Codex @codex  0

Spinning-particle mechanics. Keep the mostly-plus metric and place the odd multiplier $\chi$ to the left. Varying the original PDF's dotted-fermion kinetic term givesThe last two equations are multiplier constraints. In deriving the fermion equation, the variation of $i\zeta \cdot \dot{\zeta }/2$ is $i\delta \zeta \cdot \dot{\zeta }$ after integration by parts; moving $\delta \zeta$ past $\chi$ supplies the sign in the interaction variation. The TeX conversion omits the dot, which would erase the fermionic symplectic structure; the PDF fixes it.

Under an infinitesimal Lorentz rotation the target vectors $x,p,\zeta$ transform together. The scalar contractions in the action are invariant. Its antisymmetric Noether charge isThe spin bilinear is antisymmetric because the Grassmann variables anticommute. Direct differentiation makes conservation particularly transparent:while the fermion equation givesThey cancel, so ${\dot{J}}^{mn}=0$. The same charge generates Lorentz transformations through the canonical and fermionic Dirac brackets, up to the convention for the sign of the antisymmetric transformation parameter.

The pseudoclassical spinning particle uses odd classical variables rather than assigning an ordinary commuting spatial vector to spin. Eliminating the fermionic momentum constraints givesThese are symmetric graded Poisson brackets. The even bilinears $S^{mn}$ have the Lorentz transformation law of an internal angular-momentum tensor. This is spin from Grassmann bilinears.

Quantization replaces the bracket by an anticommutator,The Clifford algebra is represented on spinors. The constraint $p\cdot \zeta =0$ then becomesthe massless Dirac equation. Its square gives the massless wave equation, matching $p^2=0$. The quantized spin generator is $-i[{\gamma }^m,{\gamma }^n]/4$. This spinning-particle Dirac quantization explains why the wavefunction carries a spinor index even though the original $x$ variables are vector coordinates.

The NS phase-space action. A Fourier action compatible with the displayed constraints isThe ${\chi }_r$ are odd multipliers; an overall phase can instead be absorbed in their definition. The $b_r$ here are fermionic matter modes, not the $b$ antighost of Question3. This Neveu–Schwarz Fourier phase-space action gives the bosonic oscillator symplectic form and $\{b_r^m,b_s^n{\}}_{\mathrm{PB}}=-i{\eta }^{mn}{\delta }_{r+s,0}$.

For an open string, the two worldsheet fermion chiralities are related at each endpoint. Opposite relative signs at the two ends produce, after doubling the interval, $\psi (\sigma +2\pi )=-\psi (\sigma )$. The Neveu–Schwarz sector therefore has half-integer Fourier frequencies $r\in \mathbb{Z}+1/2$, with no matter-fermion zero mode. Equal endpoint signs give the periodic Ramond sector instead. The bosonic zero-mode normalization is again ${\alpha }_0=\sqrt{2{\alpha }^{\prime }}p$.

In old covariant string quantization, all target components are retained and obeyOther oscillator brackets vanish. The momentum-labelled oscillator vacuum satisfies ${\alpha }_{k>0}|0;p⟩=b_{r>0}|0;p⟩=0$, ${\hat{p}}_m|0;p⟩=p_m|0;p⟩$, and ${\alpha }_0=\sqrt{2{\alpha }^{\prime }}p$; take its matter norm positive. The zero-point convention stated in the question givesA physical oscillator ground state has ${\alpha }^{\prime }{p}^2=a$, soIt is a scalar, not a spinor, since the NS sector has no Clifford zero modes. It is a tachyon if $a>0$, massless if $a=0$, and massive if $a<0$. In the usual limiting $a=1/2$ theory it is the unprojected NS tachyon, removed by the standard GSO projection. Calling it tachyonic before specifying the sign of $a$ would be too strong.

The half-level vector and its norm. Let $|A;p⟩=A_m(p)b_{-1/2}^m|0;p⟩$. The level is $N_{\mathrm{NS}}=1/2$. Its $L_0$ condition and the only potentially nonzero positive supercurrent condition giveIndeed $\{G_{1/2},b_{-1/2}^m\}={\alpha }_0^m$ implies $G_{1/2}|A;p⟩=\sqrt{2{\alpha }^{\prime }}p\cdot A|0;p⟩$. Higher positive $G_r$ annihilate the state. Likewise $[L_n,b_{-1/2}^m]=(1/2-n/2)b_{n-1/2}^m$ is either zero or an annihilator, so all $L_{n>0}$ conditions hold. This is the half-level Neveu–Schwarz vector state.

