Expectation of Complex Operators Given an operator $\hat{\alpha}$, how do we obtain,
$$
\sqrt{
\left\langle
    \left(
        \hat{\alpha} - \left\langle\hat{\alpha}\right\rangle
    \right)^2
\right\rangle
}
=
\sqrt{
        \left\langle\hat{\alpha}^2\right\rangle
        -
        \left\langle\hat{\alpha}\right\rangle^2
}
$$
My thoughts... To simplify, I define $A$ as,
$$
A\equiv
    \left(
        \hat{\alpha} - \left\langle\hat{\alpha}\right\rangle
    \right)^2
=
        \hat{\alpha}^2 
        +
        \left\langle\hat{\alpha}\right\rangle^2
        -
        2\hat{\alpha}\left\langle\hat{\alpha}\right\rangle
$$
so what do I do with the extra $-2
\left\langle
        \hat{\alpha}\left\langle\hat{\alpha}\right\rangle
\right\rangle$ in $\langle A\rangle$?
$$
\implies
\left\langle
        \hat{\alpha}^2 
\right\rangle
        +
\left\langle
        \left\langle\hat{\alpha}\right\rangle^2
\right\rangle
        -2
\left\langle
        \hat{\alpha}\left\langle\hat{\alpha}\right\rangle
\right\rangle
$$
 A: For the ease of notation I will write $T$ for the operator instead of $\hat \alpha$ 
Remember that $\langle T \rangle$ is defined as $\langle T \rangle = \langle \psi ,T \psi\rangle$ for some state $\psi$ (in the domain of $T$ with $\langle \psi,\psi \rangle = 1$). Then $\langle T \rangle$ is indeed real if $T$ is hermetian.
In order to calculate $\left\langle (T - \langle T \rangle)^2\right\rangle$ note that $\langle T \rangle$ is a number, but $T$ is an operator, so we're acutally interested in calculating $\left \langle (T - \langle T \rangle I )^2 \right\rangle$, where $I$ is the identity operator.
For $A= (T - \langle T \rangle I )^2$, we have
$$A=(T - \langle T \rangle I )^2 = T^2 - \langle T \rangle T I -  \langle T \rangle I T  + \langle T \rangle^2 = T^2 -  (2\langle T \rangle) T +  \langle T \rangle ^2$$
Thus, 
$$\langle A \rangle = \left \langle (T - \langle T \rangle I )^2 \right\rangle = \left\langle T^2 +  (-2\langle T \rangle) T +  \langle T \rangle ^2 \right\rangle= \langle T^2 \rangle + \left\langle \left(-2\langle T \rangle \right) T \right\rangle + \left\langle \langle T \rangle ^2 \right\rangle.$$
Now, since $\langle T \rangle$ is a number, we get $\left\langle\left(-2\langle T \rangle \right) T \right\rangle = -2 \langle T\rangle \langle T \rangle$ and $\langle \langle T \rangle^2 \rangle = \langle T \rangle^2  \langle \psi,\psi \rangle = \langle T \rangle^2 \cdot 1= \langle T \rangle\langle T \rangle$.
Finally we get $$\langle A \rangle = \langle T^2 \rangle - 2 \langle T\rangle \langle T \rangle + \langle T \rangle\langle T \rangle = \langle T^2 \rangle - \langle T\rangle \langle T \rangle = \langle T^2 \rangle - \langle T\rangle ^2.$$
This number is real if $T$ is hermetian.
A: $$
         -2 \left\langle
         \hat{\alpha}\left\langle\hat{\alpha}\right\rangle \right\rangle
 =
         -2 \left\langle
         \hat{\alpha} \right\rangle \left\langle\hat{\alpha}\right\rangle $$
 and that is true because the $\langle \ \ \rangle $ "sees" $\hat{\alpha}$, since $\langle \hat{\alpha} \rangle $ has already been acted upon, and thus, is a number. In general, this is complex and in the case of hermitian $\hat{\alpha}$, real.
So, we get,
$$ \left\langle
         \hat{\alpha}^2  \right\rangle
         + \left\langle
         \left\langle\hat{\alpha}\right\rangle^2 \right\rangle
         -2 \left\langle
         \hat{\alpha}\left\langle\hat{\alpha}\right\rangle \right\rangle
 = \left\langle
         \hat{\alpha}^2  \right\rangle
         +
         \left\langle\hat{\alpha}\right\rangle^2
         -2 \left\langle
         \hat{\alpha} \right\rangle^2
 = \left\langle
         \hat{\alpha}^2  \right\rangle
         -
         \left\langle\hat{\alpha}\right\rangle^2 $$ and the rest follows.
