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.

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 $$b = 1/\sqrt{18}$$.

• Seeing as normalization and orthogonality take care of a and b, must the excited states of the nucleon be orthogonal to these states by way of spacial wave functions? Or are there other possible symmetric spin-flavour combinations I'm not seeing? Jun 25, 2019 at 21:45

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.

• You should make a comment of this since it isn't an answer 😊.
– dan
Oct 28, 2023 at 9:14