0
$\begingroup$

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 $\frac{1}{\sqrt2}$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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 $|1 0\rangle$

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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,$\downarrow\downarrow=1$ $A^2=0+1+1+0$ $\endgroup$
    – Outrageous
    Commented Dec 5, 2013 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.