Proving properties of Hermitian conjugate I have three properties:

*

*If $\hat{A}$ and $\hat{B}$ are Hermitian operators. Then $\hat{A}\hat{B}$ is Hermitian provided $\hat{A}$ and $\hat{B}$ also commute $[\hat{A},\hat{B}]=0$

*If $\hat{A}$ and $\hat{B}$ are Hermitian operators and  $\hat{A}$ and $\hat{B}$ also commute, then $\hat{A}+\hat{B}$ is Hermitian

*If $\hat{A}$ and $\hat{B}$ are Hermitian operators, and  $\hat{A}$ and $\hat{B}$ do not commute, then $\hat{A}\hat{B}+\hat{B}\hat{A}$ is Hermitian

I am trying to prove all these properties.

1st one:


For the second one I'm struggling, as I do not know how to expand $(\hat{A}+\hat{B})^\dagger$
 A: As Jakob commented, to prove identities of that kind it is often good to go back to the definition of the adjoint operator as arising from an inner product. Given an inner product $(\cdot,\cdot)$ and an operator $\hat{A}$, one defines the adjoint operator $\hat{A}^\dagger$ to be the operator that satisfies
$$(v,\hat{A}w) = (\hat{A}^\dagger v,w)$$
for all vectors $v,w$ (on a more technical note, one might have to restrict the condition from "all vectors" to "those vectors where the quantities are defined", but that is typically omitted in introductory QM lectures). With that, you can prove $(\hat{A}\hat{B})^\dagger = \hat{B}^\dagger \hat{A}^\dagger$ by
$$((\hat{A}\hat{B})^\dagger v,w) \stackrel{\textrm{def}}{=} (v,\hat{A}\hat{B}w) = (v,\hat{A}(\hat{B}w)) \stackrel{\textrm{def}}{=} (\hat{A}^\dagger v,\hat{B}w)
\stackrel{\textrm{def}}{=} (\hat{B}^\dagger(\hat{A}^\dagger v),w)
= (\hat{B}^\dagger\hat{A}^\dagger v,w).$$
Analogous to that, and just using the linearity of the inner product, i.e.
$$(v, w + \lambda u) = (v, w) + \lambda(v,u)$$
with vectors $v,w,u$ and a scalar $\lambda \in \mathbb{C}$, you can figure it out. Try it and comment if that works, otherwise I'll add another edit.

If the inner product notation is unfamiliar, replace braces with bras and kets and write greek letters,
$$\langle \psi | \hat{A} \phi \rangle\ \sim (v,\hat{A}w)$$
A: Taking Hermitian Conjugate is simply taking Complex Conjugate and then Transpose, both operations are linear so their composition is also linear:
$$(\hat{A}+\hat{B} )^{\dagger}$$
$$=((\hat{A}+\hat{B} )^{T})^{*}$$
$$=(\hat{A}^{T}+\hat{B}^{T} )^{*}$$
$$=(\hat{A}^{T})^{*}+(\hat{B}^{T} )^{*}$$
$$=\hat{A}^{\dagger}+\hat{B}^{\dagger}$$
$$=\hat{A}+\hat{B} $$
In fact it doesn't matter that $\hat{A} \text{ and } \hat{B} $ commute, their sum is Hermitian in general.
For the last one there is a similar process except when you do the transpose you have to the order of multiplication, but then by switching the order of addition you obtain the same thing back again.
