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For a one-component superfluid with nonzero order-parameter magnitude on a closed loop, the phase defines a map from the loop into the circle. Its winding number isThe fundamental group ${\pi }_1(S^1)=\mathbb{Z}$ means that smooth deformations preserving nonzero amplitude cannot change this integer. With superfluid velocity ${\mathbf{v}}_s=\hslash \nabla \theta /m_B$, it gives quantized circulation $\oint {\mathbf{v}}_s\cdot d\mathbf{l}=nh/m_B$. A vortex core permits the order parameter to vanish and carries the winding; a superfluid phase slip changes a loop's winding when a core crosses it. For paired neutral fermions, the relevant condensed-object mass is the pair mass.

For charged pairs, gauge-covariant phase rigidity gives fluxoid and flux quantization in a superconductor:The magnitude of the pair flux quantum is $h/(2e)$. Magnetic flux alone approaches an integer multiple of this value when the circulating-velocity term on a bulk contour is negligible. The topological condition is not a claim that an arbitrary finite loop has exactly quantized magnetic flux independent of its screening current.

There are also momentum-space invariants, distinct from real-space vortex winding. A fully gapped two-dimensional Bogoliubov--de Gennes Hamiltonian can have a Chern number obtained from its occupied quasiparticle bands' Berry curvature. In a chiral paired model it can protect edge modes while the bulk gap and relevant symmetry class are maintained. Such an invariant can change only when those assumptions fail, for example through a bulk gap closing. A nodal d-wave state requires a different analysis; it is not automatically a fully gapped Chern superconductor.

The electron-phonon interaction lets one electron polarize the ionic lattice and another interact with that delayed deformation. Phonon-mediated electron attraction acts over energy transfers below a characteristic phonon scale. Pairing additionally requires a net attractive interaction in some allowed pairing channel after screened Coulomb repulsion is included. In the conventional weak-coupling situation, retardation separates the low phonon scale from the much larger electronic scale, reducing the effective repulsion. The Cooper instability then makes a Fermi surface unstable to an arbitrarily weak net attraction in the idealized channel; finite temperature cuts off its logarithmic enhancement.

In BCS theory, a momentum-independent pairing amplitude within the cutoff obeys a self-consistency equation containing $\tanh (E/2k_{B}T)/(2E)$, with quasiparticle energy $E=\sqrt{{\xi }^2+|\Delta {|}^2}$. At a continuous transition, take $\Delta \to 0$. A constant effective density of states and coupling combine into $\lambda$, while the band bottom truncates the negative-energy side at $-\sigma$, with $\sigma =\min (\mu ,\theta )$. This gives the supplied linearized gap equation. Its integrand is even, so withit becomes $1/\lambda =[I(\theta ,T_c)+I(\sigma ,T_c)]/2$. If both cutoffs are much larger than $k_B{T}_c$, the given large-cutoff integral gives $I(a,T_c)\simeq \ln (1.13a/(k_B{T}_c))$. Consequently the asymmetric-cutoff BCS transition temperature isFor $\mu >\theta$, this reduces to $1.13\theta e^{-1/\lambda }$; for $0<\mu <\theta$, it is $1.13\sqrt{\theta \mu }{e}^{-1/\lambda }$. The second limit requires $k_B{T}_c\ll \mu$ as well as $k_B{T}_c\ll \theta$. Weak $\lambda$ at fixed nonzero $\mu /\theta$ supplies that limit, but taking $\mu \to 0$ afterward need not preserve it.

For a three-dimensional spin-degenerate parabolic electron band in the degenerate regime,DefineHolding the other model parameters fixed givesThus the log-log slope is $1/3$ below the cutoff-density scale and zero above it. For the specified parameters,This density crossover of a fixed-coupling BCS cutoff model is the requested prescribed model. A real three-dimensional parabolic band also has a changing density of states, so keeping $\lambda$ constant as density varies is an additional imposed assumption, not a consequence of that dispersion.

