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Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

In a bosonic Nambu spinor expression, the second diagonal entry commonly contains $a_{-k}{a}_{-k}^{†}$. The canonical commutation relations add one constant per mode when this is normal ordered. For an antiferromagnetic chain coefficient $JS$, rewriting the quadratic Hamiltonian as a full Nambu sum therefore changes the exterior constant from $-NJ{S}^2$ to $-NJS(S+1)$. Omitting the shift corrupts the ground-state energy even when the dispersion is unchanged.

An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ up to a global phase. At $\theta =\pi /4$ the trial energy is $1$; at $\theta =\pi /2$ it is $2$. These are upper bounds on the ground-state energy, even though neither is a minimizing trial.

The real trial family contains the ground state, so it attains the exact minimum despite not covering every complex quantum state. More generally, a family covering all normalized vectors in a two-dimensional Hilbert space gives the exact lowest eigenvalue of any Hermitian matrix by minimizing its Rayleigh quotient. For a complex trial $c_0|0⟩+c_1|1⟩$ with $|c_0{|}^2+|c_1{|}^2=1$, the same calculation gives $E=2|c_1{|}^2$: a relative quantum phase cannot change this diagonal Hamiltonian energy. This finite example is the weighted mean proof of the variational method with just two energy eigenstates.

In the stationary-state terminology used here, an excited state is a normalized energy eigenstate with energy eigenvalue $E>E_0$, where $E_0$ is an attained ground-state energy. Its excitation cost above the ground state is $\Delta E=E-E_0>0$. Several energy eigenstates can share one excited energy eigenvalue through degenerate energy levels. An arbitrary quantum superposition with energy expectation value above $E_0$ need not be a single stationary excited state; its energy content is described by its Born rule probabilities or, for a continuum, its scalar spectral measure.

An excited state can be a stationary state: the second infinite square well mode is an immediate example. Spontaneous photon emission is a transition caused by coupling matter to the electromagnetic field; it is absent from an isolated matter-only Hamiltonian operator. The relevant energy eigenstate of that isolated model is therefore not automatically an energy eigenstate of the full interacting system. This distinction explains how an ideal stationary excited state can approximate a state with a finite experimental lifetime. See electric-dipole interaction for the coupling used in the simplest optical model.

For a one-photon transition between stationary internal states, neglecting recoil, conservation of energy requiresHere $h$ is the Planck constant and $\nu$ is the photon frequency. Energy matching alone does not guarantee a transition: in the electric-dipole interaction approximation its strength also involves the matrix element $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$, where $\mathbf{d}$ is the electric dipole moment operator and $\mathbf{ϵ}$ specifies photon polarization. A zero matrix element gives an electric-dipole selection rule. Thus an absent optical line does not prove the corresponding energy eigenvalue is absent; a different interaction can couple different states. ^[ref]. The Franck-Hertz experiment gives a particularly instructive comparison between collision excitation and photon emission.

The first excited state terminology assumes a least energy eigenvalue above $E_0$. Such a next level is not guaranteed by quantum mechanics: a continuous spectrum can start immediately above the spectral bottom, or energy eigenvalues can accumulate there. For example, a Hamiltonian operator with energy eigenvalues $0,\epsilon ,\epsilon /2,\epsilon /3,\dots$, where $\epsilon >0$, has a ground-state energy zero but no least positive energy eigenvalue: every candidate $\epsilon /n$ is exceeded downward by $\epsilon /(n+1)$. A free particle on the whole line has neither a ground state nor a first stationary excited state. Each use of a discrete excitation ladder therefore needs a specified Hamiltonian operator and its boundary conditions.

The Franck-Hertz experiment connects a reproducible loss of Electron kinetic energy with excitation of a chemical atom. The original mercury-vapor experiments were performed during 1912–1914 while investigating ionization; they were not designed as a test of the Bohr model. Their later interpretation distinguished internal excitation from removal of an Electron. ^[ref].

In a modern mercury tube, a heated cathode emits Electrons and a positive perforated grid accelerates them through Hg vapor. A collector slightly below the grid's electric potential rejects slow Electrons. The oven controls vapor density and hence the mean free path. Keep heating and retarding settings fixed, sweep the accelerating electric potential difference $V$, and record collector electric current $I(V)$. These measured quantities indirectly probe atomic excitation. ^[ref].

