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The free kinetic terms of left- and right-handed Weyl fields are separately invariant under unitary group rotations in flavour space. For $N$ massless Dirac fields, the subgroup preserving common fermion number is $U(N{)}_L\times U(N{)}_R$. A common nonzero Dirac mass term couples the two chirality sectors and leaves its diagonal $U(N)$ subgroup. Gauge interactions can obstruct some of the classical axial currents through a chiral anomaly.

Without fixed gauge charges or interactions distinguishing particles from antiparticles, $N$ free Dirac fields give $2N$ left-handed Weyl fields after charge conjugation of the right-handed fields. Their massless kinetic term has unitary group $U(2N)$. Equal nonzero Dirac masses have equal positive singular values and can be written as $2N$ equal Majorana mass terms; their preserving group is the orthogonal group $O(2N)$. Explicitly, in a basis with symmetric mass matrix $m1$, invariance requires $V^{†}V=V^{T}V=1$, hence real $V$. Restricting to transformations preserving a chosen common fermion number gives the usual $U(N{)}_L\times U(N{)}_R$ at zero mass and $U(N)$ at nonzero mass.

For a Dirac field with a global phase symmetry, fermion number is the vacuum-subtracted charge of the Dirac current. In a mode expansion it counts particles minus antiparticles, with the overall sign fixed by the chosen generator convention. It is distinct from the total number of fermionic excitations and from fermion parity, which can survive even when interactions violate the continuous charge.

For a normalized spinless paired ground state $u|0⟩+iv{c}_p^{†}{c}_{-p}^{†}|0⟩$, its mean fermion number is $2v^2$. The canonical Bogoliubov quasiparticle creation operator produces $c_p^{†}|0⟩$, containing exactly one fermion. With BCS coherence factors $v^2=(1-ϵ/E)/2$, the change in mean fermion number is $1-2v^2=ϵ/E$. This need not be an integer because the pairing Hamiltonian preserves fermion parity while failing to conserve the original fermion number operator. Every measurement of that number operator still has integer outcomes.

Collect the two Dirac fields into $\Psi =({\psi }_1,{\psi }_2{)}^T$, with a common nonzero mass $m$. ThenAmong complex-linear rotations of the Dirac fields, the internal symmetry of a classical field theory is $\Psi \mapsto U\Psi$, $U\in U(2)$, with $\bar{\Psi }\mapsto \bar{\Psi }{U}^{†}$. Unitarity preserves the kinetic term, and the mass matrix commutes with every $U$. This unitary flavour symmetry of degenerate Dirac fields has group $(SU(2)\times U(1))/{\mathbb{Z}}_2$, where the quotient avoids counting the common central element twice.

Choose the four Hermitian matrices $T^0=1_2/2$, $T^a={\sigma }^a/2$ for $a=1,2,3$, with ${\sigma }^a$ the Pauli matrices. Promoting the four parameters to functions gives $\delta S=-\int d^{4}x({\partial }_{\mu }{\alpha }^A)\bar{\Psi }{\gamma }^{\mu }{T}^A\Psi$. The four Noether currents are thereforeTheir divergences vanish by the equal-mass Dirac equations. The singlet is half the total Dirac current; the other three currents rotate the two species.

Free-theory qualification. If “global internal symmetry” allows mixing the Dirac fields with charge-conjugate fields, the completely free theory written here has a larger enhanced internal symmetry of free Dirac fields. Write ${\psi }_i=({\lambda }_{2i-1}+i{\lambda }_{2i})/\sqrt{2}$ using four Majorana fermions. Up to a total derivative,The full orthogonal group $O(4)$ acts on the four real species; its connected part is $SO(4)$. Its six Noether currents can be chosen as $J_{AB}^{\mu }=i{\bar{\lambda }}_A{\gamma }^{\mu }{\lambda }_B$ for $A<B$, up to the common sign convention. The antisymmetry of this Majorana fermion vector bilinear makes these the six independent rotations. A concrete rotation outside the stated complex-linear $U(2)$ is ${\lambda }_2\mapsto \cos \theta {\lambda }_2+\sin \theta {\lambda }_3$, ${\lambda }_3\mapsto -\sin \theta {\lambda }_2+\cos \theta {\lambda }_3$: it preserves the displayed Lagrangian density while mixing ${\psi }_i$ with their charge-conjugate fields. The usual $U(2)$ answer specifies the subgroup preserving the chosen common fermion number.

