Interaction picture derivation of the interaction Hamiltonian and density matrix I am learning about interaction picture and I am not satisfied with the bare definition of the interaction Hamiltonian in the interaction picture, because it just seems like an ansatz. So I am trying to derive it differently. I will present my thought process for the interaction Hamiltonian and the density matrix.
1) Interaction Hamiltonian
From books Interaction Hamiltonian definition in the Interaction picture reads:
\begin{equation}
    H_I(t)=e^{\frac{i}{\hbar}H_0 t}H_I(0)e^{-\frac{i}{\hbar} H_0 t}
\end{equation}
It is easy to see that it actually comes from solving a Heisenberg equation of motion by treating interaction Hamiltonian as a Heisenberg operator:
\begin{equation}
    \dot{H}_I(t) = \frac{i}{\hbar} [H(t),H_I(t)] =\frac{i}{\hbar} [H_0+H_I(t),H_I(t)] =\frac{i}{\hbar} [H_0,H_I(t)] 
\end{equation}
We can write the solution as:
\begin{equation}
    H_I(t)=H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt' \, [H_0,H_I(t')]
\end{equation}
and we plug it iteratively again:
\begin{equation}
    H_I(t)=H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt' \, [H_0,H_I(0)]-\frac{1}{\hbar^2}\int\limits_0^t \int\limits_0^{t'}  dt' dt'' \, [H_0,[H_0,H_I(t'')]]
\end{equation}
Eventually we plug it infinite times and get an infinite series as a solution, which we can transform to exponents using Baker–Campbell–Hausdorff formula:
\begin{equation}
    H_I(t)=e^{\frac{i}{\hbar}\int\limits_0^t dt' \,H_0}H_I(0)e^{-\frac{i}{\hbar}\int\limits_0^t dt' \, H_0} =e^{\frac{i}{\hbar}H_0 t}H_I(0)e^{-\frac{i}{\hbar} H_0 t}
\end{equation}
where if the Hamiltonian is time independent, we can carry out the integral.
So far it seems to work, but my question is - where does the last term
$\int\limits_0^t dt' ...\int\limits_0^{t^{n-2}}dt^{n-1} \int\limits_0^{t^{n-1}} dt^n\, [...,[H_0,[H_0,H_I(t^n)]]$
in the infinite expansion go, because here the Hamiltonian is still dependent on time and we can not carry out the integral? I was thinking that we can throw it away because it is infinite series, but I am not convinced because the last term might also give infinite contribution to the series.
2) Density Matrix
I also try to treat the density matrix similarly as the interaction Hamiltonian. From books it is defined as:
\begin{equation}
   \rho_I=e^{\frac{i}{\hbar}H_0 t}\rho_S e^{-\frac{i}{\hbar} H_0 t}
\end{equation}
where $\rho_S$ is density matrix in the Schrödinger picture. So I think that one could also derive this definition from the density matrix equation:
\begin{equation}
\dot{\rho}_S = \frac{i}{\hbar} [H_0,\rho_S] 
\end{equation}
where the solution of $\rho_S$ would be the interaction picture density matrix $\rho_I$. Is my thought process correct?
 A: No need to iterate
\begin{equation}
    H_I(t)=H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt' \, [H_0,H_I(t')] .
\end{equation}
Just observe that it is readily solved by the Ansatz
\begin{equation}
    H_I(t)=e^{\frac{i}{\hbar}tH_0 }H_I(0)e^{-\frac{i}{\hbar} t H_0  }
\end{equation}
which yields the r.h.side
$$
H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt'  ~~ [H_0,~e^{\frac{i}{\hbar}t' H_0  }H_I(0)e^{-\frac{i}{\hbar}t' H_0  }  ]\\ 
= H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt'  ~~e^{\frac{i}{\hbar}t'[H_0  }~~[H_0,H_I(0)]\\  =  H_I(0)+ \left ( e^{\frac{i}{\hbar}t[H_0  } ~H_I (0) - H_I(0)  \right )\\  =  e^{\frac{i}{\hbar}t[H_0  }~~ H_I (0)   \equiv  e^{\frac{i}{\hbar}t H_0 } H_I (0)  e^{-\frac{i}{\hbar}t H_I }.
$$
This last line is a famous Lemma (4), namely that
$$
e^A B e^{-A}= B+ [A,B]+ [A,[A,B]]/2!+ [A,[A,[A,B]]]/3!+...
$$
where I have used the notation
$$
e^{[A} \equiv e^{\operatorname{ad}_A } = \operatorname{Ad}_{e^A}
$$
since most physicists are unfamiliar with the mathematics operators involved. The stunt then is integration of a simple exponential of the ad operator.
The very same integration of ad maneuver will net you the density matrix.