The curve shows repeated current minima, typically about $4.9\mathrm{V}$ apart. In the collision model, an Electron gains approximately $eV$ of kinetic energy, with $e$ the positive elementary charge. An inelastic collision transfers about $4.9\mathrm{eV}$ to internal excitation. Near the grid, this can prevent the Electron crossing the retarding region, lowering collected electric current. Further acceleration restores collection; repeated excitation losses produce further minima. These represent multiple collisions, rather than equally spaced atomic energy eigenvalues. Contact potentials shift the first minimum, so compare several successive spacings instead of identifying the first dip's voltage with the excitation cost. ^[ref].

The optical comparison supplies a second observable: mercury radiation at a wavelength near $253.6\mathrm{nm}$ was detected through a window in the tube. Its inferred photon energy $hc/\lambda \approx 4.89\mathrm{eV}$ agrees with the characteristic collision loss, with $c$ the speed of light. This agreement supports the interpretation as a transition between internal atomic energies. It does not make the collector trace a complete spectrum. In particular, the lowest excited mercury level, labeled $6{}^3{P}_0$, is metastable because an electric-dipole selection rule forbids its direct transition to the ground state; the nearby $6{}^3{P}_1$ level emits the quoted ultraviolet line. Collision excitation and optical detection therefore have different sensitivities to the available excited states. ^[ref].

Initial Electron kinetic energy spread and varying collision positions broaden real minima. Repeat traces at different retarding settings and oven temperatures to distinguish excitation from collection effects. This probes a discrete energy-transfer channel, rather than every energy eigenspace, its dimension, or an absolute ground-state energy. ^[ref].

James Franck and Gustav Ludwig Hertz presented their paper on 24 April 1914. Its heated platinum wire sat inside a cylindrical platinum mesh, with a collector 1–2 mm beyond it. A heated bath and pumping prepared the vapor; guard rings suppressed leakage. They differentiated retarding-electric current scans to study outgoing energies.^[ref]

Their 1925 Nobel Prize preceded their lectures of 11 December 1926. James Franck described choosing pressure for many collisions before the mesh but few afterward.^[ref] Gustav Ludwig Hertz explained how a field-free collision region narrowed excitation thresholds.^[ref]

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.

A ground state is a normalized energy eigenstate that attains the lowest possible energy of a quantum system. Precisely, let the Hamiltonian operator $H$ be a self-adjoint operator bounded below on a Hilbert space, and write $E_{\ast }=\inf \sigma (H)$ for the infimum of its spectrum. A ground state is a unit vector ${\psi }_0$ in the operator domain $D(H)$ satisfyingThus a ground state exists exactly when the spectral bottom $E_{\ast }$ is an energy eigenvalue. The lowest energy eigenvalue among some listed bound states is insufficient if other parts of the spectrum lie below it. The ground-state subspace collects all minimizing energy eigenstates; it can have ground-state degeneracy. The quantum variational principle characterizes $E_{\ast }$ as the infimum of normalized energy expectation values. ^[ref].

For the one-dimensional quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ above the minimum of the potential energy. This zero-point energy is compatible with being the lowest allowed energy: a localized wavefunction cannot simultaneously have zero kinetic energy. For a free particle on the whole line, the spectral bottom is zero but is unattained by any normalizable wavefunction; see ground-state energy. A lower bound therefore need not describe a physically preparable energy eigenstate.

The ground state supplies a reference for excitation costs and for the quantum variational principle. A ground state is also a stationary state under a time-independent Hamiltonian operator: its quantum phase changes, while its position probability density stays fixed. A nonzero ground-state energy does not imply motion along a classical trajectory, and ground-state degeneracy means there need not be a unique lowest quantum state.

When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

The ground-state energy $E_0$ is the energy eigenvalue attained by a ground state. For a self-adjoint operator $H$ bounded below, distinguish this attained value from the always-defined spectral bottom $E_{\ast }=\inf \sigma (H)$. When a ground state exists, $E_0=E_{\ast }$. Some usage calls $E_{\ast }$ the ground-state energy even when it is unattained; here an unattained bottom is explicitly identified as an infimum. The quantum variational principle statesApproaching this infimum with trial wavefunctions does not ensure that a minimizing wavefunction exists. ^[ref].

