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?


2 Answers 2


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.


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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