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: $$$$H_I(t)=e^{\frac{i}{\hbar}H_0 t}H_I(0)e^{-\frac{i}{\hbar} H_0 t}$$$$ It is easy to see that it actually comes from solving a Heisenberg equation of motion by treating interaction Hamiltonian as a Heisenberg operator:

$$$$\dot{H}_I(t) = \frac{i}{\hbar} [H(t),H_I(t)] =\frac{i}{\hbar} [H_0+H_I(t),H_I(t)] =\frac{i}{\hbar} [H_0,H_I(t)]$$$$ We can write the solution as: $$$$H_I(t)=H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt' \, [H_0,H_I(t')]$$$$ and we plug it iteratively again:

$$$$H_I(t)=H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt' \, [H_0,H_I(0)]-\frac{1}{\hbar^2}\int\limits_0^t \int\limits_0^{t'} dt' dt'' \, [H_0,[H_0,H_I(t'')]]$$$$

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:

$$$$H_I(t)=e^{\frac{i}{\hbar}\int\limits_0^t dt' \,H_0}H_I(0)e^{-\frac{i}{\hbar}\int\limits_0^t dt' \, H_0} =e^{\frac{i}{\hbar}H_0 t}H_I(0)e^{-\frac{i}{\hbar} H_0 t}$$$$ 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:

$$$$\rho_I=e^{\frac{i}{\hbar}H_0 t}\rho_S e^{-\frac{i}{\hbar} H_0 t}$$$$

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:

$$$$\dot{\rho}_S = \frac{i}{\hbar} [H_0,\rho_S]$$$$ where the solution of $$\rho_S$$ would be the interaction picture density matrix $$\rho_I$$. Is my thought process correct?

• Are you on the same page as this summary? It would be easier to connect to the Schroedinger picture. Feb 5 '21 at 15:48
• Yes, I am aware of the differences, I am just wondering how to obtain this transformation from differential equation. Because, for example, if one wants to transform to the interaction picture when Lindblad terms are present (which will contribute to the decay of operators in the interaction picture), then simple exponential transform does not work anymore. Feb 5 '21 at 16:11
• You do take $H_0(t)=H_0(0)=H_0$, right? If not your first equation does not yield the second one. Feb 5 '21 at 16:22
• Yes, of course. Sorry about the typo, I corrected it. Feb 5 '21 at 16:29

No need to iterate $$$$H_I(t)=H_I(0)+\frac{i}{\hbar}\int\limits_0^t dt' \, [H_0,H_I(t')] .$$$$ Just observe that it is readily solved by the Ansatz $$$$H_I(t)=e^{\frac{i}{\hbar}tH_0 }H_I(0)e^{-\frac{i}{\hbar} t H_0 }$$$$ which yields the r.h.side $$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.