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Spin-1/2 Chain

A spin-1/21/2 chain is a one-dimensional tensor product of two-level local spin spaces coupled by specified onsite and finite-range operators; it is a model family whose symmetry, integrability, phases, and numerics depend decisively on normalization, couplings, fields, graph, and boundary conditions.

This dossier supplies the shared family record and validation language. It does not make every spin chain one model. The Transverse-Field Ising Model dossier, Heisenberg Chain dossier, and XXZ Chain dossier own those families’ convention-complete records. Their Ising, Heisenberg, and XXZ teaching articles own derivations and physical interpretation.

The umbrella record used here is a uniform nearest-neighbor XYZ chain in a uniform field.

FieldReference choice
sitesL≥3L\geq3 sites labeled j=1,…,Lj=1,\ldots,L
local spaceHj≃C2\mathcal H_j\simeq\mathbb C^2
local operatorsdimensionless sjα=σjα/2s_j^\alpha=\sigma_j^\alpha/2
geometryone-dimensional chain
exchangereal diagonal couplings Jx,Jy,JzJ_x,J_y,J_z
fieldreal energy vector h=(hx,hy,hz)\mathbf h=(h_x,h_y,h_z)
boundariesopen or periodic, stated explicitly
energy offsetnone
thermodynamic limitL→∞L\to\infty at fixed local couplings
disorder and longer rangeabsent unless declared

For periodic boundaries, the reference Hamiltonian is

H=Hex+Hh,Hex=∑α=x,y,zJα∑j=1Lsjαsj+1α,Hh=−∑α=x,y,zhα∑j=1Lsjα.\begin{aligned} H &= H_{\mathrm{ex}} + H_h, \\ H_{\mathrm{ex}} &= \sum_{\alpha=x,y,z} J_\alpha \sum_{j=1}^{L} s_j^\alpha s_{j+1}^\alpha, \\ H_h &= - \sum_{\alpha=x,y,z} h_\alpha \sum_{j=1}^{L} s_j^\alpha. \end{aligned}

with

sL+1α=s1α.s_{L+1}^\alpha = s_1^\alpha.

This reference family contains several canonical chains as parameter restrictions, but an arbitrary point with fields is not generically integrable. Cross-component exchange, Dzyaloshinskii–Moriya terms, bond alternation, disorder, and multi-spin interactions are variants, not unspoken parts of the record.

The local basis is chosen as

sz∣↑⟩=12∣↑⟩,sz∣↓⟩=−12∣↓⟩.s^z\lvert\uparrow\rangle = \frac12\lvert\uparrow\rangle, \qquad s^z\lvert\downarrow\rangle = -\frac12\lvert\downarrow\rangle.

Equivalently,

∣0⟩=∣↑⟩,∣1⟩=∣↓⟩.\lvert0\rangle = \lvert\uparrow\rangle, \qquad \lvert1\rangle = \lvert\downarrow\rangle.

The many-body Hilbert space is

HL=⨂j=1LC2,\mathcal H_L = \bigotimes_{j=1}^{L} \mathbb C^2,

with dimension

dim⁡HL=2L.\dim\mathcal H_L = 2^L.

The site operator is embedded as

sjα=I⊗(j−1)⊗σα2⊗I⊗(L−j).s_j^\alpha = I^{\otimes(j-1)} \otimes \frac{\sigma^\alpha}{2} \otimes I^{\otimes(L-j)}.

Operators on distinct sites commute:

[siα,sjβ]=0(i≠j),[s_i^\alpha,s_j^\beta] = 0 \qquad (i\ne j),

whereas the onsite algebra is

[sjα,sjβ]=iϵαβγsjγ.[s_j^\alpha,s_j^\beta] = i\epsilon_{\alpha\beta\gamma} s_j^\gamma.

For reproducible bit-basis examples, this dossier orders kets as

∣b1b2⋯bL⟩,\lvert b_1b_2\cdots b_L\rangle,

and maps them to integers by

x=∑j=1Lbj2L−j.x = \sum_{j=1}^{L} b_j2^{L-j}.

Thus site 11 is the most significant bit. Another ordering is equally valid if all operators, basis labels, and reported vectors use it consistently.

