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When deriving the generalized Langevin equation with Mori-Zwanzig formalism, I was taught that one identity should be used, that is, $$e^{it\hat X}=e^{it\hat Y}+\int_0^te^{i(t-\tau)\hat X}i(\hat X-\hat Y)e^{i\tau\hat Y}d\tau$$ where $\hat X$ and $\hat Y$ are super-operator acts on operator of the Hillbert space of wave functions(e.g. the Liouville operator in Heisenberg picture). What I want to know is, how to prove this identity, is there a name for this identity? I hope someone can help me or give me some references on it.

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  • $\begingroup$ This operator identity depends on a continuous parameter, $t$. If you want to prove that $f(t) = g(t)$, one way to do it is to prove that $f(0) = g(0)$ and that $f'(t) = g'(t)$ for all $t$. $\endgroup$ Mar 21 '19 at 15:37
  • $\begingroup$ Related. $\endgroup$ Mar 21 '19 at 16:35
  • $\begingroup$ Thanks a lot! It is my first time to hear about Duhamel's formula, $\endgroup$
    – FaDA
    Mar 22 '19 at 0:10
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This is one of the 37 ways to re-express Duhamel's formula. Define $$Z(t)\equiv e^{-it X} e^{it Y}, $$ so that $$\partial_t Z(t)= -i e^{-it X}( X-Y )e^{it Y}. $$ Now, since $Z(t)- 1\!\! 1= \int_0^t d\tau~~ \partial_\tau Z(\tau) $, you have $$ Z(t)-1\!\! 1 = -i\int_0^t d\tau ~~e^{-i\tau X}(X-Y) e^{i\tau Y}, $$
hence $$ e^{it Y}- e^{it X } +i\int_0^t d\tau ~~e^{-i(\tau -t) X}(X-Y) e^{i\tau Y}=0, $$

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  • $\begingroup$ Reference needed for the other 36 :-) $\endgroup$
    – DanielSank
    Mar 21 '19 at 17:52
  • $\begingroup$ Only for 31.5 available.... $\endgroup$ Mar 21 '19 at 18:33
  • $\begingroup$ Agree. Though the most general version is proved via the "interaction picture" $\endgroup$
    – lcv
    Mar 21 '19 at 18:34

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