Answer to Question #282969 in Quantum Mechanics for BIGIRIMANA Elie

Question

 Show that if the linear operators Aˆ and Bˆ do not commute, the operators (AˆBˆ + BˆAˆ) and i[A, ˆ Bˆ] are Hermitian. 

Answer 1

Operator $S$ is Hermitian if $S^+=S$, where $+$ means operation of complex conjugate and transpose.

1. Let's check $AB + BA$:

$(AB + BA)^+=(AB)^++(BA)^+=B^+A^++ A^+B^+$

The last expression is equal to $AB + BA$ only if $A$ and $B$ are themselfs Hermitian.

2. Let's check $i[A, B]$:

$(i[A, B])^+ = (i(AB-AB))^+ = -i((AB)^+-(BA)^+)=i(A^+B^+-B^+A^+)$

which is, again, equal to the initial operator $i[A, B]$ only if $A$ and $B$ are Hermitian.

Answer. The operators (AˆBˆ + BˆAˆ) and i[A, ˆ Bˆ] are Hermitian only if A and B are Hermitian as well.