# Derivative of the product of operators and Derivative of exponential

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: $$\frac{d}{d\lambda}\exp(\hat{A}(\lambda) )~?$$

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. $$\Box$$

• Related perturbative formulas: $$e^{-t\hat{A}}e^{t(\hat{A}+\hat{B})}-{\bf 1} ~=~\int_0^t\!ds~e^{-s\hat{A}}\hat{B}e^{s(\hat{A}+\hat{B})}\qquad \Leftrightarrow$$ $$e^{t(\hat{A}+\hat{B})}~=~e^{t\hat{A}} +\int_0^t\!ds~e^{(t-s)\hat{A}}\hat{B}e^{s(\hat{A}+\hat{B})}~=~\ldots .$$ Apr 26, 2015 at 20:07
• $(e^{t\hat{A}+t[\hat{B},\cdot]}1) ~=~e^{t\hat{A}+t\hat{B}}e^{-t\hat{B}}$. Sep 5, 2017 at 18:02
• $\quad \frac{d\hat{U}(t)}{dt}~=~\hat{A}(t)\hat{U}(t);$ $\quad \hat{U}(t\!=\!0)~=~{\bf 1};$ $\quad \hat{U}(t)~=~\left\{\begin{array}{rcl} T\exp\left(\int_0^t\! ds~\hat{A}(s)\right)&{\rm for}& t\geq 0,\cr AT\exp\left(\int_0^t\! ds~\hat{A}(s)\right)&{\rm for}& t\leq 0,\end{array}\right.$ $\quad \hat{U}(t)^{-1}\frac{d\hat{U}(t)}{d\lambda}~=~\int_0^t\! ds~\hat{U}(s)^{-1}\frac{d\hat{A}(s)}{d\lambda}\hat{U}(s).$ Nov 2, 2019 at 13:32
• $\quad \frac{d\hat{U}(t_2,t_1)}{dt_2}~=~\hat{A}(t_2)\hat{U}(t_2,t_1);$$\quad \frac{d\hat{U}(t_2,t_1)}{dt_1}~=~-\hat{U}(t_2,t_1)\hat{A}(t_1); \quad \hat{U}(t,t)~=~{\bf 1}; \quad \hat{U}(t_2,t_1)~=~\left\{\begin{array}{rcl} T\exp\left(\int_{t_1}^{t_2}\! dt~\hat{A}(t)\right)&{\rm for}& t_1\leq t_2,\cr AT\exp\left(\int_{t_1}^{t_2}\! dt~\hat{A}(t)\right)&{\rm for}& t_2\leq t_1,\end{array}\right. Jan 11, 2022 at 14:38 • \quad\frac{d\hat{U}(t_2,t_1)}{d\lambda}~=~\frac{dt_2}{d\lambda}\hat{A}(t_2)\hat{U}(t_2,t_1) +\int_{t_1}^{t_2}\! dt~\hat{U}(t_2,t)\frac{d\hat{A}(t)}{d\lambda}\hat{U}(t,t_1) -\hat{U}(t_2,t_1)\hat{A}(t_1)\frac{dt_1}{d\lambda}. Jan 12, 2022 at 13:26$$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)$$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.