Baker-Hausdorff for normal ordering exponential

Question

Let $A=A^+ +A^-$ where $A^+,A^-$ denote the creation and annihilation portion of the field. Then in Eduardo Fradkin, Field Theories of Condensed Matter Physics, equation (5.284), it states that
$$
:e^A::e^B: ~=~ e^{[A^+,B^-]}:e^{A+B}:tag{5.284}
$$
where $::$ denotes normal-ordering of $A^+,A^-$. I'm familiar with the regular Baker-Hausdorff formula, but I'm not sure why this identity is true.

EDIT: Here's my attempt.
begin{align}
:A^n: &= text{He}_n(A) 
:e^A: &= sum_{n=0}^infty frac{1}{n!}text{He}_n(A) =e^{A-1/2}
:e^A::e^B: &= e^{-1} e^A e^B
&= e^{-1} e^{A+B} e^{frac{1}{2} [A,B]} 
&= :e^{A+B}:e^{-1/2} e^{i Im[A^+,B^-]}
end{align}
where I implicitly assumed that $[A,B]$ is a complex multiple of the identity. However, you can see that my result doesn't quite match the equation.

Qmechanic · Accepted Answer

Ref. 1 contains several$^1$  typos, e.g. the aforementioned eq. (5.284) if we use$^1$ the definition above eq. (5.262):

Let $phi^+(x)$ ($phi^-(x)$) denote the piece of $phi(x)$ which depends on the creation (annihilation) operators only,
$$phi(x) ~=~phi^+(x)+phi^-(x).tag{5.262}$$

The corrected eq. (5.284) is derived as follows:
$$begin{align}   :e^A::e^B:~=~&e^{A^+}e^{A^-}e^{B^+}e^{B^-}cr
~=~&e^{A^+}e^{[A^-,B^+]}e^{B^+}e^{A^-}e^{B^-}cr
~=~&e^{[A^-,B^+]}e^{A^++B^+}e^{A^-+B^-}cr
~=~&e^{[A^-,B^+]}:e^{A+B}:end{align} tag{5.284'}$$
References:

E. Fradkin, Field Theories of Condensed Matter Physics, 2nd ed. (2013).

--
$^1$ Independently, there is a wrong sign in the truncated BCH formula (5.269).
$^2$ Alternatively, eq. (5.284) is correct in its printed form if we use the opposite notation for creation & annihilation operators.

Baker-Hausdorff for normal ordering exponential

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