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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?

Qmechanic
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haha
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  • hint: apply derivative of products rule to derive the polynomial case, and think how to apply that to your problem – lurscher Dec 26 '18 at 04:04
  • @lurscher Ok so this is what I tried. $\frac{d A^3}{d t} = \frac{dA}{dt} A^2 + A \frac{dA}{dt}A + A^2 \frac{dA}{dt}$, $\frac{d A^4}{d t} = \frac{dA}{dt} A^3 + A \frac{dA}{dt} A^2 + A^2 \frac{dA}{dt} A + A^3 \frac{dA}{dt}$, so $\frac{d A^n}{dt} = \frac{dA}{dt} A^{n-1} + A \frac{dA}{dt} A^{n-2} + ... + A^{n-2} \frac{dA}{dt} A^2 + A^{n-1} \frac{dA}{dt}$. Is that what you mean? – haha Dec 26 '18 at 07:13
  • at this point you need to add an explicit knowledge of the commutator $\lbrace A, \frac{dA}{dt} \rbrace$ (in QM basic courses, this is a scalar operator) and use it to simplify the factors. Once you have the tidiest expression for $A^n$, use the Taylor expansion of the function you are interested (exp in this case) – lurscher Dec 26 '18 at 07:18
  • Possible duplicate: https://physics.stackexchange.com/q/15742/2451 – Qmechanic Dec 27 '18 at 14:25
  • @Qmechanic uh I don’t think they answered the question I asked in *EDIT* – haha Dec 27 '18 at 23:53
  • Consider double-checking the derivation leading to the flawed eq. (5). – Qmechanic Dec 28 '18 at 00:26
  • Being new to the community, I'd like to answer the OP's question in *EDIT*, but am not sure how to do it: the question is closed. I've submitted suggested edits to the only answer here to include my answer, but was rejected by the author of the answer because "the edit is not relevant to the answer". – terraregia Oct 30 '20 at 04:19

1 Answers1

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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.

Sunyam
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