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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.

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 $j\ge 0$.

Recover $L_1$ from the given angular momentum raising operator and angular momentum lowering operator:This is the generator of rotations about the first coordinate axis. A natural candidate is the angular function pointing along that axis, $f_x=\sin \theta \cos \varphi$. Direct substitution givesAlso $f_x=(f_{+1}+f_{-1})/2$, so linearity and part (ii) give $L^2{f}_x=2{\hslash }^2{f}_x$. Thus an answer is $\sin \theta \cos \varphi$. Its squared norm on the sphere isso the normalized quantum state isThe supplied starting state gives an alternative derivation with a changed quantization axis. Put $g=\cos \theta +i\sin \theta \sin \varphi$ and $K_{+}=L_2+i{L}_3$. The angular momentum commutation relations give $[L_1,K_{+}]=\hslash K_{+}$ and $[L^2,K_{+}]=0$, so $K_{+}$ raises the $L_1$ projection from $-\hslash$ to zero. Direct differentiation giveswhich is nonzero and proportional to the same ${\psi }_x$. This also connects the construction directly to the given state.

Starting from $f_{-1}=\sin \theta e^{-i\varphi }$, part (i) gives $L^2{f}_{-1}=2{\hslash }^2{f}_{-1}$ and $L_3{f}_{-1}=-\hslash f_{-1}$. Apply the angular momentum raising operator explicitly:A second application givesSince $L_{+}$ commutes with $L^2$, both raised functions retain the eigenvalue $2{\hslash }^2$. Their $L_3$ eigenvalues follow either by differentiating their $\varphi$ dependence or from $[L_3,L_{+}]=\hslash L_{+}$. Thus three valid unnormalized functions, in the requested order, areFor later Born rule calculations, normalize with the solid angle measure $d\Omega =\sin \theta d\theta d\varphi$. Their squared norms are $8\pi /3$, $4\pi /3$, $8\pi /3$, respectively. For example ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. A convenient orthonormal basis is thereforeOrthogonality follows from the distinct Fourier factors in $\varphi$. An overall phase of each basis function is arbitrary; these choices need not impose the usual phase convention for the spherical harmonics.