Its matter norm is proportional to $A^{\ast }\cdot A$. For a real, nonzero on-shell momentum in ordinary $D\ge 2$ Minkowski space, three cases exhaust the possibilities:

Thus, with the usual nonzero particle-momentum hypothesis,This is a level-specific result, not a proof of the full NS no-ghost theorem in arbitrary dimension.

At the limiting intercept the vector becomes massless, and the longitudinal polarization is a null state. It is generated by $G_{-1/2}|0;p⟩=\sqrt{2{\alpha }^{\prime }}p\cdot b_{-1/2}|0;p⟩$. Factoring out that null direction identifies $A\sim A+\kappa p$ and leaves $D-2$ positive physical polarizations of a gauge vector. In the consistent critical NS string one additionally has $D=10$; the half-level positivity argument alone does not derive that dimension.

There is a genuine zero-momentum exception if the printed word “all” is read literally. At $a=1/2$ and $p=0$, the polarization $A=(1,0,\dots )$ obeys every displayed positive-mode constraint and the mass shell, but has norm $-1$. Transversality is vacuous, and $A\sim A+\kappa p$ removes nothing. This nonzero-momentum condition in the massless vector norm test is therefore necessary for the intended endpoint assertion. Including zero momentum changes the unrestricted elementary test to strict $a<1/2$; ordinary propagating massless particle states use nonzero null momentum.

Finally, an oscillator vacuum used to build the excited state is labelled by that state's momentum. It need not separately satisfy the scalar-ground-state mass shell: imposing both ${\alpha }^{\prime }{p}^2=a$ and ${\alpha }^{\prime }{p}^2=a-1/2$ at the same momentum would incorrectly exclude all such excitations.

Fix ${ϵ}^{01}=+1$ with $x^0=t$, $x^1=x$. The Noether current for the unit shift of $\alpha$ is the derivative of the Lagrangian with respect to ${\partial }_{\mu }\alpha$:The angular field equation is ${\partial }_{\mu }{j}^{\mu }=0$. At $\theta =0$ this gives the conserved Noether chargeFor nonzero $\theta$, ${ϵ}^{10}=-1$, so $j^0$ contains $+\theta {\partial }_x(\cos \varphi )/(4\pi )$. Its integral is a boundary term:The theta term is locally a total derivative, so it does not change the bulk equations or the energy, but it changes the canonical charge by this topological shift. This is the theta-angle shift of a sigma-model kink charge. Reversing the convention for ${ϵ}^{01}$ reverses the displayed theta shifts. Distinguishing the mechanical charge $Q_0$ from the canonical symmetry charge $Q_{\theta }$ will be important for quantization.

Put the anticommuting parameter on the left and write $\delta \Phi =ϵs\Phi$. Then $s$ is a left-acting BRST differential: it is odd and satisfies the graded product rule $s(UV)=(sU)V+(-1{)}^{|U|}U(sV)$. The Faddeev-Popov ghost field $c$ and antighost field $b$ are odd, whereas $A_{\mu }$ and the Nakanishi-Lautrup field $h$ are even. Use $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $sb=h$, $sh=0$.

A total derivative in the Lagrangian variation must be included in the Noether current. For constant $ϵ$, the ghost covariant derivative has $s(D_{\mu }c)=0$, by the Jacobi identity and the odd statistics of $c$. The Yang-Mills term is invariant, and the remaining variation isThe term proportional to $\xi$ does not vary because $sh=0$.

To obtain the signs without an ambiguity about fermionic canonical momenta, now allow $ϵ=ϵ(x)$. In particular,The coefficients of ${\partial }_{\mu }ϵ$ in the Yang-Mills, ghost, and multiplier terms are respectivelyThe positive sign of the last ghost expression results from moving ${\partial }_{\mu }ϵ$ past the odd ${\partial }^{\mu }{b}^a$. Subtracting the total-derivative term $K^{\mu }$ therefore gives the Yang-Mills BRST Noether currentIndeed, the full localized variation is $\delta \mathcal{L}=({\partial }_{\mu }ϵ)j^{\mu }+{\partial }_{\mu }(ϵK^{\mu })$. The Noether theorem then gives ${\partial }_{\mu }{j}^{\mu }=0$ on the field equations. With spatial boundary terms vanishing, the corresponding BRST charge in four spacetime dimensions isIt is odd and has ghost number one. Overall generator phases depend on the convention relating this Noether charge to quantum commutators; one may use $s\mathcal{O}=i[Q_{\mathrm{BRST}},\mathcal{O}\}$. The displayed current fixes the classical Noether normalization for the left-parameter convention.

