? If \hat{A} and \hat{B} are two commuting operator. proof that i) [\hat{A}, \hat{B}] + [\hat{B}, \hat{A}] = 0 ii) [\hat{A}, \hat{A}] = 0 iii) [\hat{A}, \hat{B} + \hat{C}] = [\hat{A}, \hat{B}] + [\hat{A}, \hat{C}] iv) [\hat{A} + \hat{B}, \hat{C}] = [\hat{A}, \hat{B}] + [\hat{B}, \hat{C}] v) [\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{C}]\hat{B} + \hat{A}[\hat{B}, \hat{C}]
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[ABJ+[3A]=0 To prove this, we can use the commutation relation [A,B] = AB - BA. So, [ABJ+[3A] = ABJ + 3A - (BAJ + 3A) = ABJ + 3A - BAJ - 3A = ABJ - BAJ = (AB - BA)J = [A,B]J = 0. Show more…
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