I am trying to understand the last equation from page 2 of this pdf http://physics.gu.se/~frtbm/joomla/media/mydocs/LennartSjogren/kap7.pdf but I am not being able to develop as here it says. Could someone help me?

This is what I have tried:

$$ \frac{d}{dt}e^{(-L_0\cdot t)} = e^{(-L_0\cdot t)}\frac{d}{dt}(L_0\cdot t) $$

$$ p = e^{-L_0\cdot t} \sigma $$

$$ \frac{d}{dt} p= -e^{-L_0t}\frac{d}{dt}(L_0t) \sigma + e^{-L_0t}\frac{d}{dt}\sigma $$

\begin{align} \frac{d}{dt} \sigma&= e^{L_0t}\frac{d}{dt}p + \frac{d}{dt}(L_0t)\sigma=-e^{L_0t}\left( L_0p+L_1p\right)+\frac{d}{dt}(L_0t)\sigma=\\ &=\left[-e^{L_0t}\left( L_0e^{-L_0t}+L_1e^{-L_0t} \right)+\frac{d}{dt}(L_0t)\right]\sigma=\\ &=\left[ -e^{L_0t}L_0e^{-L_0t}-e^{L_0t}L_1e^{-L_0t}+\frac{d}{dt}(L_0t)\right] \sigma \end{align}


1 Answer 1



  • As $L_0$ does not depend on $t$, we have that $(d/dt) (L_0 t) = L_0$.

  • Also $e^{-L_0 t}$ commutes with $L_0$, so $$e^{L_0 t} L_0 e^{-L_0 t} = e^{L_0 t} e^{-L_0 t} L_0 = L_0.$$

So we can simply your last expression: $$ \frac{d \sigma}{dt} =\left[ - \underbrace{e^{L_0t}L_0e^{-L_0t}}_{L_0}-e^{L_0t}L_1e^{-L_0t}+ \underbrace{\frac{d}{dt}(L_0t)}_{L_0}\right] \sigma = - V(t) \sigma$$ with $$ V(t) = e^{L_0t}L_1e^{-L_0t}.$$

This derivation is quite standard in physics and corresponds (with imaginary time) to the interaction (Dyson) representation in quantum mechanics; related is also the notion of the Dyson series.


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