Combining spins of two particles of spin |1,0> and |1/2, 1/2> I am trying to combine the spin of two particles.
Their individual spins are:
$|1,0\rangle $ and $\left|\frac{1}{2},\frac{1}{2} \right>$
Now I am told that they combine to give a total spin state of:
$$\left|\frac{1}{2},\frac{1}{2} \right>$$
However, I am confused as to how this works. Thinking physically I am confused as:

I understand that this doesn't really make sense as spin is quantized, however, I am confused as to how the states combine if this is not the case.
 A: $
\newcommand{\ket}[1]{{\textstyle\left|{#1}\right>}}
\newcommand{\sqrtfrac}[2]{{\color{lightblue}{\textstyle\sqrt{\frac{#1}{#2}}}}}
%%%
$There is no way to combine $\ket{1,0}$ and $\ket{\frac12, \frac12}$ to get a pure $j=\frac12$ state.  The only possible value for the $z$-axis projection $m$ is $0 + \frac12 = \frac12$, but the total angular momentum can take either of $\frac12$ or $\frac32$.
The orthonormal combinations are given by the Clebsch-Gordan coefficients, which are usually presented in horrible tables like



The way to read this horrible table is that, if you wanted to construct the composite state like $\ket{\frac32,\frac12}$ or $\ket{\frac12,\frac12}$, you would read down the columns of the second table:
\begin{align}
\ket{\frac32,\frac12} &=
\sqrtfrac13\ \ket{1,1}\ket{\frac12,{-\frac12}}
+ \sqrtfrac23\ \ket{1,0}\ket{\frac12,{+\frac12}}
\\
\ket{\frac12,\frac12} &=
\sqrtfrac23\ \ket{1,1}\ket{\frac12,{-\frac12}}
- \sqrtfrac13\ \ket{1,0}\ket{\frac12,{+\frac12}}
\end{align}
If your constituent particles are in a pure state $\ket{1,0}\ket{\frac12,\frac12}$, your composite system is in a superposition of $\ket{\frac12,\frac12}$ and $\ket{\frac32,\frac12}$.  You should convince yourself that you can find its coefficients either by solving the system of equations above for $\ket{1,0}\ket{\frac12, \frac12}$, or by reading across the Clebsch-Gordan table.
A: This can be done as follows:
$$1\otimes \frac{1}{2}=\frac{3}{2}\oplus\frac{1}{2}$$
Using the notation $|j_1m_1,j_2m_2\rangle $ for product ket and  $|jm\rangle$ for sum ket.
$$|j_1m_1,j_2m_2\rangle =\sum_{j,m}|jm\rangle \langle jm|j_1m_1,j_2m_2\rangle $$
$$\left|10,\frac{1}{2}\frac{1}{2}\right\rangle=\left\langle \frac{1}{2}\frac{1}{2}\left|10,\frac{1}{2}\frac{1}{2}\right.\right\rangle \left|\frac{1}{2}\frac{1}{2}\right\rangle+
\left\langle \frac{3}{2}\frac{1}{2}\left|10,\frac{1}{2}\frac{1}{2}\right.\right\rangle \left|\frac{3}{2}\frac{1}{2}\right\rangle$$
Putting the value of Clebsch-Gordan coefficient from here.
$$\left|10,\frac{1}{2}\frac{1}{2}\right\rangle=\sqrt{\frac{2}{3}}\left|\frac{3}{2}\frac{1}{2}\right\rangle-\sqrt{\frac{1}{3}}\left|\frac{1}{2}\frac{1}{2}\right\rangle$$