This dossier uses dimensionless spin operators

sα=σα2,s^\alpha = \frac{\sigma^\alpha}{2},

whose eigenvalues along any axis are ±1/2\pm1/2. Physical angular momentum is

Sα=ℏsα.S^\alpha = \hbar s^\alpha.

A bond and field transform as

σiασjα=4siαsjα,σiα=2siα.\sigma_i^\alpha\sigma_j^\alpha = 4s_i^\alpha s_j^\alpha, \qquad \sigma_i^\alpha = 2s_i^\alpha.

Therefore a Pauli-form Hamiltonian

H=∑⟨i,j⟩,αKασiασjα−∑i,αgασiαH = \sum_{\langle i,j\rangle,\alpha} K_\alpha \sigma_i^\alpha\sigma_j^\alpha - \sum_{i,\alpha} g_\alpha\sigma_i^\alpha

uses

Jα=4Kα,hα=2gαJ_\alpha = 4K_\alpha, \qquad h_\alpha = 2g_\alpha

in the present spin convention. Spectra, critical fields, and trace moments can differ by factors of two or four when this conversion is missed.

Define

sj±=sjx±isjy.s_j^\pm = s_j^x \pm is_j^y.

The anisotropic transverse bond is

Jxsjxsj+1x+Jysjysj+1y=Jx+Jy4(sj+sj+1−+sj−sj+1+)+Jx−Jy4(sj+sj+1++sj−sj+1−).\begin{aligned} J_xs_j^xs_{j+1}^x &+ J_ys_j^ys_{j+1}^y \\ ={}& \frac{J_x+J_y}{4} \left( s_j^+s_{j+1}^- + s_j^-s_{j+1}^+ \right) \\ &+ \frac{J_x-J_y}{4} \left( s_j^+s_{j+1}^+ + s_j^-s_{j+1}^- \right). \end{aligned}

The first line of processes exchanges an up and a down spin and preserves total zz magnetization. The second creates or removes two down spins relative to the all-up reference and changes total zz magnetization by two units.

This operator decomposition immediately diagnoses a central symmetry condition:

[H,Stotz]=0,Stotz=∑jsjz,[H,S_{\mathrm{tot}}^z] = 0, \qquad S_{\mathrm{tot}}^z = \sum_j s_j^z,

when

Jx=Jy,hx=hy=0.J_x = J_y, \qquad h_x = h_y = 0.

In that case, a sector with N↓N_\downarrow down spins has

Stotz=L2−N↓S_{\mathrm{tot}}^z = \frac L2-N_\downarrow

and dimension

dim⁡HN↓=(LN↓).\dim\mathcal H_{N_\downarrow} = \binom{L}{N_\downarrow}.

The parameter restrictions below identify standard nearest-neighbor descendants in the present spin convention.

FamilyParameter restrictionCharacteristic symmetryStandard solution status
longitudinal IsingJx=Jy=0J_x=J_y=0, hx=hy=0h_x=h_y=0all sjzs_j^z commute with HHproduct-basis exact
transverse-field Isingone Ising exchange plus a perpendicular fieldglobal Z2\mathbb Z_2quadratic fermions in the standard chain
XYJz=0J_z=0, field normal to the exchange planeusually Z2\mathbb Z_2quadratic after Jordan–Wigner
XXJx=JyJ_x=J_y, Jz=0J_z=0, longitudinal fieldU(1)U(1)free spinless fermions
XXZJx=JyJ_x=J_y, JzJ_z independent, longitudinal fieldU(1)U(1)Bethe-ansatz integrable in the uniform chain
HeisenbergJx=Jy=JzJ_x=J_y=J_z, zero fieldSU(2)SU(2)Bethe-ansatz integrable for the uniform spin-1/21/2 chain
XYZunequal Jx,Jy,JzJ_x,J_y,J_z, zero fielddiscrete spin rotationsintegrable in the uniform zero-field chain via the eight-vertex structure
generic field XYZunequal exchange and arbitrary fieldoften only lattice symmetriesgenerally nonintegrable

Axis labels are conventional. A global spin rotation can turn a zzzz Ising exchange with an xx field into an xxxx exchange with a zz field. The transformed Hamiltonian is physically equivalent only when every operator, observable, boundary term, and state is rotated consistently.

