Spin half for the value of $|1 0\rangle$?

Spin-1/2 The eigenspinor , $X=aX_++bX_-$ $$X_+=\left( \begin{array}{cc} 1\\ 0\end{array} \right)$$$$X_-=\left( \begin{array}{cc} 0\\ 1\end{array} \right)$$ They are define like this because they work well in the following? $S_zX_+={\hbar}/2X_+$ and $S^2X_+={\frac{3}{4}}{\hbar}X_+$.

But for $|s m \rangle$, I don't understand why do we need to put $|1 0\rangle = \frac{1}{\sqrt2} | \uparrow \downarrow + \downarrow \uparrow\rangle$ Because without $\frac{1}{\sqrt2}$ we can prove the eigenvalue of $S^2$ is 2har. Why $\frac{1}{\sqrt2}$?

That $1/\sqrt{2}$ factor is for the normalization. i.e. to ensure that $\langle 1 0 | 1 0\rangle = 1$ where $\langle 1 0 |$ is the conjugate transpose of $|1 0\rangle$
• Thanks. so easy,no wonder my book didn't explain. <10|10>=1 <10|10>=$A^2(\uparrow\downarrow + \downarrow\uparrow)(\uparrow\downarrow + \downarrow\uparrow)$ $A^2(\uparrow\downarrow\uparrow\downarrow+\uparrow\downarrow \downarrow\uparrow+\downarrow\uparrow\uparrow\downarrow+ \downarrow\uparrow\downarrow\uparrow)=1$ The arrow can be expand like this? Then use $\downarrow\uparrow$=o,$\downarrow\downarrow=1$ $A^2=0+1+1+0$ Dec 5 '13 at 7:02