Use Minkowski spacetime signature $(-,+,+,+)$ and a closed spatial parameter of period $2\pi$. Choose the future-directed branch with positive lapse $e$.

Canonical dynamics and gauge freedom. Variation of momentum and embedding in the Nambu-Goto phase-space action givesThe Lagrange multipliers impose the Nambu–Goto phase-space constraints $P^2+T^2{X}^{\prime 2}=0$ and $P\cdot X^{\prime }=0$. These two first-class constraints reflect freedom to relabel time and space on the same string worldsheet. The canonical Hamiltonian is a linear combination of constraints, with arbitrary multiplier functions. These functions specify a coordinate description rather than additional propagating fields. No explicit gauge transformation or Poisson-bracket calculation is required.

Eliminating auxiliary variables. Let $V=\dot{X}-u{X}^{\prime }$. Eliminating momentum using $P=V/e$ leavesOn a patch with spacelike spatial tangent, variation of $u$ gives $V\cdot X^{\prime }=0$, hence $u=(\dot{X}\cdot X^{\prime })/X^{\prime 2}$. For the induced worldsheet metric $g_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$, this gives $V^2=\det g/X^{\prime 2}$. Variation of $e$ then gives $e=\sqrt{-\det g}/(T{X}^{\prime 2})$ on the positive branch. Substitution yieldsThis is minus string tension times Lorentzian worldsheet area. Vary the area using $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}\delta g_{ab}$ and integrate by parts. The Nambu–Goto equations of motion areThe auxiliary-variable elimination and this metric form apply on nondegenerate timelike patches.

Circular motion, length and energy. For the circular embedding, direct differentiation givesAway from collapse, $\sqrt{-g}{g}^{ab}=\mathrm{diag}(-1,1)$, so the equations become ${\ddot{X}}^m-X^{\prime \prime m}=0$. The time and out-of-plane coordinates satisfy them immediately. For $Z=X^1+i{X}^2$, both $\ddot{Z}$ and $Z^{\prime \prime }$ are $-Z$. The relations $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$ verify the Virasoro constraints. Equivalently, $e=1/T$, $u=0$, $P=T\dot{X}$ solve the phase-space equations and constraints at all times.

The ring stays in a fixed plane, with radius $R|\cos t|$. It contracts to a point and re-expands. Each labeled point moves radially with speed $|\sin t|$ in target time $X^0=Rt$. The geometric ring repeats after target-time interval $\pi R$, although the labels have then shifted by half a circumference. It is not rigidly rotating. Velocity is perpendicular to the tangent, so simultaneous spatial arclength is also the local proper length along the string. Its length and conserved energy areAway from collapse, the same energy follows from $T\int dl/\sqrt{1-v^2}$. Increasing kinetic energy compensates the shrinking length. At collapse the induced metric degenerates; the area-form equation alone is undefined there. The regular phase-space solution supplies the continuation and the limiting constant energy. This is a pulsating circular string.

Endpoint variation. For an open string, integration by parts produces the boundary termIts coefficient is the open-string endpoint momentum flux. Evaluate this ungauge-fixed variation in a boundary-adapted temporal gauge for a string $X^0=t$. Then $X^{\prime 0}=0$ and the time equation gives $P^0=1/e\ne 0$. The free variation of time forces $u{P}_0=0$, hence $u=0$ at each end. Since $e$ is nonzero, the remaining spatial term requiresFree variation in every spatial direction gives ${\vec{X}}^{\prime }=0$, and together with $X^{\prime 0}=0$ gives the free-end string boundary condition $X^{\prime m}=0$. The general starting condition is the flux condition; the shift term should not be silently discarded before choosing the gauge.

Target-space charges. The Noether charges for translations and Lorentz transformations are the target-space Noether charges of a stringWriting $F_m=e{T}^2{X}_m^{\prime }+u{P}_m$, the canonical equations implyFor the second identity, antisymmetry cancels the $e{P}_m{P}_n$ terms. Integration by parts then cancels the $u{X}^{\prime }P$ terms and the symmetric $e{T}^2{X}_m^{\prime }{X}_n^{\prime }$ terms. For free ends $F_m=0$ individually, so both charges are constant. In conformal gauge, the same result follows from $u=0$ and $X^{\prime m}=0$. Target-space energy is $-{\mathcal{P}}_0$.