For noninteracting spinless bosonic pairs, mass $m_B=2m$ and density $n_B=n/2$, homogeneous ideal Bose gas condensation givesUsing the supplied integral $I_B={\int }_0^{\infty }{x}^{1/2}/(e^x-1)dx\simeq 2.3$ gives $C_B=\frac{1}{2}[2/(3I_B){]}^{2/3}\simeq 0.219$. Its slight difference from the more precise $0.218$ is consistent with the rounded integral. Write $a=T_0/(\theta /k_B)=1.13e^{-1/\lambda }\simeq 0.0403$. On the low-density branch,Since $a/C_B<1$, no part of the high-density plateau exceeds $T_B$. The formal crossing condition isThe crossing occurs near $T_c\simeq 7.4\mathrm{K}$ and $T_F\simeq 34\mathrm{K}$.

The apparent low-density excess should not be read as a reliable transition above the ideal condensation scale of preformed pairs. On this extrapolation $k_B{T}_c/\mu =a{u}^{-1/3}$ grows without bound, violating the degenerate-gas and large-lower-cutoff approximations. It reaches one already near $7.5\times 10^{21}{\mathrm{m}}^{-3}$; well before using an arbitrarily small density, the chemical potential, pairing, density of states and pair fluctuations need a self-consistent treatment. The asymptotic formula and ideal-boson curve compare two stipulated models, rather than solving the full BCS-BEC crossover.

A complex order parameter captures both the amplitude and phase rigidity of a superfluid. A local isotropic Ginzburg-Landau theory writes its free-energy functional, to fourth order in amplitude and second order in gradients, asThe uniform minimum has $\Psi =0$ for $\alpha >0$ and $|{\Psi }_0{|}^2=-\alpha /\beta$ for $\alpha <0$. Its phase remains arbitrary under a global $U(1)$ transformation. Minimizing with respect to ${\Psi }^{\ast }$ gives $-K{\nabla }^2\Psi +\alpha \Psi +\beta |\Psi {|}^2\Psi =0$. The phenomenological description specifies the allowed structures and coefficients; microscopic input is needed to identify the normalization of $\Psi$ and calculate those coefficients. A static functional by itself does not fix the dynamical equation or mode frequencies.

For a dilute three-dimensional Bose gas with short-range repulsion and particle mass $m_B$, the microscopic low-energy Hamiltonian isThe interaction parameter is expressed through the low-energy s-wave scattering length $a_s$, rather than an unrenormalized bare delta potential. In a symmetry-selected description, $\Psi =⟨\hat{\psi }⟩$ and $|\Psi {|}^2=n_0$ is condensate density. Equivalently, fixed-number off-diagonal long-range order identifies the macroscopically occupied natural orbital and its amplitude without a nonzero one-point expectation.

Evaluating $\hat{H}-\mu \hat{N}$ in a coherent-field condensate variational state gives the Ginzburg-Landau functional of a dilute Bose gas, with the mean-field microscopic identificationsIts stationary variational equation is the Gross–Pitaevskii equation. The time-dependent variational action additionally contains $i\hslash {\Psi }^{\ast }{\partial }_t\Psi$, givingThis term expresses density-phase conjugacy. It is the missing microscopic dynamical input; a purely dissipative relaxation law would describe a different approximation.

Bogoliubov theory writes $\hat{\psi }=\Psi +\delta \hat{\psi }$ and retains quadratic fluctuations. In a uniform condensate, $\mu =g_B{n}_0$ and the fluctuation matrix isIts eigenvalues are $\pm \sqrt{{ϵ}_k({ϵ}_k+2g_B{n}_0)}$. The positive branch is the gapless sound mode, continuously crossing over to particle-like excitations at larger $k$. The same result follows by linearizing the time-dependent Gross–Pitaevskii equation about its uniform solution. Thus the microscopic condensate amplitude, scattering interaction and mass determine the mean-field static coefficients and the collective mode. Quadratic quantum fluctuations also produce quantum depletion of a Bose condensate, which a pure coherent-state product omits.

These identifications are controlled for a dilute weakly interacting gas away from a fluctuation-dominated critical region. They are not a universal statement that the exact finite-temperature GL coefficient is always $-\mu$ or that its quartic coefficient never receives corrections. Integrating out noncondensed and short-wavelength modes produces effective temperature-dependent coefficients and can change their near-critical behavior. The phase rigidity described by $K$ and the thermodynamic order-parameter expansion remain the shared organizing structure.

