Answer to Question #282969 in Quantum Mechanics for BIGIRIMANA Elie

Question #282969

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


1
Expert's answer
2022-02-06T14:34:52-0500

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.


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