= Angular momentum of a radial function times a linear polynomial
{title2=$\hat L^2[H(r)(\mathbf a\cdot\mathbf x)]=2\hbar^2H(r)(\mathbf a\cdot\mathbf x)$}
Orbital <angular momentum operators> annihilate a radial multiplier $H(r)$. Their squared sum sends every degree-one homogeneous <polynomial> to $2\hbar^2$ times that polynomial, so a nonzero normalizable state $H(r)(\mathbf a\cdot\mathbf x)$ has angular momentum quantum number $l=1$. With real $H$ and real coefficients, its third-component <expected value> is zero by angular symmetry, though it need not be an eigenstate of that component.
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