Codex @codex  0

The first two probability densities areBoth vanish at the walls. The ground state has one maximum at $L/2$ and no interior node. The first excited state has a node at $L/2$ and equal maxima at $L/4$ and $3L/4$.

For a finite square well whose exterior potential is $V_0>E_n$, the exterior wavefunction decays as $e^{-\kappa d}$, where $\kappa =\sqrt{2m(V_0-E_n)}/\hslash$ and $d$ is distance from the relevant edge. Wavefunction matching at a finite potential step makes the wavefunction and, for constant mass, its first derivative continuous. Thus the probability density is nonzero at the edges and has evanescent tails outside, the interior wavelength increases, and the energy levels are lower than for infinite walls at the same positions. Reflection symmetry still gives an even ground state and odd first excited state about the center, when both are bound. A shallow well may not support that excited bound state at all.

The commutators implyThus any nonzero resulting vector is the adjacent energy eigenstate. Since $a^{†}a|n⟩=n|n⟩$ and $a{a}^{†}=a^{†}a+1$, their squared norms areChoosing consistent phases for the eigenstates yieldsThe vanishing last expression terminates the lowering sequence at the ground state, rather than introducing a negative-energy oscillator state.

The Born rule gives the required probability by integrating the probability density over the central interval. Put $d=0.1\mathrm{nm}$. For this infinite square well ground state,Substitution of $d/L=0.1$ givesThis exceeds the interval's $20\%$ fraction of the box length because the ground state probability density is largest at the centre; it is not a uniform position distribution.

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.

Solving the time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

Identical spin one-half particles are fermions. By fermionic exchange symmetry, the complete spatial-and-spin wavefunction must change sign when the two particle labels are exchanged. Define the normalized antisymmetric spin-one-half singlet stateand the three symmetric spin-one-half triplet statesThe Hamiltonian is independent of spin, so these spin functions carry no additional energy.

Both particles can occupy $u_0$ if their spin factor is antisymmetric. The unique ground state wavefunction isTwo equal-spin particles could not share this orbital, by the Pauli exclusion principle.

At the first excited energy, $E=ϵ={\hslash }^2/(2m{R}^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functionsFor each $s$, all allowed wavefunctions are spanned byThere are $2(1+3)$ orthonormal states, hence first-excited-state degeneracy $g=8$. In each product, exactly one factor is antisymmetric, ensuring the required total fermionic exchange symmetry.

At fixed positive pressure, the extrapolated term $C_{mp}\ln (T/T_0)$ tends to $-\infty$ as $T\to 0^{+}$. That prediction is not physically realistic for an equilibrium sample. The classical ideal-gas description with constant heat capacity cannot extend unchanged to absolute zero. A real substance generally undergoes phase changes, and quantum statistics and thermal freeze-out change its heat capacity and entropy. For a nondegenerate equilibrium ground state, the third law gives entropy approaching zero; residual ground-state degeneracy can give a finite residual value rather than this negative divergence. This is a breakdown of the extrapolation, not a violation of the third law by a real gas.

The zeroth law of thermodynamics states the transitivity of thermal equilibrium: if two systems each equilibrate thermally with a third, they equilibrate with one another. A reproducible thermometric property can therefore label their temperature. Its empirical numerical scale is not itself the complete thermodynamic content; reversible heat-engine relations establish an absolute temperature scale.

The first law of thermodynamics expresses energy conservation. For a closed system, $dU=\delta Q-\delta W$ when $\delta W$ is work done by the system. internal energy is a state function, while heat and work depend on the path. In reversible simple compressible motion, $\delta W=pdV$ and $dU=TdS-pdV$; an open system additionally needs the energies carried by transferred matter. An ideal gas can expand freely with $Q=W=0$ and unchanged internal energy. The first law alone does not decide whether the reverse spontaneous process occurs.

The Second law of thermodynamics introduces entropy and restricts direction. For a reversible path, $dS=\delta Q_{\mathrm{rev}}/T$; for an irreversible process the entropy balance includes nonnegative production. The entropy of an isolated system cannot decrease. In a cycle, the Clausius inequality is $\oint \delta Q/T\le 0$, using the appropriate boundary/reservoir temperatures. The Kelvin–Planck statement forbids a cyclic device whose sole effect is extracting heat from one reservoir and turning all of it into work; the Clausius statement forbids moving heat from cold to hot with no compensating effect. A refrigerator is allowed because it consumes work. A two-reservoir engine has $\eta \le 1-T_c/T_h$, with equality in the reversible limit. Work may be fully dissipated as heat without contradicting this one-way restriction.

