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A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.

The energy needed to remove specified nucleons from a bound atomic nucleus in its ground state. It compares neighboring nuclear masses, and so can reveal a nuclear shell closure more clearly than a smoothly varying total nuclear binding energy.

An odd periodic nearest-neighbour Ising model with antiferromagnetic coupling has unavoidable geometrical frustration. For $k$ parallel bonds, $H=-NJ+2Jk$, while ${\prod }_j{S}_j{S}_{j+1}=1$ forces $N-k$ to be even. Thus odd $N$ requires at least one parallel bond. A ground state has exactly one, with $N$ possible locations and two global spin inversion symmetry choices, proving the displayed ground-state degeneracy and ground-state energy. The zero-temperature canonical ensemble distributes that defect uniformly. A directed arc of $n$ bonds contains it with probability $n/N$, giving two-point correlation function $(-1{)}^n(1-2n/N)$. The finite antiferromagnetic Ising-ring correlation gives the corresponding positive-temperature result.

A normalized ground state initially supported on $(0,a)$ is $\sqrt{2/a}\sin (\pi x/a)$. Expanding the infinite square well to $(0,2a)$ leaves this wavefunction unchanged during an ideal sudden switch. Its overlap with the new ground state $a^{-1/2}\sin (\pi x/(2a))$ is $4\sqrt{2}/(3\pi )$, so the Born rule gives $32/(9{\pi }^2)$. This differs from ground-state overlap after sudden expansion of a square well with a fixed midpoint, because translation of the new well changes the overlap. Under a fixed final Hamiltonian operator, all final energy eigenstate probabilities remain constant during subsequent unitary time evolution.

Optical pumping redistributes populations among internal quantum states by selective photon absorption followed by spontaneous emission. It can prepare atoms for measurement and participate in sub-Doppler cooling.

The practical aim is to make otherwise almost equally populated magnetic sublevels behave differently. Repeated absorption and emission can store a population imbalance in the long-lived ground state; a change in photon absorption or fluorescence then reveals a perturbation of that prepared distribution. This combines preparation and measurement: a weak radio wave can reorganize the populations, while abundant optical photons supply the readout.

An imbalance among ground state magnetic sublevels need not be a population inversion between optical energy levels. Optical pumping also supplies excitation in some lasers, where an appropriate population inversion matters for gain. The preparation problem here concerns the distribution within the ground state. A two-level system with only one lower quantum state cannot describe this redistribution.

The uniform paramagnetic three-dimensional electron-gas determinant is unstable within Hartree-Fock theory to a spin-density-wave deformation, even at high density. Comparing only spatially uniform polarized and unpolarized determinants therefore does not find the unrestricted Hartree-Fock ground state.^[ref]

A pairing gap is an energy scale for fermionic excitations above a paired superfluid or superconductor ground state. In a simple isotropic paired system the quasiparticle dispersion has the form $E_{\mathbf{k}}=\sqrt{{\xi }_{\mathbf{k}}^2+|\Delta {|}^2}$, where ${\xi }_{\mathbf{k}}$ measures the unpaired single-particle energy relative to the chemical potential. A nonzero minimum gap gives exponentially small fermionic thermal occupations when $k_{B}T\ll |\Delta |$. If the gap varies with direction and has nodes, gapless directions can instead produce power-law thermodynamics. Collective excitations must be counted separately from the gapped fermionic quasiparticles.

An even one-dimensional confining potential preserves even and odd parity. Restricting a normalized trial state to the odd sector makes its Rayleigh quotient an upper bound on the lowest odd energy, which is the first excited energy for the usual nonsingular one-dimensional bound-state problem. This avoids contamination by the even ground state. For an adjustable-width odd Gaussian, optimize its own width rather than borrowing the ground-state optimum.