In the Weyl representation of the gamma matrices, write each Dirac field as ${\psi }_i=({\ell }_{i\alpha },r_i^{\dot{\alpha }}{)}^T$, where ${\ell }_i$ and $r_i$ have opposite chirality. With the Weyl sigma matrices ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(1,-\mathbf{\sigma })$,The Dirac mass term couples the two Weyl fields. Under $\ell \mapsto U_L\ell$, $r\mapsto U_{R}r$, its first term acquires $U_L^{†}{U}_R$, so a common nonzero mass requires $U_L=U_R$. At zero mass there is no such constraint:This is the chiral flavour symmetry of massless Dirac fields. Its eight Noether currents are $j_{L,A}^{\mu }={\ell }^{†}{\bar{\sigma }}^{\mu }{T}^A\ell$ and $j_{R,A}^{\mu }=r^{†}{\sigma }^{\mu }{T}^{A}r$. The sum $j_R+j_L$ is the vector current, and $j_R-j_L$ is the axial current, for each flavour generator. This statement concerns the free theory in flat spacetime; an interacting theory can have a chiral anomaly.

For completeness, allowing the extra transformations described in part e gives four left-handed Weyl fields, $q=({\ell }_1,{\ell }_2,r_1^c,r_2^c{)}^T$, where $r_i^c$ is the left-handed charge-conjugate field of $r_i$. After integration by parts their kinetic term is $i{q}^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }q$, whose maximal linear unitary group is $U(4)$. The symmetric mass matrix for equal Dirac masses iswith the antisymmetric spinor contraction implicit. A unitary change of basis makes $M=m{1}_4$; preserving both the kinetic term and $M$ requires $V^{†}V=V^{T}V=1_4$, hence real $V\in O(4)$. Thus the unrestricted free-theory enhancement is $U(4)$ at zero mass and $O(4)$ at equal nonzero mass; the familiar $U(2{)}_L\times U(2{)}_R$ and diagonal $U(2)$ are their common-fermion number-preserving subgroups.

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.

At the parton level, the dominant valence reactions are $e^{-}u\to {\nu }_{e}d$ through W boson exchange and $e^{+}d\to {\bar{\nu }}_{e}u$ through W boson exchange. The first also acts on $c$, while the second acts on $s$ with the appropriate Cabibbo-Kobayashi-Maskawa matrix factors; crossing supplies the antiquark reactions. The lines in the figure show fermion number flow; the Positron and antineutrino arrows point opposite to their momenta.

For unpolarized beams, the leading charged current cross sections, suppressing Cabibbo-Kobayashi-Maskawa matrix factors, have the formHere each parton distribution function is evaluated at $(x,Q^2)$. The proton has two valence up quarks and one valence down quark; moreover, its valence contribution in the Positron reaction carries the chiral angular factor $(1-y{)}^2$. Both effects favour $e^{-}p$. Increasing $Q^2$ at fixed collider energy enforces $x\ge Q^2/s$, progressively suppressing the sea relative to the valence quarks; large $y$ further suppresses the Positron valence term. There is no universal factor of two valid at every $(x,y)$.

The escaping neutrino prevents direct reconstruction from a scattered charged lepton. Instead use the Jacquet-Blondel reconstruction. With the $+z$ axis along the proton beam, measure the complete hadronic final state and formThese relations follow from four-momentum conservation, with the unobserved neutrino transverse momentum opposite to the measured hadronic sum. A nearly hermetic hadronic calorimeter, supplemented by tracking detectors, must measure jets, their energies and their directions. Forward losses, initial-state radiation and small $y_h$ cause significant reconstruction uncertainties.

Use incoming fermion $p_1$ and antiparticle $p_2$, outgoing fermion $p_3$ and antiparticle $p_4$. The two tree-level Feynman diagrams are the $t$ channel, with $q_t=p_1-p_3=p_4-p_2$, and the $s$ channel, with $q_s=p_1+p_2=p_3+p_4$. They occupy the lower row of the figure. In the $t$ channel the fermion flow follows $1\to 3$ and $4\to 2$; in the $s$ channel it follows $1\to 2$ at the incoming interaction vertex and $4\to 3$ at the outgoing interaction vertex.

