The uncertainty relation
Contents:
# Dispersion
Given $\hat A$, define
$$ \Delta A \equiv A - \langle A\rangle $$
where $\langle A\rangle$ is the expectation value $\langle\alpha|\hat A|\alpha\rangle$. (And for a constant $c$: $\langle c\rangle = c\langle\alpha|\alpha\rangle = c$ if $|\alpha\rangle$ is normalized.)
The expectation of $(\Delta A)^2$:
$$ \langle(\Delta A)^2\rangle = \left\langle A^2 + \langle A\rangle^2 - 2A\langle A\rangle\right\rangle $$
Note that $\langle A\rangle$ is a number, hence $A\langle A\rangle = \langle A\rangle A$.
$$ \begin{aligned} \implies \langle(\Delta A)^2\rangle &= \langle A^2\rangle + \left\langle\langle A\rangle^2\right\rangle - \left\langle 2A\langle A\rangle\right\rangle \\ &= \langle A^2\rangle + \langle A\rangle^2 - 2\langle A\rangle\cdot\langle A\rangle \end{aligned} $$
$$ \boxed{\langle(\Delta A)^2\rangle = \langle A^2\rangle - \langle A\rangle^2} $$
$\to$ dispersion (the variance / mean-square deviation).
# Dispersion of Sz in the S-G experiment
Let the state ket be
$$ |\alpha\rangle = |S_x;+\rangle = \frac{1}{\sqrt{2}}\left(|+\rangle + |-\rangle\right) $$
and $\hat A = \hat S_z$.
$$ \langle\hat S_z\rangle = \langle\alpha|\hat S_z|\alpha\rangle = \left(\langle S_x;+|\right)\left(\hat S_z|S_x;+\rangle\right) $$
First, $S_z$ in terms of the base kets:
$$ S_z = S_z\cdot I = S_z\cdot\sum_{a'}|a'\rangle\langle a'| $$
$$ \implies S_z = \sum_{a''}\sum_{a'}|a''\rangle\langle a''|S_z|a'\rangle\langle a'| $$
For 2-states, $S_z = \sum_{a'}a'|a'\rangle\langle a'|$ if the basis vectors are eigenkets. We don’t need 2 insertions:
$$ \implies S_z = \sum_{a'}S_z|a'\rangle\langle a'| = \sum_{a'}a'|a'\rangle\langle a'| $$
$$ \boxed{S_z = \frac{\hbar}{2}\left(|+\rangle\langle+| - |-\rangle\langle-|\right)} $$
Done.
Now with that, the variance / mean-square deviation is $\langle A^2\rangle - \langle A\rangle^2$.
Eg. dispersion of $S_z$ for a spin $\frac{1}{2}$ system $= \langle S_z^2\rangle - \langle S_z\rangle^2$.
For $|\alpha\rangle = |S_x;+\rangle$:
$$ \langle S_z\rangle = \langle S_x;+|\hat S_z|S_x;+\rangle = 0 $$
(since $\hat S_z|S_x;+\rangle = \frac{\hbar}{2}\cdot\frac{1}{\sqrt{2}}\left(|+\rangle - |-\rangle\right) = \frac{\hbar}{2}|S_x;-\rangle$, which is orthogonal to $|S_x;+\rangle$)
$$ S_z^2 = S_z\cdot S_z = \frac{\hbar^2}{4}\left(|+\rangle\langle+| - |-\rangle\langle-|\right)\left(|+\rangle\langle+| - |-\rangle\langle-|\right) $$
$$ S_z^2 = \frac{\hbar^2}{4}\underbrace{\left(|+\rangle\langle+| + |-\rangle\langle-|\right)}_{\text{closure}} = \frac{\hbar^2}{4}I $$
$$ \therefore \langle S_z^2\rangle = \frac{\hbar^2}{4} $$
$\implies$ the dispersion of $S_z$:
$$ \boxed{\langle(\Delta S_z)^2\rangle = \frac{\hbar^2}{4} - 0 = \frac{\hbar^2}{4}} $$
$\implies$ the state in x ($S_x$) has an undefined $\pm S_z$.
If instead the state was $|S_z;+\rangle$, then the dispersion would be 0 $\implies$ no spread in $S_z$. We know it exactly.