Interaction picture equation from Heisenberg equation

[copernicus1]

The standard Heisenberg picture equation of motion is $$i\hbar\frac d{dt}A_H=[A_H,H],$$ assuming no explicit $t$-dependence on the Heisenberg-picture operator $A_H$. I've been trying to go directly from this equation to the corresponding interaction-picture equation, $$i\hbar\frac d{dt}A_I=[A_I,H_0],$$ (see Sakurai 5.5.12) which I thought at first would be simple, but I keep coming up with $$i\hbar\frac d{dt}A_I=[A_I,H_0]+[A_I,V_I],$$ where $V_I$ is the interaction part of the hamiltonian in the interaction picture. The basic problem is that in the original equation $H$ contains both $H_0$ and $V$ and I don't know how to get rid of the $V$ part. Has anyone been through this calculation?

Thanks!

P.S. I know I could just start with $A_I(t)=e^{iH_0t/\hbar}A_Se^{-iH_0t/\hbar}$, where $A_S$ is in the Schrödinger picture, and I can derive the equation this way, but I feel like it should work the other way too.

 

[DrClaude]

Could you please show your derivation?

 

[copernicus1]

I start with $$i\hbar\frac{}d{dt}A_H(t)=[A_H(t),H_H(t)]$$ (subscript means Heisenberg picture) and plug in $A_H(t)=e^{iH_St/\hbar}A_Se^{-iH_St/\hbar}$ and $H_H(t)=e^{iH_St/\hbar}H_Se^{-iH_St/\hbar}$. (I then replace $H_S=H_{0,S}+V_S$ everywhere and transform both sides of the original Heisenberg equation using $$i\hbar\frac{}d{dt}e^{-iV_St/\hbar}A_H(t)e^{iV_St/\hbar}=e^{-iV_St/\hbar}[A_H(t),H_H(t)]e^{iV_St/\hbar}.$$ Simplify and I'm left with $$i\hbar\frac{d}{dt}A_I(t)=A_I(t)e^{iH_{0,S}t/\hbar}H_Se^{-iH_{0,S}t/\hbar}-e^{iH_{0,S}t/\hbar}H_Se^{-iH_{0,S}t/\hbar}A_I(t),$$ but $H_S\neq H_{0,S}$. If it did I would be done. Instead it's $H_S=H_{0,S}+V_S$. This is where I get stuck.

 

[DrClaude]

I then replace $H_S=H_{0,S}+V_S$ everywhere

I guess this is where it doesn't work. Since $[H_0, V] \neq 0$, $e^{i (H_0 + V) t/ \hbar} \neq e^{i H_0 t/ \hbar} e^{i V t/ \hbar}$.

 

[copernicus1]

Ah, interesting! Thanks for pointing that out.