The third law of thermodynamics constrains low-temperature entropy. An equilibrium perfect crystal with a nondegenerate ground state approaches zero entropy as $T\to 0$, fixing the conventional absolute entropy reference. Degenerate ground states or frozen nonequilibrium disorder require care when discussing residual entropy. Under regular equilibrium conditions the heat capacity must decrease sufficiently for $\int C/TdT$ to remain finite; a constant positive classical heat capacity cannot be extrapolated to zero temperature. The unattainability formulation states that absolute zero cannot be reached by a finite sequence of admissible cooling operations. These statements concern the limiting thermodynamic regime, rather than predicting that every substance stays in one classical phase during cooling.

The laws work together: conservation sets a device's energy budget, entropy production restricts the attainable budget, and constitutive equations give the properties of its working material. For fixed temperature and volume, Helmholtz free energy decreases toward equilibrium; at fixed temperature and pressure the corresponding potential is Gibbs free energy. Stable equilibrium, phase coexistence and Maxwell relations follow from these potentials with appropriate constraints. Calorimetry, mechanical-work measurements, thermometry and reversible-path comparisons test different parts of this framework. None of the laws by itself supplies a particular gas equation of state or microscopic heat capacity.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

The third law of thermodynamics, in its perfect-crystal form, states that the entropy of a pure, perfectly ordered crystal with a unique ground state tends to zero as $T\to 0$. Its more general entropy-difference formulation says that the entropy difference between equilibrium states related by a finite change of a regular external control parameter tends to zero at absolute zero:The common limiting entropy is independent of $X$. Ground-state degeneracy or frozen disorder can produce residual entropy, so zero entropy is not an unconditional assertion about every sample.

In the usual regular thermodynamic setting, absolute zero cannot be attained from positive temperature by a finite sequence of cooling operations with finite resources. An entropy-temperature plot shows why a finite reversible adiabatic step cannot reach the common zero-temperature endpoint from a state with positive entropy above that endpoint: an isentropic process follows a horizontal line, while all the regular entropy curves meet only at $T=0$.

The figure illustrates this with two possible settings having $S-S_0=a_{1}T$ and $S-S_0=a_{2}T$, where $0<a_1<a_2$. First change the setting isothermally from the upper to the lower curve, rejecting heat; then change it reversibly and adiabatically back to the upper curve. Entropy conservation on the latter step givesfor every finite $n$. The staircase approaches the origin but reaches it only after infinitely many such stages. Irreversibility generates entropy and cannot improve on the reversible endpoint.

This is a thermodynamic illustration of unattainability of absolute zero, not a claim that every cooling apparatus has this particular linear entropy law. Its conclusion presumes regular equilibrium states and bounded control parameters; dynamical statements about finite cooling time additionally require physical assumptions about the available interactions and cooling rates.

Direct differentiation gives ${\psi }^{\prime \prime }/\psi =x^2/a^4-1/a^2$. Inserting this into the Schrödinger equation and separately matching the $x^2$ and constant terms yieldsFor $V_0=m{\omega }_0^2/2$, $a^2=\hslash /(m{\omega }_0)$ and $E=\hslash {\omega }_0/2$. The normalizable Gaussian has no nodes, as expected for the nondegenerate one-dimensional ground state of this confining potential.

The normalization integral is $|C{|}^{2}a\sqrt{\pi }=1$. Choose its irrelevant global phase so thatFor the oscillator ground-state density in a comparison wavelength, choose a free comparison particle with the same total energy. It has $p=\sqrt{2mE}=\hslash /a$, hence comparison wavelength $\lambda =h/p=2\pi a$. ConsequentlyThe bound density is localized on scale $a=\lambda /(2\pi )$, with standard deviation $\lambda /(2\pi \sqrt{2})$, rather than being spatially uniform like the density of a single free plane wave. The comparison does not assign constant free-particle momentum to the bound state; its total energy also includes potential energy.

Here $V(x)$ is potential energy, $E$ is the stationary-state total energy eigenvalue, and $\psi (x)$ is its spatial wavefunction. Wavefunction normalization gives $\int |\psi {|}^{2}dx=1$, and $|\psi {|}^{2}dx$ is the probability of finding the particle in $dx$.

A bound state requires $-V_0<E<0$. Inside the well the equation is ${\psi }^{\prime \prime }=-k^2\psi$, whereas outside it is ${\psi }^{\prime \prime }=q^2\psi$, withThe ground state is even and has no node, selecting $\cos (kx)$ inside. Normalizability selects decaying, not growing, exterior exponentials. Continuity of $\psi$ and ${\psi }^{\prime }$ givesThe first excited state, if present, is odd: a sine inside and opposite-sign decaying tails outside. Matching gives $-k_1\cot (k_{1}a/2)=q_1>0$. On its first odd branch, $\pi /2<k_{1}a/2<\pi$, hence $k_{1}a>\pi$. Since $k_1^2<2m{V}_0/{\hslash }^2$, existence requiresThis is the first odd bound-state threshold of a square well. There is no excited bound state below that threshold; equality gives a non-normalizable zero-energy limiting state, not an extra bound state.

