[Physics] n identity for anti-commutator $\{ A B, \, C D \}$ in terms of commutators $[\, , \,]$ only

anticommutatorclifford-algebracommutatoroperatorsquantum mechanics

I'm looking for an identity that could express the anti-commutator
$$\tag{1}
\{ A B , \, C D \} \equiv A B C D + C D A B
$$

expressed as a combination of commutators only: $[A,\, C]$, $[A, \, D]$, etc.

Is there such an identity? I suspect there is none.

I'm thinking of something similar to the following identity (with only commutators on the right side, no anti-commutators):
$$\tag{2}
[A B, \, C D] = A C \, [B, \, D] + A \, [B, \, C] \, D + C \, [A, \, D] \, B + [A, \, C] \, D B.
$$


EDIT: I'm more specifically interested in evaluating the following:
$$\tag{3}\{ a_i^{\dagger} \, a_j, \, a_k^{\dagger} \, a_l \},$$
where $a_i$ and $a_i^{\dagger}$ are bosonic operators satisfying
\begin{align}
[ a_i, \, a_j^{\dagger}] &= \delta_{ij}, \tag{4}\\[1ex]
[a_i, \, a_j] = [a_i^{\dagger}, \, a_j^{\dagger}] &= 0. \tag{5}
\end{align}

Best Answer

Let me put on my math hat by providing a counterexample:

Let's say $$ A = a_1, \\B= a_2, \\C= a_3, \\D = a_4 $$ where $a_i$ are regular bosonic anhiliation operators, therefore $$ [a_i, a_j] \equiv 0 $$

If OP's proposition works: $$ \{ A B , \, C D \} \equiv A B C D + C D A B = 2a_1a_2a_3a_4 $$ will be reduced to zero, since any combination of $[a_i, a_j] $ is zero. But $2a_1a_2a_3a_4$ is evidently not identical to zero.

Thus there is no such construct.

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