- #1

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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.

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- Thread starter copernicus1
- Start date

- #1

- 99

- 0

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.

- #2

DrClaude

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Could you please show your derivation?

- #3

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- #4

DrClaude

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

- #5

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Ah, interesting! Thanks for pointing that out.

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