For uniform periodic couplings, the Hamiltonian commutes with one-site translation. It also has spatial reflection because the exchange is symmetric under interchanging the two sites of a bond.

At zero field, real bilinear exchange is time-reversal invariant. For an odd number of spin-1/21/2 sites,

T2=−1,\mathcal T^2 = -1,

so every finite-size energy level is at least Kramers degenerate. For even LL, T2=+1\mathcal T^2=+1 on the many-spin space and time reversal alone does not enforce pairwise degeneracy.

Internal symmetry depends on parameters:

  • Jx=JyJ_x=J_y with a longitudinal field has axial U(1)U(1) symmetry generated by StotzS_{\mathrm{tot}}^z.
  • Jx=Jy=JzJ_x=J_y=J_z and h=0\mathbf h=0 has global SU(2)SU(2) spin-rotation symmetry.
  • A field at the isotropic point reduces SU(2)SU(2) to rotations about the field axis.
  • At zero field, diagonal XYZ exchange is invariant under global π\pi rotations about each principal axis.
  • The transverse-field Ising restriction has a global spin-flip Z2\mathbb Z_2 symmetry about the field axis.

Translation, reflection, spin flip, magnetization, and total spin are distinct labels. They should not be assumed simultaneously. A numerical block decomposition is valid only after checking

[H,Q]=0[H,Q] = 0

for each claimed sector operator QQ.

For open boundaries,

HO=∑j=1L−1hj,j+1+∑j=1Lhj.\begin{aligned} H_{\mathrm O} ={}& \sum_{j=1}^{L-1} h_{j,j+1} + \sum_{j=1}^{L} h_j. \end{aligned}

There are L−1L-1 nearest-neighbor bonds. For periodic boundaries,

HP=HO+hL,1,H_{\mathrm P} = H_{\mathrm O} + h_{L,1},

so a ring with L≥3L\geq3 has LL undirected bonds.

For L=2L=2, modulo indexing is ambiguous: the formal terms j=1j=1 and j=2j=2 may describe the same undirected pair twice. A two-site benchmark must state whether it uses one bond, two parallel bonds, or a coupling rescaled to compensate. The minimal example below is an open one-bond dimer.

Open boundaries break translation symmetry but may preserve reflection. Periodic boundaries remove physical ends and support crystal-momentum sectors. Twisted boundaries are meaningful when a continuous spin component is conserved; in an axial chain one may replace the wrapping flip-flop terms by

sL+s1−⟼eiϕsL+s1−,s_L^+s_1^- \longmapsto e^{i\phi} s_L^+s_1^-,

and its Hermitian conjugate by the opposite phase.

After a Jordan–Wigner transformation, periodic spin boundaries do not become one universal fermion boundary condition. The fermionic boundary sign depends on parity. Boundary Conditions on Lattices owns that full audit.

A convenient local energy scale is

J∗=max⁡(∣Jx∣,∣Jy∣,∣Jz∣,∣hx∣,∣hy∣,∣hz∣).J_* = \max \left( |J_x|, |J_y|, |J_z|, |h_x|, |h_y|, |h_z| \right).

When J∗>0J_*>0, useful dimensionless controls include

JyJ∗,JzJ∗,hαJ∗,kBTJ∗,L.\frac{J_y}{J_*}, \quad \frac{J_z}{J_*}, \quad \frac{h_\alpha}{J_*}, \quad \frac{k_{\mathrm B}T}{J_*}, \quad L.

For the XXZ restriction, one usually writes

Jx=Jy=J,Jz=JΔ,J_x = J_y = J, \qquad J_z = J\Delta,

so Δ\Delta is the exchange anisotropy. For a transverse-field Ising restriction, a field-to-exchange ratio organizes the competition, but its numerical critical value depends on whether Pauli or spin operators define those coefficients.

Finite-size control requires comparing several LL values, boundary choices, and symmetry sectors. A small gap can represent a true bulk gap, a critical 1/L1/L level spacing, tunneling between finite-size symmetry partners, or a crossing between different conserved sectors.

“Spin-1/21/2 chain” does not name one exactness class.

