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For $S_n^{+}=P_n{c}_n$ with $P_n={\prod }_{m<n}(1-2c_m^{†}{c}_m)$, neighbouring Jordan–Wigner transformation strings differ by only the local parity factor. The canonical anticommutation relations give $(c_n+c_n^{†})(1-2c_n^{†}{c}_n)=c_n^{†}-c_n$, proving the displayed local spin identity. The closing bond of a periodic chain instead retains total fermion parity; cancellation on bulk bonds does not justify discarding that global sector.

A linear change of creation operators and annihilation operators is canonical when it preserves their defining brackets. For finitely many bosonic modes, the canonical commutation relations require $U{U}^{†}-V{V}^{†}=I$ and $U{V}^T-V{U}^T=0$. For fermions, the canonical anticommutation relations instead require $U{U}^{†}+V{V}^{†}=I$ and $U{V}^T+V{U}^T=0$. A Bogoliubov transformation uses such mixing to diagonalize a quadratic Hamiltonian or relate different vacuum descriptions. Infinite-mode unitary implementation has further conditions, as in bosonic mode mixing implementability.

Exclusion constrains complete one-particle states, not a collection of little spatial boxes. In particular, two opposite-spin electrons in the same spatial orbital haveThe spatial factor is symmetric; the spin-one-half singlet state supplies the required antisymmetry. At ${\mathbf{r}}_1={\mathbf{r}}_2$, this wavefunction need not vanish for opposite spin labels. For equal spin labels it does vanish. In a continuous position description, an exact coincident point has zero measure anyway; the useful comparison concerns joint probability densities or detection in finite regions. Equal positions do not imply equal complete quantum states.

The fermionic occupation numbers $n_{\alpha }=0,1$ refer to a chosen orthonormal basis of one-particle states. A quantum superposition need not have definite occupation numbers in that basis, and an expectation value $⟨n_{\alpha }⟩$ can be fractional. For example, one electron in $(u+v)/\sqrt{2}$ has mean occupation number $1/2$ in each of two orthogonal spin orbitals, without splitting into two particles. The canonical anticommutation relations give $(a_{\alpha }^{†}{)}^2=0$ and make $a_{\alpha }^{†}{a}_{\alpha }$ a projector, so its expectation value stays between zero and one in every normalized quantum state.

For constant internal matrices acting on Dirac fields, the equal-time canonical anticommutation relations givewith the corresponding spatial delta functions for fields. The quartic terms cancel after fermion reordering. Integrating proves the displayed algebra for well-defined global Noether charges, with normal ordering and a regulator that preserves the nonanomalous internal symmetry. Orthogonal chiral projectors make the left/right mixed commutators zero. This reduces current-charge closure to a finite matrix calculation.

For $d$ fermionic creation operators with the canonical anticommutation relations, the Hilbert space $L^2({\mathbb{R}}^d)\otimes {\Lambda }^{\bullet }{\mathbb{C}}^d$ splits into ${\mathcal{H}}_n=L^2({\mathbb{R}}^d)\otimes {\Lambda }^n{\mathbb{C}}^d$, $0\le n\le d$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$; its internal multiplicity is $\left(\genfrac{}{}{0ex}{}{d}{n}\right)$. An odd supercharge $Q=q+q^{†}$, with $q$ raising $n$ and $q^2=0$, gives $H=\{q,q^{†}\}$ and preserves these sectors.

On an eigenspace with $E>0$, $Q/\sqrt{E}$ is an odd involution pairing bosonic and fermionic states. More precisely,gives an orthogonal decomposition of each state into the images of $q$ and $q^{†}$: their inner product vanishes because $(q^{†}{)}^2=0$. If $u\in \mathrm{im}{q}^{†}$, then $q^{†}u=0$, $q^{†}qu=Eu$ and $\Vert qu{\Vert }^2=E\Vert u{\Vert }^2$. Thus $q/\sqrt{E}$ pairs such a state in degree $n$ with one in degree $n+1$, and $q^{†}/\sqrt{E}$ is the inverse. States with zero energy are killed by both operators and may remain unpaired. The pairing applies to positive spectral subspaces as well; discreteness is not implied.

For a real smooth function $\chi$ on Euclidean space, put $A_a={\partial }_a-{\partial }_a\chi$ and $q={\sum }_a{\psi }_a^{†}{A}_a$. The adjoint operator for the usual $L^2$ inner product is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. Since $[A_a,A_b]=0$, the canonical anticommutation relations give $q^2=0$. Expanding $H=\{q,q^{†}\}$ and using $[A_a,A_b^{†}]=-2{\partial }_a{\partial }_b\chi$ givesIn the empty fermionic Fock space sector the last term vanishes; in the filled sector ${\psi }_a^{†}{\psi }_b={\delta }_{ab}$. Consequently the two scalar Hamiltonian operators are $H_0=-\Delta +|\nabla \chi {|}^2+\Delta \chi$ and $H_d=-\Delta +|\nabla \chi {|}^2-\Delta \chi$. On the compatible self-adjoint operator domain the Hamiltonian operator is nonnegative because $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2$.

