Interaction picture equation from Heisenberg equation

  • #1
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 Schrodinger picture, and I can derive the equation this way, but I feel like it should work the other way too.
 

Answers and Replies

  • #2
DrClaude
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Could you please show your derivation?
 
  • #3
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.
 
  • #4
DrClaude
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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}##.
 
  • #5
Ah, interesting! Thanks for pointing that out.
 

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