What is the time derivative $\frac{d}{dt}(\exp(\hat A))$ of operator exponential $\exp(\hat{A})$?

Question

Assuming that the operator $\hat{A}$ does not necessarily commute with $d \hat A/dt$, what is $$\frac{d}{dt}(\exp(\hat A))~?\tag{1}$$ Is it $\exp(\hat A(t)) \frac{d \hat A}{dt}$ or $\frac {d\hat A}{dt} \exp(\hat A(t))$ and why?

EDIT: Actually here is my original question: I was asked to show that

$$\frac{d(\exp{A})}{dt} \exp{(-A)} = \sum_{n=0}^{\infty} \frac{1}{(n+1)!} L^n_A(\frac{dA}{dt}),\tag{2}$$ with $$L_A(X) = [A,X], \qquad L^2_A(X) = [A,[A,X]],\qquad \ldots \tag{3}$$

I thought the Baker–Campbell–Hausdorff formula said

$$\exp{(A)} B \exp{(-A)} = \sum_{n=0}^{\infty} \frac{1}{n!} L^n_A(B),\tag{4}$$

and so comparing the two equations above, I would get

$$\frac{d\exp(A)}{dt} = \exp{(A)} \frac{dA}{dt}\tag{5}$$ (by plugging in $B = \frac{dA}{dt}$). Now that it seems that this is not true, does it mean that the identity I was asked to prove is also not true?

Answer 1

Note : This is a not so rigorous answer. Here non-rigorous proof of the following identity is given : $$\frac{d}{dt}e_{}^{\hat A(t)}=\int_{0}^{1}dz e_{}^{\hat A(t) z}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]e_{}^{\hat A(t) (1-z)}.$$

OP's question is about the derivative (with respect to a parameter) of exponential of an operator which depends on a parameter. i.e., $$\frac{d}{dt}e_{}^{\hat A(t)}.$$

This can be deduced using non-commutatitve Liebnitz rule as follows :

(i) Expand the exponential using Taylor-Maclaurin formula as : $$\frac{d}{dt}e_{}^{\hat A(t)}= \frac{d}{dt}\sum_{n=0}^{\infty}\frac{1}{\Gamma[n+1]}\hat A(t)_{}^{n}.$$ (ii) Apply non-commutatitve Liebnitz rule to each term of the sum as: $$\frac{d}{dt}e_{}^{\hat A(t)}= \sum_{n=1}^{\infty}\frac{1}{\Gamma[n+1]}\sum_{k=0}^{n-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}.$$ (iii) Use the following identity : $$\frac{1}{\Gamma[k+1]\Gamma[n-k]}\int_{0}^{1}dz z_{}^{k}(1-z_{}^{})_{}^{n-k-1}=\frac{1}{\Gamma[n+1]}$$ to get : $$\frac{d}{dt}e_{}^{\hat A(t)}= \frac{d}{dt}\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{1}{\Gamma[k+1]\Gamma[n-k]}\int_{0}^{1}dz z_{}^{k}(1-z_{}^{})_{}^{n-k-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}$$ which gives : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{1}{\Gamma[k+1]\Gamma[n-k]} z_{}^{k}(1-z_{}^{})_{}^{n-k-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}$$ which upon change of summation variable $n \rightarrow n-1$ gives : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{1}{\Gamma[k+1]\Gamma[n-k+1]} z_{}^{k}(1-z_{}^{})_{}^{n-k}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k}.$$ (iv) Finally using the change of double summation formula , it follows : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{\Gamma[k+1]\Gamma[n+1]} z_{}^{k}(1-z_{}^{})_{}^{n}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n}.$$ which when rearranged gives : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{k=0}^{\infty}\frac{z_{}^{k}\hat A(t)_{}^{k}}{\Gamma[k+1]} \left[\frac{d}{dt}\hat A(t)_{}^{}\right]\sum_{n=0}^{\infty}\frac{(1-z_{}^{})_{}^{n}\hat A(t)_{}^{k}}{\Gamma[n+1]}.$$ (v) Upon taking the exponential again, gives the desired result : $$\boxed{\frac{d}{dt}e_{}^{\hat A(t)}=\int_{0}^{1}dz e_{}^{\hat A(t) z}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]e_{}^{\hat A(t) (1-z)}}.$$

Note : For a short and elegant proof see this answer by Qmechanic.