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I'm reading the paper (eq.(14) and eq.(10)) and got curious how the paper uses this equation:

$\frac{\partial}{\partial c}\exp(-i\Delta t (X+cY)) = \exp(-i\Delta t (X+cY))(-iY\Delta t + \frac{\Delta t^2}{2}[X+cY, Y] + \frac{i\Delta t^3}{6}[X+cY, [X+cY,Y]]+ \cdots )$

Can anybody help me deriving the equation?

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2 Answers 2

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The standard identity for the derivative of the exponential map is $$ \partial_c e^{M(c)}= e^{M} \left (1-\frac{1}{2}[M,\bullet]+ \frac{1}{6}[M,[M,\bullet]]+... \right ) \partial_c M, $$ where $\bullet$ pipes the argument on the right in case you were not familiar with the adjoint map.

So, just plug in, $M= -i\Delta t (X+cY)$, $$ \partial_c e^{-i\Delta t (X+cY)} \\ = e^{-i\Delta t (X+cY)} \Delta t \Bigl (-i Y +[\Delta t (X+cY), Y ]/2 + i[\Delta t (X+cY) , [\Delta t (X+cY),Y]]/6 +...\Bigr ) , $$ amounting to your result.

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OP's sought-for identity is

$$\begin{align}e^{-\hat{A}}\frac{d}{d\lambda}e^{\hat{A}} ~=~& \int_0^1\!ds~e^{-s\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}}\cr ~\stackrel{(3)}{=}~& \int_0^1\!ds~e^{-s~{\rm ad}\hat{A}}\frac{d\hat{A}}{d\lambda} \cr ~=~& \int_0^1\!ds\sum_{n=0}^{\infty}\frac{(-s~{\rm ad}\hat{A})^n}{n!}\frac{d\hat{A}}{d\lambda}\cr ~\stackrel{(4)}{=}~& \sum_{n=0}^{\infty}\frac{(-{\rm ad}\hat{A})^n}{(n+1)!}\frac{d\hat{A}}{d\lambda} ,\end{align}\tag{1}$$

where we have defined the adjoint map

$${\rm ad}\hat{A}~\equiv ~[\hat{A},~\cdot~],\tag{2}$$

used the identity

$$ e^\hat{X} \hat{Y} e^{-\hat{X}}~=~e^{{\rm ad}\hat{X}}\hat{Y},\tag{3}$$

and used the integral

$$ \int_0^1\!ds~s^n~=~\frac{1}{n+1}.\tag{4}$$

The first equality in eq. (1) is proven in my Phys.SE answer here.

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  • $\begingroup$ This is very helpful, thanks. How did you go from 2nd to 3rd, and 3rd to 4th line in eq. (1)? (And I assume that from eq. (1), we can just put $e^{A}$ on each side and plug A = M to reproduce the equation of interest since $e^{A}e^{-A} = I$.) $\endgroup$
    – Jon Megan
    Mar 12, 2022 at 22:14
  • $\begingroup$ (Oops, the transition from 2nd to 3rd line is just a Taylor expansion!) $\endgroup$
    – Jon Megan
    Mar 12, 2022 at 23:11
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Mar 13, 2022 at 2:29
  • $\begingroup$ Thanks! I'm not familiar with adjoint map (and Lie algebra/group). How is the adjoint map supposed to appear here? What is the mathematical/physical intuition of connecting adjoint map in the calculation? $\endgroup$
    – Jon Megan
    Mar 13, 2022 at 3:27
  • $\begingroup$ While the adjoint map is certainly an interesting topic, only the definition (2) itself is needed for the sought-for formula (1). $\endgroup$
    – Qmechanic
    Mar 14, 2022 at 16:37

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