Hermiticity supplies the reverse matrix elements, also equal to the real number $t$. In an orthonormal basis for the degenerate eigenspace, the perturbation iswhere ${\mathbf{J}}_3$ is the all-ones matrix. Its symmetric vector $(1,1,1)/\sqrt{3}$ has eigenvalue $2t$. Every vector whose components sum to zero has eigenvalue $-t$; that subspace has dimension two. For example, $(1,-1,0)/\sqrt{2}$ and $(1,1,-2)/\sqrt{6}$ form an orthonormal basis for it.

The first-order energies are thereforeThe trace of the shifts is $2t-t-t=0$, agreeing with the zero diagonal. The ordering depends on the sign of $t$; the formula does not assume that the symmetric state is the lowest. Diagonalizing the perturbation inside the degenerate space is essential: the original diagonal entries alone would incorrectly predict no splitting.

The open shell contains one $f$ electron, so $L=3$ and $S=1/2$. For a less-than-half-filled shell, Hund's rules select the smaller total angular momentum, $J=L-S=5/2$, giving the free-ion ground level ${}^2{F}_{5/2}$. The Landé g-factor isIn a magnetic field weak compared with the spin-orbit splitting, the electronic Zeeman shifts are $\delta E_{m_J}=g_J{\mu }_{B}B{m}_J$, for $m_J=-5/2,-3/2,\dots ,5/2$. Thus six equally spaced magnetic substates appear. At $B=1\mathrm{T}$, the adjacent-level spacing isThe complete lowest-to-highest multiplet spread is five spacings,Giving both quantities removes the ambiguity in the word splitting. This is the free-ion, weak-field result; no crystal-field doublet or hyperfine structure is specified.

For $-z$, use a resonant half-turn pulse,The up amplitude vanishes because $\cos (\Omega {\tau }_{\pi }/2)=0$, while the down amplitude has unit magnitude. Its phase is a common phase of the final state, so any RF phase givesThe final population inversion is insensitive to a free-precession phase, unlike the orientation of a transverse superposition. This is a Rabi pi pulse with the drive-amplitude convention specified above.

The $+y$ eigenstate is $|+y⟩=(|\uparrow ⟩+i|\downarrow ⟩)/\sqrt{2}$, with ${\sigma }_y(1,i{)}^T=(1,i{)}^T$. Use the same duration ${\tau }_{\pi /2}=\pi /(2\Omega )$ but phaseThe relative amplitude is now $i$, givingThe two transverse preparations differ by a quarter-cycle of RF phase, not by changing the pulse area. Reversing the desired transverse direction adds $\pi$ to the chosen phase.

The $+x$ eigenstate is $|+x⟩=(|\uparrow ⟩+|\downarrow ⟩)/\sqrt{2}$, since ${\sigma }_x(1,1{)}^T=(1,1{)}^T$. Use a resonant quarter-turn pulse, ${\tau }_{\pi /2}=\pi /(2\Omega )$, and choose its phase so thatThe laboratory-frame relative amplitude then is $i{e}^{-i\pi /2}=1$, givingIn rotating axes the equivalent target is obtained with $\phi =\pi /2$; the extra laboratory phase above compensates Larmor precession during the pulse. After switching the RF field off, the spin continues to precess in the static field, so this statement specifies the preparation instant rather than a state fixed along laboratory $x$ forever.

A helium atom has two electrons bound to nuclear charge $+2e$. In the nonrelativistic fixed-nucleus approximation, using atomic units,The last term couples the electrons, so a product of independent hydrogenic orbitals is not an exact solution. The dominant ground configuration is $1s^2$. Its spatial wavefunction is symmetric, so Fermi antisymmetry requires the spin singletA triplet cannot put both electrons in the same $1s$ spatial orbital. Correlation changes the exact spatial function, including its dependence on $r_{12}$, while preserving this ground-state symmetry.

A useful helium variational effective charge takes normalized orbitals ${\varphi }_{\zeta }(\mathbf{r})=({\zeta }^3/\pi {)}^{1/2}{e}^{-\zeta r}$ for both electrons and multiplies their product by ${\chi }_{00}$. The energy in Hartree units isThe reduced effective charge expresses screening. Neglecting repulsion gives $-4E_h$, while its expectation in the unscreened $\zeta =2$ product gives $-2.75E_h$; optimizing the orbital improves that trial estimate. The exact ground energy is lower than the optimized product energy, since the product omits explicit electron correlation. This calculation keeps nuclear charge two in $H$; only the trial orbital parameter is varied.

