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For $N$ identical fermions restricted to one angular-momentum multiplet $j$, with all other one-particle labels fixed, the Pauli exclusion principle permits distinct angular momentum projections $j,j-1,\dots ,j-N+1$. Their sum is $Nj-N(N-1)/2$, provided $N\le 2j+1$. The resulting Slater determinant is killed by the total raising angular momentum ladder operator because raising any occupied state either exceeds $j$ or duplicates an occupied state. It is therefore a highest-weight vector of precisely this total angular momentum. The formula is invariant under replacing particles by holes in the full shell.

A spherical tensor operator of rank $k$ transforms under rotation like an angular momentum $k$. The Wigner-Eckart theorem writes its matrix elements asThe reduced matrix element contains the radial and other dynamical information; the Clebsch-Gordan coefficients determine the dependence on the projections. Ratios within a fixed pair of multiplets therefore follow without a radial integral. The angular selection rules include $m^{\prime }=m+q$ and $|j-k|\le j^{\prime }\le j+k$.

For the position vector, choose the rank-one componentsLet $R=⟨{\lambda }^{\prime },3/2\Vert T^{(1)}\Vert \lambda ,1/2⟩$. The stretched Clebsch-Gordan coefficients and one lowering step giveFor example, apply the total angular momentum ladder operator to the stretched state. The result has components with coefficients $1/\sqrt{3}$ and $\sqrt{2/3}$, corresponding to lowering the initial angular momentum and the vector component, respectively.

Only the $q=+1$ component of $x$ contributes to $M_1$. The $q=0$ operator $z$ cannot connect the projections in $M_2$, so that matrix element vanishes. With the convention above,Thus the requested relative magnitudes and their squared values areThese ratios presume a common nonzero reduced matrix element. The position operator has odd parity, so an electric dipole transition requires opposite-parity multiplets; for one electron the electric-dipole selection rule also gives $\Delta \ell =\pm 1$. The displayed $j,m$ labels alone do not establish that condition. If the common reduced matrix element vanishes, all three matrix elements vanish and no nonzero ratio can be inferred.

The Zeeman effect comes from $-\mathbf{\mu }\cdot \mathbf{B}$. For an electron, the orbital and spin magnetic moments are approximately $-{\mu }_B\mathbf{L}/\hslash$ and $-2{\mu }_B\mathbf{S}/\hslash$, respectively. This explains both the sign and the relative factor two in the term linear in $B$. The second term is the spin-orbit interaction: the electron's motion through the atomic electric field produces a magnetic interaction in its rest frame. The relativistic correction, including Thomas precession, and the radial expectation value are incorporated into $K$. This effective Hamiltonian retains the linear Zeeman splitting and spin-orbit coupling, while omitting diamagnetic and hyperfine structure corrections.

Define the angular momentum ladder operators by $L_{\pm }=L_x\pm i{L}_y$ and similarly for $S_{\pm }$. Direct expansion givesso adding $L_z{S}_z$ gives the required scalar product. The off-diagonal terms change $m_{\ell }$ and $m_s$ in opposite directions. Consequently $J_z=L_z+S_z$ commutes with $H$, and $m_j$ remains a good quantum number at every magnetic field.

In the given product eigenbasis, the diagonal matrix elements areFor $\ell =1$ and $s=1/2$, the only off-diagonal couplings are between $(1,-1/2)$ and $(0,1/2)$, and between $(0,-1/2)$ and $(-1,1/2)$. Each relevant orbital ladder factor is $\hslash \sqrt{2}$ and each spin ladder factor is $\hslash$. Multiplying by $K/{\hslash }^2$ gives $\sqrt{2}K$. ThereforeAll other off-diagonal entries vanish either by the ladder endpoints or by conservation of $m_j$. This derives every entry of the displayed matrix.

The stretched states have no mixing and yieldThe $m_j=1/2$ block is $\left(\begin{array}{cc}-K& \sqrt{2}K\\ \sqrt{2}K& a\end{array}\right)$, while the $m_j=-1/2$ block is $\left(\begin{array}{cc}-a& \sqrt{2}K\\ \sqrt{2}K& -K\end{array}\right)$. Taking the trace and determinant of each $2\times 2$ block gives the remaining four energy eigenvalues:The subscripts indicate $m_j=\pm 1/2$ and the superscripts indicate the upper or lower root. These formulas are the Zeeman spectrum for orbital angular momentum one and remain valid for either sign of $K$.

