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The Berezinskii–Kosterlitz–Thouless transition in the two-dimensional XY model separates a low-temperature phase with algebraically decaying correlations from a high-temperature phase with unbound vortices and exponentially decaying correlations. It does not produce conventional long-range order.

How can something resist a twist when its orientation never settles down? The distinction is between long-range order, a nonzero limiting correlation function, and phase stiffness, a free energy cost for a slow spatial twist. In a uniform two-dimensional XY model at positive temperature, smooth spin waves can erase the first while leaving the second finite. Vortex-antivortex binding then decides whether that resistance survives on arbitrarily large scales.

Vadim Lvovich Berezinskii developed an earlier description of two-dimensional order. John Michael Kosterlitz and D. J. Thouless independently developed the defect mechanism; Kosterlitz's 2016 lecture explicitly acknowledges their initial ignorance of Berezinskii's work. The central step was to include rare vortices alongside smooth spin waves.^[ref]

The Mermin-Wagner theorem concerns continuous symmetry and sufficiently short-range interactions in the thermodynamic limit. It does not say that every two-dimensional system is disordered in every sense, nor that every continuous symmetry gives a Berezinskii–Kosterlitz–Thouless transition. Here the order parameter has a circular phase, permitting integer winding numbers. Finite-scale vortex renormalization explains why a finite sample can look coherent even when its infinite counterpart lacks long-range order.

In a coupled quantum rotor chain, each angle is identified modulo $2\pi$. Expanding a neighboring-angle interaction $-J\cos ({\varphi }_{j+1}-{\varphi }_j)$ to quadratic order gives a tractable harmonic phase mode of a quantum rotor chain, but that local expansion does not remove the original angular identification.

A smooth small-gradient configuration is described well by the harmonic theory. Configurations differing by a winding around a ring, or histories containing phase slips, require additional sectors of the compact theory. The uniform angle also remains a rotor, as described by the global phase zero mode of a finite rotor chain, rather than an ordinary oscillator of zero frequency. Consequently, a gapless harmonic branch alone cannot establish true long-range order or justify discarding compact fluctuations. The approximation is useful when local angle differences stay small and processes connecting distinct winding sectors are negligible on the scales studied.

An XY model orientation takes values on a circle, so a circuit around an excluded defect core can carry integer winding number. Smooth single-valued spin waves cannot change that integer without a core event or boundary crossing. The locally quadratic order-parameter phase action therefore describes a sector, rather than automatically summing every compact-field configuration. Vortex-antivortex unbinding changes the large-distance correlation function and renormalized phase stiffness, even though ordinary long-range order is already absent in two dimensions at positive temperature.

A domain wall is a topological defect separating regions whose order parameter approaches different disconnected components of a vacuum manifold. In $d$ spatial dimensions, a flat domain wall extends in $d-1$ directions: a surface in three dimensions, a line in two, and a localized defect in one. A scalar quartic domain wall provides an explicit smooth interpolation between two degenerate minima.

For the one-dimensional, zero-field Ising model with nearest-neighbor coupling $J>0$, a reversed bond costs $2J$. A single domain wall can occupy order $N$ positions on a chain of $N$ sites, giving a free-energy balance $\Delta F\simeq 2J-k_{B}T\log N$. At every positive temperature, positional entropy defeats a fixed wall cost as $N\to \infty$. Thermally excited domain walls then destroy long-range order, despite two ordered zero-temperature ground states. This argument assumes short-range interactions and finite wall cost; it does not apply unchanged to long-range interacting chains. In higher dimensions the wall cost grows with its area, allowing different ordering behavior.^[ref]

Physical Review 158, 383–386. Uses a rigorous inequality to exclude finite-temperature long-range order in homogeneous low-dimensional quantum gases under its stated assumptions. doi.org/10.1103/PhysRev.158.383

