Question

show that [A²,B]=A[A,B]+[A,B]A

Answer 1

We use the definition for the operator on a function f:

[A^2,B]f=A[A,B]f+[A,B]Af \\ [A^2,B]f=A^2Bf-BA^2f \\ [A^2,B]f=A(AB)f-(BA)Af

Following that, we add one term that is equal to zero to then rearrange and find the equivalence with the definitions for [A,B]:

[A^2,B]f=A(AB)f-(BA)Af-A(BA)f+(AB)Af \\ =A(AB)f-A(BA)f-(BA)Af+(AB)Af \\ =A[AB-BA]f+[AB-BA]Af \\ [A^2,B]f =A[A,B]f+[A,B]Af

In conclusion, we were able to show that [A²,B]=A[A,B]+[A,B]A.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.