Proving the Identity for Non-Commuting Operators A and B | Operator Algebra

[S.G.]

How do you prove the following identity for non-commuting operators A and B?

[[[A,B],B]A]=[B,[A,[A,B]]]

 

[Edgardo]

Hello S.G.,

you could first use the abbreviation C:= [A,B], so the equation
looks better:

[[[A,B],B]A]=[B,[A,[A,B]]]
<=>
[[C,B],A] = [B,[A,C]]

Then use the definition of the commutator [Y,Z] = YZ-ZY
and in the end, instead of C, use [A,B] again.

 

[R0man]

To prove this identity, we can use the properties of operator algebra, specifically the Jacobi identity. This identity states that for any three operators A, B, and C, we have [[A, B], C] + [[B, C], A] + [[C, A], B] = 0.

Now, let's apply this identity to our equation: [[[A, B], B], A] = [[B, [A, B]], A] + [[A, B], [B, A]].

Since A and B do not commute, [A, B] is not equal to [B, A], so we cannot simply swap the order of these operators. However, we can use the Jacobi identity to rewrite the second term as [[A, [A, B]], B].

Substituting this into our equation, we get: [[[A, B], B], A] = [[B, [A, B]], A] + [[A, [A, B]], B].

Now, we can use the Jacobi identity again to rewrite the first term as [[A, B], [B, A]]. Substituting this into our equation, we get: [[[A, B], B], A] = [[B, [A, B]], A] + [[A, [A, B]], B] = [[B, [A, B]], A] + [[A, B], [B, A]].

Since A and B do not commute, we cannot further simplify this expression. However, we can see that the right side of the equation is equal to the left side, therefore proving the identity: [[[A, B], B], A] = [[B, [A, B]], A] + [[A, B], [B, A]].