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My professor wants me to solve this identity in two ways. Sadly, I could only do one way and haven't figure out how to solve it another way. Here is my way,

Denote $F(t)=e^{t\hat{A}}\hat{B}e^{-t\hat{A}}$

$\frac{dF}{dt}=\hat{A}e^{t\hat{A}}\hat{B}e^{-t\hat{A}}-e^{t\hat{A}}\hat{B}e^{-t\hat{A}}\hat{A}=[\hat{A},F]=ad\hat{A}\cdot F(t)\quad\Rightarrow\frac{d^nF}{dt^n}=\left(ad\hat{A}\right)^n\cdot F(t)$

Using Taylor expansion $f(x)=\Sigma_n\frac{f^{(n)}(a)}{n!}(x-a)^n$ for $a=0$

$\begin{array}{l}F(t)&=F(0)+F'(0)t+F''(0)\frac{t}{2}+\dots+\frac{F^{(n)}(0)t^n}{n!}\\&=\hat{B}+ad\hat{A}\cdot\hat{B}+\dots\\&=e^{ad\hat{A}t}\cdot\hat{B}\end{array}$

Thus, $e^{\hat{A}}\hat{B}e^{-\hat{A}}=F(1)=e^{ad\hat{A}}\cdot\hat{B}$

Could you show me another way prove this identity?

TBBT
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    Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Pedro Apr 06 '15 at 06:47
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    Yes, I edited my question. Could you take a look, @Pedro? – TBBT Apr 06 '15 at 06:55

1 Answers1

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Here's another way: consider the ordinary, linear, constant coefficient differential equation

$\dfrac{dX(t)}{dt} = AX - XA = [A, X] = (ad A) X, \tag{1}$

$X(0) = B. \tag{2}$

(1) has a unique solution satisfying (2), and since $ad A = [A, \cdot]$ is a constant linear map, this solution is

$X(t) = e^{t (ad A)} B; \tag{3}$

we note that

$X(0) = e^{0 (ad A)} B = e^0 B = B, \tag{4}$

verifying that (3) indeed fulfils (2). Now set

$Y(t) = e^{tA} B e^{-tA}; \tag{5}$

we find

$\dfrac{dY(t)}{dt} = Ae^{tA}Be^{-tA} - e^{tA}B e^{-tA} A = AY(t) - Y(t) A = (ad A) Y(t); \tag{6}$

we see that $Y(t)$ also obeys (1) and furthermore

$Y(0) = B. \tag{7}$

Since both $X(t)$ and $Y(t)$ satisfy the same differential equation with the same initial condition, uniqueness of the solution implies $X(t) = Y(t)$ for all $t$, whence

$e^{tA} B e^{-tA} = e^{t (ad A)} B \tag{8}$

holds for all $t \in \Bbb R$. Now set $t = 1$ and you're done!

Robert Lewis
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  • Thank you very much for your help. – TBBT Apr 07 '15 at 03:33
  • Your are more than welcome. I was glad to write this one up, because I think the equation $e^{tA}Be^{-tA} = e^{t(ad A)}B$ is really important, and this gave me a chance to re-create the derivation. Thanks for the "acceptance", and, of course, Cheers! – Robert Lewis Apr 07 '15 at 03:46