For the usual ordering $K>0$, including sodium, the zero-field states follow fromThe $j=3/2$ quartet lies at $K$ and the $j=1/2$ doublet at $-2K$, separated by $3K$. In a weak magnetic field, the Landé g-factors giveThese label the curves near $B=0$. In a strong magnetic field, the Paschen-Back effect makes $(m_{\ell },m_s)$ the appropriate approximate labels:for the mixed states, with the stretched-state formulas already exact. The branch connections for $B>0$ areWithin a fixed $m_j$, the nonzero coupling prevents a crossing; this is an avoided crossing. Curves with different $m_j$ may cross because the Hamiltonian does not couple them.

For the sodium estimate, the zero-field frequency separation fixesThe lower $3s$ doublet has shifts $\pm a$. The Sodium D-line Zeeman pattern gives photon-frequency shifts $a/h$ times $\{\pm 2/3,\pm 4/3\}$ for D${}_1$ and $\{\pm 1/3,\pm 1,\pm 5/3\}$ for D${}_2$. Thus the highest-frequency D${}_1$ component and lowest-frequency D${}_2$ component approach each other by $3a/h$. Extrapolating the weak-field Zeeman splitting givesso the crossover occurs at a field of order $10\mathrm{T}$, where $a\sim K$ and the weak-field approximation is becoming marginal.

The exact eigenvalues refine the estimate within this Hamiltonian. The relevant photon energies, with a common zero-field reference omitted, are $E_{+}^{(-)}+a$ and $E_{-}^{(+)}-a$. The corresponding electric dipole transitions have $\Delta m_j=+1$ and $-1$, respectively. Equating them givesIts positive solution is $a/K=\sqrt{10}/3$, hencein the retained model. The crossing is between optical transition frequencies; the participating upper states themselves need not be degenerate.

The orbital angular momentum commutation relations giveConsequentlyThe same commutation relations imply $[L^2,L_x]=[L^2,L_y]=0$, and henceThese are the defining identities for the angular momentum ladder operators.

Ifthen, whenever $L_{\pm }\varphi \ne 0$,Thus $L_{\pm }\varphi$ has quantum numbers $(\ell ,m\pm 1)$.

orbital angular momentum generates spatial rotations and is $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. With $\mathbf{p}=-i\hslash \nabla$, its components satisfyThus $L^2$ and one component, conventionally $L_z$, have simultaneous eigenstates, while all three components cannot generally have sharp values. On ordinary single-valued spatial wavefunctions,The spherical harmonics form an orthonormal angular basis. The integer restriction is specific to orbital wavefunctions; half-integer spin is an intrinsic degree of freedom, not the angular dependence of a scalar spatial wavefunction.

The angular momentum ladder operators obeyThe ladder ends at $m=\pm \ell$, producing $2\ell +1$ states. A state with specified $m$ has a sharp projection $\hslash m$, but the angular momentum magnitude is $\hslash \sqrt{\ell (\ell +1)}$, not $\hslash \ell$. No classical picture of a vector with simultaneously fixed Cartesian components reproduces all these measurement properties.

In a central potential, rotational symmetry gives $[H,L_i]=0$. separation of variables yields $\psi (\mathbf{r})=R_{\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The radial Schrödinger equation includes a centrifugal potential ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$, which suppresses small-radius amplitude for nonzero $\ell$. The energy is independent of $m$ when rotational symmetry is intact, but need not be independent of $\ell$. A magnetic field can break that degeneracy. Spatial inversion gives $Y_{\ell m}(-\hat{\mathbf{r}})=(-1{)}^{\ell }{Y}_{\ell m}(\hat{\mathbf{r}})$, connecting orbital angular momentum to parity and to the even/odd rotational restriction for the hydrogen spin isomers.

For Hermitian $J_3$ and adjoint angular momentum ladder operators $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space, a highest-weight vector of weight $j$ obeys $J_{+}|j,j⟩=0$. The commutation relations give $J_{+}(J_{-}{)}^{\ell }|j,j⟩=\ell (2j-\ell +1)(J_{-}{)}^{\ell -1}|j,j⟩$. Taking inner products forces $2j$ to be a nonnegative integer: otherwise the first negative factor would give a negative squared norm. The norm first vanishes at $\ell =2j+1$, so the generated unitary representation has dimension $2j+1$. Without positivity an abstract highest-weight module can remain infinite-dimensional even at an integral highest weight.