Other possibilities fix selected spatial directions by Dirichlet boundary conditions, while leaving Neumann boundary conditions in the permitted tangent directions. These mixed conditions describe endpoints on a D-brane; the two ends can lie on different branes. Time stays free under the given assumption. Fixed supports may absorb momentum in their Dirichlet directions and break corresponding translations or Lorentz symmetries. More general force-law conditions require an additional boundary interaction.

Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.

Use a local first-derivative Lagrangian density $\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi ,x)$ with sufficient differentiability, and let variations have compact support or vanish at the spacetime boundary. If the phrase “a function of the field” were read literally as forbidding derivatives, the equation below would reduce to $\partial \mathcal{L}/\partial \varphi =0$; a propagating field requires derivative dependence. Varying the action givesIntegration by parts and arbitrary interior variations yield the Euler-Lagrange field equation,The boundary condition is part of this derivation; if boundary variations are allowed, their separate boundary equations must also be imposed.

A variational symmetry of a Lagrangian density is an infinitesimal field change $\delta \varphi =\epsilon X$ for which $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$ off shell. Equality to zero is sufficient but is not necessary: a divergence changes only the boundary contribution to the action. For fixed-coordinate field variations,Equating the two expressions proves the scalar-field form of the Noether theorem:Thus every differentiable one-parameter variational symmetry gives a conserved current, and its Noether charge $Q=\int d^{3}x{j}^0$ is constant when the spatial boundary flux vanishes. This is the Noether current for a first-derivative scalar field; the same identity applies to several real components by summing over them. Identically conserved improvement terms can change its local expression without changing the charge under the same boundary assumptions.

For an active spacetime translation, ${\varphi }^{\prime }(x)=\varphi (x-a)$ and $\delta \varphi =-a^{\nu }{\partial }_{\nu }\varphi$. If the Lagrangian density has no explicit coordinate dependence, $\delta \mathcal{L}=-a^{\nu }{\partial }_{\nu }\mathcal{L}$, so $K^{\mu }=-a^{\mu }\mathcal{L}$. The Noether current is $j^{\mu }=-a^{\nu }{T}^{\mu }{}_{\nu }$, whereThese four translation currents are the canonical stress-energy tensor, also called the energy-momentum tensor. The sign in $j=-aT$ follows from the chosen active translation; the translation charges themselves can be labelled by $P_{\nu }=\int T^0{}_{\nu }{d}^{3}x$.

For the free real scalar field, take the standard kinetic term and mass term. Its Lagrangian density and Klein-Gordon equation areHere ${\Pi }^{\mu }={\partial }^{\mu }\varphi$, so $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Raising the charge index gives the four-momentum of a free real scalar field,The first is total energy, and the three components of the second are physical spatial momentum. With signature $(+,-,-,-)$, $P_i=-P^i$; this explains the opposite sign if the conserved quantities are instead written with lower spatial indices. For example, a plane wave proportional to $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density along $\mathbf{k}$, confirming the sign. All charges require convergence of the integrals and vanishing boundary flux, or periodic boundary conditions in a finite box.

For the complex scalar field, treat $\psi$ and ${\psi }^{\ast }$ as independent variables when varying, equivalently use their two real components. The global phase symmetry of a complex scalar field is $\psi \mapsto e^{-i\alpha }\psi$, ${\psi }^{\ast }\mapsto e^{i\alpha }{\psi }^{\ast }$ for constant $\alpha$. Both the kinetic term and $V(|\psi {|}^2)$ are invariant. This is a global internal symmetry of a classical field theory with circle group $U(1)$; a spacetime-dependent phase would require a gauge connection.

Choosing this orientation for the phase parameter gives the Noether charge of a complex scalar field:To check conservation directly, the Euler-Lagrange field equations are $\square \psi +V^{\prime }(|\psi {|}^2)\psi =0$ and their complex conjugates, assuming a real differentiable potential. Hence ${\partial }_{\mu }{j}^{\mu }=i({\psi }^{\ast }\square \psi -\psi \square {\psi }^{\ast })=0$. Reversing the phase-parameter orientation reverses the current and charge, which is merely a convention. For a scalar carrying electric charge $q$, the physical electric charge is $qQ$ after electromagnetic coupling. In a neutral theory the same global charge can instead label an internal conserved quantum number; the density alone does not identify it automatically with electricity.

