Normalize Triplet State of Hydrogen For hydrogen, the total spin of the electron and proton is $s = 1$, while $m_s = -1,0,1$. If $m_s = 1$, one of the states can be written as
$$\left| 1\;1 \right > = \left |\uparrow \uparrow\right >$$
Applying the $S_{-}$ operator to this state, we obtain
$$\left| 1\;0 \right > = (\left |\uparrow \downarrow\right > + \left |\downarrow \uparrow\right >)$$
Then my text normalizes the state by dividing it by $\sqrt{2}$. Where does that $\sqrt{2}$ come from? I know if you have a $(1\; 0 \; 1)$ vector, you'd normalize it by dividing by $\sqrt{1^2 + 1^2}$. $\uparrow = (1\;0)$ and $\downarrow = (0\;1)$. $\uparrow\downarrow$ does not mean multiplication, I think. I'm just trying to figure out where the $\sqrt{2}$ comes from. Is it because there are two states? But where in the math is the 2?
 A: Your unnormalised state is a superposition of 2 states, $\left |\uparrow \downarrow\right>$ and $\left |\downarrow \uparrow\right>$ with probability amplitudes equal to unity. Their probability amplitudes are the coefficients in front of the 2 states. In this case they are 1. But what does normalisation mean anyway? Currently as the state is, and since the square of the probability amplitude of either state should give the probability of obtaining the said state, there is a 100% chance of obtaining either state which follows from:
$$P_{\left |\uparrow \downarrow\right>} = 1^2 = 1 = 100\%$$ and $$P_{\left |\downarrow \uparrow\right>} = 1^2 = 1 = 100\%$$
This is obviously nonsense. To avoid this problem we normalise the state to produce probabilities which sum up to unity, and thus become physical. This is done by summing the squares of each probability amplitude and taking the square root:
$N = \sqrt{1^2 + 1^2} = \sqrt{2}$
Where 1/N will re-weight the amplitudes to sum up to $1$(and thus $100\%$), without affecting the actual probability ratios themselves.