For the whole-line free particle with mass $m>0$,Its spectrum is continuous and contains no energy eigenvalues represented by normalizable wavefunctions. The Fourier transform makes $H$ multiplication by ${\hslash }^2{k}^2/(2m)$; a fixed energy restricts $k$ to a set of zero length, on which a nonzero square-integrable function cannot be supported. ^[ref]. An explicit family of normalized Gaussian wave packets demonstrates how the bottom is approached:The formula follows from ${\psi }_s^{\prime }=-x{\psi }_s/(2s^2)$, integration by parts, and $⟨x^2{⟩}_s=s^2$. Every finite width gives positive kinetic energy. At zero energy, $H\psi =0$ would require an affine function on the whole line, which is never a nonzero normalizable wavefunction. Spreading a Gaussian wave packet indefinitely produces an unattained infimum, rather than a zero-energy ground state.

An additive choice of energy origin changes the numerical ground-state energy without changing the ground-state subspace. For $H^{\prime }=H+CI$, every energy eigenvalue becomes $E^{\prime }=E+C$, while all differences $E_j-E_i$ stay fixed. Time evolution gains only the global phase $e^{-iCt/\hslash }$. This makes excitation costs and spectroscopic differences more intrinsic than the isolated number $E_0$.

The infinite square well is an ideal confinement model with potential energy $V(x)=0$ for $0<x<L$ and infinite walls outside. The walls impose Dirichlet boundary conditions $\varphi (0)=\varphi (L)=0$ and a vanishing exterior wavefunction. Inside, the time-independent Schrödinger equation isThe first wall removes the cosine term; the second requires $\sin (kL)=0$, hence $k=n\pi /L$. The $n=0$ solution is identically zero and cannot satisfy wavefunction normalization. Each energy eigenstate has $n-1$ interior nodes of a standing wave, at $x=jL/n$ for $j=1,\dots ,n-1$. More curvature costs more kinetic energy, explaining the $n^2$ growth. The positive ground-state energy is a confinement effect: a constant zero-energy wavefunction cannot satisfy both walls and remain normalized.

For an Electron with mass $m_e=9.1093837139\times 10^{-31}\mathrm{kg}$ in a box of length $L=1\mathrm{nm}$, use $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. These constants are from the 2022 CODATA adjustment. ^[ref]. Substitution givesDoubling $L$ quarters every energy eigenvalue; doubling $m$ halves them. This scaling makes the particle in a box useful for estimating confinement energies before attempting a more realistic calculation. A finite square well permits exterior tails for bound states and has only finitely many bound energy eigenstates. For an Electron, the scalar box calculation assumes fixed spin and no spin-dependent interactions.

For the ground state, the probability of finding the particle in the middle half of the box is fully computable:Thus half the available length contains about $81.8\%$ of the probability. The figure contrasts the signed wavefunctions with their nonnegative probability densities: a negative lobe is an amplitude with opposite quantum phase, not negative probability.

An equal quantum superposition of the first two energy eigenstates gives a simple example of moving density:Its density period is $T=h/(E_2-E_1)\approx 3.66608\mathrm{fs}$ for the Electron box above. At $t=0$ the cross term enhances the left side; at $T/4$ it vanishes; at $T/2$ it enhances the right side. More quantitatively,which takes the values $0.924413$, $0.5$, and $0.075587$ at those times. Orthogonality keeps the total probability one throughout. An energy measurement still returns $E_1$ or $E_2$ with equal probabilities: changing position density arises from relative quantum phase, not changing populations. A mixed state describing a statistical mixture with those populations has the constant probability density $({\varphi }_1^2+{\varphi }_2^2)/2$; it lacks this matter-wave interference term.

