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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.

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2 Answers 2

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

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  • $\begingroup$ 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? $\endgroup$
    – Craig
    Jun 25, 2019 at 21:45
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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.

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

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