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.