For BCS theory, the microscopic order parameter is instead an anomalous pair amplitude, commonly normalized as an energy gap $\Delta \propto ⟨{\hat{\psi }}_{\downarrow }{\hat{\psi }}_{\uparrow }⟩$. It is not an ordinary single-electron field expectation or automatically the square root of a pair number density. Integrating out fermions and expanding in small $\Delta$ and slow gradients near $T_c$ gives the microscopic derivation of Ginzburg-Landau theory. The BCS pairing-field effective action supplies its coefficients through normal-state pair-response functions and thermal fermion sums.

For example, a clean isotropic three-dimensional weak-coupling s-wave model, with $N(0)$ the density of states per spin at the Fermi energy and $\Delta$ measured in energy units, giveswith microscopic Ginzburg-Landau coefficients from BCS theoryThe numerical factors arise from fermionic Matsubara sums and the isotropic velocity average; they depend on the stated gap and density-of-states conventions. Rescaling the order parameter rescales all three coefficients. The expansion applies near the transition with sufficiently slow spatial variation; disorder, anisotropic pairing and strong coupling alter the coefficients or require additional components. Microscopic derivation also fixes the pair charge in the covariant gradient, producing the superconducting screening current and the Meissner effect.^[ref]

The neutral Bose and charged Fermi theories therefore share a complex order parameter and a phase-stiffness functional, while deriving that field from different microscopic objects. The Bogoliubov transformation mixes particle and hole fluctuations of the bosonic condensate in one case and paired fermionic excitations in the other. The static GL limit organizes long-distance behavior; the full microscopic theories retain excitation spectra and temperature dependence beyond that local near-transition expansion.

For a neutral condensate with order parameter $\psi =\sqrt{n_s}{e}^{i\varphi }$, the superfluid velocity away from singularities is ${\mathbf{v}}_s=\hslash \nabla \varphi /m$. Single-valuedness around a closed contour gives $\oint \nabla \varphi \cdot d\mathbf{l}=2\pi n$, hence quantization of circulation:The mass is that of the condensed object: an atom for a bosonic atomic superfluid, or a pair for a paired-fermion condensate. A nonzero winding cannot be continuously removed while the order parameter remains nonzero throughout a spanning region. A vortex allows its phase to wind while its amplitude vanishes in a core.

For a charged condensate, minimal coupling gives $m{\mathbf{v}}_s+q\mathbf{A}=\hslash \nabla \varphi$. ThusUsing ${\mathbf{j}}_s=n_{s}q{\mathbf{v}}_s$ and ${\lambda }_L^2=m/({\mu }_0{n}_s{q}^2)$ gives the fluxoid and flux quantization in a superconductor relation $\Phi +{\mu }_0{\lambda }_L^2\oint {\mathbf{j}}_s\cdot d\mathbf{l}=nh/q$. The signs of oriented flux and winding depend on the signed pair charge; the magnitude of the elementary superconducting flux quantum is ${\Phi }_0=h/(2e)$. Flux alone is quantized when the current term is negligible on a contour deep in the bulk. A thin ring or a contour near a core can instead require the full fluxoid.

In a neutral superfluid, vortex flow carries kinetic energy and angular momentum. Rotation can favor vortices because equilibrium minimizes the rotating-frame quantity $E-\mathbf{\Omega }\cdot \mathbf{L}$. The core size is set by a coherence or healing scale. In a type-II superconductor, an Abrikosov vortex carries quantized magnetic flux and circulating supercurrent; its core scale is the coherence length and its magnetic screening scale is the penetration depth. Between the lower and upper critical fields, a vortex lattice can be favored. Flux pinning affects how that structure responds to current. Type-I bulk superconductors instead favor macroscopic normal/superconducting separation rather than the usual stable dilute Abrikosov-vortex regime.

