Clebsch-Gordan coefficients and their sign

[Aleolomorfo]

TL;DR

Different sign in the combination of two $\textbf{1/2}$ isospins with opposite third component

Summary: Different sign in the combination of two $\textbf{1/2}$ isospins with opposite third component

Hello everybody!
I was doing an exercise regarding isospin and I noticed something from the Clebsch-Gordan coefficients that made me think.
For example, if I consider the combination between the two angular momenta $|\textbf{1/2}, +1/2>$ and $|\textbf{1/2}, -1/2>$, in the Clebsch-Gordan table there are two possibilities. One in which $I_z=+1/2$ is put first and the second in which $I_z=+1/2$ is put after. I have attached the table and I have highlighted what I am saying inside a red circle.
The two combinations gives different coefficients, just for a sign.
My question is: what is the difference between combining $|\textbf{1/2}, +1/2>$ $|\textbf{1/2}, -1/2>$ and $|\textbf{1/2}, -1/2>$ $|\textbf{1/2}, +1/2>$? Should not they be equal? From what comes the minus sign?
Thank you in advance!

 

[vanhees71]

There are two states with $\Sigma_z=0$. One is for $S=0$:
$$|S=0,\Sigma_z=0 \rangle=\frac{1}{\sqrt{2}} (|1/2,1/2 \rangle \otimes |1/2,-1/2 \rangle - |1/2,-1/2 \rangle \otimes |1/2,1/2 \rangle$$
and one for $S=1$:
$$|S=1,\Sigma_z=0 \rangle=\frac{1}{\sqrt{2}} (|1/2,1/2 \rangle \otimes |1/2,-1/2 \rangle +|1/2,-1/2 \rangle \otimes |1/2,1/2 \rangle.$$
Note the footnote in the Review of Particle Physics table of Clebsch-Gordon coefficients: You have to take the square root of all the number (with the minus-signs outside of the square root)!