For a singly excited helium state with distinct orthonormal orbitals $a=1s$ and $b=n\ell$, the spatial combinations areThe symmetric ${\Phi }_{+}$ pairs with the antisymmetric spin singlet, while antisymmetric ${\Phi }_{-}$ pairs with the symmetric spin triplet. With the same orbitals, their leading energies areHere $J_{ab}$ is the direct Coulomb integral and $K_{ab}$ is the exchange integral. The latter is positive for the Coulomb interaction, so the triplet is lower for this fixed-orbital comparison. singlet-triplet exchange splitting in helium arises from antisymmetry and spatial correlation in the Coulomb interaction, not from inserting a separate classical spin-spin force. Spin-singlet helium states are parahelium; spin-triplet helium states are orthohelium. The two spectra approximately form separate families because the leading electric dipole operator does not change total spin.^[ref]

The lowest $1s2s$ terms are ${}^3{S}_1$ and ${}^1{S}_0$, with the triplet lower. The $1s2p$ configuration gives ${}^3{P}_{0,1,2}^{\mathrm{o}}$ and ${}^1{P}_1^{\mathrm{o}}$, where the odd parity allows singlet $P$ to connect to the even-parity ground singlet by an electric dipole transition. Both $1s2s$ levels lack an allowed direct electric dipole decay to $1s^2$: the triplet is spin forbidden, and singlet $S\to S$ is forbidden by orbital angular momentum and parity. This explains helium metastability without implying an infinite lifetime once weaker radiative processes or collisions are allowed.

The observed first $1s2s$ excitation energies are approximately $19.82$ and $20.62\mathrm{eV}$ for the triplet and singlet, respectively. Higher singly excited Rydberg series approach the He II ground-state ionization threshold; doubly excited states above that threshold can autoionize and are not all stable bound excited levels.^[ref]

For a particle of charge $q$ and mass $m$ in a uniform magnetic field $\mathbf{B}=B\hat{z}$, minimal coupling replaces momentum by $\mathbf{p}-q\mathbf{A}$. Choose Landau gauge $\mathbf{A}=(0,Bx,0)$ and first restrict motion to the $xy$ plane. Since $p_y=\hslash k_y$ is conserved,These Landau levels quantize the classical cyclotron motion. The wavefunctions are harmonic-oscillator functions centred at $x_c$, multiplied by $e^{i{k}_{y}y}$. Changing $k_y$ shifts the guiding centre without changing the energy. Gauge-dependent labels such as $k_y$ are not themselves observables; the level spacing and degeneracy are gauge independent. The magnetic length ${\ell }_B=\sqrt{\hslash /|qB|}$ is the transverse orbital scale.

For a large rectangle of sides $L_x,L_y$, periodicity along $y$ gives $\Delta k_y=2\pi /L_y$. Requiring centres inside the sample gives, for each orbital level and each specified spin state,Thus each magnetic flux quantum supplies one guiding-centre state. The continuum bulk count neglects boundary corrections; magnetic periodic conditions on a torus require integer flux count. A finite confining edge lifts the simple flat guiding-centre spectrum near that edge.

spin adds a Zeeman term. For electrons with gyromagnetic factor $g$, its contribution is $g{\mu }_{B}B{m}_s$, with $m_s=\pm 1/2$ and ${\mu }_B=e\hslash /(2m_e)$. If $g=2$ and the orbital mass is $m_e$, the branches become $\hslash {\omega }_{c}n$ and $\hslash {\omega }_c(n+1)$: the lowest combined level has one spin branch, while positive levels coincide in pairs. In a material, effective mass and effective $g$ need not give that coincidence.

In three dimensions, free motion parallel to $B$ adds ${\hslash }^2{k}_z^2/(2m)$, so each orbital index labels a one-dimensional continuum rather than an isolated three-dimensional energy. Resolved magnetic quantization requires the spacing to exceed substantial thermal and disorder broadening. The changing occupation of these levels underlies magnetic quantum oscillations; in two dimensions their degeneracy and filling are central to the integer quantum Hall effect. These statements assume nonrelativistic, approximately independent particles, rather than arbitrary interacting or disordered spectra.

The quantum variational principle turns the ground-state problem into minimization of a Rayleigh quotient:The Hamiltonian must be a self-adjoint operator bounded below, and the nonzero trial state must have finite form energy. Expanding a normalized trial state in exact energy eigenstates gives $E[\psi ]={\sum }_n|c_n{|}^2{E}_n$, a weighted average no smaller than $E_0$. Equality requires support only in the ground-state eigenspace. The spectral version also accommodates continuous spectrum; a lowest normalizable eigenstate is not required merely to bound the spectral infimum.

The variational method chooses a trial family with the correct boundary conditions, fermionic exchange symmetry and other required quantum numbers, evaluates its energy and minimizes over its parameters. It gives an upper bound on ground-state energy. Enlarging a nested trial family can only improve or preserve its optimal bound. A low energy does not by itself establish a highly accurate wavefunction, especially near degeneracy; energy errors are second order in small admixtures orthogonal to a nondegenerate exact ground state.