Condensate depletion means particles outside a macroscopically occupied bosonic single-particle mode. Interactions can deplete that mode even at zero temperature: the Bogoliubov transformation mixes creation and annihilation operators, so its quasiparticle vacuum contains particles at nonzero momentum. A weak homogeneous three-dimensional gas with repulsive scattering length $a_s$ has leading quantum depletion of a Bose condensateThermal quasiparticles add another contribution. Depletion and normal-fluid density are distinct: a Galilean-invariant fluid can have quantum depletion while retaining its full superfluid mass response at zero temperature.

For the fermionic pairing problem, mean-field decoupling of the reduced BCS pairing Hamiltonian introduces the anomalous amplitude $\Delta$. Each paired Nambu block has diagonal entries ${\xi }_{\mathbf{k}},-{\xi }_{\mathbf{k}}$ and off-diagonal pairing entries of magnitude $\Delta$. A phase convention with negative off-diagonal entries gives eigenvalues $\pm E_{\mathbf{k}}$, where $E_{\mathbf{k}}=\sqrt{{\xi }_{\mathbf{k}}^2+{\Delta }^2}$. Its canonical Bogoliubov rotation yields the BCS coherence factorsThe partner transformation has the complementary sign required by fermionic anticommutation. At zero temperature all positive-energy BCS quasiparticle modes are empty. Expanding the supplied electron operator therefore givesSince $E_{\mathbf{k}}^2-{\xi }_{\mathbf{k}}^2={\Delta }^2$,The electron occupation is close to one far below the normal-state Fermi surface, equals one half at $\xi =0$, and decays toward zero above it. Pairing smooths the Fermi step over an energy range of order $\Delta$. The positive-energy quasiparticle occupation is instead zero everywhere at this temperature; the electron sea and the quasiparticle vacuum are different descriptions of the same state. At finite temperature, $n_{\mathbf{k}\sigma }=[1-(\xi /E)\tanh (E/(2k_{B}T))]/2$ in the same mean-field model.

For a three-dimensional parabolic normal band, the total density of states including both spins is $D(\mu )=3N/(2\mu )$. In the weak-gap limit, replace it by this constant across the narrow pairing region and use $\mu \simeq E_F$. The number of electrons above the normal-state Fermi surface isThe integral is $[\xi -\sqrt{{\xi }^2+{\Delta }^2}{]}_0^{\infty }/2=\Delta /2$. Therefore the BCS population above the normal-state Fermi surface isThe constant-density-of-states approximation and a small gap relative to the chemical potential are essential. This is the question's explicitly defined fermionic population measure; it is not the fraction of electrons outside a single bosonic condensate orbital.

For an order-of-magnitude mercury estimate, use $T_c\simeq 4.2\mathrm{K}$ and the weak-coupling scale ${\Delta }_0\simeq 1.76k_B{T}_c\simeq 0.64\mathrm{meV}$, with a representative metallic $\mu \simeq 7\mathrm{eV}$. These input approximations give $\Delta N/N\sim 7\times 10^{-5}$, of order $10^{-4}$. The use of a weak-coupling gap is an estimate, not an assertion of mercury's exact measured gap.

For a copper oxide, an illustrative gap $20\mathrm{meV}$ and effective chemical-potential scale $0.5\mathrm{eV}$ give about three per cent in this formula. Choosing $0.2$–$1\mathrm{eV}$ instead gives roughly $7.5$–$1.5$ per cent. These are explicit example inputs: the paper specifies no compound, doping or band parameters from which a unique number follows. A real cuprate's anisotropic gap, quasi-two-dimensional band and strong correlations require a band-resolved occupation integral, so the reliable contrast is the much larger gap-to-electronic-energy scale, rather than a universal cuprate percentage.

At atomic separations both massive exchange potentials are exponentially negligible. Molecular binding needs an interaction that survives at atomic distances; actual chemical bonds is governed principally by electromagnetism and quantum mechanics, with a massless photon rather than a scalar mediator. A hypothetical massless scalar would also have an unscreened $1/r$ potential, but its coupling and spin-dependent phenomenology would be different. The Higgs mechanism affects particle masses; that does not imply an appreciable direct Higgs-exchange force between atoms separated by $10^{-10}\mathrm{m}$.