For integer modes filled at indices below zero, set $E_{rs}=c_r^{†}{c}_s-{\delta }_{rs}{1}_{r<0}$ and $J_n={\sum }_r{E}_{r,r+n}$ on the finite-energy domain. The canonical anticommutation relations give a central term ${\delta }_{st}{\delta }_{ur}(1_{r<0}-1_{s<0})$ in $[E_{rs},E_{tu}]$. Summing across the filled-empty boundary gives $[J_n,J_m]=n{\delta }_{n+m,0}I$, with $J_n^{\ast }=J_{-n}$. The current's level is the coefficient of this central term.

For a paired non-self-conjugate momentum, rotate the Ising-chain Nambu spinor by an angle with $\cos \theta =(g-\cos k)/r_k$, $\sin \theta =-\sin k/r_k$ and $r_k=\sqrt{1+g^2-2g\cos k}$. Opposite angles at $k,-k$ preserve the canonical anticommutation relations. The result is $H={\sum }_k{ϵ}_k({\gamma }_k^{†}{\gamma }_k-1/2)$. Self-paired momenta have no pairing and are handled directly as number levels. Periodic spin-chain sectors additionally constrain fermion parity and allowed mode quantization.

The Jordan–Wigner transformation represents a one-dimensional spin one-half chain by fermions with a parity string preceding each site. With $P_n={\prod }_{m<n}(1-2n_m)$, one has $S_n^{+}=P_n{c}_n$ and $S_n^z=1/2-n_n$. The string converts the off-site canonical anticommutation relations into commuting spin operators. Adjacent exchange terms become quadratic fermion expressions, while a periodic end bond depends on global fermion parity.

Let $S=\bar{q}q$, $P_a=i\bar{q}{\gamma }_5{\tau }_{a}q$ and $Q_{5,a}=\int q^{†}{\gamma }_5{\tau }_{a}q/2d^{3}x$. The canonical anticommutation relations imply $[Q_{5,a},q]=-{\gamma }_5{\tau }_{a}q/2$ and $[Q_{5,a},\bar{q}]=-\bar{q}{\gamma }_5{\tau }_a/2$. The Leibniz rule for the commutator, together with $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}$ and ${\gamma }_5^2=1$, gives the displayed transformations. These are equal-time operator relations; common nonzero quark masses remove conservation of the charges without changing their canonical action on the fields.

An operator algebra is an algebra of linear operators, with multiplication given by composition. In quantum examples it can be generated by creation and annihilation operators subject to the canonical commutation relations or canonical anticommutation relations. Its analytic realization requires specifying a common domain for unbounded operators.

Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

A relativistic quantum field describes both creation and annihilation of excitations; a one-particle wavefunction alone is insufficient when interactions can change particle number. For a free complex scalar field, one conventional mode expansion of a free field isThe apparently negative-frequency term creates a positive-energy antiparticle; it is not a physical particle of unbounded negative energy. The normal-ordered Hamiltonian and conserved global phase charge have the formsParticles and antiparticles have equal mass and opposite additive internal charges. Including both frequency sectors is also essential to the local field commutator: the positive- and negative-frequency terms combine so that spacelike-separated scalar observables commute, respecting microcausality.

A Dirac field similarly contains particle annihilation operators multiplying $u$ spinors and antiparticle creation operators multiplying $v$ spinors. Its canonical anticommutation relations implement Fermi-Dirac statistics, whereas scalar modes use commutators and Bose-Einstein statistics. These assignments are tied to the Spin-statistics theorem under the usual relativistic locality and positivity assumptions. Conserved charge constrains particle creation and annihilation, but particle and antiparticle numbers need not be separately conserved in an interacting theory.

An antiparticle need not be a distinct species. A neutral real scalar field is self-conjugate; the photon is also its own antiparticle. A Majorana fermion obeys a self-conjugacy condition and cannot carry an unbroken additive charge that distinguishes it from its antiparticle. Electric neutrality alone does not force self-conjugacy when other conserved quantum numbers distinguish the states.

Let $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.

If every conserved axial charge left the vacuum invariant, the vacuum expectation of each such commutator would vanish. A nonzero quark chiral condensate instead makes $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨S⟩\ne 0$ for $a=b$. It therefore diagnoses spontaneous breaking of the nonsinglet chiral symmetry. In the massless two-flavor limit, the observed hadronic pattern is described by $SU(2{)}_L\times SU(2{)}_R\to SU(2{)}_V$, with three Goldstone boson modes identified with the pions. The vacuum condensate is a dynamical property of QCD, not something derivable from the canonical algebra alone. The light pion multiplet and absence of parity-degenerate light hadrons support this pattern; the singlet axial anomaly prevents interpreting an additional singlet mode as a fourth Goldstone boson.