Density functional theory replaces the interacting many-electron ground state wavefunction by its spatial density $n(\mathbf{r})$. For fixed interaction and a nondegenerate ground state, the Hohenberg-Kohn theorem fixes the external potential up to an additive constant. If two distinct potentials gave the same density, their strict variational energy inequalities would sum to the contradiction $E_0+E_0^{\prime }<E_0+E_0^{\prime }$.

The density functional variational principle minimizesover admissible densities of the correct particle number. The universal $F$ contains the kinetic and interaction energies. This is exact in principle; an explicit useful functional is the practical difficulty.^[ref]

The Kohn-Sham equations introduce noninteracting orbitals with the same density and decompose $F=T_s+E_H+E_{\mathrm{xc}}$. With $n={\sum }_i{f}_i|{\varphi }_i{|}^2$,The exchange-correlation functional includes exchange, correlation and the interacting kinetic-energy correction relative to $T_s$. Guess a density, construct the potentials, solve and occupy orbitals, then mix and iterate to self-consistent field convergence. The local density approximation uses $E_{\mathrm{xc}}^{\mathrm{LDA}}=\int n{ϵ}_{\mathrm{xc}}^{\mathrm{uniform}}(n)d^{3}r$; a generalized gradient approximation also depends on density gradients.

Kohn–Sham orbital energies are not generally excitation energies, and their sum double counts interactions. The Kohn-Sham energy double-counting correction iswith ion–ion energy added for a total crystal energy. Approximate functionals can miss self-interaction cancellation, band gaps and strong correlation; convergence alone does not prove the global ground state.^[ref]

Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

Use a three-dimensional electron gas with a uniform positive background, the jellium model. The Hartree-Fock method minimizes the expectation of the interacting Hamiltonian over normalized Slater determinants. Antisymmetry is exact within that trial class, but correlations requiring a superposition of determinants are omitted. For orthonormal occupied spin orbitals ${\varphi }_i$, the energy iswhere $J_{ij}$ is the direct Coulomb integral and $K_{ij}$ the exchange integral; the spin overlap makes exchange vanish between opposite-spin orbitals. The self-interaction terms $J_{ii}$ and $K_{ii}$ cancel. Variation gives a self-consistent Fock operator $\hat{f}=h_0+{\sum }_j({\hat{J}}_j-{\hat{K}}_j)$; exchange is nonlocal. Its eigenfunctions define occupied orbitals of a stationary determinant, and the determinant must be compared with competing self-consistent solutions to select a variational ground state.

The eigenvalues are orbital Lagrange multipliers. Their sum counts each interaction twice:Frozen-orbital removal energies may be estimated using Koopmans' theorem, but measured quasiparticle energies can contain relaxation and correlations absent from this approximation. Unoccupied orbital energies are not automatically an exact excitation spectrum. In particular, unscreened exchange in a uniform three-dimensional gas produces an unphysical logarithmic slope singularity at its Fermi surface; a one-determinant mean field does not reproduce the interacting Fermi liquid's screened response.

For a spatially uniform plane-wave determinant, electron-background and background-background terms cancel the uniform direct Coulomb contribution. Retain the background when making this cancellation: an isolated infinite electron charge distribution would have a divergent direct energy. The surviving contributions per particle are kinetic energy and negative same-spin exchange energy. Let $(3/(4\pi n){)}^{1/3}=r_s{a}_B^{\ast }$ define the Dimensionless Wigner-Seitz radius, with $a_B^{\ast }=4\pi \epsilon {\hslash }^2/(m^{\ast }{e}^2)$, $E_H^{\ast }={\hslash }^2/(m^{\ast }{a}_B^{\ast 2})$ and $\zeta =(n_{\uparrow }-n_{\downarrow })/n$. ThenFor the unpolarized state this is approximately $1.105/r_s^2-0.4582/r_s$. The kinetic contribution penalizes spin polarization because it enlarges an occupied Fermi sphere; exchange favors polarization because equal-spin occupation gains more exchange energy. As density falls, the $r_s^{-1}$ exchange scale overtakes the $r_s^{-2}$ kinetic scale. The uniform fully polarized and unpolarized energies cross atThis is a variational comparison within the homogeneous Hartree-Fock electron gas, not an exact physical ferromagnetic phase boundary. It demonstrates how Pauli kinetic energy and exchange compete and why the allowed trial states matter.