The Bose-gas phase-only action gives a thermal infrared integral $\int d^{d}q/q^2$ and a zero-temperature equal-time integral $\int d^{d}q/|q|$. Thus an infinite uniform short-range interacting gas has no true finite-temperature condensate in $d\le 2$, and the one-dimensional ground state also lacks true condensate long-range order. This does not exclude local density coherence, a finite-size condensate description or two-dimensional algebraic order and superfluidity.^[ref]

Long-range order means that an appropriate order-sensitive correlation function remains nonzero as the separation becomes arbitrarily large in the thermodynamic limit. In a ferromagnet, a spin-spin correlation can approach a positive constant, indicating a persistent relation between distant spins. An antiferromagnet requires an alternating sign convention, captured by staggered magnetization, to reveal its corresponding order.

The choice of average matters. A finite zero-field equilibrium sample can preserve symmetry and have zero mean magnetization, while distant spin correlations still diagnose the approach to an ordered thermodynamic phase. Within a selected symmetry-broken state, the disconnected contribution can carry the order and the connected correlation can instead decay. One must therefore state the observable, state and order of limits before applying a nonzero-correlation criterion.

This also explains a locality constraint: a circuit with bounded depth and bounded-range gates cannot generate arbitrary-distance connected correlations from a product state. Its finite propagation range limits which separated regions can share correlations.

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.

Use the same normalization for the elastic energy and electronic energy density: if $\kappa {\delta }^2$ is an energy per unit cell, divide it by $a$ first; if it is already per length, use it directly. Write its coefficient per length as $K>0$ and put $A={\pi }^2{V}_0^2/(4a{E}_{\mathrm{ZB}})>0$. The small-distortion change in total energy density isNo minimization calculation is needed. For arbitrarily small nonzero $|\delta |$, the logarithm is negative with arbitrarily large magnitude. Hence the negative electronic Peierls energy gain, proportional to ${\delta }^2\ln (1/|\delta |)$, eventually exceeds any fixed positive quadratic elastic energy. There are distorted configurations with lower energy than the undistorted configuration for every finite $K$ and nonzero coupling. The ideal zero-temperature infinite-chain model therefore hasThe source's ${\delta }_{\mathrm{eq}}>0$ is correct when $\delta$ denotes a nonnegative distortion amplitude. If $\delta$ is signed, the energy density is even and the two signs describe equally stable patterns related by a lattice translation. For example, $V(x+a/2;\delta )=V(x;-\delta )$, so a negative equilibrium distortion has the same energy as the positive one. The zero-coupling case $V_0=0$ instead has its minimum at $\delta =0$. The argument assumes a coupled chain, complete lower-energy band filling and zero temperature; it does not by itself establish finite-temperature long-range order in an isolated one-dimensional system.

The physical angle takes values on a circle, so it is defined modulo $2\pi$. Around a closed curve enclosing a phase vortex,The integer winding number cannot change through a smooth single-valued deformation unless a defect crosses the curve. Thus a globally smooth Gaussian spin wave field omits whole topological sectors of the XY model.

Outside a vortex core of radius $a$, an isolated charge-$n$ phase vortex has $\theta =n\phi$ and $|\nabla \theta |=|n|/r$. Its dimensionless energy in a sample of size $L$ isThere are of order $(L/a{)}^2$ positions for its core, so the positional entropy contributes $2\log (L/a)$. The competition for unit charge is therefore $(\pi \bar{K}-2)\log (L/a)$, apart from core terms. Opposite charges have a logarithmic attraction and form a vortex Coulomb gas.

At low temperature, vortex-antivortex binding leaves no free charges on large scales. Spin waves still destroy true long-range order, but the correlation function decays algebraically with the renormalized phase stiffness. At the Berezinskii–Kosterlitz–Thouless transition, pairs unbind; free vortices screen the interaction and produce a finite correlation length and exponential decay above the transition. The energy-entropy estimate identifies the threshold stiffness $K_R=2/\pi$, where $K_R$ is the long-distance dimensionless stiffness, not necessarily the bare $\bar{K}$. It agrees with the universal stiffness jump and the limiting algebraic exponent $1/(2\pi K_R)=1/4$. The two-dimensional transition is vortex unbinding between quasi-long-range order and disorder, rather than the onset of nonzero spontaneous magnetization.