For equal nonzero masses, add ${\mathcal{L}}_m=-m\bar{q}q$. It is invariant under local color gauge transformations and common vector flavor rotations, but not under independent left/right rotations. The exact continuous flavor symmetry becomes $SU(2{)}_V\times U(1{)}_V$. The kinetic term and canonical anticommutators are unchanged, and the two proved equal-time commutators still hold. The charges, however, are no longer conserved:The first equation follows from the massive Dirac equations, or the last from $H_m=m\int S$ and $\dot{Q}=i[H,Q]$. This is the flavor-nonsinglet axial Ward identity with equal quark masses. The pions become pseudo-Goldstone bosons with $m_{\pi }^2=O(m)$, rather than exact massless modes. To leading order, in the convention $S=\bar{u}u+\bar{d}d$, $f_{\pi }^2{m}_{\pi }^2=-m⟨S⟩+O(m^2)$. The condensate persists for small masses, but the symmetry is now explicitly broken, so the exact-symmetry Goldstone argument is replaced by an approximate one. The singlet anomaly and gauge symmetry are unaffected.

Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

For the specified $1/\sqrt{2E_{\mathbf{p}}}$ mode normalization, the canonical anticommutation relations areand every other anticommutator of the $a,b$ operators vanishes. For the Dirac field, they imply the Equal-time canonical anticommutator of a Dirac fieldTo check the normalization, the two sectors contribute $u(\mathbf{p})u^{†}(\mathbf{p})$ and $v(\mathbf{p})v^{†}(\mathbf{p})$ with opposite spatial Fourier signs. Reverse $\mathbf{p}$ in the second contribution. The completeness relations then givewhere $\tilde{p}=(E_{\mathbf{p}},-\mathbf{p})$. This cancels the mode denominator and leaves the Fourier representation of the delta function. Same-momentum $u^{†}v$ orthogonality must not be used in place of this reversed-momentum identity.

For arbitrary spacetime points,For spacelike $z$, choose a Lorentz frame with $z^0=0$; reversing spatial momentum shows $\Delta (z)=0$. Its derivatives vanish there as well, so the field anticommutators vanish at spacelike separation. This is fermionic microcausality. The anticommutation rules make the occupation of each normalized mode zero or one, implementing the Pauli exclusion principle, and make the normal-ordered free Hamiltonian positive for both particle sectors.

Use the complex inner product convention linear in its first argument. Give the exterior algebra its induced inner product, with an orthonormal wedge basis. Exterior multiplication is $e(v)\eta =v\wedge \eta$. Its adjoint is contraction:Applying this formula to $a\wedge \eta$ separates its first term from the remaining terms, giving $e(b{)}^{\ast }e(a)\eta =(a,b)\eta -e(a)e(b{)}^{\ast }\eta$. Hence the canonical anticommutation relations areFor irreducibility, choose an orthonormal basis $u_1,\dots ,u_d$. The occupancy projections are $N_j=e(u_j)e(u_j{)}^{\ast }$, and $P_0={\prod }_j(I-N_j)$ projects onto the vacuum $1\in {\Lambda }^{0}V$. For an increasing index set $J$, put $E_J=e(u_{j_1})\cdots e(u_{j_r})$. Then $E_I{P}_0{E}_J^{\ast }$ maps the wedge basis vector $u_J$ to $u_I$ and kills every other wedge basis vector. Thus these products are all matrix units and the generated star-algebra is $\mathrm{End}(\Lambda V)$. Any common invariant subspace for the creation and annihilation operators is therefore either zero or all of $W$.

The loop group here is $LU(1)=C^{\infty }(S^1,U(1))$, with pointwise multiplication. Although it is abelian, its nontrivial positive-energy quantum representations are projective: implementers can commute only up to a phase. A positive energy representation of the loop group of U(1) has rotations $R_t$ whose self-adjoint generator is bounded below and which implement $g(z)\mapsto g(e^{it}z)$ in the projective unitary group. The fermionic construction gives both that phase and an explicit bosonic description.

Choose the one-particle Hilbert space $H=L^2(S^1)$ with modes $e_r(z)=z^r$, $r\in \mathbb{Z}$, and projection $P$ onto $r\ge 0$. Use the polarized fermionic Fock spacewhere the exterior powers are Hilbert-completed and the tensor product is graded. Its vacuum fills all modes $r<0$ and leaves $r\ge 0$ empty; excitations are particles in the empty modes and holes in the filled modes. The fermionic annihilation operators $c_r$ and fermionic creation operators $c_r^{\ast }$ satisfy the canonical anticommutation relationsFinite particle-hole configurations form an orthonormal basis. As in the finite-dimensional exterior algebra construction, their creation and annihilation operators act irreducibly: an operator in the commutant sends the vacuum to a scalar vacuum, since the joint kernel of its vacuum annihilators is one-dimensional, and then acts by that scalar on the dense configuration basis.

