Prove that $(\hat A \hat B + \hat B \hat A )$ is self-adjoint if $\hat A$ and $\hat B$ are self-adjoint

Question

I'm not clear why, in case $\hat A$ and $\hat B$ are self-adjoint, then $(\hat A \hat B + \hat B \hat A )$ is also self-adjoint. In order to show that $(\hat A \hat B + \hat B \hat A )=(\hat A \hat B + \hat B \hat A )^\dagger$, I have seen that the property $(\hat a \hat b)^\dagger=\hat b^\dagger \hat a^\dagger$ is applied, so $(\hat A \hat B + \hat B \hat A )^\dagger=\hat B^\dagger \hat A^\dagger + \hat A^\dagger \hat B^\dagger$. However, I don't get with this last is equal to $\hat A \hat B + \hat B \hat A $. Could someone give me a hint?

Answer 1

$$\begin{align} \left( \hat A\hat B+\hat B\hat A\right)^\dagger &=\left(\hat A\hat B\right)^\dagger+\left( \hat B\hat A\right)^\dagger \tag{1 }\\ \tag{2} &= \hat B^\dagger\hat A^\dagger+\hat A^\dagger\hat B^\dagger \\ \tag{3} &= \hat A^\dagger\hat B^\dagger+ \hat B^\dagger\hat A^\dagger \\ \tag{4 }&= \hat A\hat B + \hat B\hat A.\end{align}$$

1. Distribute the Hermitian conjugate between terms.

2. $(\hat X\hat Y)^\dagger = \hat Y^\dagger \hat X^\dagger$ as you have correctly said.

3. Matrix addition is commutative.

4. $\hat A^\dagger=\hat A$ and $\hat B^\dagger =\hat B$ as they are both self-adjoint.