Quantity or subfamilyStatusCaveat
finite-LL matrixexactly defined once basis, graph, and coefficients are fixeddimension grows as 2L2^L
full exact diagonalizationnumerically exact up to arithmetic and eigensolver tolerancesfeasible only for limited LL
commuting longitudinal Ising limitexact product-basis spectruma transverse field removes commutativity
standard XY and transverse-field Ising chainsreducible to quadratic fermionsboundaries and parity sectors remain essential
uniform spin-1/21/2 Heisenberg and XXZ chainsBethe-ansatz integrablefinite roots, thermodynamics, and correlators are separate tasks
uniform zero-field XYZ chainintegrable through the eight-vertex correspondencegeneric fields destroy that structure
generic finite-range chainno general analytic solutiontensor networks or other controlled numerics may still be accurate
real-time generic dynamicsunitary and exactly definedthermalization, transport, and hydrodynamics are model-dependent
thermodynamic phase claimsrequire L→∞L\to\infty evidenceno finite chain has a nonanalytic partition function

Integrability is not synonymous with triviality. An exact spectral construction may still leave difficult correlation functions, quench overlaps, or finite-temperature limits. Conversely, a nonintegrable finite chain is still an exact finite-dimensional quantum problem even when no closed-form thermodynamic solution exists.

With no added identity offset, every term in the reference Hamiltonian is a nonidentity Pauli string, so

Tr⁡H=0.\operatorname{Tr}H = 0.

For a uniform periodic chain with L≥3L\geq3, Pauli-string orthogonality gives

12LTr⁡H2=L[Jx2+Jy2+Jz216+hx2+hy2+hz24].\begin{aligned} \frac{1}{2^L} \operatorname{Tr}H^2 ={}& L \left[ \frac{ J_x^2+J_y^2+J_z^2 }{16} \right. \\ &\left. \qquad + \frac{ h_x^2+h_y^2+h_z^2 }{4} \right]. \end{aligned}

This identity is independent of integrability and is a strong assembly test. It changes if bonds are counted differently, an identity shift is added, or Pauli matrices replace spin operators without rescaling.

At infinite temperature,

12LTr⁡(siαsjβ)=14δijδαβ.\frac{1}{2^L} \operatorname{Tr} (s_i^\alpha s_j^\beta) = \frac14 \delta_{ij}\delta_{\alpha\beta}.

The entropy is

S∞=LkBln⁡2.S_\infty = Lk_{\mathrm B}\ln2.

When Tr⁡H=0\operatorname{Tr}H=0, the high-temperature energy begins as

U=−βTr⁡H22L+O(β2),U = -\beta \frac{\operatorname{Tr}H^2}{2^L} +O(\beta^2),

and therefore

CV=kBβ2Tr⁡H22L+O(β3).C_V = k_{\mathrm B}\beta^2 \frac{\operatorname{Tr}H^2}{2^L} +O(\beta^3).

These are normalization checks, not phase diagnostics. They do not determine low-energy order, criticality, or transport.

LimitExact structureWhat it does not imply
all Jα=0J_\alpha=0independent spins in a uniform fieldno exchange correlations
only Jz,hzJ_z,h_z nonzerocommuting classical Ising energy in the zz basisno transverse quantum dynamics
strong fieldpolarized product state with perturbative spin flipsexact polarization only at infinite ratio unless terms commute
Jx=JyJ_x=J_y, longitudinal fieldfixed-StotzS_{\mathrm{tot}}^z sectorsnot necessarily free or integrable
Jx=Jy=JzJ_x=J_y=J_z, zero fieldSU(2)SU(2) multipletsnot classical alignment for antiferromagnetic exchange
LL finitediscrete analytic spectrumno spontaneous symmetry breaking or true critical singularity
L→∞L\to\inftypossible phases and critical scalingboundary and parity sequences must still be controlled
spin s→∞s\to\inftysemiclassical limit of a different familya fixed spin-1/21/2 chain has no tunable large-spin parameter

Changing the sign of a coupling is not always removable. On a bipartite open chain, some signs can be changed by staggered spin rotations. On an odd periodic ring, the same transformation can leave a frustrated wrapping bond. Geometry and boundary conditions are part of every sign-convention claim.

The uniform magnetization density is

mα=1L∑j⟨sjα⟩.m^\alpha = \frac1L \sum_j \langle s_j^\alpha\rangle.

