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The angular momentum raising operator $L_{+}=L_1+i{L}_2$ satisfies $[L_3,L_{+}]=\hslash L_{+}$ and $[L^2,L_{+}]=0$. It preserves the total angular momentum while increasing the third-component magnetic quantum number by one, whenever its result is nonzero. Its adjoint operator is the angular momentum lowering operator $L_{-}$. Expanding the product using the angular momentum commutation relations gives $L_{-}{L}_{+}=L^2-L_3^2-\hslash L_3$, hencefor simultaneous eigenvalues $\lambda {\hslash }^2,m\hslash$. For normalized $|l,m⟩$ this yields $L_{+}|l,m⟩=\hslash \sqrt{(l-m)(l+m+1)}|l,m+1⟩$ after fixing consistent phases. At $m=l$ the result is zero, the upper angular momentum ladder endpoint. For a different quantization axis, cyclically permute the components; for example $L_2+i{L}_3$ raises the $L_1$ projection.

Expanding with the angular momentum commutators givesFor ${\hat{L}}_z|\psi ⟩=\alpha \hslash |\psi ⟩$, commuting ${\hat{L}}_z$ past a ladder operator givesThus the raised or lowered vector, if nonzero, is an eigenvector with the shifted eigenvalue; an angular momentum ladder endpoint can instead be annihilated. Also $[{\hat{L}}^2,{\hat{L}}_{\pm }]=0$, so ${\hat{L}}^2{\hat{L}}_{\pm }|\psi ⟩=\Lambda {\hslash }^2{\hat{L}}_{\pm }|\psi ⟩$ for every nonzero ladder result.

For orbital angular momentum, ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$ gives an azimuthal factor $e^{i\alpha \varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ requires integer $\alpha$. Positivity of ladder norms givesfor normalized states. A fixed finite $\Lambda$ therefore has upper and lower endpoints. Writing the upper endpoint as $m=\ell$, its vanishing raising norm gives $\Lambda =\ell (\ell +1)$; the lower endpoint is $m=-\ell$. Consequentlywith $\ell =0,1,2,\dots$. This integer conclusion is for orbital angular momentum; spin representations can also have half-integer labels.

For the specified Gaussian-radial state, use a Cartesian p-orbital decomposition: write $x+y+z=r(\sin \theta \cos \varphi +\sin \theta \sin \varphi +\cos \theta )$. Its angular factor isOnly $\ell =1$ occurs. The squared angular coefficients are $2,2,2$ before common normalization, so each allowed $m$ has probability one third. Therefore the simultaneous outcome isThe common normalized radial factor cancels from this ratio. The standard symmetric $m$ sequence used above resolves the duplicated central positive entry in the printed list.

The given angular momentum commutation relations imply $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. By linearity,Similarly,For the remaining commutators, the commutator product rule gives, for each $j$,The last sum vanishes because the Levi-Civita symbol is antisymmetric under $i\leftrightarrow k$, while $L_i{L}_k+L_k{L}_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves

If $L^2\psi =\lambda {\hslash }^2\psi$ and $L_3\psi =m\hslash \psi$, these identities yieldOne must also prove that this candidate eigenstate is nonzero. Since the angular momentum operators are self-adjoint operators, $L_{+}^{\ast }=L_{-}$, and expanding the product givesThereforeThis is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The angular momentum raising operator really does produce a nonzero simultaneous eigenstate with the claimed eigenvalues.

To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity givesso $q^2\le \lambda$. Thus the sequence cannot remain nonzero for every integer $r\ge 0$. At its last nonzero state $v=(L_{+}{)}^k\psi$, one has $L_{+}v=0$, and the squared norm identity givesUnder the paper's initial inequality the first step is nonzero, so in fact $k\ge 1$. An angular momentum ladder endpoint must therefore be reached: bounded projection follows from positivity, and the final raising norm fixes the total angular momentum eigenvalue. Writing $j=m+k$, the complementary product $L_{+}{L}_{-}=L^2-L_3^2+\hslash L_3$ gives $\Vert L_{-}v{\Vert }^2=2j{\hslash }^2\Vert v{\Vert }^2$, also showing $j\ge 0$.

For $\lambda =3/4$, the adjoint operator relation $J_{+}^{\ast }=J_{-}$ and the given product identity implyA vector of zero norm is zero, so $J_{+}\psi =0$. Part (i) givesThe commutator $[J^2,J_{-}]=0$ also gives $J^2\phi =(3/4){\hslash }^2\phi$, while $J_3\phi =-(\hslash /2)\phi$. Applying the other product identity to this unnormalized vector,ThereforeThese are the two angular momentum ladder endpoints of a spin one-half pair. In particular, ${\psi }_2=\phi /\hslash$ has unit norm, and $J_{+}{\psi }_2=\hslash {\psi }_1$ because $J_{+}{J}_{-}\psi ={\hslash }^2\psi$.