How do we decide the proton wave function? The fully symmetric spin-up proton spin-flavour wave function in the constituent quark model is usually presented as follows:
$$\begin{align}
\frac{1}{\sqrt{18}} ~ (
&2 |u\uparrow ~  u\uparrow ~ d\downarrow \rangle
- |u\uparrow ~  u\downarrow ~ d\uparrow \rangle
- |u\downarrow ~  u\uparrow ~ d\uparrow \rangle \\
+ & 2 |u\uparrow ~ d\downarrow ~  u\uparrow \rangle
- |u\downarrow ~ d\uparrow ~  u\uparrow \rangle
- |u\uparrow ~ d\uparrow ~  u\downarrow \rangle \\
+ & 2|d\downarrow ~u\uparrow ~  u\uparrow \rangle
- |d\uparrow ~u\downarrow ~  u\uparrow \rangle
- |d\uparrow ~u\uparrow ~  u\downarrow \rangle).
\end{align}$$
However the following also satisfies all the symmetry properties desired:
$$\begin{align}
\propto (
 &a\> |u\uparrow ~  u\uparrow ~ d\downarrow \rangle
- b\> |u\uparrow ~  u\downarrow ~ d\uparrow \rangle
- b\> |u\downarrow ~  u\uparrow ~ d\uparrow \rangle \\
+&a\> |u\uparrow ~ d\downarrow ~  u\uparrow \rangle
- b\> |u\downarrow ~ d\uparrow ~  u\uparrow \rangle
- b\> |u\uparrow ~ d\uparrow ~  u\downarrow \rangle \\
+&a\> |d\downarrow ~u\uparrow ~  u\uparrow \rangle
- b\> |d\uparrow ~u\downarrow ~  u\uparrow \rangle
- b\> |d\uparrow ~u\uparrow ~u\downarrow \rangle ).
\end{align}$$
How does one "narrow" it down so to speak. A group theoretic aproach to this would be preferred.
 A: I'm not an expert in this, and this isn't explicitly group-theoretical, but here's my understanding of it:
If you set $a = -b = 1$ in the ansatz you've provided, you get something that factors out nicely:
\begin{align}
(
 &\> |u\uparrow ~  u\uparrow ~ d\downarrow \rangle
+ \> |u\uparrow ~  u\downarrow ~ d\uparrow \rangle
+ \> |u\downarrow ~  u\uparrow ~ d\uparrow \rangle \\
+&\> |u\uparrow ~ d\downarrow ~  u\uparrow \rangle
+ \> |u\downarrow ~ d\uparrow ~  u\uparrow \rangle
+ \> |u\uparrow ~ d\uparrow ~  u\downarrow \rangle \\
+&\> |d\downarrow ~u\uparrow ~  u\uparrow \rangle
+ \> |d\uparrow ~u\downarrow ~  u\uparrow \rangle
+ \> |d\uparrow ~u\uparrow ~u\downarrow \rangle )\\
& \qquad = \left( | uud \rangle + | udu \rangle + | duu \rangle \right) \left( | \uparrow\uparrow\downarrow \rangle + |\uparrow\downarrow\uparrow \rangle + | \downarrow\uparrow\uparrow \rangle \right) 
\end{align}
Since this is expressible as a completely symmetric state in both flavor and spin, it is part of the baryon decuplet, not the baryon octet.  (Specifically, I believe it would be a $\Delta^+$ baryon in a $m = \frac{1}{2}$ state.)  
The proton state must be orthogonal to this decuplet state;  and if you take the inner product of this decuplet state with your state, you find that you must have $a - 2b = 0$.  Normalizing the state, this then implies that $a = 2/\sqrt{18}$ and $b = 1/\sqrt{18}$.
A: Michael Seifert's answer is exactly what I needed. An additional piece of information is also helpful: most configurations of $a$ and $b$ will not be eigenvectors of total square spin and isospin, $\hat{S}^2$ and $\hat{I}^2$. Physical states need to be simultaneous eigenstates of both. 