Fill the lower energy band of the dimerized tight-binding chain with both spin states. Take zero on-site energy and measure ground-state energy per doubled primitive unit cell. State counting in a Bloch band and $q=ka$ giveEquivalently, for $|\delta |\le t$, $\mathcal{E}(\delta )=-(8t/\pi )E(\sqrt{1-{\delta }^2/t^2})$, where $E(k)$ is the complete elliptic integral of the second kind in the modulus convention.

For fixed $t>0$ and $|\delta |/t\to 0$, the change isThe logarithm comes from states near the edge of the Brillouin zone. Put $u=\pi /2-q$. To leading order the square root there is $\sqrt{t^2{u}^2+{\delta }^2}$. For a fixed small cutoff $\eta >0$, direct integration givesThe rest of the band, and the higher-order corrections near its edge, contribute only $O({\delta }^2/t)$; multiplying by $-8/\pi$ gives the displayed asymptotic expansion. Here the remainder uses Big O notation. The negative logarithmic term exceeds any fixed quadratic elastic cost $K{\delta }^2$, $K>0$, for sufficiently small nonzero $\delta$. This explains the zero-temperature instability to a Peierls distortion in the ideal chain model.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

For physical spin ladder operators, $S_j^{\pm }=S_j^x\pm i{S}_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and givesAdding the remaining component provesFor spin one-half, $S^{+}|\downarrow ⟩=\hslash |\uparrow ⟩$, $S^{-}|\uparrow ⟩=\hslash |\downarrow ⟩$, and raising an up spin or lowering a down spin gives zero. These formulas restore the physical $\hslash$ factors implicit in the dimensionless ladder hint.

Let $|n⟩$ abbreviate the state with the only down spin at site $n$, keeping $|{\psi }_0⟩$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up spins. The ladder terms vanish, while $S_j^z{S}_{j+1}^z$ gives ${\hslash }^2/4$. ThusFor a down spin at $j$, the $z$ product is instead $-{\hslash }^2/4$, and $S_j^{+}{S}_{j+1}^{-}$ moves it to $j+1$ with amplitude ${\hslash }^2$. For a down spin at $j+1$, the other ladder product moves it back. ThereforeEvery bond in $|{\psi }_0⟩$ is aligned, so summing its $N$ bond energies givesThis is indeed a ground-state energy: a pair of spin-one-half particles has ${\mathbf{S}}_j\cdot {\mathbf{S}}_{j+1}$ eigenvalues ${\hslash }^2/4$ and $-3{\hslash }^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic Heisenberg model has a whole maximum-total-spin ground multiplet, not a unique all-up ground state.

In the single-excitation subspace, only the two bonds adjacent to the down spin differ from the aligned background. Their diagonal energy increase is $J$, and each neighboring hop has amplitude $-J/2$. Hence, with periodic site indices,This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n⟩=|n+1⟩$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, soThere are $N$ independent such wavevectors, for example one set of $N$ consecutive integers $\ell$ in a Brillouin zone. The normalized one-magnon Bloch states on a ring are $|q⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{iqna}|n⟩$. Reindexing with the periodic coefficient condition gives $T|q⟩=e^{-iqa}|q⟩$ and hence $T^N|q⟩=|q⟩$.

There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation eigenstate, equivalently a consistently periodic plane wave. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed momentum modes.

Apply the hopping Hamiltonian operator and shift the summation index using periodic boundary condition. The coefficients from $|n-1⟩$ and $|n+1⟩$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thusand the nearest-neighbour ferromagnetic magnon dispersion isThe unnormalized state in the question has the same energy. Distinct allowed momenta are orthogonal, since $N^{-1}{\sum }_n{e}^{i(q-q^{\prime })na}={\delta }_{q{q}^{\prime }}$ over one set of representatives.

For large $N$ the sampled wavevectors become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos (qa)$ over $-2\pi <qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm \pi /a$. Near any equivalent zero,using $q$ measured from that zero. The low-momentum ferromagnetic magnon is quadratic, with group velocity ${\hslash }^{-1}dE/dq=(Ja/\hslash )\sin (qa)$. The $q=0$ state is proportional to the total spin lowering operator acting on $|{\psi }_0⟩$, so it belongs to the same ground multiplet and has exactly zero excitation energy. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.

