# Is quantum entanglement beyond Jacobi identity?

If the state of some particles can't be derived by the tensor product of each particle's state, those are called entangled, which is beyond the simple pairwise additive classical correlation. What about Jacobi identity? In classical physics, magnetism is not pairwise additive but it can be described by satisfying Jacobi identity. I mean Bell's inequality should hold for a 2-body problem not a 3-body one. Let's assume $$S_A$$ , $$S_B$$ , $$S_M$$ corresponds to Alice, Bob and measurement states respectively, Why can't we explain the violation of Bell's inequality using the Jacobi identity classically?

$$S_A \times (S_B \times S_M ) + S_B \times (S_M \times S_A ) + S_M \times (S_A \times S_B ) = 0$$
Or: $$[S_A,[S_B,S_M]]+ [S_B,[S_M,S_A]]+ [S_M,[S_A,S_B]]= 0$$

Please edit my question to be more accurate.

• What do you mean with the sentence " magnetism is not pairwise additive but satisfies Jacobi identity"? Jun 18 at 19:21
• @Davius You are right, In formal classical physics we only talk about magnetism under field-particle approximation, I'll edit the question. Jun 18 at 19:36