Allowing spatial symmetry breaking changes the result: the Overhauser instability lowers the paramagnetic determinant's Hartree-Fock energy through a spin density wave, even in the high-density limit. A calculation restricted to uniform Fermi spheres therefore does not give the unrestricted Hartree-Fock ground state.^[ref] At still lower density, localization and a Wigner crystal compete with fluid states. electron-gas correlation energy beyond a single determinant is important in assessing their competition. Quantum Monte Carlo provides a concrete route to comparing correlated fluid and crystal energies; the original stochastic calculation demonstrates why exchange-only energy differences are insufficient.^[ref]

Hartree-Fock is a useful variational baseline with exact exchange, not a controlled complete low-density phase diagram. At high density the kinetic scale dominates, but even a small symmetry-breaking energy gain can invalidate a restricted determinant as the exact Hartree-Fock minimizer. At low density neglected correlations, screening and localization can change both transition locations and the ordering of candidate phases. A numerical phase boundary should therefore state dimension, spin constraints, background model and whether correlation and spatially modulated states were included.

For any fixed $r\ge 0$, the source contribution obeys $c{\varphi }_1\ge cr$, because $c<0$ and ${\varphi }_1\le r$. Equality occurs only at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(r,0,0)$ when $r>0$. Thus the global ground state lies on the positive first axis. Minimize there:For $0\le r\le v=\mu /\sqrt{\lambda }$, $f^{\prime }(r)<0$. For $r>v$, $f^{\prime \prime }(r)=3\lambda r^2-{\mu }^2>0$, so $f^{\prime }$ increases strictly from $c$ to $+\infty$. There is exactly one root $\tilde{v}>v$, and it is the unique global minimum. ThereforeThis argument selects the true ground state even if the full cubic has additional negative stationary roots.

Expand ${\varphi }_1=\tilde{v}+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The source is linear, so it shifts the expansion point but has zero Hessian matrix. The scalar mass matrix is diagonal at the new vacuum expectation value:Eliminating $\lambda$ with the stationary equation givesAll three squared masses are positive. The masses themselves are the positive square roots of these expressions.

The source explicitly reduces the stated SO(3) symmetry to SO(2); the selected vacuum state preserves that remaining SO(2). There is consequently no spontaneously broken exact continuous global symmetry in field theory to which the Goldstone theorem applies. For small $|c|$, the two angular excitations are pseudo-Goldstone bosons. Their squared masses vanish as $c\to 0^{-}$, while $\tilde{v}\to \mu /\sqrt{\lambda }$ and $m_h^2\to 2{\mu }^2$.

Put $r^2={\varphi }_i{\varphi }_i$. Completing the square in the scalar potential givesSince $\lambda >0$, its global minima form the vacuum manifoldHere $\mu$ denotes the positive square root of the parameter ${\mu }^2$. The origin is a local maximum, not a ground state. Choose $⟨\mathbf{\varphi }⟩=(v,0,0)$. Its stabilizer subgroup is the SO(2) subgroup of rotations in the $(2,3)$ plane, so the stated SO(3) global symmetry in field theory is spontaneously broken to SO(2).

Write ${\varphi }_1=v+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The kinetic terms stay canonically normalized, while the scalar mass matrix isEquivalently, the quadratic scalar potential is ${\mu }^2{h}^2$. Therefore the particle masses areThe two angular Goldstone bosons match $\dim SO(3)-\dim SO(2)=3-1=2$. The radial displacement changes $r$ and costs potential energy, giving the massive Higgs mode. These are tree-level masses; the Goldstone theorem protects the angular zero-mass modes in the exact global theory.