The two contractions again have a relative fermionic sign. With the same harmless common-phase convention as i, their sum isThe auxiliary-field propagator is the same constant $-i$ in both channels. There is no $u$ channel preserving the external fermion number assignments.

Label the incoming fermion momenta $p_1,p_2$ and outgoing fermion momenta $p_3,p_4$, with $p_1+p_2=p_3+p_4$. There are two tree-level Feynman diagrams: the $t$ channel connects $1$ to $3$ and $2$ to $4$, with exchanged momentum $q_t=p_1-p_3=p_4-p_2$; the $u$ channel connects $1$ to $4$ and $2$ to $3$, with $q_u=p_1-p_4=p_3-p_2$. Both internal lines are auxiliary-field propagators. The upper row in the figure gives these labelled connections; the outgoing labels are exchanged in its second panel to avoid a visually unnecessary crossing.

Each tree-level Feynman diagram contains two factors $ig$ and one factor $-i$, giving $i{g}^2$. The exchange of the two identical outgoing fermions gives the relative fermionic sign. Up to a common external-state phase, the corresponding scattering amplitude isThere is no annihilation channel: both incoming particles carry the same fermion number, whereas a $\sigma \bar{\psi }\psi$ interaction vertex couples one fermion to one antiparticle when both are incoming.

For arbitrary linear operators, expansion of the anticommutators givesNow use the commutator product rule $[VX,W]=V[X,W]+[V,W]X$, with $W=YZ$, twice. The result isNo grading assumption is needed for these algebraic identities.

For the Dirac field, supplement the supplied equal-time canonical anticommutation relations with $\{{\psi }_a(x),{\psi }_b(y)\}=\{{\psi }_a^{†}(x),{\psi }_b^{†}(y)\}=0$. Define the fermion number density $n(\mathbf{x})={\sum }_a{\psi }_a^{†}(\mathbf{x}){\psi }_a(\mathbf{x})$. In the preceding identity take $V={\psi }_a^{†}(\mathbf{x})$, $X={\psi }_a(\mathbf{x})$, $Y={\psi }_b^{†}(\mathbf{y})$, $Z={\psi }_c(\mathbf{z})$ and sum over $a$. Only its first and last anticommutators survive, givingThis is an identity of operator-valued distributions. Point splitting, or smearing with test functions, gives a precise meaning to coincident bilinears. Normal ordering only subtracts a constant from the density and does not alter this commutator.

Combine the two equal-mass Dirac fields into $\Psi =({\psi }^1,{\psi }^2{)}^T$. A constant unitary matrix $U$ acting on this flavour index commutes with the common Dirac Hamiltonian, so $\Psi \mapsto U\Psi$ gives the unitary flavour symmetry of degenerate Dirac fields $U(2)$. Its real Lie algebra has dimension four. A convenient Hermitian generator basis is $T^0=1_2/2$, $T^a={\tau }^a/2$, $a=1,2,3$, where ${\tau }^a$ are the Pauli matrices.

For each generator the canonical anticommutation relations giveThe calculation is exactly the one in b, since $T^A$ commutes with both the Dirac gamma matrices and the common fermion mass matrix. ThusExplicitly their densities areThe two off-diagonal conserved currents are why counting only the individual fermion numbers misses two generators. The singlet generates $U(1)$; the other three generate special unitary group $SU(2)$. Unequal masses would break the off-diagonal conservation laws.

Scope of the count. Four is the count of currents of the displayed ${\bar{\psi }}^A{\gamma }^{\mu }{T}_{AB}{\psi }^B$ form. The completely free system also has an enhanced internal symmetry of free Dirac fields if mixing with charge-conjugate fields is admitted. For $m\ne 0$, write its two Dirac fields as four equal-mass Majorana spinors ${\chi }^I$. Their orthogonal group $O(4)$ rotations have six continuous generators, with conserved currents proportional to $i{\bar{\chi }}^I{\gamma }^{\mu }{\chi }^J$ for $I<J$; the four $U(2)$ currents form a subgroup. This larger count is relevant if “similar” means all local internal currents of the free system, without preserving a chosen fermion number. If $m=0$, the usual number-preserving chiral flavour symmetry of massless Dirac fields is $U(2{)}_L\times U(2{)}_R$ with eight currents; allowing charge conjugation mixing enlarges it to $U(4)$ with sixteen. These counts concern internal currents, excluding spacetime Noether currents.

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.