Can someone help me with a proof of the expectation value of the dispersion of an observable A only using Dirac Notation? Unfortunately I learned Quantum Mechanics from Griffiths and I struggle greatly with Dirac notation. I'm working my way through Sakurai learning it and tried to do a very trivial proof using only Dirac notation and can't quite get it. For those with the text I'm trying to prove equation 1.4.51. I'll put my work on the problem below and comment on it afterwards.

We know the expectation value of an observable $A$ is
$ \left< A \right> = \left< \alpha \right| A \left| \alpha \right> $
Define 
$ \Delta A \equiv A - \left< A \right> $
Theorem:
$\left< \left( \Delta A \right)^{2} \right> = \left< A^{2} \right> - \left< A \right>^{2} $
Proof:
\begin{align}
\left< \left( \Delta A \right)^{2} \right> &= \left< \left( A - \left< A \right> \right)^{2} \right>\\
&=\left< A^{2} - 2 A \left< A \right> + \left< A \right>^{2} \right>\\
&=\left< \alpha \right| \left( A^{2} - 2 A \left< A \right> + \left< A \right>^{2} \right) \left| \alpha \right>\\
&=\left( \left< \alpha \right| A^{2} + \left< \alpha \right| \left( - 2 A \left< A \right> \right) + \left< \alpha \right| \left< A \right>^{2} \right) \left| \alpha \right> \, , \qquad \text{Bras are distributive}\\
&=\left< \alpha \right| A^{2} \left| \alpha \right> + \left< \alpha \right| \left( - 2 A \left< A \right> \right) \left| \alpha \right> + \left< \alpha \right| \left< A \right>^{2} \left| \alpha \right> \, , \qquad \text{Kets are distributive} \\
&=\left< \alpha \right| A^{2} \left| \alpha \right> - 2 \left< \alpha \right| \left( A \left< A \right> \right) \left| \alpha \right> + \left< \alpha \right| \left< A \right>^{2} \left| \alpha \right>  \, , \qquad \text{-2 is constant and factors out}\\
\end{align}
Then $\left< \alpha \right| A^{2} \left| \alpha \right> = \left< A^{2} \right>$

Now this is the part I'm stuck on. I want to be able to say "An expectation value is just a number and factors out", but it isn't totally obvious to me that for some observable that 
$$
\left< \alpha \right| \left< \alpha \right| A \left| \alpha \right> \left| \alpha \right> = \left< \alpha \right| A \left| \alpha \right> \left< \alpha | \alpha \right> = \left< \alpha \right| A \left| \alpha \right> = \left< A \right>
$$
For a few different reasons. 


*

*We have to assume that our space is normalized to say $\left< \alpha | \alpha \right> = 1$. 

*What if we consider $\left< \beta \right| \left< \alpha \right| A \left| \alpha \right> \left| \beta \right>$. Is this still true?

*Without magically "factoring the entire term out" I can't think of how to properly "commute" the bras and kets and operators to get from $\left< \alpha \right| \left< \alpha \right| A \left| \alpha \right> \left| \alpha \right> $ to $ \left< \alpha \right| A \left| \alpha \right> \left< \alpha | \alpha \right>$


Not to mention that 
$$
\left< \alpha \right| \left< A \right>^{2} \left| \alpha \right> = 
\left< \alpha \right| \left< \alpha \right| A \left| \alpha \right> \left< \alpha \right| A \left| \alpha \right> \left| \alpha \right> = \left< \alpha \right| A \left| \alpha \right> \left< \alpha \right| A \left| \alpha \right> = \left< A \right> \left< A \right> = \left< A \right>^{2}
$$
Or worse still because we have adjacent kets in the case of
$$
\left< \alpha \right| \left( A \left< A \right> \right) \left| \alpha \right> = 
\left< \alpha \right| A \left< \alpha \right| A \left| \alpha \right> \left| \alpha \right> = \left< \alpha \right| A \left| \alpha \right> \left< \alpha \right| A \left| \alpha \right> = \left< A \right> \left< A \right> = \left< A \right>^{2}
$$
If I was to word this as a list of questions:


*

*How do I properly "commute" the bras and kets to fill in the intermediate steps I've listed above, when bras and kets don't generally commute? (potentially using transposes perhaps?)

