= Orthohydrogen
{title2=$I_{\rm n}=1,\quad J=1,3,5,\ldots$}
<Orthohydrogen> uses the three symmetric <spin-one-half triplet states>,
$$
\chi_{11}=|\uparrow\uparrow\rangle,\qquad
\chi_{10}=\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2},\qquad
\chi_{1,-1}=|\downarrow\downarrow\rangle.
$$
These three states give a nuclear-spin multiplicity of three, in addition to the rotational multiplicity $2J+1$. Thus the lowest <orthohydrogen> level, $J=1$, has nine states, whereas the lowest <parahydrogen> level, $J=0$, has one. The familiar high-<temperature> $3:1$ ratio is not the ratio $9:1$ of these two lowest levels: at high <temperature>, many even and odd rotational levels contribute.
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