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In a bosonic Nambu spinor expression, the second diagonal entry commonly contains $a_{-k}{a}_{-k}^{†}$. The canonical commutation relations add one constant per mode when this is normal ordered. For an antiferromagnetic chain coefficient $JS$, rewriting the quadratic Hamiltonian as a full Nambu sum therefore changes the exterior constant from $-NJ{S}^2$ to $-NJS(S+1)$. Omitting the shift corrupts the ground-state energy even when the dispersion is unchanged.

A two-component Nambu spinor $({\psi }_{k\uparrow },{\bar{\psi }}_{-k\downarrow })$ for every k accounts for both spin species once after relabelling. Its Grassmann integral contributes one functional determinant. A fully doubled spin-particle-hole basis instead has redundancy and requires one half in the trace logarithm or a corresponding Pfaffian convention. A reliable check is the zero-gap limit: the grand potential must become $-2{\beta }^{-1}{\sum }_k\ln (1+e^{-\beta {\xi }_k})$, with exactly two spin states and no invented vacuum offset.

If the traceless matrix $h_k$ has eigenvalues $\pm d_k$ and the Nambu spinor includes an opposite-momentum creation operator, a full-zone sum contains both partners. Its physical quasiparticle energy is $2J{d}_k$, not $J{d}_k$. The negative term becomes a vacuum constant only after normal ordering. In the Ising-chain Nambu spinor, $d_k=\sqrt{1+g^2-2g\cos k}$. A half-zone calculation instead groups one complete two-mode block; changing the momentum range without changing the normalization is inconsistent.

A Nambu spinor groups an annihilation operator with an opposite-momentum creation operator to express pairing in a quadratic Hamiltonian as a matrix problem. Its components are not independent: the spinor at $-k$ is related to conjugated components at $k$. This redundancy must be accounted for when summing energies and normal ordering constants. Component phases are a convention and change the displayed off-diagonal matrix entries.

For a spinless Bogoliubov transformation $c_p=\cos {\theta }_p{\alpha }_p+i\sin {\theta }_p{\alpha }_{-p}^{†}$, consistency of the doubled Nambu spinor requires $\cos {\theta }_{-p}=\cos {\theta }_p$ and $\sin {\theta }_{-p}=-\sin {\theta }_p$. Thus choose ${\theta }_{-p}=-{\theta }_p$ modulo $2\pi$. The cross anticommutator from the first-row formulas is $\{c_p,c_{-p}\}=i\sin ({\theta }_p+{\theta }_{-p})$; a nontrivial even angle would violate the canonical anticommutation relations. A self-paired zero mode must instead have vanishing mixing in this convention. Real-angle unitarity within each two-by-two block alone does not enforce consistency between duplicated momentum labels.

Group the nonzero modes into independent pairs $p>0$. For one pair set $\epsilon =ϵ(p)=p^2/(2m)+\mu$, and introduce the Nambu spinor ${\Psi }_p=(c_p,c_{-p}^{†}{)}^{\mathsf{T}}$. Normal ordering using $c_{-p}{c}_{-p}^{†}=1-c_{-p}^{†}{c}_{-p}$ givesThis exactly contains both kinetic terms and the pairing terms once; there is no additional factor of one half because only $p>0$ is summed.

Insert ${\Psi }_p=U_p{\mathcal{A}}_p$, where ${\mathcal{A}}_p=({\alpha }_p,{\alpha }_{-p}^{†}{)}^{\mathsf{T}}$ and $U_p$ is the Bogoliubov transformation matrix. Direct matrix multiplication yieldsChoose the diagonalizing branchThen the off-diagonal terms vanish and the upper diagonal entry is $E_p>0$. Extend the angle to negative momenta as in part b; the sine relation just written applies to the positive representative of each pair. The BCS coherence factors can be chosen asHere $u_p$ denotes $\cos {\theta }_p$, and $v_p$ denotes $\sin {\theta }_p$; these expressions are nonsingular because $\mu >0$ and $m>0$. The minus sign of $v_p$ for positive $\Delta$ is fixed by the pairing phase in this Hamiltonian.

A second use of the canonical anticommutation relations givesThe zero-momentum term was never paired and is simply $\mu {\alpha }_0^{†}{\alpha }_0$. HencewithThe constant is the ground-state energy. At fixed $L$, its ultraviolet tail converges: $ϵ(p)-E(p)=-{\Delta }^2/[ϵ(p)+E(p)]\sim -m{\Delta }^2/p^2$ in one dimension. Replacing $E(0)$ by $\sqrt{{\mu }^2+{\Delta }^2}$ would incorrectly pair a spinless mode with itself.

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere is one fermionic Bogoliubov quasiparticle of energy $E$ per occupied transformed orbital. The two Nambu spinor matrix eigenvalues $\pm E$ should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.

Use the reduced attractive reduced BCS pairing Hamiltonian with volume V, ${\xi }_k={\epsilon }_k-\mu$ and attraction $\lambda >0$:A specified pairing shell or an ultraviolet renormalization is part of the model. The trace is grand canonical. Insert fermionic coherent states into its time-sliced trace. The resulting Grassmann variables are antiperiodic, and with inverse-energy imaginary time $0\le \tau \le \beta$ the dimensionless action isThis boundary condition is essential for Fermi-Dirac statistics; replacing it by a bosonic periodic trace changes the answer. A complex Hubbard–Stratonovich transformation uses the local identitywith time-slice factors understood. The pairing field renders the fermion action quadratic. For a uniform static field the independent Nambu spinor block $({\psi }_{k\uparrow },{\bar{\psi }}_{-k\downarrow }{)}^T$ hasIts determinant is $-({\omega }_n^2+E_k^2)$. The phase-independent determinant sign cancels under consistent reference normalization. Integrating the fermions gives the BCS pairing-field effective action, ${\mathcal{S}}_{\mathrm{eff}}={\int }_0^{\beta }V|\Delta {|}^2/\lambda d\tau -\mathrm{Tr}\ln {\mathcal{G}}^{-1}$, with the normal-ordering/reference term retained. The fermionic pairing-block determinant normalization uses one independent pairing block for every k; a fully doubled spin-particle-hole trace instead needs one half. The subtracted grand potential of the reduced BCS model is particularly unambiguous:Performing the Matsubara frequency sum yieldsup to a common constant. At zero gap this reduces exactly to the two-spin free-gas result. The Matsubara derivation of the BCS saddle follows by differentiating with respect to ${\Delta }^{\ast }$ and using ${\beta }^{-1}{\sum }_n({\omega }_n^2+E^2{)}^{-1}=\tanh (\beta E/2)/(2E)$ givesThe second equation fixes the chemical potential at specified number. This stationary action is BCS mean-field theory, not an exact integration of pairing-field fluctuations. With constant single-spin density of states ${\nu }_F$ per volume and a symmetric energy cutoff $\Lambda$, weak coupling gives $\Delta (0)\simeq 2\Lambda e^{-1/(\lambda {\nu }_F)}$ and $k_B{T}_c\simeq (2e^{{\gamma }_E}/\pi )\Lambda e^{-1/(\lambda {\nu }_F)}$. The zero-temperature condensation grand-potential density is $-{\nu }_F{\Delta }^2/2$ to leading weak-coupling order; using a spin-summed density would change these displayed factors. The BCS coherence factors are $u_k^2=(1+{\xi }_k/E_k)/2$ and $v_k^2=(1-{\xi }_k/E_k)/2$, and the associated BCS ground state is the normalized product of empty and occupied pair states. ^[ref]

Allowing a slowly varying complex pairing field generates phase stiffness and, near the transition, the Ginzburg-Landau theory of superconductivity. A neutral paired system has a gapless phase sound mode; in a charged bulk superconductor electromagnetic coupling produces the Anderson-Higgs mechanism in a superconductor and the Meissner effect. These collective fluctuations are distinct from the gapped fermionic excitations. A microscopic dynamical calculation is needed to fix collective-mode dispersion and damping; the static saddle alone does not determine them. ^[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 occupations small compared with $S$, use $S^z=S-n$, $S^{-}\simeq \sqrt{2S}{a}^{†}$ and $S^{+}\simeq \sqrt{2S}a$. The longitudinal bond contributes $-J\Delta S^2+JS\Delta (n_m+n_{m+1})+O(S^0)$; the transverse pair terms contribute $JS(a_m{a}_{m+1}+a_m^{†}{a}_{m+1}^{†})+O(S^0)$. Thus the direct real-space expansion isTake $a_m=N^{-1/2}{\sum }_k{e}^{ikm}{a}_k$, with $k=2\pi j/N$ over one full Brillouin zone and the lattice spacing set to one in $k$. Symmetrizing $k$ and $-k$ changes the phase in the pair sum to $\cos k$, givingnormal ordering the second Nambu spinor component then yieldsThe printed matrix has a factor-of-two error on its diagonal. With a full-zone sum its printed $2\Delta$ would produce $4\Delta a_k^{†}{a}_k$, and even its vacuum expectation would differ from the direct expansion by $J\Delta NS$. A half-zone convention cannot repair both coefficients: one must then group each $k,-k$ pair and double its off-diagonal pair coefficient as well. The formula above follows from the specified Hamiltonian and has the expected gapless isotropic limit; subsequent physical results use this consistent expansion. Terms quartic in bosons first enter at order $S^0$.

Use the contact-interaction Bose-gas field integral for a homogeneous three-dimensional weakly interacting Bose gas with repulsive short-range interaction as the standard model. Let particle mass be $m_b$, scattering length $a_s>0$, physical coupling $g=4\pi {\hslash }^2{a}_s/m_b$ and gas parameter $n{a}_s^3\ll 1$. These are symbolic model definitions, not supplied numerical parameters. The grand-canonical coherent-state integral has periodic fields and actionThe contact term is the normal symbol of $g{\hat{\psi }}^{†2}{\hat{\psi }}^2/2$. In three dimensions a contact interaction needs a regulator and coupling matched to the measured scattering length; the displayed $g$ is the low-energy physical coupling. This field integral is useful because an ordered saddle and its quantum/thermal fluctuations follow from the same action.

Stationarity with respect to $\bar{\psi }$ gives the imaginary-time Gross–Pitaevskii equation. A uniform saddle exists at $\mu >0$ with ${\psi }_0=\sqrt{n_0}{e}^{i{\theta }_0}$ and condensate density $n_0=\mu /g$. Its arbitrary phase reflects global $U(1)$ symmetry; a selected condensate fixes a phase in the thermodynamic symmetry-breaking description. Write $\psi =\sqrt{n_0}+\eta$ after choosing that phase. Linear terms vanish, and the quadratic part isWith ${\epsilon }_k={\hslash }^2{k}^2/(2m_b)$ and the Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}-i{\omega }_n\tau }$, the paired quadratic kernel isThe one-half in the Nambu spinor action removes pair double counting. Analytic continuation of the inverse kernel gives the Bogoliubov dispersionAt high momentum it approaches the free-particle kinetic energy plus its mean-field shift. The crossover scale is the healing length $\xi =\hslash /\sqrt{2m_{b}g{n}_0}$. Gaussian integration adds a fluctuation determinant to the saddle free energy. Expressed using the physical three-dimensional coupling, its zero-temperature integrand can be regularized as $\frac{1}{2}[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$; the subtraction follows from renormalization of a three-dimensional dilute-gas contact coupling, not discarding an infinite answer arbitrarily. The thermal contribution is $k_{B}T\int d^{3}k/(2\pi {)}^3\log (1-e^{-\beta E_k})$. A calculation beyond mean field must keep that matching and its thermodynamic derivatives consistent.

The density-phase action of a Bose gas makes the physical content clear, $\psi =\sqrt{n_0+\delta n}{e}^{i\theta }$. Up to a periodic density total derivative, the long-wavelength quadratic action containsnumber-phase conjugacy pairs the density fluctuation and phase in the quadratic density-phase action. Dropping the density-gradient term only in the long-wave limit and completing the density Gaussian square gives the Bose-gas phase-only actionThis has the same sound pole. A term $i\hslash n_0{\partial }_{\tau }\theta$ is also present before restricting to the smooth zero-winding sector; although locally a total derivative, it cannot in general be ignored in topologically nontrivial sectors. The density and phase variables describe the same coupled Bogoliubov branch, not a second independent massive particle simply because the density has a static restoring cost.^[ref]

A phase gradient carries superflow ${\mathbf{v}}_s=(\hslash /m_b)\nabla \theta$. Its energy cost defines the superfluid number density. At leading mean-field order the stiffness uses $n_0$, but condensate density and superfluid number density are different observables: for a Galilean-invariant continuum gas at zero temperature the latter is the full density even when interactions deplete the condensate. Bogoliubov coefficients give $v_k^2=[({\epsilon }_k+g{n}_0)/E_k-1]/2$ and a small zero-temperature depletion of relative order $\sqrt{n{a}_s^3}$ in three dimensions. Large depletion or strong long-wavelength phase fluctuations would invalidate the small-fluctuation saddle approximation.

The Landau criterion for creating excitations is $v_L={\inf }_{k>0}{E}_k/(\hslash k)=c_s$ for this homogeneous dispersion. Thus repulsion produces a linear sound mode and a positive critical velocity from the Landau criterion, whereas an ideal quadratic gas has zero value of this criterion. Vortices, obstacles and nucleation can reduce an actual flow threshold below $c_s$; the criterion alone is not a complete theory of metastable flow. At finite temperature thermal quasiparticles reduce the superfluid number density. Dimensionality and finite size also matter: the uniform three-dimensional weak-depletion condensate expansion must not be transferred unchanged to a one- or two-dimensional thermodynamic system. The field integral provides systematic corrections only within its dilute, stable and sufficiently weak-fluctuation regime.

Group the nonzero modes into independent pairs $p>0$. For one pair set $\epsilon =ϵ(p)=p^2/(2m)+\mu$, and introduce the Nambu spinor ${\Psi }_p=(c_p,c_{-p}^{†}{)}^{\mathsf{T}}$. Normal ordering using $c_{-p}{c}_{-p}^{†}=1-c_{-p}^{†}{c}_{-p}$ givesThis exactly contains both kinetic terms and the pairing terms once; there is no additional factor of one half because only $p>0$ is summed.

Insert ${\Psi }_p=U_p{\mathcal{A}}_p$, where ${\mathcal{A}}_p=({\alpha }_p,{\alpha }_{-p}^{†}{)}^{\mathsf{T}}$ and $U_p$ is the Bogoliubov transformation matrix. Direct matrix multiplication yieldsChoose the diagonalizing branchThen the off-diagonal terms vanish and the upper diagonal entry is $E_p>0$. Extend the angle to negative momenta as in part b; the sine relation just written applies to the positive representative of each pair. The BCS coherence factors can be chosen asHere $u_p$ denotes $\cos {\theta }_p$, and $v_p$ denotes $\sin {\theta }_p$; these expressions are nonsingular because $\mu >0$ and $m>0$. The minus sign of $v_p$ for positive $\Delta$ is fixed by the pairing phase in this Hamiltonian.

A second use of the canonical anticommutation relations givesThe zero-momentum term was never paired and is simply $\mu {\alpha }_0^{†}{\alpha }_0$. HencewithThe constant is the ground-state energy. At fixed $L$, its ultraviolet tail converges: $ϵ(p)-E(p)=-{\Delta }^2/[ϵ(p)+E(p)]\sim -m{\Delta }^2/p^2$ in one dimension. Replacing $E(0)$ by $\sqrt{{\mu }^2+{\Delta }^2}$ would incorrectly pair a spinless mode with itself.