Consider a depth-$D$ circuit whose gates have range at most $r$. Under backward Heisenberg evolution, a one-site observable has a backward light cone of a local quantum circuit of radius at most $rD$. Choose sites $i,j$ separated by $L>2rD$. Their two backward light cones are disjoint. Since the input is a product state, expectations factorize, and hence every connected correlation function between the two output observables vanishes.

For the GHZ statehowever,at every separation. The light cones must therefore overlap, which forcesFor opposite ends of a one-dimensional chain, $L=\Theta (N)$, so $D=\Omega (N)$ and no constant-depth local circuit can prepare the GHZ state. The continuous-time version follows directly from the Lieb-Robinson bound, with preparation time at least $L/(2v_{\mathrm{LR}})$ up to exponentially small tails.

This is the GHZ-state circuit-depth lower bound. Finite-depth local circuits define equivalence within a gapped phase, so a state with this long-range order cannot lie in the same circuit phase as a product state. The GHZ state is the finite-size cat state associated with spontaneous symmetry breaking; it is not a unique short-range-entangled ground state. The persistent distant correlation is precisely the obstruction.

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]

For a uniform time-independent saddle, extremizing the action gives $(-\mu +g|{\psi }_0{|}^2){\psi }_0=0$. Its grand-potential density is $f(\rho )=-\mu \rho +g{\rho }^2/2$, with $\rho =|{\psi }_0{|}^2\ge 0$. Since $g>0$, the minimum at positive density for $\mu >0$ isFor $\mu \le 0$ the uniform mean-field minimum is the vacuum, rather than a positive-density condensate. At fixed nonzero density in the weakly interacting condensed regime, the saddle relation sets $\mu \simeq g{\rho }_0$. The mean-field one-body density matrix factorizes as $⟨a^{†}(\mathbf{r})a({\mathbf{r}}^{\prime })⟩\simeq {\rho }_0$, giving macroscopic occupation of the uniform mode. Depletion corrections are omitted at this stage.

The action has the global particle-number symmetry $\psi \mapsto e^{i\alpha }\psi$, $\bar{\psi }\mapsto e^{-i\alpha }\bar{\psi }$. Choosing one saddle phase breaks this $U(1)$ symmetry; the degenerate mean-field minima form the circle $U(1)\simeq S^1$. There is no energetic preference for a uniform value of ${\varphi }_0$, but spatially varying phase costs gradient energy. Consequently a neutral condensate has a gapless Goldstone mode and a nonzero phase stiffness. The mode becomes sound once density dynamics is included. With $\hslash =1$ the superfluid velocity is $\nabla \varphi /m$, so a single-valued field allows circulation in units $2\pi /m$ around a closed loop that avoids zeros of the amplitude. This phase coherence and rigidity are the physical content of the broken-symmetry description.

The claim is a mean-field saddle statement, not an assertion that every $d$ and temperature in the question supports true long-range order. In a finite number-conserving system $⟨a⟩=0$; the symmetry-breaking order parameter is defined with the thermodynamic limit taken before an infinitesimal source is removed. In a homogeneous infinite short-range gas at nonzero temperature, infrared phase fluctuations prohibit true condensation in $d\le 2$. In one dimension they also destroy true condensate long-range order at zero temperature. A two-dimensional gas can nevertheless have superfluidity and algebraically decaying correlations below a Berezinskii–Kosterlitz–Thouless transition. A finite trap can have substantial coherence without contradicting these thermodynamic statements.^[ref]

At wavelengths with $q^2/(4m{\rho }_0)\ll g$, neglect the density-gradient term in the quadratic density-phase action. Complete the square in its remaining density terms:The convergent Gaussian integral can be performed by a contour shift or by its Fourier-transform identity. Its determinant is independent of $\phi$ at this order. Combining that normalization with the saddle action into a constant $S_0$ givesThe saddle contribution to $S_0$ is $-\beta V{\mu }^2/(2g)$; the additional measure-dependent constant need not be assigned an invented numerical value. The density Gaussian extends the fluctuation integration beyond its exact positivity domain, which is justified only while fluctuations around ${\rho }_0>0$ are small. The smooth zero-winding assumption is also required as discussed above. A constant global phase zero mode gives a finite compact-group volume, handled separately from nonzero phonon modes.

