= Solution
Denote the two complex spatial <atomic orbitals> by $\psi_+$ and $\psi_-$, retaining the same <spin> state. With the phases specified by the supplied <spherical harmonics>, choose
$$
\boxed{\psi_x=\frac{\psi_- -\psi_+}{\sqrt2},\qquad
\psi_y=\frac{i(\psi_-+\psi_+)}{\sqrt2}.}
$$
The two original <atomic orbitals> are <orthonormal>, so the squared coefficients add to one and these combinations are normalized and mutually <orthogonal>. Explicitly,
$$
\psi_x=\frac{1}{\sqrt{32\pi}}\,x\,e^{-r/2},\qquad
\psi_y=\frac{1}{\sqrt{32\pi}}\,y\,e^{-r/2},
$$
which are real spatial <wavefunctions>. An overall sign on either orbital has no physical consequence.
These are the familiar \b[$2p_x$ and $2p_y$ <p orbitals>]. Each has two lobes of opposite <wavefunction> sign, with an <atomic-orbital nodal plane> through the <atomic nucleus>: $x=0$ for $2p_x$ and $y=0$ for $2p_y$. The sign is a phase convention, not a positive or negative <electric charge>; the <probability density> is nonnegative on both lobes.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1-real-2p.png]
{title=Original sections through the real hydrogen 2p <atomic orbitals>, showing opposite phases and their nodal planes}
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