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The orbital angular momentum operators are $\mathbf{L}=\mathbf{x}\times \mathbf{p}$, soThe canonical commutation relations are $[x_i,x_j]=[p_i,p_j]=0$ and $[x_i,p_j]=i\hslash {\delta }_{ij}$. On a common domain of smooth wavefunctions, the product commutator identity givesContract this with ${ϵ}_{iab}{ϵ}_{jcd}$. The first term contributes $i\hslash ({\delta }_{ij}\mathbf{x}\cdot \mathbf{p}-x_j{p}_i)$ and the second contributes $i\hslash (x_i{p}_j-{\delta }_{ij}\mathbf{x}\cdot \mathbf{p})$. Their sum isThis verifies the angular momentum commutation relations explicitly. Similarly,

In the position representation $\mathbf{p}=-i\hslash \nabla$. For a radial wavefunction, $\nabla {\Psi }_0={\Psi }_0^{\prime }(r)\mathbf{x}/r$, parallel to $\mathbf{x}$, soWrite ${\Phi }_i=h(r)x_i$ with $h(r)={\Psi }_0^{\prime }(r)/r$. The operator $L_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$ annihilates the radial factor. It gives $L_3{x}_1=i\hslash x_2$, $L_3{x}_2=-i\hslash x_1$, and $L_3{x}_3=0$, henceThus these components, when nonzero, are the specified eigenfunctions with eigenvalues $0,+\hslash ,-\hslash$. This is an example of angular momentum of a radial function times a linear polynomial.

Use the radial exponential $e^{-\alpha r}$ shown in the original PDF; the TeX's $e^{-\alpha x}$ is a transcription defect. Every orbital angular momentum operator annihilates a radial function, so it acts only on $P=x_1+x_2+2x_3$. The angular momentum of a radial function times a linear polynomial is determined by its degree-one angular factor: direct application of the rotation derivatives givesEquivalently its restriction to the sphere is a degree-one spherical harmonic. ThusFurthermore ${\hat{L}}_3\psi =-i\hslash K(x_1-x_2)e^{-\alpha r}$. Its expected value is proportional to the integral of $(x_1+x_2+2x_3)(x_1-x_2)e^{-2\alpha r}$. The cross terms are odd in a coordinate, and the integrals of $x_1^2$ and $x_2^2$ agree by rotational symmetry. HenceThe state need not be an eigenstate of ${\hat{L}}_3$. If normalization is desired, $\int P^2{e}^{-2\alpha r}{d}^{3}x=6\pi /{\alpha }^5$, so $K={\alpha }^{5/2}/\sqrt{6\pi }$; the zero expectation holds after normalization for any nonzero initial $K$.