Wavefunctions – Finding ? Wavefunctions from Proton Wavefunction

Question

Knowing the isospin part of the wavefunction of the proton, it is possible to find that of the neutron by applying the isospin lowering operator $I_-$ which sits horizontally to the left of the proton in the baryon octet. Is there any method (actually, an operator) by which knowing the isospin wavefunction of the proton, we can obtain those of the $\Sigma^{-}, \Sigma^{0}, \Sigma^{+}$ particles (which sit below neutron and proton in the isospin $I=1$ multiplet in the baryon octet)?

Answer 1

U and V spins move you diagonally in SU(3), so that $U_- ~d= s$ and $V_- ~u= s$.

It then follows that $$U_- ~p = \Sigma^+ ~;\\ I_- U_- ~p=V_- ~p = \Sigma^0 ~; \\ I_- V_- ~p =V_- I_- ~p = \Sigma^- ~. $$

For example, the first line amounts to $$ |\Sigma^+ _\uparrow\rangle= \frac{U_-}{\sqrt {18}}\bigl [2| u_\uparrow d_\downarrow u_\uparrow \rangle + 2| u_\uparrow u_\uparrow d_\downarrow \rangle +2| d_\downarrow u_\uparrow u_\uparrow \rangle \\ - | u_\uparrow u_\downarrow d_\uparrow\rangle -| u_\uparrow d_\uparrow u_\downarrow\rangle -| u_\downarrow d_\uparrow u_\uparrow\rangle -| d_\uparrow u_\downarrow u_\uparrow\rangle -|d_\uparrow u_\uparrow u_\downarrow\rangle-| u_\downarrow u_\uparrow d_\uparrow\rangle \bigr ]\\ = \frac{1}{\sqrt {18}} \bigl [2| u_\uparrow s_\downarrow u_\uparrow \rangle + 2| u_\uparrow u_\uparrow s_\downarrow \rangle +2| s_\downarrow u_\uparrow u_\uparrow \rangle \\ - | u_\uparrow u_\downarrow s_\uparrow\rangle -| u_\uparrow s_\uparrow u_\downarrow\rangle -| u_\downarrow d_\uparrow u_\uparrow\rangle -| d_\uparrow u_\downarrow u_\uparrow\rangle -|d_\uparrow u_\uparrow u_\downarrow\rangle-| u_\downarrow u_\uparrow d_\uparrow\rangle\bigr ]. $$