a^{\dagger})$

where

$H_0 = \hbar \omega \left(a^{\dagger}a + \frac{1}{2}\right)$

and $x_0 = \sqrt{\frac{\hbar}{2 m \omega}}$. I am interested in transforming the perturbation term into the interaction picture with respect to $H_0$ (ignoring fast oscillating terms). So first I do:

$V_I(t) = x_0 e^{i H_0 t /\hbar} (a + a^{\dagger}) e^{-i H_0 t /\hbar}$

I don't remember how to apply exponential operators. I think you would do something like:

$e^{i H_0 t /\hbar} \approx 1 + \frac{it}{\hbar}H_0$

But I am not so sure about this as it introduces a linear time component. But there is also the Baker-Hausdorff relationship:

$e^{i\hat{G}\lambda}\hat{A}e^{-i\hat{G}\lambda} \approx \hat{A} + i \lambda [\hat{G},\hat{A}]$

Doing it the second way I get:

2$x_0(a + a^{\dagger})$

And doing it the first way I get something clearly wrong. Apparently, from what I am looking at here, the answer should be:

$x_0 (a e^{-i\omega t} + a^{\dagger}e^{i\omega t})$

• In this specific case can resum the whole series arising from the full expansion of $e^{iG\lambda} A e^{-iG\lambda}$ and you will recover the full answer. – ZeroTheHero Oct 2 '17 at 13:49

So, consider the interaction picture creation operator $$\hat{a}(t)=e^{i\hat{\mathcal{H}}_{0}t/\hbar}\hat{a}e^{-i\hat{\mathcal{H}}_{0}t/\hbar}$$ Then $$i\hbar\frac{d\hat{a}(t)}{dt}=[\hat{a}(t), \hat{\mathcal{H}}_{0}]=\hbar\omega\hat{a}(t)$$ Thus $$\hat{a}(t)=ae^{-i\omega{t}}$$ Analoguasly $$i\hbar\frac{d\hat{a}^{\dagger}(t)}{dt}=[\hat{a}^{\dagger}(t), \hat{\mathcal{H}}_{0}]=-\hbar\omega\hat{a}^{\dagger}(t)$$ So that $$\hat{a}^{\dagger}(t)=\hat{a}^{\dagger}e^{i\omega{t}}$$ So, the potential becomes $$V_{I}(t)=x_{0}(\hat{a}(t)+\hat{a}^{\dagger}(t))=x_{0}(\hat{a}e^{-i\omega{t}}+\hat{a}^{\dagger}e^{i\omega{t}})$$