Forms That Are Always Hermitian for an Arbitrary Operator
Formula
For an arbitrary operator $A$, the following forms are always Hermitian operators.
$$ A + A^{\dagger} \tag{1} $$
$$ \i (A - A^{\dagger}) \tag{2} $$
$$ A A^{\dagger} \tag{3} $$
Proof
The proof is complete if we show that taking the conjugate transpose $^{\dagger}$ of the original expression yields the original form again.
(1)
$$ (A+A^{\dagger})^{\dagger}=A^{\dagger}+{(A^{\dagger})}^{\dagger}=A^{\dagger}+A=A+A^{\dagger} $$
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(2)
$$ [\i(A-A^{\dagger}) ]^{\dagger} = -\i\left[ A^{\dagger}-{(A^{\dagger})}^{\dagger} \right] = -\i(A^{\dagger}-A) = \i(A-A^{\dagger}) $$
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(3)
$$ (AA^{\dagger})^{\dagger}={(A^{\dagger})}^{\dagger} A^{\dagger}=AA^{\dagger} $$
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