A relativistic two-component spinor is an element of a two-dimensional complex representation of the special linear group $SL(2,\mathbb{C})$, the double cover of the proper orthochronous Lorentz group. A Weyl spinor transforms in either $(1/2,0)$ or $(0,1/2)$; parity interchanges the two chiralities. It describes a chiral spin-$1/2$ degree of freedom. A free massless Weyl field has one particle helicity and the opposite antiparticle helicity.

The number of components alone does not prove zero mass. A massive Dirac field has two coupled two-component chiral fields, and a nonrelativistic Pauli spinor describes the two spin states of a massive spin-$1/2$ particle. A Majorana mass term can also be written using a two-component field and its complex conjugate. Thus a single independent free massless Weyl equation is a dynamical assumption, not part of the bare algebraic definition of a two-component spinor.

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,The positive spring term in $L_2$ means the spring lowers its energy when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

Put $G=g/\ell$ and $s=\kappa /m$. Applying the Euler-Lagrange equations to the quadratic Lagrangian givesA nonzero motion satisfying the specified equal-and-opposite initial displacements can have one frequency only if $(1,-1)$ is a normal-mode vector. Its acceleration terms are proportional to $(1,0)$, while the restoring terms are proportional to $(2G-2s,2s-G)$. HenceandThe rods swing oppositely. The hinge mass moves sideways, while the final mass has no first-order displacement: the two horizontal contributions cancel and its vertical change is quadratic in angle. In the corresponding geometric ansatz it moves vertically at twice the small-oscillation frequency to leading nonzero order, but the displayed single-frequency solution is a linearized result. The trivial case $\xi =0$ imposes no parameter condition.

Removing the hinge mass gives the exact LagrangianThe stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomesConsequentlyFor unequal squared frequencies, $\nu =0$: the rods remain aligned to first order and the end mass swings like a pendulum of length $2\ell$. This degenerate Lagrangian gives the difference coordinate no quadratic kinetic term, so its equation is an algebraic constraint, not a second oscillator. If ${\omega }_0={\omega }_1$, the quadratic action loses all dependence on $\nu$. Its motion is then undetermined at this order; nonlinear terms or a small hinge mass must resolve the degeneracy. For larger spring stiffness the straight constrained branch is not a minimum of the bending potential, despite the absence of a finite-inertia growing mode in this strictly massless quadratic description.

Differentiate $\mu \dot{\mathbf{r}}=\mathbf{p}+e^{\ast }\mathbf{B}\times \mathbf{r}$ and substitute the previous momentum equation:For a planar vector, $\mathbf{B}\times \dot{\mathbf{r}}=B(-\dot{y},\dot{x})$, which gives the two component equations with the printed signs.

For circular relative motion of a neutral Coulomb pair of separation radius $R$, $\dot{\mathbf{r}}=\omega \hat{\mathbf{z}}\times \mathbf{r}$ and $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. ThusSince $(e^{\ast }{)}^2+\mu e^2/M=e^2/4$, the two signed frequencies areOne is positive and the other negative. At $B=0$ they give the two senses of an ordinary Coulomb circular orbit; equal masses remove the term linear in $B$ from their average. With negligible Coulomb attraction and $eB>0$, the branches approach $eB/m_2$ and $-eB/m_1$, the appropriate signed free cyclotron limits.

The actual particle paths follow by integrating the centre-of-mass velocity:Both particles move on circles about the same fixed guiding centre with a constant separation and common signed frequency, with a possible relative phase reversal encoded by a negative coefficient. Their mechanical centre of mass also generally circles. This is consistent with magnetic deflection plus a central mutual force; central interaction alone does not eliminate the external Lorentz forces.

For a real scalar triplet, with metric signature $(+---)$, differentiating the Lagrangian density gives $\partial \mathcal{L}/\partial {\varphi }_a=-\lambda {\varphi }_a$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)={\partial }^{\mu }{\varphi }_a$. HenceFor a constant internal rotation parameter, the variation isEach bracketed tensor is symmetric in $a,c$, so its contraction with the antisymmetric symbol vanishes. This rotates the three field components, not the space coordinates; both the kinetic term and common mass term are invariant.

