Interaction picture derivation of the interaction Hamiltonian and density matrix

Question

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?

Answer 1

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.