For the numerical well,The exterior probability of a symmetric finite-well state includes both tails:The threshold is $9.40\mathrm{eV}$, exceeding the well depth, so there is no excited bound state. The rounded supplied coefficients integrate to $0.9973$ rather than exactly one; this rounding does not change the quoted probability precision materially.

Equation and interpretation. The supplied equation is the time-independent Schrödinger equation for a nonrelativistic particle with mass $m$ in a prescribed potential energy $V(x)$. Equivalently,It is an eigenvalue problem for the Hamiltonian operator. A stationary energy state has full wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. By the Born rule, $|\psi (x){|}^{2}dx$ is the probability of finding the particle in a small interval $dx$, with wavefunction normalization ${\int }_{-\infty }^{\infty }|\psi {|}^{2}dx=1$.

Infinite barriers. Infinite exterior potential energy excludes any finite-energy probability outside the interval. The infinite square well consequently has $\psi =0$ outside and the hard-wall boundary conditionsThe interior wavefunction must meet the zero exterior amplitude continuously. Its derivative need not be continuous across an infinite wall; imposing both zero value and zero derivative would eliminate every nonzero interior solution.

Ground state. Inside the well $V=0$, so ${\psi }^{\prime \prime }+k^2\psi =0$, with $k=\sqrt{2mE}/\hslash$. Its general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$, and the right wall gives $ka=n\pi$ for $n=1,2,\dots$. Thus the discrete energy eigenvalues are $E_n=n^2{\pi }^2{\hslash }^2/(2m{a}^2)$. The smallest, $n=1$, is the ground state. Sinceits normalized wavefunction, choosing an irrelevant overall phase to make it positive inside, is

Finite barriers. For a finite step and constant particle mass, both $\psi$ and ${\psi }^{\prime }$ are continuous at $x=0,a$. Wavefunction matching at a finite potential step follows by integrating the Schrödinger equation across a vanishingly small neighborhood of the step: a bounded potential cannot generate a finite jump in ${\psi }^{\prime }$. The bound-state exterior solutions decay rather than vanish identically:Normalizability excludes the growing exterior exponentials. The equivalent interior boundary conditions are ${\psi }^{\prime }(0)=\kappa \psi (0)$ and ${\psi }^{\prime }(a)=-\kappa \psi (a)$; neither boundary value is zero for the nontrivial ground state.

The ground state of a finite square well remains symmetric about $a/2$ and has no nodes. It bends smoothly into exponential tails at both walls, extending over a larger region than the infinite-well state. Its maximum after normalization is lower for the deep well considered here. When $V_w\gg E$, the penetration length $1/\kappa$ is short, so the two shapes are close except near the walls and in the exterior tails.

The finite curve is an illustrative exact matched solution with $V_w=10E_{1,\infty }$; no numerical wall height is specified by the question.

Energy comparison. The finite barriers relax the confinement, allowing a less rapidly varying interior wavefunction and therefore smaller kinetic energy. More explicitly, the even interior solution is proportional to $\cos [k(x-a/2)]$, and matching to the tails gives $k\tan (ka/2)=\kappa$. The lowest root lies at $0<ka<\pi$, while the infinite well has $ka=\pi$. HenceAs $V_w\to \infty$, the tails disappear and the finite-well energy approaches the infinite-well value from below. This comparison uses the same zero of potential energy inside both wells.

The infinite square well levels for width $L$ are $E_n=n^2{h}^2/(8m_e{L}^2)$. Starting in the ground state, the smallest excitation is $1\to 2$, and its electric-dipole matrix element is nonzero. It therefore sets the longest absorption wavelength:The corresponding energy separation is about $12.4\mathrm{eV}$. Elastic scattering of light is a different process from the stipulated absorption into an excited well state.

For an infinite cubic potential well of side $L$, the one-particle energy eigenvalues are $E_{n_x{n}_y{n}_z}=E_0(n_x^2+n_y^2+n_z^2)$, where $E_0={\pi }^2{\hslash }^2/(2m{L}^2)$ and each quantum number is a positive integer. The lowest spatial orbital is $(1,1,1)$, with energy $3E_0$ and orbital degeneracy one. The next energy is $6E_0$, with three orbitals given by permutations of $(2,1,1)$.

Spin-zero identical particles are bosons. Both can occupy the lowest orbital, so the two-boson ground state is nondegenerate. The first excited total energy is $9E_0$: one particle occupies the lowest orbital and one an excited orbital. Symmetrizing the two-particle wavefunction gives one state for each of the three excited orbitals, so the first-excited degeneracy is three.