For a first-principles Noether current derivation, temporarily let $\theta$ depend on spacetime. The terms proportional to $\theta$ still cancel, leavingOn a solution of the field equations, the action variation for a compactly supported $\theta$ vanishes after integration by parts. Thus ${\partial }_{\mu }{j}^{\mu }=0$. To match the charges printed in the question, use the equivalent overall sign convention for the SO(3) Noether current of a real scalar tripletArbitrariness of $\mathbf{n}$ gives three continuity equations. Their charges are exactlyAn overall reversal of all currents is immaterial; the minus sign above keeps the generator convention distinct from the supplied charge convention.

Direct verification is also short:The mass term and the products of equal-type derivatives cancel by antisymmetry; integration by parts produces the boundary term. The charges are conserved when these integrals exist and the net boundary flux vanishes: sufficiently decaying fields on all space, periodic boundary conditions, or suitable symmetry-preserving no-flux boundaries suffice. Arbitrary open boundaries do not. A common constant $\lambda$ is necessary for this internal symmetry; its positivity is a stability issue, not needed for the algebraic conservation proof.

Use metric signature $(+,-,-,-)$ and units $\hslash =c=1$. Varying ${\varphi }^{\ast }$ and $\varphi$ independently givesIn the divergence of the stated current, the products of first derivatives cancel, leavingThe two identical mass terms cancel on the field equations.

An infinitesimal constant phase change has $\delta \varphi =-iq\alpha \varphi$ and $\delta {\varphi }^{\ast }=iq\alpha {\varphi }^{\ast }$. Both the derivative product and the mass term are invariant. Noether's theorem gives, per unit $\alpha$,Thus the global phase symmetry supplies the conserved current in part a.

Write the bead position in Cartesian coordinates as ${\mathbf{x}}_B=(a\cos {\theta }_B,a\sin {\theta }_B,b{\theta }_B)$. Differentiation gives $|{\dot{\mathbf{x}}}_B{|}^2=(a^2+b^2){\dot{\theta }}_B^2$, while the free particle has $|\dot{\mathbf{x}}{|}^2={\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. Thus the kinetic energy is the sum of these squared speeds weighted by half the respective mass. The squared spring length isA spring obeying Hooke's law with zero natural length has potential energy $k{\ell }^2/2$. Consequently $L=T-k{\ell }^2/2$, and deleting the additive constant $-k{a}^2/2$ givesThe spring length is the straight separation of its endpoints, not distance measured along the helix. The fixed frictionless constraint is already incorporated by using the generalized coordinate ${\theta }_B$; it supplies no extra potential energy.

Let $v=\sqrt{-\ln \lambda /\lambda }>0$ and write $\varphi =v+\chi$. ThenSince $V^{\prime }(v^2)=0$ and $V^{\prime \prime }(v^2)=\lambda$, the quadratic potential isThereforeThe constant can be dropped from the fluctuation equations. For canonical real fields, use $\chi =(h+i\pi )/\sqrt{2}$:These are the radial and angular masses of a phase-invariant scalar potential. The radial Higgs mode changes the field magnitude and costs energy to second order. The angular Goldstone boson changes the phase along the vacuum circle without a quadratic restoring force. Its linear dispersion is ${\omega }^2=|\mathbf{k}{|}^2$, whereas the radial mode has ${\omega }^2=|\mathbf{k}{|}^2+m_h^2$. The zero angular mass follows the continuous broken global symmetry, not from an accidental cancellation in a particular coordinate choice.

Write $\varphi =x+iy$. The added term is $-(x^2-y^2)$, so the full potential simplifies toThe phase-rotation subgroup is reduced from U(1) to the two-element symmetry $\varphi \mapsto -\varphi$. More fully, the kinetic term and potential are invariant under independent reflections $x\mapsto -x$ and $y\mapsto -y$, a ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$ internal symmetry generated by $\varphi \mapsto -\varphi$ and complex conjugation. There is no continuous internal rotation symmetry for the explicit twofold anisotropy of a complex scalar field.

