# Baker-Hausdorff for normal ordering exponential

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.

## 1 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:

$$:e^A::e^B:~=~e^{A^+}e^{A^-}e^{B^+}e^{B^-}~=~e^{A^+}e^{[A^-,B^+]}e^{B^+}e^{A^-}e^{B^-}$$ $$~=~e^{[A^-,B^+]}e^{A^++B^+}e^{A^-+B^-}~=~e^{[A^-,B^+]}:e^{A+B}:\tag{5.284'}$$

References:

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

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$$^1$$ 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.