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Orbital angular momentum operators annihilate a radial multiplier $H(r)$. Their squared sum sends every degree-one homogeneous polynomial to $2{\hslash }^2$ 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.

Applying the angular momentum operator directly givesFor an eigenvalue $\lambda$, comparison with $\lambda (f(x)+iy)z$ gives $\hslash x=\lambda f(x)$ and $\hslash f^{\prime }(x)=\lambda$. The first identity excludes $\lambda =0$; differentiating it and using the second gives ${\lambda }^2={\hslash }^2$. No additive constant in $f$ is permitted. The possibilities are $f(x)=x$, $\lambda =\hslash$, and $f(x)=-x$, $\lambda =-\hslash$.

For $f=x$, the polynomial $\psi =(x+iy)z$ is homogeneous of degree two and satisfies $\Delta \psi =0$. Expanding the three squared orbital angular momentum operators gives, with $D=x{\partial }_x+y{\partial }_y+z{\partial }_z$,Here $D\psi =2\psi$ and $D^2\psi =4\psi$. Therefore ${\hat{L}}^2\psi =6{\hslash }^2\psi$. This is the degree-two case of angular momentum of a homogeneous harmonic polynomial; its angular labels are $l=2$, $m=1$.

The product rule for a commutator gives $[A,B^2]=[A,B]B+B[A,B]$. Using the permitted cyclic angular momentum commutation relations,The first two expressions cancel, even though the factors need not commute. HenceThe squared angular momentum operator commutes with each of its components by the same cyclic calculation.

Distinct eigenvalues of the self-adjoint angular momentum operator imply $⟨{\psi }_1,{\psi }_2⟩=0$. The given combination is not normalized: its squared norm is $({\cos }^2\omega t+3{\sin }^2\omega t)/4$, which is always positive. Therefore its properly normalized expectation value isIt vanishes when ${\tan }^2\omega t=1/3$. The first positive solution has $\omega t=\pi /6$, soNormalization changes the value of the expectation value away from zero, but not its zero times.

The orbital angular momentum operators areIn particular ${\hat{L}}_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$. For smooth compactly supported functions $\varphi ,\psi$, integration by parts produces no boundary term, and ${\partial }_2{x}_1={\partial }_1{x}_2=0$. HenceThis proves Hermiticity on the common test-function domain, and the same calculation applies on suitable decaying domains. As an unbounded operator, it is not defined on every $L^2$ function: the usual self-adjoint operator realization is the generator of unitary rotations about the third axis. Thus the integration statement includes its necessary domain/boundary assumptions.

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 $K$.

The total angular momentum operator isUsing the cyclic orbital angular momentum commutation relations,while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,{\mathbf{L}}^2]=0}$. In Cartesian coordinates,Because $L_z(x+iy)=\hslash (x+iy)$ and it annihilates $z$ and every radial function,Replacing $i$ by $-i$ gives eigenvalue $-m\hslash$.

The six-dimensional space splits into the five trace-free quadratic spherical harmonics with $\ell =2$ and one radial state with $\ell =0$. A simultaneous eigenbasis for $L_z$ and ${\mathbf{L}}^2$ iswhere the eigenvalue $\ell (\ell +1){\hslash }^2$ of ${\mathbf{L}}^2$ was used. Every displayed state is a complex linear combination of the original six energy eigenstates, so it is still an energy eigenstate at the same level.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.

Since $O$ commutes with every angular momentum operator, it commutes with $J^2$. Taking a matrix element of $[O,J^2]=0$ givesThe allowed nonnegative values of $j$ make $j(j+1)$ injective, so the matrix element vanishes unless $j=j^{\prime }$.

In the ordered spin-one basis $(m=1,0,-1)$, the second angular momentum operator satisfies $J_2^3=J_2$. Therefore its matrix exponential reduces to $I-i\sin \theta J_2+(\cos \theta -1)J_2^2$. A half-turn exchanges the two outer weights and negates the middle weight. Applied to the pion isospin triplet, this fixes the rotation factor in G parity.