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  1. I'm asked to show that $$\frac{d(\hat{A}\hat{B})}{d\lambda} ~=~ \frac{d\hat{A}}{d\lambda}\hat{B} + \hat{A}\frac{d\hat{b}}{d\lambda}$$ With $\lambda$ a continuous parameter. Should I use the definition $$\frac{d\hat{A}}{d\lambda} ~=~ \lim_{\epsilon \to 0} \frac{\hat{A}(\lambda + \epsilon) - \hat{A}(\lambda)}{\epsilon}$$ applied to $\hat{A}\hat{B}$ like $$\frac{d(\hat{A}\hat{B})}{d\lambda} ~=~ \lim_{\epsilon \to 0} \frac{\hat{A}(\lambda + \epsilon)\hat{B}(\lambda + \epsilon) - \hat{A}(\lambda)\hat{B}(\lambda)}{\epsilon}$$ and do some algebra to get the RHS of the first equation, or I'm missing something?

  2. Another interesting derivative to pay attention to is: $$\frac{d}{d\lambda}\exp(\hat{A}(\lambda) )~?$$

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$$A(\lambda+\epsilon)B(\lambda+\epsilon) = (A(\lambda) + \epsilon \dot{A} )(B(\lambda) +\epsilon \dot B ) = A(\lambda)B(\lambda) + \epsilon(\dot AB+A\dot B) + o(\epsilon^2)$$

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Here we will only consider the added last subquestion (v4):

$$ \frac{d}{d\lambda}e^{\hat{A}} ~=~ \int_0^1\!ds~e^{(1-s)\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}} .\tag{1}$$

The identity (1) follows by setting $t=1$ in the following identity

$$ e^{-t\hat{A}} \frac{d}{d\lambda}e^{t\hat{A}} ~=~ \int_0^t\!ds~e^{-s\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}} .\tag{2}$$

To prove equation (2), first note that (2) is trivially true for $t=0$. Secondly, note that a differentiation wrt. $t$ on both sides of (2) produces the same expression

$$e^{-t\hat{A}}[\frac{d}{d\lambda},\hat{A}]e^{t\hat{A}}~=~e^{-t\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t\hat{A}},\tag{3}$$

where we use the fact that

$$\frac{d}{dt}e^{t\hat{A}}~=~\hat{A}e^{t\hat{A}}~=~e^{t\hat{A}}\hat{A}.\tag{4}$$

So the two sides of eq.(2) must be equal. $\Box$

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  • $\begingroup$ Related perturbative formulas: $$e^{-t\hat{A}}e^{t(\hat{A}+\hat{B})}-{\bf 1} ~=~\int_0^t\!ds~e^{-s\hat{A}}\hat{B}e^{s(\hat{A}+\hat{B})}\qquad \Leftrightarrow $$ $$e^{t(\hat{A}+\hat{B})}~=~e^{t\hat{A}} +\int_0^t\!ds~e^{(t-s)\hat{A}}\hat{B}e^{s(\hat{A}+\hat{B})}~=~\ldots .$$ $\endgroup$ – Qmechanic Apr 26 '15 at 20:07
  • $\begingroup$ $(e^{t\hat{A}+t[\hat{B},\cdot]}1) ~=~e^{t\hat{A}+t\hat{B}}e^{-t\hat{B}}$. $\endgroup$ – Qmechanic Sep 5 '17 at 18:02
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Here are hints for 2: Note that $e^{\hat{A}(\lambda)}=\sum_{n=1}^\infty \frac{\hat{A}(\lambda)^n}{n!}$ and in the interest of preserving order that $\frac{d}{d\lambda} \hat{A}(\lambda)^n=\sum_{k=0}^{n-1} \hat{A}(\lambda)^k \frac{d\hat{A}}{d\lambda} \hat{A}(\lambda)^{n-k-1}$. Using these facts, and the beta function fact $B(n,k) = \frac{n!(k+1)!}{(n+k+1)!}=\int_0^1 x^{n-1}(1-x)^k dx$, through some sum swapping, index changing, and factorial rewriting you can arrive at "Sneddon's formula" for $\frac{d}{d\lambda}e^{\hat{A}(\lambda)}$. All these details can be found in "Mathematical Methods of quantum optics" by Ravinder Rupchand Puri, the beginning of chapter 2.

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