For example, a normalized Gaussian trial for a one-dimensional quantum harmonic oscillator is ${\psi }_{\beta }=(\beta /\pi {)}^{1/4}{e}^{-\beta x^2/2}$, with $\beta >0$. Its energy isThe two terms balance kinetic localization cost against the potential cost of spreading out. This Gaussian variational energy for the harmonic oscillator happens to attain the exact answer because the family contains the exact ground state.

A trial expansion $\psi ={\sum }_j{c}_j{\chi }_j$ in a finite, possibly nonorthogonal basis gives the generalized eigenvalue problem $H\mathbf{c}=ES\mathbf{c}$, with $H_{ij}=⟨{\chi }_i|H|{\chi }_j⟩$ and $S_{ij}=⟨{\chi }_i|{\chi }_j⟩$. Its lowest Ritz value minimizes the quotient in that subspace. For excited states, the excited-state variational constraint imposes exact orthogonality to the lower eigenstates to obtain a corresponding upper bound; the Courant–Fischer min-max principle yields upper bounds for ordered Ritz eigenvalues from a common subspace. orthogonality only to an independently approximate ground state does not automatically provide an excited-state bound: residual true-ground-state contamination can lower the trial energy below the desired excited energy. Symmetry can impose exact orthogonality when the relevant states have different symmetry sectors.

Closed $1s^2$ and $2s^2$ subshells have $L=S=0$. In the open $2p^4$ subshell, three electrons can have parallel spins in $m_{\ell }=-1,0,1$, and the fourth must have the opposite spin, giving maximum $S=1$. Treating the shell as two $p$ holes gives the same allowed terms as $p^2$; at this spin, antisymmetry permits $L=1$, rather than $L=2$. One can also see $M_L=1$ by placing the fourth electron in $m_{\ell }=1$.

Hund's rules therefore select the ${}^{3}P$ term. Since four electrons exceed half filling of the six-state $p$ subshell, its lowest atomic fine-structure level has the largest $J=L+S=2$:Its zero-field magnetic degeneracy is $2J+1=5$. The experimentally tabulated neutral-oxygen ground level has precisely this term and $J$.^[ref]

At incident wave number $k$, low kinetic energy means $ka\ll 1$: the de Broglie wavelength is much larger than the scattering region. A generic finite-range scatterer can then be described to leading order without resolving its angular structure. In partial-wave scattering, this is leading S-wave scattering. The $\ell =0$ spherical harmonic is constant, whereas the centrifugal term ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$ impedes penetration to the potential for $\ell >0$.

For a central potential with regular threshold behaviour, the finite-range partial-wave threshold law gives ${\delta }_{\ell }=O(k^{2\ell +1})$. Thuswhich is independent of scattering angle. For finite nonzero scattering length $a_s$, $f\to -a_s$ and $d\sigma /d\Omega \to a_s^2$ is isotropic. The potential range $a$ and scattering length $a_s$ are different quantities. The claim describes the generic leading S-wave scattering limit; tuned vanishing S-wave scattering amplitude or exceptional threshold resonances require separate analysis rather than a universal angular conclusion.

The Pauli exclusion principle allows each one-particle magnetic substate only once, since the fermions share the other quantum numbers. To maximize the total angular momentum projection, occupy $m_j=5/2,3/2,1/2,-1/2$, givingThe corresponding Slater determinant is a highest-weight vector: every attempted raising operation either tries to exceed $j$ or duplicates an occupied substate and vanishes. Therefore it belongs to a multiplet with $\boxed{J_{\max }=4}$, not the distinguishable-particle value $4j=10$. Equivalently, four particles in a six-state shell have two holes whose maximum angular momentum projection is $5/2+3/2=4$. The general maximum total angular momentum of equivalent fermions is $J_{\max }=Nj-N(N-1)/2$ for $N\le 2j+1$.

For $|w|\gg |\Delta |$, the mixed states approach opposite-spin product states andThe omitted mixed-branch correction is of order ${\Delta }^2/|w|$. The four levels are generically nondegenerate, apart from special moment choices or crossings.

For equal-and-opposite signed moments, $g_2=-g_1$, $u=0$ and the parallel-spin level remains exactly twofold degenerate at $\Delta /4$; the other two approach $-\Delta /4\pm |g_1{\mu }_{0}B|$. Equal moment magnitudes are an extra assumption, not implied by opposite charges alone. If $g_1=g_2$, the differential Zeeman term vanishes and the singlet/triplet-zero block stays unmixed by the field. This is the strong-field limit of a two-spin exchange spectrum.

At zero field, the spin-one-half triplet state has energy $\Delta /4$ and degeneracy three; the spin-one-half singlet state has energy $-3\Delta /4$ and degeneracy one:For $\Delta >0$ the singlet is lower; for $\Delta <0$ the triplet is lower. If $\Delta =0$, all four states coincide.