Spin-half identical particles are fermions, so the total state must be antisymmetric. Two particles in the same lowest orbital have a symmetric spatial wavefunction and must form the unique spin singlet state. Thus the fermion ground state is also nondegenerate. At the first excited energy, each choice of excited spatial orbital permits a symmetric spatial state with a singlet state, or an antisymmetric spatial state with a three-dimensional triplet state. ThereforeThe spin multiplicity is essential: counting only the three orbital choices would omit allowed triplet states. The result assumes ordinary spin-statistics, an unsplit spin degeneracy and the hard-wall cube model.

Adopt the usual orthogonal tight-binding approximation, so the three localized spatial states are orthonormal. Each site has two neighbors, with the same real hopping matrix element. For coefficients $c_n$, the Schrödinger equation at site $n$ isThe constant-coefficient state has $c_1=c_2=c_3$, giving energy $E_0+2t$. Its normalization therefore yieldsTo verify that it is the ground state, the three discrete ring momenta are $\theta =0,\pm 2\pi /3$, with $c_n\propto e^{i\theta (n-1)}$ and energies $E_0+2t\cos \theta$. The other two states both have energy $E_0-t$. Since $t<0$, $E_0+2t<E_0-t$ and the gap is $3|t|$. This is the three-site tight-binding ring spectrum.

The spatial ground state is unique and has equal probability $1/3$ on every site; its common overall phase is arbitrary. Spin would supply a separate twofold degeneracy when no spin-dependent terms are included, but does not change the requested spatial state. If the localized functions were instead nonorthogonal raw atomic orbitals, their overlap matrix would be additional data needed for normalization and a generalized eigenvalue problem.

For the Transverse-field Ising model, use dimensionless spin one-half operators, so $S^x$ has eigenvalues $\pm 1/2$. At zero transverse field every aligned bond contributes $-J$. The two ground states areThey span an exactly two-dimensional ground space on the finite periodic ferromagnetic chain. Any linear combination is also a ground state; parity eigenstates are the symmetric and antisymmetric combinations. Choosing one ordered product state in the thermodynamic limit is spontaneous symmetry breaking.

For large positive field the leading term is $-2Jg{\sum }_n{S}_n^z$. Its unique minimizing state isThe exchange has zero first-order expectation in this state and produces virtual spin-pair fluctuations at finite g. Thus the polarized product is the limiting wavefunction, not an exact finite-field ground state. Bulk perturbation theory gives $E_0/N=-Jg-J/(4g)+O(J/g^3)$. The field-polarized phase has no twofold ordered-state degeneracy.

In a fixed fermion-parity sector use lattice spacing one and the unitary discrete Fourier transformThe allowed momenta satisfy $e^{ikN}=-P$: they are $k=2\pi (\ell +1/2)/N$ for $P=+1$ and $k=2\pi \ell /N$ for $P=-1$, with $N$ distinct values modulo $2\pi$. The hopping and number sums become ${\sum }_{k}2\cos k{c}_k^{†}{c}_k$ and ${\sum }_k{c}_k^{†}{c}_k$. ThusHere $E$ is the energy of the all-up spin configuration, not generally the ground-state energy. The dispersion has minimum $h-J$ at $k=0$, maximum $h+J$ at $k=\pi$, and positive curvature at its minimum. Since the fermions do not interact, each mode can be occupied zero or one time. A negative-energy occupation lowers the many-body energy, so in the thermodynamic limitThis is a filled Fermi sea containing many fermions; it is not obtained by putting a single particle into the band minimum. For $h>J$ every mode has positive energy and the vacuum is the ground state. Its spin interpretation is complete longitudinal polarization. At $h=J$ the zero-momentum one-particle excitation first becomes gapless.

For an exact finite ring use parity-consistent filling of a free-fermion band. In each momentum grid fill negative-energy levels, check that their particle number has the chosen parity, and if necessary toggle a mode with the smallest $|{ϵ}_k|$. Compare the two sector energies. A zero-energy mode allows degenerate occupations. This restores the physical spin Hilbert space without changing the stated band.

Because $S_{\mathrm{tot}}^z=N/2-N_f$, the exact finite-size answer is $m_N=1/2-N_f^{\mathrm{gs}}/N$, with $N_f^{\mathrm{gs}}$ determined by the parity-consistent filling above. For the thermodynamic magnetization of the quantum XX chain, the occupied fraction is the length of the negative-energy interval divided by the Brillouin-zone length:ThereforeThe zero-field limit has zero magnetization and the high-field limit has all spins up. The finite-ring curve consists of magnetization steps approaching this continuous limit. Strictly above $h_c$ the fully polarized ground state is unique; exactly at $h_c$ a zero-energy spin-flip state can be degenerate with it. This finite-size qualification does not alter the saturation field or limiting magnetization density.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]