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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 orbital angular momentum of a particle is the operator $\mathbf{L}=\mathbf{r}\times \mathbf{p}$, with $\mathbf{p}=-i\hslash \nabla$ in the position representation. Its angular momentum commutation relations areThus one may specify $L^2$ and one component, conventionally $L_z$, simultaneously. One generally cannot assign sharp values to all three components, because they do not commute. The common angular eigenfunctions are the spherical harmonics:For a scalar orbital wavefunction, single-valuedness in the azimuthal angle gives integer $m$, and regularity on the sphere and termination of the ladder give integer $l\ge |m|$. Half-integer angular momenta describe intrinsic spin rather than scalar orbital motion.

The ladder operators $L_{\pm }=L_x\pm i{L}_y$ obey $[L_z,L_{\pm }]=\pm \hslash L_{\pm }$. Their normalized action isThe coefficient vanishes at the top or bottom of the multiplet. Its squared magnitude follows from $L_{\mp }{L}_{\pm }=L^2-L_z^2\mp \hslash L_z$, and positivity of the resulting norm constrains the allowed $m$ values.

For a central potential, the Hamiltonian commutes with $L^2$ and every component of $\mathbf{L}$. Stationary states can be separated as $R_{nl}(r)Y_l^m(\theta ,\varphi )$, and rotational symmetry makes the energy independent of $m$, giving a $2l+1$ quantum degeneracy at fixed radial label and $l$. Additional degeneracy between distinct $l$ values requires additional structure, as in the Coulomb potential. The parity of $Y_l^m$ is $(-1{)}^l$. These results organize atomic orbitals and their selection rules. They describe quantized rotation and symmetry, without assigning the electron a classical orbit or a simultaneously definite three-component angular momentum vector.

For the quantum harmonic oscillator, using $[\hat{x},\hat{p}]=i\hslash$, direct multiplication givesThe commutators areThus each ladder operation shifts energy by one quantum. The ladder operator norms $\Vert a|n⟩{\Vert }^2=n$ and $\Vert a^{†}|n⟩{\Vert }^2=n+1$, with conventional phases, giveLet $\ell =\sqrt{\hslash /(m\omega )}$. In position space $\hat{p}=-i\hslash {\partial }_x$, so $a{\psi }_0=0$ becomes $(x+{\ell }^2{\partial }_x){\psi }_0=0$. Normalizing and then applying $a^{†}$ givesTheir overlap vanishes because ${\psi }_0^{\ast }{\psi }_1$ is odd on the full real line; the Gaussian second moment verifies $\int |{\psi }_1{|}^{2}dx=1$.

Invert the given ladder operators:Because the Energy eigenstate are orthonormal, $⟨n|{\hat{a}}_{\pm }|n⟩=0$: each operator changes the state to one orthogonal to $|n⟩$. HenceThe result also follows from the definite parity of every oscillator energy eigenfunction, but here it follows directly from the specified operators.

Expanding the total angular momentum and replacing transverse components by ladder operators givesThe particle exchange operator obeys $P_{12}^2=I$ and $P_{12}{s}_{1\alpha }{P}_{12}=s_{2\alpha }$, with the reverse relation for index $2$. These statements also apply to the ladder operators. Exchanging the indices preserves $s_1^2+s_2^2$ and $s_{1z}+s_{2z}$; it interchanges the last two terms in $S^2$. Operators on different tensor factors commute. HenceThe full particle exchange includes the spin coordinates as well as the spatial coordinates.

In the ordered orthonormal basis $(|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩)$, the spin one-half angular momentum operators are $J_i=(\hslash /2){\sigma }_i$, with Pauli matricesThe ladder operators areExplicit multiplication gives $J_{+}{J}_{-}={\hslash }^2\mathrm{diag}(1,0)$ and $J_{-}{J}_{+}={\hslash }^2\mathrm{diag}(0,1)$. Their commutator is therefore

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

In the ordered angular momentum eigenstate basis $(|1,1⟩,|1,0⟩,|1,-1⟩)$, with positive raising coefficients,To derive them, $[J_3,J_{+}]=\hslash J_{+}$ restricts $J_{+}$ to its two adjacent superdiagonal entries. Write these real entries as $A,B\ge 0$. Since $J_{-}=J_{+}^{†}$, the relation $[J_{+},J_{-}]=2\hslash J_3$ reads $\mathrm{diag}(A^2,B^2-A^2,-B^2)=2{\hslash }^2\mathrm{diag}(1,0,-1)$, forcing $A=B=\sqrt{2}\hslash$. Adding and subtracting the ladder operators gives the displayed matrices. Direct multiplication verifies the angular momentum commutation relations and $J^2=2{\hslash }^{2}I$.

In the product basis $|m_1,m_2⟩$ for two spin-one particles, total magnetic quantum number zero allows only $|1,-1⟩$, $|0,0⟩$ and $|-1,1⟩$. Applying the total ladder operators to $\alpha |1,-1⟩+\beta |0,0⟩+\gamma |-1,1⟩$ givesA singlet state is annihilated by $S_{+}$ and $S_{-}$, so $\alpha =\gamma =-\beta$. Wavefunction normalization gives the displayed state, up to an overall phase. Here the ket on the left of the title formula has total-spin labels, while its right-hand kets have individual magnetic labels. The state is symmetric under exchanging the particles, although no exchange constraint is needed for distinguishable particles.