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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 $l=2$, $m=1$.

For the separated stationary state, restore its time factor $e^{-iEt/\hslash }$. Away from radial nodes, the quantum phase is $S=k\varphi -Et/\hslash$, up to a constant $0$ or $\pi$ where the real radial function has fixed sign. The guidance equation in plane polar coordinates therefore gives the Bohmian circulation of an angular-momentum eigenstateThe direction is $\hat{\mathbf{\varphi }}$ for $k>0$ and $-\hat{\mathbf{\varphi }}$ for $k<0$; the velocity is zero for $k=0$. Each admissible trajectory is a circle:The orbital angular momentum along the trajectory is $m{r}_0{v}_{\varphi }=\hslash k$. The origin or any zero-amplitude circle is excluded from this local formula. For the real degenerate superposition constructed above, the spatial quantum phase is constant on each nodal sector, so its Bohmian mechanics velocity is instead zero. Thus the circular motion is a conclusion about the separated angular-momentum eigenstate, not an arbitrary energy eigenstate.

The rotational invariance of a central-potential Hamiltonian allows a simultaneous eigenstate of energy and the two-dimensional orbital angular momentum $L_z=-i\hslash {\partial }_{\varphi }$. For a separated wavefunction $F(r)G(\varphi )$, the Laplacian in polar coordinates givesThe $\varphi$ term must be a constant; write $G^{\prime \prime }/G=-k^2$. The angular eigenfunctions can be chosen as $e^{ik\varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ imposes $e^{2\pi ik}=1$, hence $k\in \mathbb{Z}$. The radial equation in the separation of a two-dimensional central-potential eigenstate isFor a real central potential and the usual real self-adjoint radial boundary conditions, the radial equation admits a basis of real solutions: real and imaginary parts of a complex solution obey the same equation and boundary conditions. Choose a real normalized radial eigenfunction. Since the plane area element in plane polar coordinates is $rdrd\varphi$, the normalization isThis establishes the intended separated simultaneous eigenstate form. It is not the form of every stationary state. The radial equation depends on $k^2$, so the $k$ and $-k$ sectors have the same energy. For $k\ge 1$, their normalized superpositionis a single-valued stationary state with that energy but is not a single angular exponential. Quantum degeneracy is precisely why separation of variables selects a convenient eigenstate basis rather than all vectors in an energy eigenspace. The printed assertion needs this qualification. The printed polar-coordinate aid also labels a gradient component tuple as a divergence; the gradient used below is the two-dimensional vector $\hat{\mathbf{r}}{\partial }_r+\hat{\mathbf{\varphi }}{r}^{-1}{\partial }_{\varphi }$.

With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\ge 3/2$ vanish.

For the decay, the final total intrinsic spin is $1/2$. Combining it with integer orbital angular momentum $\ell$ to obtain $J=3/2$ permits only $\ell =1,2$. Parity conservation gives ${\eta }_X={\eta }_A{\eta }_B(-1{)}^{\ell }=(-1{)}^{\ell }$. Thus $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

For $\ell =2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell =1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.

The agreement is in the energy levels, not in electron trajectories. The Bohr model assumes a particle on a circular orbit, with orbital angular momentum quantized as $N\hslash$. The Coulomb bound states here are spatially distributed wavefunctions with no definite classical orbit. Their probability density is spherically symmetric, and their orbital angular momentum is zero because $\ell =0$. Their radial distributions also depend on $N$ and can have nodes. A real radial wavefunction times a global stationary phase has zero probability current, unlike the circulating point charge in the Bohr picture.

The orbital angular momentum linear operator is the cross product of the position operator and the momentum operator:In position space, $\hat{\mathbf{p}}=-i\hslash \nabla$, so $\hat{\mathbf{L}}=-i\hslash \mathbf{r}\times \nabla$. The Levi-Civita symbol gives ${\hat{L}}_1={\hat{x}}_2{\hat{p}}_3-{\hat{x}}_3{\hat{p}}_2$ and its cyclic analogues.

In atomic units the Time-independent Schrodinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the Gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

The intrinsic parity $\eta$ of a particle is the parity eigenvalue of its one-particle state at rest:For the particles in this question, $\eta =\pm 1$. A two-particle state has multiparticle parity ${\eta }_1{\eta }_2(-1{)}^{\ell }$, where $\ell$ is its relative orbital angular momentum.

The pionic-deuterium ground state has atomic $\ell =0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is ${\eta }_d{\eta }_{\pi }={\eta }_{\pi }$. The two neutrons are identical fermions. Their antisymmetric spin-one-half singlet state would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell =1$. Hence they must occupy the symmetric spin-one-half triplet state, so the orbital state is antisymmetric and $\ell$ is odd. The addition of angular momentum with $S=J=1$ permits $\ell =0,1,2$, leaving $\ell =1$. Thus parity conservation givesso the pion has intrinsic parity $\boxed{{\eta }_{\pi }=-1}$.

In ${\Delta }^{-}\to {\pi }^{-}+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ givesso the allowed orbital angular momenta are $\boxed{\ell =1,2}$. The observed angular probability is ${\cos }^2\theta =|P_1(\cos \theta ){|}^2$, where $P_1$ is a Legendre polynomial; it therefore selects the $\ell =1$ partial wave rather than $\ell =2$. The parity selection rule for a two-body decay now givesTherefore the delta particle has intrinsic parity $\boxed{{\eta }_{{\Delta }^{-}}=+1}$.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

The Hamiltonian is the sum of three commuting one-dimensional harmonic-oscillator Hamiltonians. Its product eigenstates arewith energiesEquivalently, the level with $N=n_x+n_y+n_z$ has energy $E_N=\hslash \omega (N+3/2)$, as in the three-dimensional isotropic harmonic oscillator.

Up to normalization, the ground-state wavefunction isIt is radial, so $\nabla {\Psi }_{000}$ is parallel to the position vector. Since the orbital angular momentum is $L=-i\hslash r\times \nabla$, every component of $L$ annihilates this state. Thereforeand the ground state has $\ell =m=0$.

The first excited level has $N=1$ and is spanned byThe statesatisfies $L_z{\Phi }_0=0$. Applying the ladder operators givesHence convenient $m=\pm 1$ eigenstates areup to normalization and irrelevant overall phases.

Finally, the isotropic Hamiltonian is rotationally invariant. The rotational invariance of a central-potential Hamiltonian givesThese commuting self-adjoint operators can be simultaneously diagonalized within each energy eigenspace, which is why joint eigenstates of $H,L^2,L_z$ must exist.

Parity conservation and the negative parity of $X$ require odd orbital angular momentum $L$. Since the two $Y$ particles are identical bosons, their spin state must then be antisymmetric, so their total spin is $S=1$. Total angular momentum zero forces $L=S=1$.

Conditioning on spin projection $m_S=1$ leaves orbital projection $m_L=-1$. The angular density is therefore proportional to $|Y_{1,-1}(\theta ,\phi ){|}^2\propto {\sin }^2\theta$. Including the solid-angle measure, the requested probability is

The planar orbital angular momentum isThe canonical commutation relations giveConsequentlyThus angular momentum preserves each energy eigenspace. More explicitly,soWithin a degenerate energy eigenspace, this argument places no further restriction on the matrix of $L$.

An energy and orbital angular momentum eigenstate for a real two-dimensional central potential can be chosen with integer $k$ and real radial function $f$. Its radial equation isQuantum degeneracy permits other stationary states, including $f(r)\cos (k\varphi )/\sqrt{\pi }$, which are not single angular exponentials. Separation of variables supplies a basis rather than describing every superposition in a degenerate eigenspace.