Fourier transforming the Bose-gas phase-only action gives a kernel ${\omega }_n^2/g+({\rho }_0/m)q^2$. Analytically continuing ${\omega }_n\mapsto -i(\omega +i{0}^{+})$ gives its real-frequency poles,In the present units the excitation energy equals $\omega$. Each real Fourier coordinate is a harmonic oscillator with inertia coefficient $1/g$ and spring coefficient ${\rho }_0{q}^2/m$, hence frequency equal to the square root of their ratio. This is the direct relation to a quantum harmonic chain. The mode is a gapless sound phonon; its finite speed reflects both the compressibility $\partial \rho /\partial \mu =1/g$ and the phase stiffness ${\rho }_0/m$.

Retaining the density-gradient term provides a useful check. Its integration instead gives a temporal coefficient $1/A_q$, soThis is the Bogoliubov spectrum. Its small-$q$ limit is the derived sound branch; at large $q$ within the contact model it crosses towards particle-like behaviour. The crossover in the squared dispersion is $q\sim 2\sqrt{mg{\rho }_0}$, precisely the scale beyond which the discarded density gradient is not small.

The same action checks the dimensional restriction in the saddle analysis. At nonzero temperature its zero Matsubara mode gives phase fluctuations proportional to $\int d^{d}q/q^2$; at zero temperature integrating the frequency gives $\int d^{d}q/|q|$. Their infrared divergences invalidate uniform true long-range order for $d\le 2$ at nonzero temperature and for $d=1$ at zero temperature. The long-wave collective description can remain useful for correlations even where a global condensate saddle is not a literal symmetry-broken state. The contact coupling here is an effective low-energy parameter; this calculation does not determine ultraviolet renormalization or trap-specific surface modes.

A theoretical condensed matter physicist who, with Walter Kohn, established a foundation of density functional theory in 1964. Their variational principle expresses the ground-state energy of interacting electrons through the electron density, making a three-dimensional quantity central to a many-particle problem.^[ref]

His 1967 work also ruled out the relevant finite-temperature long-range order in one- and two-dimensional bosonic and fermionic systems under the theorem's assumptions. This helps explain why dimensionality matters when discussing Bose-Einstein condensation.^[ref]

Quasi-long-range order means that correlations decay algebraically rather than approaching a nonzero constant. It occurs in the low-temperature phase of the two-dimensional XY model, where spin waves destroy true long-range order but bound vortex pairs preserve power-law correlations.

For a dimensionless Gaussian action $F/(k_{B}T)=(K_R/2)\int d^{2}r|\nabla \theta {|}^2$, the phase-difference variance at $a\ll r\ll L$ is $\log (r/a)/(\pi K_R)$ up to bounded cutoff terms. Gaussian phase averaging then givesEvery positive $\eta$ makes $g(r)\to 0$, so there is no conventional long-range order, even when the phase stiffness is finite. This is how an accumulation of gentle, cheap twists destroys global alignment without making every local twist free.

For a two-species uniform boson number state, set $S_i^{+}=a_i^{†}{b}_i$. A single spin ladder operator changes the species occupation numbers, so its expectation value vanishes. At distinct sites, $⟨S_i^{+}{S}_j^{-}⟩=n_A{n}_B/L^2$, proving the displayed transverse two-point correlation function. For $n_A/L,n_B/L$ tending to nonzero constants, this correlation function has a nonzero large-distance limit despite the global U(1) symmetry. This diagnoses long-range order in the auxiliary state; its interpretation as a state of fixed physical spins still requires the Schwinger boson constraint.