# [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}

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.