Adjacent nodes of a wavefunction are separated by roughly $a/n$, so a locally sinusoidal wavefunction has typical wavevector $k\sim \pi n/a$. The kinetic-energy estimate from wavefunction nodes is thereforeEquivalently, $⟨K⟩={\hslash }^2\int |{\psi }^{\prime }{|}^{2}dx/(2m\int |\psi {|}^{2}dx)$. The robust scaling is ${\hslash }^2{n}^2/(m{a}^2)$; the precise coefficient depends on how the extent is defined and how the oscillations vary near classical turning points. For few nodes, use an order-one offset to $n$: the nodeless ground state still has confinement energy bound from position support of order ${\hslash }^2/(m{a}^2)$, rather than zero.

Centre the infinite square well at zero. Its ground state wavefunction is $\psi (x)=\sqrt{2/L}\cos (\pi x/L)$ inside the well and zero outside. The momentum representation in quantum mechanics gives the wavevector amplitudeConsequently the unnormalised wavevector distribution of an infinite-well ground state isThe apparent singularities at $k=\pm \pi /L$ are removable. The normalized probability density is $4\pi L$ times the boxed expression and has limit $L/(4\pi )$ there. Translating the well changes only the Fourier transform's phase. The wavevector is continuously distributed; a box energy eigenstate is not a free momentum eigenstate.

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]

With $y=x-x_0$ and oscillator length $a_0=\sqrt{\hslash /(m\omega )}$, the first two harmonic oscillator eigenfunctions and energies areThe ground state is even and nodeless; the first excited state is odd with one node of a wavefunction. Their rms widths are $a_0/\sqrt{2}$ and $\sqrt{3/2}{a}_0$. Also, $m{\omega }^2{a}_0^2\sim \hslash \omega$ gives the characteristic extent directly.

For the shaken harmonic trap, expanding about the unshifted Hamiltonian givesThe scalar term gives only a common phase and no transitions. In time-dependent perturbation theory, initially $c_0=1$, $c_{n\ne 0}=0$ and the oscillatory factor for $0\to 1$ is $e^{i(E_1-E_0)t/\hslash }=e^{i\omega t}$. ThusThe odd operator $x$ makes transitions to even $n>0$ vanish by parity. In fact the first-order selection rule for a linearly driven oscillator is stronger: since $x=a_0(a+a^{†})/\sqrt{2}$,All excited $n\ne 1$ amplitudes vanish at first order, including odd $n>1$. This assertion does not apply to the surviving $c_0$ or to higher perturbative orders.

Put $\delta =\omega -\Omega$, $\Sigma =\omega +\Omega$. The exact first-order time integral isNear resonance, the second term oscillates rapidly at approximately $2\omega$ and is bounded by $1/\Sigma$, whereas the first grows as $T/2$ on resonance. Thus $\omega T\gg 1$ justifies the rotating-wave approximation, $\cos \Omega t\mapsto e^{-i\Omega t}/2$. At zeros of the resonant term, the omitted small contribution can dominate the residual; the approximation has no uniform relative-error bound there.

Using the requested order-one approximation $⟨1|x|0⟩\simeq a_0$ and $m{\omega }^2{a}_0/\hslash =\omega /a_0$ gives the square-pulse resonance profileThe exact harmonic oscillator matrix element instead gives half this probability. The finite central maximum is at $\Omega =\omega$; first zeros are at $|\Omega -\omega |=2\pi /T$, with smaller side lobes and a $1/{\delta }^2$ envelope. Increasing $T$ narrows the peak as $1/T$ and increases its height as $T^2$, as long as weak resonant driving keeps $p_1\ll 1$. Small displacement alone does not ensure this for arbitrarily long $T$; near resonance one also needs $\epsilon \omega T/a_0\ll 1$.

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 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.

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.