A translation-averaged connected correlator is

Cαβ(r)=1L∑j[⟨sjαsj+rβ⟩−⟨sjα⟩⟨sj+rβ⟩].\begin{aligned} C^{\alpha\beta}(r) ={}& \frac1L \sum_j \left[ \langle s_j^\alpha s_{j+r}^\beta \rangle \right. \\ &\left. \qquad - \langle s_j^\alpha\rangle \langle s_{j+r}^\beta\rangle \right]. \end{aligned}

The static structure factor is

Sαβ(q)=1L∑j,ke−iq(j−k)⟨sjαskβ⟩.S^{\alpha\beta}(q) = \frac1L \sum_{j,k} e^{-iq(j-k)} \langle s_j^\alpha s_k^\beta \rangle.

Other standard observables include:

  • symmetry-resolved excitation gaps;
  • uniform and staggered susceptibilities;
  • dynamical structure factors;
  • spin currents when a component is conserved;
  • domain-wall, chirality, or string observables in suitable variants;
  • bipartite entanglement entropy and entanglement spectra;
  • fidelity and response to boundary twists.

The phrase “the gap” is incomplete. A calculation should identify the reference state, target symmetry sector, momentum, parity, and boundary condition. Equal-Time Correlations and Structure Factors own the general normalization conventions.

Consider one open XYZ bond with no field:

H2=Jxs1xs2x+Jys1ys2y+Jzs1zs2z.H_2 = J_xs_1^xs_2^x + J_ys_1^ys_2^y + J_zs_1^zs_2^z.

The Bell basis is

∣Φ±⟩=∣↑↑⟩±∣↓↓⟩2,∣Ψ±⟩=∣↑↓⟩±∣↓↑⟩2.\begin{aligned} \lvert\Phi_\pm\rangle &= \frac{ \lvert\uparrow\uparrow\rangle \pm \lvert\downarrow\downarrow\rangle }{\sqrt2}, \\ \lvert\Psi_\pm\rangle &= \frac{ \lvert\uparrow\downarrow\rangle \pm \lvert\downarrow\uparrow\rangle }{\sqrt2}. \end{aligned}

These states diagonalize all three commuting two-site products

σ1xσ2x,σ1yσ2y,σ1zσ2z.\sigma_1^x\sigma_2^x, \qquad \sigma_1^y\sigma_2^y, \qquad \sigma_1^z\sigma_2^z.

Their energies are

EΦ+=Jx−Jy+Jz4,EΦ−=−Jx+Jy+Jz4,EΨ+=Jx+Jy−Jz4,EΨ−=−Jx−Jy−Jz4.\begin{aligned} E_{\Phi_+} &= \frac{ J_x-J_y+J_z }{4}, \\ E_{\Phi_-} &= \frac{ -J_x+J_y+J_z }{4}, \\ E_{\Psi_+} &= \frac{ J_x+J_y-J_z }{4}, \\ E_{\Psi_-} &= \frac{ -J_x-J_y-J_z }{4}. \end{aligned}

The energies sum to zero, as required by Tr⁡H2=0\operatorname{Tr}H_2=0. At the isotropic point,

Jx=Jy=Jz=J,J_x = J_y = J_z = J,

the first three states form the triplet with energy J/4J/4, while ∣Ψ−⟩\lvert\Psi_-\rangle is the singlet with energy −3J/4-3J/4. This dimer fixes the spin normalization but does not represent a two-site periodic ring unless its bond convention is stated separately.

Use the periodic L=4L=4 reference chain with

Jx=1,Jy=2,Jz=3,J_x = 1, \qquad J_y = 2, \qquad J_z = 3,

and

hx=1,hy=hz=0.h_x = 1, \qquad h_y = h_z = 0.

The four undirected bonds are

(1,2),(2,3),(3,4),(4,1).(1,2), \quad (2,3), \quad (3,4), \quad (4,1).

The Hilbert-space dimension is

24=16.2^4 = 16.

The basis-independent trace targets are

Tr⁡H=0,\operatorname{Tr}H = 0,

and

116Tr⁡H2=92.\frac1{16} \operatorname{Tr}H^2 = \frac92.

Equivalently,

Tr⁡H2=72.\operatorname{Tr}H^2 = 72.

Although StotzS_{\mathrm{tot}}^z is not conserved, the global π\pi rotation about xx is:

Px=∏j=14σjx,[H,Px]=0.P_x = \prod_{j=1}^{4} \sigma_j^x, \qquad [H,P_x] = 0.

A valid implementation should:

  1. reproduce dimension 1616, Hermiticity, and the four-bond graph;
  2. verify the trace and second moment without using eigenvectors;
  3. check Px2=IP_x^2=I and [H,Px]=0[H,P_x]=0;
  4. confirm that [H,Stotz]≠0[H,S_{\mathrm{tot}}^z]\ne0 for these parameters;
  5. reconstruct the full spectrum from both PxP_x sectors if block diagonalization is used;
  6. compare the one-bond dimer spectrum before trusting the ring assembly;
  7. test specializations against the stable benchmark suite.

Two specializations connect directly to canonical contracts. For the Pauli-form transverse-field Ising Hamiltonian

HTFIM=−JI∑jσjzσj+1z−hI∑jσjx,H_{\mathrm{TFIM}} = -J_{\mathrm I} \sum_j \sigma_j^z\sigma_{j+1}^z - h_{\mathrm I} \sum_j \sigma_j^x,

use

Jz=−4JI,hx=2hI,Jx=Jy=0.\begin{aligned} J_z &= -4J_{\mathrm I}, & h_x &= 2h_{\mathrm I}, \\ J_x &= J_y = 0. \end{aligned}

Then apply MB-B001. For the spin-form Heisenberg ring, set

Jx=Jy=Jz=J,h=0,J_x = J_y = J_z = J, \qquad \mathbf h = 0,

and apply MB-B002.

Passing the family trace test validates operator normalization and bond assembly. It does not validate a thermodynamic phase diagram, a Jordan–Wigner parity choice, or long-time dynamics.

For a finite chain:

  • the spectrum is discrete and analytic in generic parameters away from exact crossings;
  • a symmetry-preserving eigenstate has zero expectation value for an odd order parameter;
  • nearly degenerate symmetry partners can precede bulk symmetry breaking;
  • momenta depend on boundary conditions and, after nonlocal mappings, parity sectors;
  • even and odd rings can have different frustration and Kramers structure;
  • the lowest excitation in the full Hilbert space need not be the gap relevant to a chosen response operator.

If StotzS_{\mathrm{tot}}^z is conserved, fixed-magnetization blocks must reconstruct the full dimension:

∑N↓=0L(LN↓)=2L.\sum_{N_\downarrow=0}^{L} \binom{L}{N_\downarrow} = 2^L.

If translation sectors are used, basis states fall into orbits whose lengths divide LL. Momentum-block dimensions are therefore not generally equal. Symmetry Sectors in Many-Body Numerics owns the full reconstruction audit.

Finite-size evidence should report at least:

  • LL and its parity;
  • open, periodic, or twisted boundaries;
  • operator normalization;
  • exact symmetry sector;
  • bond list;
  • energy offset;
  • convergence or extrapolation sequence.

A general bilinear bond uses a real exchange tensor:

Hij=∑α,βJijαβsiαsjβ.H_{ij} = \sum_{\alpha,\beta} J_{ij}^{\alpha\beta} s_i^\alpha s_j^\beta.

Its antisymmetric part is equivalent to a Dzyaloshinskii–Moriya vector. Such terms can break inversion, change conserved spin components, and shift spiral correlations.

Next-nearest-neighbor exchange introduces competing paths. The J1J_1–J2J_2 chain can dimerize and frustrate simple antiferromagnetic order. The ratio J2/J1J_2/J_1 becomes new model data.

Random fields or bonds break translation symmetry and require sample ensembles. Localization and rare-region claims need disorder-size and realization convergence, not one spectrum.

Adding measurement, noise, or dissipation requires a master equation or quantum channel. A non-Hermitian effective Hamiltonian alone does not specify the unconditional dynamics.

Hard-core bosons share the local two-state algebra. Spinless fermions in one dimension require Jordan–Wigner strings, and periodic boundaries require parity-sector care. Equal local dimensions do not erase exchange statistics or boundary data.

This dossier owns:

  • the common spin-1/21/2 chain family record;
  • spin-versus-Pauli normalization and basis ordering;
  • the XYZ parameter map and shared symmetry conditions;
  • generic boundary, finite-size, and sector audits;
  • the Bell-basis XYZ dimer;
  • the family-wide trace-moment benchmark and benchmark handoffs.

It does not rederive:

  • Naming a “spin chain” without its local spin, operator normalization, Hamiltonian, graph, and boundaries.
  • Substituting Pauli matrices for sα=σα/2s^\alpha=\sigma^\alpha/2 without rescaling couplings.
  • Counting the periodic wrapping bond twice.
  • Treating a two-site modulo sum as an unambiguous ring.
  • Claiming StotzS_{\mathrm{tot}}^z conservation when Jx≠JyJ_x\ne J_y or a transverse field is present.
  • Assuming translation, reflection, spin flip, and total spin are always simultaneous symmetries.
  • Calling every one-dimensional spin chain integrable.
  • Treating a finite-size avoided crossing as a bulk phase transition.
  • Calling a symmetry-preserving finite-size cat state a broken-symmetry state.
  • Comparing gaps from different sectors or boundary conditions.
  • Ignoring parity-dependent fermion boundaries after Jordan–Wigner transformation.
  • Inferring long-range order from a short-distance correlator.
  • Reporting eigenvectors inside a degenerate subspace without a phase or basis convention.
  • Treating an effective spin flip as a literal microscopic particle.

For LL spin-1/21/2 sites, show that the sector with N↓N_\downarrow down spins has dimension (LN↓)\binom{L}{N_\downarrow} and verify that all sectors reconstruct the full Hilbert space.

Solution

A computational-basis state in the sector is fixed by choosing which N↓N_\downarrow of the LL sites carry bit 11. Therefore

dim⁡HN↓=(LN↓).\dim\mathcal H_{N_\downarrow} = \binom{L}{N_\downarrow}.

Each bit string belongs to exactly one such sector. Summing over all allowed down-spin counts gives the binomial theorem:

∑N↓=0L(LN↓)=(1+1)L=2L.\sum_{N_\downarrow=0}^{L} \binom{L}{N_\downarrow} = (1+1)^L = 2^L.

The magnetization eigenvalue in this sector is

Stotz=12(L−N↓)−12N↓=L2−N↓.S_{\mathrm{tot}}^z = \frac12(L-N_\downarrow) - \frac12N_\downarrow = \frac L2-N_\downarrow.

Exercise 2: Derive the magnetization condition

Section titled “Exercise 2: Derive the magnetization condition”

Use the raising-and-lowering decomposition to determine when the reference Hamiltonian commutes with StotzS_{\mathrm{tot}}^z.

Solution

The flip-flop operators

sj+sj+1−,sj−sj+1+s_j^+s_{j+1}^-, \qquad s_j^-s_{j+1}^+

raise one site and lower the other, so their net change in total zz magnetization is zero.

The pair operators

sj+sj+1+,sj−sj+1−s_j^+s_{j+1}^+, \qquad s_j^-s_{j+1}^-

change total magnetization by +2+2 and −2-2. Their coefficient is (Jx−Jy)/4(J_x-J_y)/4, so they vanish precisely when Jx=JyJ_x=J_y.

A longitudinal field hzh_z commutes with StotzS_{\mathrm{tot}}^z, while hxsjx+hysjyh_xs_j^x+h_ys_j^y contains single-spin raising and lowering terms. Thus

[H,Stotz]=0[H,S_{\mathrm{tot}}^z] = 0

for the reference family when

Jx=Jy,hx=hy=0.J_x = J_y, \qquad h_x = h_y = 0.

Derive the four Bell-basis energies of the one-bond XYZ dimer and recover the singlet–triplet splitting at the isotropic point.

Solution

The Bell states are simultaneous eigenstates of the three Pauli-pair operators. Their eigenvalue triples for

(σxσx,σyσy,σzσz)(\sigma^x\sigma^x,\sigma^y\sigma^y,\sigma^z\sigma^z)

are

xxyyzzΦ++1−1+1Φ−−1+1+1Ψ++1+1−1Ψ−−1−1−1.\begin{array}{c|ccc} & xx & yy & zz \\ \hline \Phi_+ & +1 & -1 & +1 \\ \Phi_- & -1 & +1 & +1 \\ \Psi_+ & +1 & +1 & -1 \\ \Psi_- & -1 & -1 & -1 \end{array}.

Since sαsα=σασα/4s^\alpha s^\alpha=\sigma^\alpha\sigma^\alpha/4, multiplying by the couplings gives

EΦ+=Jx−Jy+Jz4,EΦ−=−Jx+Jy+Jz4,EΨ+=Jx+Jy−Jz4,EΨ−=−Jx−Jy−Jz4.\begin{aligned} E_{\Phi_+} &= \frac{J_x-J_y+J_z}{4}, \\ E_{\Phi_-} &= \frac{-J_x+J_y+J_z}{4}, \\ E_{\Psi_+} &= \frac{J_x+J_y-J_z}{4}, \\ E_{\Psi_-} &= \frac{-J_x-J_y-J_z}{4}. \end{aligned}

At Jx=Jy=Jz=JJ_x=J_y=J_z=J, the first three energies equal J/4J/4, while

EΨ−=−3J4.E_{\Psi_-} = -\frac{3J}{4}.

The Bell state Ψ−\Psi_- is the spin singlet; the other three form the triplet.

Exercise 4: Prove the trace-moment formula

Section titled “Exercise 4: Prove the trace-moment formula”

Use Pauli-string orthogonality to derive the normalized second moment of the uniform periodic reference chain for L≥3L\geq3.

Solution

Distinct Pauli strings are orthogonal under the Hilbert–Schmidt inner product:

12LTr⁡(PaPb)=δab.\frac1{2^L} \operatorname{Tr}(P_aP_b) = \delta_{ab}.

A spin bond is a Pauli string divided by four:

sjαsj+1α=14σjασj+1α.s_j^\alpha s_{j+1}^\alpha = \frac14 \sigma_j^\alpha\sigma_{j+1}^\alpha.

Its normalized squared trace is therefore 1/161/16. A field operator is a Pauli string divided by two, so its normalized squared trace is 1/41/4.

For L≥3L\geq3, all bond and onsite strings in the stated uniform sum are distinct. Cross terms vanish, and there are LL copies of each component. Hence

12LTr⁡H2=L[Jx2+Jy2+Jz216+hx2+hy2+hz24].\begin{aligned} \frac{1}{2^L} \operatorname{Tr}H^2 ={}& L \left[ \frac{J_x^2+J_y^2+J_z^2}{16} \right. \\ &\left. \qquad + \frac{h_x^2+h_y^2+h_z^2}{4} \right]. \end{aligned}

For the numerical target, this is

4(1+4+916+14)=92.4 \left( \frac{1+4+9}{16} + \frac14 \right) = \frac92.

Multiplying by 24=162^4=16 gives Tr⁡H2=72\operatorname{Tr}H^2=72.

Exercise 5: Translate benchmark conventions

Section titled “Exercise 5: Translate benchmark conventions”

Convert the Pauli-form transverse-field Ising chain to the dimensionless-spin convention, and explain why the Heisenberg specialization needs no further rescaling.

Solution

Using σα=2sα\sigma^\alpha=2s^\alpha,

−JIσjzσj+1z=−4JIsjzsj+1z,-J_{\mathrm I} \sigma_j^z\sigma_{j+1}^z = -4J_{\mathrm I} s_j^zs_{j+1}^z,

and

−hIσjx=−2hIsjx.-h_{\mathrm I}\sigma_j^x = -2h_{\mathrm I}s_j^x.

Comparing with

Jzsjzsj+1z−hxsjxJ_zs_j^zs_{j+1}^z - h_xs_j^x

gives

Jz=−4JI,hx=2hI.J_z = -4J_{\mathrm I}, \qquad h_x = 2h_{\mathrm I}.

The Heisenberg benchmark is already written as

J∑jsj⋅sj+1J \sum_j \mathbf s_j\cdot\mathbf s_{j+1}

with s=σ/2\mathbf s=\boldsymbol\sigma/2. Therefore its specialization is simply

Jx=Jy=Jz=J,h=0.J_x = J_y = J_z = J, \qquad \mathbf h = 0.

The remaining issue is graph data: the L=4L=4 ring has four undirected bonds, whereas a one-bond L=2L=2 dimer must not be generated by an ambiguous modulo sum.

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