*Under what conditions it is valid to do those operations?

*Does my space have to be normalized so $\left< \alpha | \alpha \right> = 1$? 

*Are my operators required to be Hermitian (or self adjoint) so $A = A^\dagger$? 

*Is my notation correct when I use $\alpha$ for everything, or should I be writing $\left< A \right> \left< A \right> = \left< \alpha \right| A \left| \alpha \right> \left< \beta \right| A \left| \beta \right>$, where $\beta$ is just a different "dummy variable" than $\alpha$? (I think back to dummy indexes when using the Einstein Summation Notation).


If you can think of anything I perhaps should have asked feel free to address that too. I can do this proof using the integral definitions but Dirac notation unfortunately doesn't seem to be clicking for me.
 A: (Sorry I can't comment!) You can see that indeed the expectation value of an expectation value is indeed a number: here are two ways to see it.
You know that operators act on kets (or bras) and their output is another ket (or bra):
$$ \hat{O}\left| \psi \right> = \left| \chi \right>
$$
Now, in its most general sense, the contraction of a bra and a ket gives a complex number - always:
$$
\left<\chi | \psi \right> \in \mathbb{C} = \{\text{The set of complex numbers} \; a+bi, \text{where}\; a,b \; \text{real}, \; i =\sqrt{-1}\}.
$$
One of the properties of bras and kets is they follow most of the formalism you are familiar with from algebra (as you've shown above!). The one to see is that complex numbers can be distributed inside or outside the bra-kets. So if you have an object like $$\left<\alpha\right|\left<\beta\right|\hat{O}\left|\gamma \right>\left|\delta\right>$$
You can use that $\hat{O}\left|\gamma\right>$ is just some other ket, say $\left|\epsilon\right>$; hence $\left<\beta|\epsilon\right>$ is a complex number you can move around freely!:
$$\left<\alpha\right|\left<\beta\right|\hat{O}\left|\gamma \right>\left|\delta\right>=\left<\beta\right|\hat{O}\left|\gamma \right>\left<\alpha|\delta\right>$$
Which is simply the multiplication of two complex numbers. One would say that bras and kets are duals (of course the algebra of quantum mechanics isn't finite linear/multivariate algebra, but much of the techniques do cross over).
In respect then to your result, you can see that
\begin{align}
\left< \left( \Delta A \right)^{2} \right> &= \left< \left( A - \left< A \right> \right)^{2} \right>\\
&=\left< A^{2} - 2 A \left< A \right> + \left< A \right>^{2} \right>\\
&=\left< \alpha \right| \left( A^{2} - 2 A \left< A \right> + \left< A \right>^{2} \right) \left| \alpha \right>\\
&=\left( \left< \alpha \right| A^{2} + \left< \alpha \right| \left( - 2 A \left< A \right> \right) + \left< \alpha \right| \left< A \right>^{2} \right) \left| \alpha \right> \, , \qquad \text{Bras are distributive}\\
&=\left< \alpha \right| A^{2} \left| \alpha \right> + \left< \alpha \right| \left( - 2 A \left< A \right> \right) \left| \alpha \right> + \left< \alpha \right| \left< A \right>^{2} \left| \alpha \right> \, , \qquad \text{Kets are distributive} \\
&=\left< \alpha \right| A^{2} \left| \alpha \right> - 2 \left< \alpha \right| \left( A \left< A \right> \right) \left| \alpha \right> + \left< \alpha \right| \left< A \right>^{2} \left| \alpha \right>  \, , \qquad \text{-2 is constant and factors out}\\
&= \left< \alpha \right| A^{2} \left| \alpha \right> - 2\left< \alpha \right| A \left| \alpha \right> \left< A \right> + \left< A \right>^2\, , \qquad \text{Expectation values are also constant and $\left| \alpha \right> $ is normalised}\\
&=\left< A^{2} \right> - \left< A \right>^{2}
\end{align}
As required!