For fixed $x$, positive $\lambda$ makes the exponential grow with $y^2$, so every global minimum has $y=0$. On the real axis,ThusFor $0<\lambda <2$, the nonzero stationary points are the two global minima; the origin is unstable along the real direction. For $\lambda \ge 2$ the origin is the unique minimum, with zero real-direction quadratic curvature at the critical value $2$. Selecting $+v$ breaks the sign reversal that exchanges the two vacua, while complex conjugation remains unbroken.

Set $\varphi =v+\chi$, $\chi =u+iw$. Since $e^{\lambda v^2}=2/\lambda$, expansion givesEquivalently, in canonical fields $h=\sqrt{2}u$, $\pi =\sqrt{2}w$,Both masses are positive throughout the broken phase $0<\lambda <2$. A broken discrete symmetry does not demand a massless Goldstone boson. At $\lambda =2$ the radial mass tends to zero at a phase boundary, so a claim of strictly positive mass there would be incorrect. For completeness, expanding around the symmetric origin gives squared masses $\lambda -2$ and $\lambda$ for the canonical real and imaginary fields.

Assume a positive-energy incoming particle, $E>m{c}^2$, and a nonzero upward step $V_0>0$. Set $M=m{c}^2$ and $k=\sqrt{E^2-M^2}/(\hslash c)>0$. With the common time factor $e^{-iEt/\hslash }$ removed, take ${\varphi }_{<}(x)=e^{ikx}+r{e}^{-ikx}$ and ${\varphi }_{>}(x)=t{e}^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0{)}^2-M^2]/(\hslash c{)}^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\varphi$ and ${\varphi }^{\prime }$, giving $1+r=t$ and $k(1-r)=qt$. ThusThese are wave amplitudes, not flux probabilities. The Noether current of scalar electrodynamics, in a convenient common normalization for a stationary mode, is ${\rho }_Q=(E-V)|\varphi {|}^2/(m{c}^2)$ and $j_Q=(\hslash /m)\mathrm{Im}({\varphi }^{\ast }{\partial }_x\varphi )$. For a real $q$, its charge-current ratios are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(q/k)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The same statement uses $\mathrm{Re}q$ for a decaying solution. The electrostatic potential couples through the time derivative; it is not a Lorentz-scalar mass shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.

The angular action is $S=\int I{\dot{\theta }}^{2}ds/2$. In each winding sector, write $\theta ={\theta }_p+\xi$ with $\xi$ zero at both endpoints. The cross term vanishes and the Gaussian fluctuation integral is the free-particle prefactor with mass replaced by $I$. ThusEquivalently, multiply the line kernel for displacement $R(\Delta +2\pi p)$ by $R$ to change its measure from arc length to angle. The quadratic action makes the classical-path sum exact, rather than merely a stationary phase approximation. Real-time sums are understood with the usual negative imaginary part of $T$ and its boundary limit.

First trace the state explicitly specified in (c). orthogonality in the $y$ oscillator removes terms with different occupation numbers:For a thermal oscillator of frequency $\omega$, the ratio of successive Boltzmann weights is $e^{-\hslash \omega /(k_{B}T)}$. Consequently the conditional answer for that two-mode squeezed state isAt $\theta =0$ this is interpreted by the limit $T_{\mathrm{sq}}=0$. The reduced state is mixed for nonzero squeezing although the joint state is pure; its effective thermal statistics arise from entanglement, not from heating by a bath.

The ground-state identification must still be qualified. For the true coordinate-coupled ground state, direct normal-mode averages giveTheir ratio is $m^2{\omega }_{+}{\omega }_{-}$, rather than $m^2{\omega }^2$. Therefore this reduced state is not thermal with respect to the original bare frequency $\omega$ for $k\ne 0$. It does have an exact thermal representation for an effective oscillator of mass $m$ and frequencyThe corresponding occupation basis is that of $\nu$, not that of the original $a$. This follows by matching the Gaussian variances to $⟨x^2⟩=\hslash \coth [\hslash \nu /(2k_{B}T)]/(2m\nu )$ and its momentum counterpart. A direct reduced-Gaussian eigenfunction calculation gives the same effective frequency and geometric eigenvalues.^[ref]