For a self-adjoint Hamiltonian operator bounded below with ground-state energy $E_0$, every admissible nonzero trial wavefunction has a Rayleigh quotient at least $E_0$. To see this explicitly in a complete orthonormal eigenbasis, write $|\psi ⟩={\sum }_n{c}_n|n⟩$ with $H|n⟩=E_n|n⟩$. ThensoFor continuous spectral components, the spectral theorem replaces the sum by an integral with the same nonnegative integrand. Equality requires the trial quantum state to lie entirely in the ground-state subspace. The variational method chooses a tractable family, evaluates the quotient and minimizes over its parameters. This gives an upper bound; enlarging the family can only improve that bound. The trial must belong to the operator domain for the operator expression above, or have finite energy in the corresponding quadratic-form version of the quantum variational principle.

For the oscillator, use the standard kinetic operator $T=-{\hslash }^2{d}^2/(2md{x}^2)$: the original PDF omits the superscript $2$ on the derivative in its numerator. Take $a>0$, and normalize the scaled exact ground state asThe factor $\sqrt{\lambda }$ is optional in a quotient but makes the calculation transparent. Changing variable to $u=\lambda x$ and differentiating the scaled wavefunction givesAt $\lambda =1$ this trial is the exact minimizing state, so its energy $\mathcal{E}(\lambda )={\lambda }^2{T}_0+{\lambda }^{-2}{V}_0$ has zero derivative there. Hence $2T_0-2V_0=0$. Since $E_0=T_0+V_0$,This is the quantum virial identity from scaling for a quadratic potential, and hence the oscillator form of the virial theorem.

For the inverse-power attraction, take $a>0$ and suppose, for contradiction, that an exact ground state exists with separately finite kinetic energy $T_0>0$ and potential mean $U_0<0$, admitting the same dilation variations. Homogeneity now givesStationarity at $\lambda =1$ would require $2T_0+p{U}_0=0$. Its second derivative would then beFor $p>2$ this is negative, contradicting a ground-state minimum. More strongly, one does not need a presumed exact state: choose any normalized smooth test function supported away from the singular origin. Its $T$ and $U$ are finite, and its scaled quotient satisfiesThe unregularized attraction therefore exhibits fall to the centre for a supercritical inverse-power potential and has no stable ground state.

There is a necessary qualification to the PDF's broader wording about “no bound state”. Under the usual finite-energy virial theorem assumptions, a putative negative-energy bound state would satisfywhich contradicts being below the continuum threshold at zero. Thus no negative-energy bound state satisfying those regularity assumptions is possible. But this singular differential expression needs a specified domain at $x=0$; separately divergent kinetic and potential means or boundary conditions not preserved by dilation invalidate that ordinary calculation. The domain assumptions in singular-potential virial arguments must therefore be stated: the scaling proof establishes absence of a semibounded ground-state problem, rather than a domain-independent theorem about every formal square-integrable solution. A short-distance regularization would be a different potential.

Periodic boundary conditions give wavevectors $(2\pi /L)(n_x,n_y)$. In the continuum, large-box limit each state occupies area $(2\pi /L{)}^2$ in wavevector space. A circular annulus therefore contains $[L^2/(2\pi {)}^2]2\pi kdk$ spinless states. Since $ϵ={\hslash }^2{k}^2/(2m)$, the density of states isAt finite $L$ the exact spectrum is discrete; this formula is its continuum density.

With electron spin the two Zeeman branches are ${ϵ}_{\mathrm{kin}}\mp \mu H$. The requested calculation includes this spin coupling and neglects orbital coupling to the field; for a perpendicular field, Landau levels would change the result. It is appropriate for a field parallel to the ideal two-dimensional plane, or for the stated spin-only model. Taking $\mu H\ge 0$, the lower branch starts at $-\mu H$, and the upper one at $\mu H$, soConstruct the ground state by filling the lowest one-electron levels subject to the Pauli exclusion principle. If the Fermi energy ${ϵ}_F$ exceeds $\mu H$, the two populations are $N_{+}=g_0({ϵ}_F+\mu H)$ and $N_{-}=g_0({ϵ}_F-\mu H)$. ConsequentlyThe ground-state energy, including the Zeeman term, follows by integrating the energy over each occupied branch: