Codex @codex  0

Use the canonical anticommutation relations $\{c_n,c_m^{†}\}={\delta }_{nm}$ and $\{c_n,c_m\}=0$. In particular, $n_m=c_m^{†}{c}_m$ is a projector, so $(1-2n_m{)}^2=1$. Define $K_n={\prod }_{m<n}(1-2n_m)$. Then $K_n^{†}=K_n$, $K_n^2=1$, and $K_n$ commutes with $c_n$ and $c_n^{†}$. Taking the adjoint of the given raising operator gives $S_n^{-}=c_n^{†}{K}_n$. HenceAlso $[n_n,c_n]=-c_n$ and $[n_n,c_n^{†}]=c_n^{†}$ imply $[S_n^z,S_n^{\pm }]=\pm S_n^{\pm }$. For different sites, exchanging two fermion operators contributes one minus sign and commuting an operator through the other site's parity string contributes another; their product is positive. Thus distinct spin algebras commute. The string supplies precisely the signs required for the Jordan–Wigner transformation. The fermion vacuum corresponds to spin up, while one occupied site corresponds to spin down; this convention fixes subsequent signs.

For a bulk bond, $K_{n+1}=K_n(1-2n_n)$ and $c_n(1-2n_n)=-c_n$. Reordering the remaining different-site operators then givesThe periodic spin boundary condition requires separate attention. Set $P={\prod }_{n=1}^N(1-2n_n)=(-1{)}^{N_f}$. The end bond becomes $S_N^{+}{S}_1^{-}=-P{c}_1^{†}{c}_N$, with the conjugate relation for its adjoint. Consequently the exact finite-ring Hamiltonian isfermion parity is conserved, so diagonalize in a sector $P=\pm 1$. There the end bond can be written like every other bond provided $c_{N+1}=-P{c}_1$. This parity-dependent boundary condition of the Jordan-Wigner chain is antiperiodic for even fermion number and periodic for odd fermion number. Simply assuming periodic fermions in both sectors changes the finite-size spectrum.

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]

Consider a conventional spin-singlet, weak-coupling, three-dimensional superconductor. Set $\hslash =k_B=1$ in the microscopic action and restore electromagnetic units below. An attraction $\lambda >0$ in an energy shell about the Fermi surface gives a Grassmann actionThe attractive sign permits pairing even when the single-particle Fermi gas is otherwise stable. A complex Hubbard–Stratonovich transformation introduces a pair field $\Delta$ proportional at the saddle to $\lambda ⟨{\psi }_{\downarrow }{\psi }_{\uparrow }⟩$. The action is then quadratic in fermions. Integrating them out gives the BCS pairing-field effective actionHere the trace uses one independent Nambu spinor pairing block; a fully doubled particle-hole representation instead carries its corresponding one-half factor. At a uniform saddle the Bogoliubov--de Gennes Hamiltonian can be written asThe negative branch describes the paired vacuum; excitations are Bogoliubov quasiparticles mixing particles and holes. Extremizing the effective action givesThe physical cutoff or a renormalized interaction is part of the microscopic model; an unregulated contact integral must not be used indiscriminately. With constant density of states $N(0)$ per spin and shell half-width ${\Omega }_c$, the zero-temperature weak-coupling solution is ${\Delta }_0\simeq 2{\Omega }_c{e}^{-1/[N(0)\lambda ]}$. The exponentially small scale follows from the logarithmic pairing susceptibility of states near the Fermi surface. For this ideal weak-coupling model, $2{\Delta }_0/(k_B{T}_c)\simeq 3.53$. The microscopic theory relates the gap, pairing and thermodynamic transition.^[ref]

Close to a continuous transition, for fields varying slowly compared with microscopic scales, expand the effective action in powers and gradients of $\Delta$. Its static part is the Ginzburg-Landau theory of superconductivity:The normalization of $\Delta$ fixes the coefficients; here $a(T)$ changes sign at $T_c$. The uniform minimum has $|{\Delta }_0{|}^2=-a/b$ for $a<0$, condensation free-energy density $-a^2/(2b)$, and characteristic coherence-length scale $\xi =\sqrt{K/|a|}$. Above the transition the minimum is at zero. The quartic stabilization and gradient cost are precisely the ingredients of Landau theory for a complex order parameter. This controlled near-transition expansion supplies a bridge between microscopic BCS theory and macroscopic magnetic response; its coefficients and validity depend on the material and fluctuation regime.^[ref]

Write $\Delta =(|{\Delta }_0|+\eta )e^{i\varphi }$. The radial fluctuation has positive local restoring curvature, whereas the phase has no local potential in a neutral theory with global $U(1)$ symmetry. Freezing the amplitude givesThe neutral case supports a gapless collective phase mode. In the charged case the electromagnetic field couples to this phase rigidity. Under $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi +q\chi /\hslash$, the displayed combination is invariant. Locally choosing a gauge with constant phase leaves a quadratic term in $\mathbf{A}$. Varying the free energy gives the transverse London current and magnetic screening scale,Combining this relation with static Maxwell equations gives exponential field penetration, the Meissner effect. Single-valuedness of the pair field around a loop with negligible current gives flux in units ${\Phi }_0=2\pi \hslash /|q|=\pi \hslash /e$. These are gauge-invariant physical consequences of the condensate stiffness.

For the dynamics, the neutral phase and density mode is reorganized by the Coulomb interaction and the dynamical electromagnetic field. In a bulk three-dimensional system the long-wave longitudinal mode has a plasma gap, rather than remaining the neutral sound mode. This is the Anderson-Higgs mechanism in a superconductor: a phase coordinate becomes part of the electromagnetic collective response instead of surviving as an independent massless physical particle. Local gauge transformations are redundancies of the description, so a literal breaking of a local redundancy is not an observable. Phase coherence, current response and magnetic screening provide the physical statements.^[ref]

A static Landau functional alone cannot determine that full dynamical spectrum. A two-dimensional film, for example, need not have the same long-wave plasma dispersion as a bulk charged system. Nor does a positive radial curvature automatically guarantee a sharp Higgs amplitude resonance: coupling to quasiparticles and other modes can damp it. The chosen mean-field phase is also a thermodynamic symmetry-breaking description; an isolated finite system can be formulated with definite particle number. The microscopic pairing saddle, the near-transition Landau expansion and the electromagnetic Anderson-Higgs response describe successive scales of the same superconducting state, with distinct assumptions at each step.

For the XXZ spin chain, write $\Delta =1+\alpha$ and replace each spin by a classical vector of length $S$. A bond is $J{S}^2{\mathbf{n}}_m^{\mathsf{T}}\mathrm{diag}(1,1,\Delta ){\mathbf{n}}_{m+1}$, whose minimum is $-J{S}^2\max (1,|\Delta |)$. For an even periodic chain all bonds can attain that bound simultaneously. The classical ground states of an XXZ spin chain therefore give $\alpha >0$ gives the two easy-axis Néel states ${\mathbf{S}}_m=(-1{)}^{m}S\hat{\mathbf{z}}$ and its reversal, with energy $-JN{S}^2(1+\alpha )$.

For $-2<\alpha <0$ the ground states are planar Néel configurations, ${\mathbf{S}}_m=(-1{)}^{m}S(\cos \phi ,\sin \phi ,0)$, with arbitrary $\phi$: their degeneracy is a circle and their energy is $-JN{S}^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha <-2$, $\Delta <-1$ and the lowest states are instead the two uniform $z$-polarized ferromagnetic configurations, with energy $JN\Delta S^2$. At $\alpha =-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic ferromagnet: ${\mathbf{S}}_m=S((-1{)}^m{n}_x,(-1{)}^m{n}_y,n_z)$ for any unit vector $\mathbf{n}$ is a ground state. This special degeneracy is a sphere. At the other boundary, $\alpha =0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum ground state.

Let $[a,a^{†}]=1$, $n=a^{†}a$, and take $S^{+}=(S^{-}{)}^{†}=\sqrt{2S-n}a$. The physical Holstein–Primakoff occupation constraint limits the Hilbert space, which contains only $n=0,1,\dots ,2S$. Acting in this number basis givesThe upper and lower endpoints are annihilated as required. ConsequentlyThe change in $n$ also gives $[S^z,S^{\pm }]=\pm S^{\pm }$. These matrix elements prove the spin algebra on the full physical spin-$S$ space. Extending the oscillator space to arbitrary occupation would not give a physical finite spin: the occupation constraint and operator ordering in the square root matter. bosons on distinct sites have $[a_m,a_n^{†}]={\delta }_{mn}$ and all other site commutators zero.

Rotate the spin on one sublattice through $\pi$ about its $x$ axis, using the unitary product of the corresponding spin-rotation operators. On a rotated site the old components obey $S_{\mathrm{old}}^x=S^x$, $S_{\mathrm{old}}^y=-S^y$, $S_{\mathrm{old}}^z=-S^z$. Every bond joins opposite sublattices, soThe longitudinal product changes sign. Hence the rotated bond is $-J\Delta S_m^z{S}_{m+1}^z+\frac{J}{2}(S_m^{+}{S}_{m+1}^{+}+S_m^{-}{S}_{m+1}^{-})$, giving the requested form. The canonical spin algebra is unchanged because the transformation is unitary. A staggered easy-axis reference configuration is now the uniform all-up state. At $\alpha =0$ one first chooses a Néel direction, here $z$, before performing the same rotation.

For $\alpha >0$, the dispersion is periodic and has minima at $k=0,\pi$ in the rotated full-zone convention, withThe maximum is $2JS(1+\alpha )$ at $k=\pm \pi /2$. The easy-axis state has no continuously broken symmetry, so a gap is allowed. Near a minimum the energy behaves as $2JS\sqrt{\alpha (2+\alpha )+(\delta k{)}^2}$. At $\alpha =0$ it becomes ${ϵ}_k=2JS|\sin k|$, linear near its zeros, with small-$k$ slope $2JS$ in the chosen units. Restoring lattice spacing $a$ and $\hslash$ gives spin-wave speed $2JSa/\hslash$. Zone folding reorganizes the same modes into the magnetic Brillouin zone.

These are linear spin-wave predictions, not a statement that the exact one-dimensional quantum chain always has long-range Néel order. At the isotropic point the zero-point spin reduction contains $N^{-1}{\sum }_k(|\sin k{|}^{-1}-1)/2$, which diverges logarithmically as the infrared cutoff vanishes. This failure of the assumed ordered reference must not be mistaken for a calculation of the exact interacting spectrum.

In this part $0<\Delta =1+\alpha <1$, so the ordered reference lies in the $xy$ plane. Choosing one direction breaks the continuous rotation symmetry within that plane. A gapless, linearly dispersing phase mode is therefore expected in the spin-wave approximation, while tilting the Néel axis out of the plane costs anisotropy energy. It is incorrect to continue the easy-axis expansion about an unstable $z$-ordered reference and interpret its imaginary frequencies as physical modes.

For the easy-plane XXZ antiferromagnetic spin-wave spectrum, take the Néel direction along old $x$, then rotate alternating local spins so that all longitudinal components point up. One such local frame gives the full-zone quadratic coefficientsand thereforeNear $k=0$, ${ϵ}_k\simeq JS\sqrt{2(1+\Delta )}|k|$. The zone-edge mode has gap ${ϵ}_{\pi }=2JS\sqrt{2(1-\Delta )}$. A different sublattice phase convention shifts the zero by $\pi$; folding into the magnetic zone displays an acoustic and a gapped polarization. The figure uses this explicitly chosen frame. One-dimensional long-wavelength fluctuations again limit the ordered-reference interpretation.

In the convention of iii, the invariant decay amplitude is ${\mathcal{M}}_{s_1{s}_2}=-g{\bar{u}}_{\chi }^{s_2}(p_2)u_{\Psi }^{s_1}(p_1)$. Including the common overall $i$ in the definition changes no squared decay amplitude. An unpolarized ensemble is an incoherent equal mixture of the two initial spin states; hence take a spin average over $s_1$ and a spin sum over the unobserved final $s_2$. Different final spin states are orthogonal outcomes, so their probabilities, rather than their decay amplitudes, add.

Using the Dirac adjoint and the fermion spin sum givesThe gamma-matrix trace identities used here are $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. The mass-shell conditions $p_1^2=M^2$ and $p_2^2=p_3^2=m^2$, together with four-momentum conservation, implyThereforeIn the parent's rest frame this also says $E_2=E_3=M/2$.

Read the unsuperscripted $a_{\mathbf{p}}$ in the printed Hamiltonian as $a_{\mathbf{p}}^s$: the repeated spin label is otherwise missing. Relabel $\mathbf{p}\mapsto -\mathbf{p}$ in its antiparticle term, and insert both projections from c. Keeping the original operator ordering givesThe difference of the two orthogonal projectors in a is $h(\mathbf{p})/E$, explaining the cancellation of the factors $E$. Under inverse Fourier transformation, $\mathbf{p}$ becomes $-i\nabla$ acting on $\psi$. HenceSpatial indices run from $1$ to $3$. This is the local Hamiltonian density of the free Dirac field. It also follows from $\mathcal{H}=\pi \dot{\psi }-\mathcal{L}$ with $\pi =i{\psi }^{†}$.

The original antiparticle term is $-b{b}^{†}$, so the result retains its infinite additive vacuum energy. If the Hamiltonian is normal-ordered, its occupation-number expression becomes $\int d^{3}p/(2\pi {)}^{3}E{\sum }_s(a^{s†}{a}^s+b^{s†}{b}^s)$ and its local expression is $:\mathcal{H}:$, with the same subtraction. The displayed unsymmetrized Hamiltonian density is Hermitian after spatial integration under vanishing boundary terms; a manifestly Hermitian local form replaces the derivative term by $-\frac{i}{2}{\psi }^{†}\mathbf{\alpha }\cdot \overleftrightarrow{\nabla }\psi$ and differs by a spatial divergence.

Every positive-energy Dirac spinor obeys $(/p-m)u^s(\mathbf{p})=0$. Therefore the fermion spin sum satisfiesSince ${\bar{u}}^s=u^{s†}{\gamma }^0$ and $({\gamma }^0{)}^2=I$, the trace of each outer product is its contracted scalar. The prescribed normalization and the two spin states yieldNow make exactly the stated polynomial assumption and apply polynomial reduction of a slashed vector. By a, on the mass shell $p^2=m^2$ write $P=A/p+BI$. ThenThe matrices $I$ and $/p$ are linearly independent for a physical nonzero $p$, so $B=mA$. The gamma matrix trace identities, $\mathrm{tr}{\gamma }^0=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^0)=4{\eta }^{\mu 0}$, giveThus $A=1$ andNo explicit representation of the Dirac spinors has been used. The same reasoning covers $m=0$ provided $p^0>0$.

Interpret the repeated species label in the Yukawa interaction as summed over $i=1,2$: each Feynman vertex is diagonal in species and equals $-ig$. Define the invariant scattering amplitude by $S-1=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. There is one tree-level Feynman diagram, an $s$-channel scalar exchange. In the left panel, $j=2$ for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Writing $k=p_1+p_2=p_3+p_4$ and $s=k^2$, the momentum-space Feynman rules giveThe scalar carries no spin label. Average over the $2\times 2$ incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of ${\bar{v}}_2{u}_1$ is ${\bar{u}}_1{v}_2$. The fermion spin sums then yieldThe bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities giveand the second trace has the same value because $p_3+p_4=p_1+p_2$ and all external masses are $m$. ThereforeThis is the fixed-order expression away from $s=M^2$. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

The canonical anticommutation relations give $\{a_1^{†},a_2^{†}\}=0$, where each label includes both momentum and spin. ThereforeThus exchanging the complete labels of the two identical fermions changes the sign of their state. If the labels coincide, $(a_1^{†}{)}^2=0$; the state vanishes, which is the Pauli exclusion principle. A negative state vector describes the same isolated physical ray, but its relative sign matters when exchange alternatives interfere.

The corrected expression in d is a vacuum two-point function, or Wick contraction, of the free Dirac field. It equals the negative spinor transpose of the usual Dirac propagator with reversed spacetime arguments. It supplies the internal fermion lines in Feynman diagrams, rather than a position-space detection probability.

For $x^0>y^0$, $\psi (y)$ creates an antiparticle and $\bar{\psi }(x)$ annihilates it. The $v\bar{v}$ term describes positive-energy antiparticle propagation from $y$ to $x$. For $y^0>x^0$, $\bar{\psi }(x)$ creates a particle and $\psi (y)$ annihilates it. The $u\bar{u}$ term describes positive-energy particle propagation from $x$ to $y$, with the minus sign required by exchanging the odd operators in fermionic time ordering. Neither term describes a physical negative-energy excitation.

The Heaviside step functions select these chronological alternatives. The factors $e^{\mp ip\cdot z}$ are on-shell plane waves with $p^0=E_{\mathbf{p}}>0$, and $d{\Pi }_p$ is the positive-energy Lorentz-invariant phase-space measure. The sum over $s$ includes the two spin states. With standard relativistic spinor normalization, the fermion spin sums areThey encode the spinor structure and solve the corresponding on-shell Dirac equations. In particular, the corrected answer can also be writtenThe Feynman i-epsilon prescription specifies the temporal boundary conditions of the Green function; its positive- and negative-frequency residues reproduce the two terms and their relative minus sign. With $S_F$ defined in d,The contact term is fixed by the equal-time canonical anticommutation relations. Choosing a value for $\theta (0)$ does not change the unequal-time sign demonstrated above.

Finally, microcausality requires field anticommutators to vanish at spacelike separation; the time-ordered vacuum two-point function itself need not vanish there. These facts are compatible with relativistic causality. The plus-plus expression printed in the paper lacks the fermionic sign needed for this standard Dirac propagator interpretation.

Include Alice's detector and environment before taking a partial trace. Ideal measurement produces the global stateOrthogonality of the two record states removes their cross terms from the reduced state of the particles:This nonselective quantum measurement has changed the initially entangled singlet into a separable, classically anticorrelated reduced state. Its entropy is one bit, although the full particles-plus-records state is still pure under the assumed unitary dynamics. The reduced density matrix is an improper mixture; tracing the environment does not by itself assert a literal collapse of that global state.

The Born rule gives Alice outcomes zero and one with equal probability. Conditional on a recorded result, the selective quantum measurement assigns $|01⟩$ or $|10⟩$ to the particles. The unitary measurement coupling and orthogonal records explain decoherence in this basis; they do not supply an additional rule selecting a unique outcome. That is part of the measurement problem. In a collapse postulate the individual outcome is stochastic; other interpretations account for outcome experience differently while retaining these operational probabilities.

When Bob makes the same kind of measurement, the global state becomesTracing both laboratories gives the same reduced matrix, ${\rho }_{AB}^{(2)}={\rho }_{AB}^{(1)}$. Bob has added a correlated record, not restored the lost reduced coherence. If Alice's outcome is known, Bob's opposite outcome is certain in this ideal common basis. Without that record, his marginal is $I/2$, so his outcome remains equiprobable. Alice cannot choose her random result or use this correlation for faster-than-light signalling, consistently with quantum no-signalling.

These common-basis measurements alone do not distinguish quantum predictions from a local source sending opposite preassigned bits. To test that explanation, prepare fresh singlets and let the observers independently choose among incompatible measurement axes before recording each result. For singlet spin measurements along unit vectors $\mathbf{a},\mathbf{b}$, quantum mechanics predicts $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. The CHSH inequality bounds $|S|\le 2$ for Bell-local models with measurement-independent settings. Choosinggives $S=E(\mathbf{a},\mathbf{b})+E(\mathbf{a},{\mathbf{b}}^{\prime })+E({\mathbf{a}}^{\prime },\mathbf{b})-E({\mathbf{a}}^{\prime },{\mathbf{b}}^{\prime })=-2\sqrt{2}$. A suitable Bell test constrains that class of explanations. Independently chosen settings, control of detection bias and separation of the measurement events matter when interpreting an actual experiment.

If Alice first performs the fixed-axis irreversible measurement described above, the resulting mixed state has $E(\mathbf{a},\mathbf{b})=-a_z{b}_z$ and cannot supply that singlet violation. The incompatible settings must belong to the Bell measurement itself. Moreover, Bell violations do not choose among all interpretations of quantum mechanics: interpretations sharing the Born predictions remain observationally equivalent for this experiment. Objective-collapse modifications could instead require additional interference tests with specified departures from unitary dynamics.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.

At $N=M$, the two spin occupations per site are $x$ and $1-x$. With $\tau =|t|>0$, the trial energy per site isIts derivative is $e^{\prime }(x)=-4\tau \cos (\pi x)+V[4x-2-(2/\pi )\sin (2\pi x)]$. Since $x=0$ is a boundary point, its local minimum test is one-sided: $e^{\prime }(0)=-4\tau -2V$. At $V=-2\tau$, where this derivative vanishes, its Taylor series begins $e(x)-e(0)=(2{\pi }^2\tau /3)x^3+O(x^4)>0$ for small positive $x$. HenceThe same applies to $x=1$ by spin interchange.