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In Griffith's text, they apply the lowering operator on the |$11\rangle$ state to get the |$10\rangle$ state. They show this result in two forms on pg. 185:

$S_{-}\left(\uparrow\uparrow\right) = \hbar \left(\uparrow \downarrow + \downarrow \uparrow \right)$

and

$|10\rangle = \frac{1}{\sqrt 2} \left(\uparrow \downarrow + \downarrow \uparrow \right)$

Where does the $ \hbar \text{ go and how does the} \frac{1}{\sqrt 2} \text{ come in?} $

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The first equation is not normalized, which is the result of the lowering operator. While the second equation is the normalized state, you can check it easily by using $\langle 10|10 \rangle$.

Or you can use the normalization condition $\langle \psi|\psi \rangle = 1$ where $|\psi \rangle = Z \hbar | \uparrow \downarrow + \downarrow \uparrow \rangle$.

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  • $\begingroup$ How do you normalize something with arrows like that? $\endgroup$
    – Logan
    Commented Nov 30, 2014 at 5:24
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    $\begingroup$ @Logan Use orthogonality. there are 4 possible states (arrow set) here, whenever they are the same, the result is , otherwise, 0. Say $\langle \uparrow \downarrow | \uparrow \downarrow \rangle = 0$. So, the expansion $\langle \psi|\psi \rangle = Z^2 * \hbar^2 * (1 + 1) = 1$ and you will get the normalized states. $\endgroup$
    – unsym
    Commented Nov 30, 2014 at 6:00

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