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I want to simplify a product of normally ordered exponentials that are in the following form

$$:e^{x(\hat{a}^\dagger+\alpha_x^*)(\hat{a}+\alpha_x)}:\times :e^{y(\hat{a}^\dagger+\alpha_y^*)(\hat{a}+\alpha_y)}:$$

where $\times$ is the usual multiplication, $\hat{a}$ is a bosonic annihilation operator, $\alpha_i$ are complex numbers, and $:\bullet:$ represents normal ordering where you rewrite the all $\hat{a}^\dagger$ to the left of $\hat{a}$.

Is the solution to this simply $::$ of the product? It looks to me that it shouldn't. Is there a trick or an identity to solve this, in case it is not trivial?

Qmechanic
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sash
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2 Answers2

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Comments to the post (v3):

  1. OP is right that a product $~:\hat{A}::\hat{B}:~$ of normal-ordered operators is not necessarily the same as normal-ordering $~:\hat{A}\hat{B}:~$ the product of operators. Instead the 2 expressions are related via a pertinent nested Wick theorem/star product, such as e.g. $$:\hat{A}\hat{B}:~=~:\hat{A}:\exp\left(-\frac{\stackrel{\longleftarrow}{\partial}}{\partial \hat{a}}\frac{\stackrel{\longrightarrow}{\partial}}{\partial \hat{a}^{\dagger}}\right):\hat{B}:$$ similar to my Phys.SE answer here.

  2. For what it's worth, a single factor in OP's product can be rewritten as $$:e^{x(\hat{a}^{\dagger}+\alpha^{\ast})(\hat{a}+\alpha)}:~=~ e^{x(\hat{a}^{\dagger}+\alpha^{\ast})\alpha} :e^{x \hat{a}^{\dagger} \hat{a}}:e^{x\alpha^{\ast}\hat{a}} $$

Qmechanic
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2

Consider a simpler example, where $f=a a^\dagger a a^\dagger$.

Then $$ :f: \times :f: = \left(a^\dagger a^\dagger a a\right) \left(a^\dagger a^\dagger a a\right) = a^\dagger a^\dagger a a a^\dagger a^\dagger a a $$ whereas $$ :f\times f: = :\left(a a^\dagger a a^\dagger\right) \left(a a^\dagger a a^\dagger\right): = a^\dagger a^\dagger a^\dagger a^\dagger a a a a $$ Therefore $$ :f: \times :f: \neq :f\times f: $$

Andrew
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