The angular momentum of a hydrogen atom in its ground state is entirely due to the spins of the electron and proton. The atom is in the state $|1,0\rangle$ in which it has one unit of angular momentum but none of it is parallel to the $z$-axis. Express this state as a linear combination of products of the spin states $|\pm, \mathrm{e}\rangle$ and $|\pm, \mathrm{p}\rangle$ of the proton and electron. Show that the states $|x \pm, \mathrm{e}\rangle$ in which the electron has well-defined spin along the $x$-axis are
$$
|x \pm, \mathrm{e}\rangle=\frac{1}{\sqrt{2}}(|+, \mathrm{e}\rangle \pm|-, \mathrm{e}\rangle)
$$
By writing
$$
|1,0\rangle=|x+, \mathrm{e}\rangle\langle x+, \mathrm{e} \mid 1,0\rangle+|x-, \mathrm{e}\rangle\langle x-, \mathrm{e} \mid 1,0\rangle
$$
express $|1,0\rangle$ as a linear combination of the products $|x \pm, \mathrm{e}\rangle|x \pm, \mathrm{p}\rangle .$ Explain the physical significance of your result.