Segal's quantization criterion for fermions says that a unitary one-particle operator $u$ has a unitary implementer $\Gamma (u)$ on this Fock space, satisfying $\Gamma (u)c^{\ast }(f)\Gamma (u{)}^{\ast }=c^{\ast }(uf)$, exactly whenEquivalently $P-uP{u}^{\ast }$ is Hilbert-Schmidt. Implementers are unique up to phase, hence form a projective unitary representation of the restricted unitary group. This is the fermionic, polarized criterion; it is not a condition that the entire operator $u-I$ be Hilbert-Schmidt. The criterion and the corresponding continuity topology are stated in Theorem III.2.1.1 of the fermionic construction of loop-group representations. For the compact principal-angle part of a pair of polarizations, the threshold has a concrete interpretation. Its two-dimensional blocks are rotations through angles ${\theta }_j$; finitely many unmatched modes can be handled separately. Each angle ${\theta }_j$ rotates the vacuum into $\cos {\theta }_j$ times its original component plus $\sin {\theta }_j$ times a particle-hole pair. After discarding finitely many exceptional blocks, the successive tensor products are Cauchy precisely when ${\prod }_j\cos {\theta }_j$ converges to a nonzero value, equivalently ${\sum }_j{\sin }^2{\theta }_j<\infty$. In this block model the sum is exactly half the squared Hilbert-Schmidt norm of the projection difference; finitely many unmatched modes account for a change of charge. Infinitely many unmatched modes already violate the criterion. This explains why infinitely many arbitrarily small mixings can fail to be implementable.

For $g(z)={\sum }_n{g}_n{z}^n\in LU(1)$, multiplication $M_g$ is a unitary one-particle operator. Its matrix entry from $e_s$ to $e_r$ is $g_{r-s}$. ThereforeFor a fixed difference $n$, exactly $|n|$ pairs cross the polarization boundary. Smoothness makes the sum finite, so Segal's quantization criterion for fermions constructs a continuous projective unitary representation of every smooth loop. Smooth convergence gives operator-norm convergence of $M_g$ and Hilbert-Schmidt convergence of its commutator, which is sufficient for the strong projective continuity of the implementers.

Normal ordering fixes the charge and rotation energy. On the dense finite-configuration domain putThe sums are finite on each configuration. A particle at $r\ge 0$ adds charge one and energy $r$; a hole at $r<0$ adds charge minus one and energy $-r$. Consequently $L_0$ is a nonnegative self-adjoint diagonal operator with integer spectrum and finite-dimensional eigenspaces. The unitaries $R_t=e^{it{L}_0}$ are strongly continuous, have period $2\pi$, and implement $e_r\mapsto e^{itr}{e}_r$. Conjugating multiplication by this one-particle rotation gives $M_{g(e^{it}\cdot )}$, so their Fock implementers rotate the loop representation projectively. This is an actual positive-energy circle action.

The scalar anomaly is visible in the Heisenberg current algebra of a fermion on the circle. Define on finite-energy vectorsFor fixed $n$ the sum has only finitely many nonzero actions on any configuration. Direct use of the canonical anticommutation relations givesOn summing, the two noncentral sums cancel. If $m=-n$ the central sum is ${\sum }_r(1_{r<0}-1_{r+n<0})=n$: for $n>0$ the terms are precisely $r=-n,\dots ,-1$. ThusIn particular $J_n\Omega =0$ for $n>0$. The coefficient one in the central term is the level of this basic representation.

Let $h(\theta )={\sum }_n{h}_n{e}^{in\theta }$ be smooth and real. The normal-ordered infinitesimal implementer of multiplication by $h$ isThe sign of the current index matters: multiplication by $z^n$ raises the one-particle mode by $n$, so it corresponds to $J_{-n}$. It follows either directly from the bilinear commutator or from the fermionic currents on the circle that $[A(h),c_r^{\ast }]={\sum }_n{h}_n{c}_{r+n}^{\ast }$. For trigonometric $h$ these are oscillator fields on the finite-energy domain; their self-adjoint closures exponentiate to the implementers of $e^{ih}$. For general smooth $h$, their extensions are obtained by the oscillator-field construction below: ${\sum }_{n>0}n|h_n{|}^2<\infty$ guarantees the square-summable field coefficients, and Fourier truncation converges in the corresponding Weyl relations implementers. Thus no unjustified infinite operator product is required.