A laser combines pumping, an amplifying medium and optical feedback. Stimulated emission adds photons to an occupied optical mode; net amplification requires population inversion between the relevant levels after accounting for their degeneracies. Pumping through additional levels and rapid lower-level relaxation can sustain inversion; a closed two-level system under steady incoherent resonant pumping cannot sustain it.

An optical resonator selects modes and feeds light back through the gain medium. For intensity gain coefficient $g$, internal intensity loss ${\alpha }_i$, length $L$ and mirror intensity reflectivities $R_1,R_2$, the round-trip laser threshold condition isAbove threshold, saturation limits gain and a partially transmitting mirror provides output. Typical properties are strong directionality, substantial coherence and narrow bandwidth; multiple modes and pulsed operation are possible, so perfect monochromaticity is not universal.^[ref]

The central field approximation replaces the anisotropic many-electron environment by an average spherical potential, giving one-electron orbitals labelled by $n,l,m$. Screening changes the effective radial attraction and removes the pure Coulomb degeneracy between different $l$.

The Hartree method improves the potential self-consistently. For normalized occupied orbitals,Guess orbitals, construct their mean fields, solve the Hartree equations, and iterate to convergence. The $j\ne i$ sum excludes self-interaction in this product-state model. A spherical average can enforce the central-field specialization. A Hartree product state omits fermionic exchange and dynamical correlation; Hartree-Fock method adds antisymmetry and exchange through a Slater determinant. Converged orbital eigenvalues must not simply be added as the total energy, because that double counts pair interactions.^[ref]

The postulates of quantum mechanics assign a pure state to a normalized ray in a complex Hilbert space, and an observable to a self-adjoint operator. A spectral outcome projector $P_a$ has Born rule probability $p_a=⟨\psi |P_a|\psi ⟩$. An ideal projective measurement updates the state to $P_a|\psi ⟩/\sqrt{p_a}$ when $p_a>0$; degeneracy requires the full outcome projector.

Between measurements, an isolated system obeys $i\hslash {\partial }_t|\psi ⟩=H|\psi ⟩$, giving unitary evolution. Composite systems use tensor products; identical particles additionally obey the appropriate exchange symmetry. Mixed preparations use a density matrix, with $p_a=\mathrm{tr}(\rho P_a)$. These postulates require a specified Hamiltonian and measurement model; they do not assign those automatically.^[ref]

For two equivalent $p$ electrons, antisymmetry permitsTheir $9+5+1=15$ states exhaust $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund's rules first maximize spin, then orbital angular momentum; the less-than-half-filled $p$ shell puts the smallest $J$ lowest. Hence the ground level is ${}^3{P}_0$. This is the allowed LS terms of a p-squared configuration.^[ref]

An impulse integrated across its full support produces the impulsive quantum kick $K=e^{-i\hat{U}/\hslash }$. Thus the leading transition amplitude is $-i{U}_{21}/\hslash$ andHere $U_{21}$ has units of action. The off-diagonal two-state impulse is exactly $P={\sin }^2(|U_{21}|/\hslash )$ for a Hermitian two-state impulse with zero diagonal. No energy denominator occurs: the impulse is instantaneous.

The normalized quantum harmonic oscillator ground state has $|A{|}^2=\sqrt{m\omega /(\pi \hslash )}$ and energy $E_0=\hslash \omega /2$. Its classical turning points are $x=\pm \sqrt{\hslash /(m\omega )}$. With $y=x\sqrt{m\omega /\hslash }$, the ground-state oscillator probability beyond the turning points isThe stated rounded error function gives $0.16$; the more precise Gaussian integral is $0.15730$.

The Stark effect changes atomic energy levels in a static electric field. For $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$, the electric-dipole interaction is $H^{\prime }=-d_z\mathcal{E}=e\mathcal{E}z$ for an electron, with $e>0$. A nondegenerate state of definite parity has $⟨z⟩=0$, so its leading quadratic Stark effect isThe ground state shift is negative, conventionally $-\alpha {\mathcal{E}}^2/2$ with positive electric polarizability.^[ref]

Degenerate opposite-parity states require degenerate perturbation theory. In ideal Coulomb hydrogen, the $2s,2p_0$ subspace has an off-diagonal matrix element of magnitude $3e{a}_{\mathrm{B}}\mathcal{E}$, where $a_{\mathrm{B}}$ is the Bohr radius. Its mixed states shift byThis Linear Stark effect in the first excited hydrogen eigenspace assumes fine-structure and Lamb splittings are negligible compared with the mixing. In an isolated parity eigenstate with such degeneracies resolved, sufficiently weak fields restore quadratic behaviour. Strong fields require further mixing and can ionize the atom.^[ref]