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Hint: OP's formula follows from a Wick-type theorem $$\tag{1} T(f(\hat{A})) ~=~ \exp\left(\frac{1}{2}\hat{C}\frac{\partial}{\partial\hat{A}}\frac{\partial}{\partial\hat{A}} \right):f(\hat{A}):$$ between time-ordering $T$ and normal ordering $::$. Here $$\tag{2} \hat{C}~=~T(\hat{A}\hat{A})~-~:\hat{A}\hat{A}:$$ is a contraction. See e.g. this Phys.SE post ...

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I found a paper that helped me a bit in understanding how things work. So here is what I understood. Given a random variable we define a formal power series and a formal derivation such that: $$\frac{\partial}{\partial U}\left(\sum_{n=0}^\infty a_nU^n\right)=\sum_{n=0}^\infty (n+1)a_{n+1}U^n$$ The "usual" Wick product $:\ :$ is ...

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I expect this to be a result of the Wick theorem. If you are considering an equilibrium situation with a quadratic Hamiltonian all odd moments will vanish. Thus you remain with even powers of your operator. If you now count the number of possible contractions you should get the correct result. P.S.: I just tried and it works indeed. Note the helpful ...

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Comments to the question (v3): The main fields assumption that goes into the proof of the Wick's theorem for fields $\hat{\phi}^i\in{\cal A}$ is that their (super)commutators $$[\hat{\phi}^i,\hat{\phi}^j]~\in~ Z({\cal A}) \tag{1}$$ are central elements of the operator algebra ${\cal A}$, cf. e.g. this Phys.SE post. For free fields $\hat{\phi}^i\in{\cal A}$,...

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