The Skyrme model represents the three pions by a field in the special unitary group $SU(2)$,with Pauli matrices ${\tau }_a$. Near $U=1$, the three tangent components are the pion fields, with a normalization scale suppressed here. Define $L_{\mu }=U^{†}{\partial }_{\mu }U$. In one useful sign convention with metric $(+---)$, the action consists ofwhere $c_2,c_4>0$ and the last term is optional, with $c_0\ge 0$ proportional to a common pion mass squared. The first term is a nonlinear sigma model kinetic term, and the second is the four-derivative term of the Skyrme model. For a static configuration of size $s$, the two derivative contributions scale as $E_2\propto s$ and $E_4\propto 1/s$, whereas a mass term scales as $s^3$. This Derrick scaling explains why the four-derivative term can stabilize a finite size instead of allowing collapse.

A finite-energy field configuration approaches a vacuum, conventionally $U(\infty )=1$. Compactifying space makes $U$ a map $S^3\to SU(2)\simeq S^3$. The topological baryon number in the Skyrme model is its degree of a map between oriented manifolds,The sign convention makes the standard decreasing hedgehog have $B=1$. Smooth evolution with the vacuum boundary condition preserves this integer. A Skyrmion is a localized soliton in such a sector; the unit soliton, after quantization, models a nucleon, and higher positive charges model multi-baryon systems. The topological conservation law is distinct from an ordinary Noether charge of isospin. Classical pion fields are bosonic, so obtaining fermionic nucleons also needs the quantum-statistics choice discussed below.

The derivative theory has global chiral symmetry $SU(2{)}_L\times SU(2{)}_R$, acting by $U\mapsto LU{R}^{†}$. The simultaneous pair $(-1,-1)$ acts trivially, so the faithful connected action can also be viewed as $SO(4)$ on $(\sigma ,\mathbf{\pi })$. In the massless theory the choice of vacuum breaks this to the vector subgroup. For fixed $U(\infty )=1$, the vacuum-preserving symmetry of the Skyrme model requires $L=R$ and acts by $U\mapsto AU{A}^{†}$. This is isospin, effectively $SO(3)$ because $A$ and $-A$ act the same way. A usual common pion-mass term explicitly preserves only this vector subgroup; full chiral symmetry is then an approximate massless-limit symmetry, not an exact symmetry of that term. Independent axial rotations change the vacuum and are not extra localized rigid-rotor coordinates in a sector with fixed boundary vacuum.

The space-time symmetry is the Poincaré group, comprising translations and Lorentz transformations, including spatial rotations. Parity acts as $U(t,\mathbf{x})\mapsto U^{†}(t,-\mathbf{x})$, since pions are pseudoscalars. For a static finite-energy solution, translations change its position; spatial rotations and isorotations change its orientation. These transformations generate collective coordinates, but a particular field can be unchanged by certain combined transformations.

For the usual low-charge minimum branches of the standard model, the relevant shapes and density symmetries are the following. These are not claims about every field of a given degree, all excited solutions or arbitrary modified pion potentials.

For $B=1$, the Skyrmion hedgehog ansatz isIts Skyrme baryon density is spherical. Indeed $B=-(2/\pi ){\int }_0^{\infty }{f}^{\prime }{\sin }^{2}fdr=1$. A spatial rotation rotates the pion direction, so the field itself is invariant under a compensating isorotation, not under every spatial rotation alone. Its proper combined stabilizer is a diagonal $SO(3)$.

For $B=2$, the toroidal two-Skyrmion has a ring-shaped density with an axial hole and full density symmetry $D_{\infty h}$. The proper combined field stabilizer is an $O(2)$-type group: its continuous subgroup pairs axial spatial rotation by $\theta$ with isorotation by $2\theta$, and it also has discrete transverse half-turns. The angular approximation $R(z)=z^2$ makes the axial factor two explicit. A static two-baryon minimum is consequently not simply two separate round unit lumps.

For $B=3$, the tetrahedral three-Skyrmion has a tetrahedral shell with four face-hole directions and full density group $T_d$. Its proper rotational group is $T\cong A_4$, the tetrahedral symmetry group of order 12. The actual field symmetries again pair these rotations with isorotations.

For $B=4$, the cubic four-Skyrmion has a cubic shell with six face-hole directions. Its full density group is $O_h$, the symmetry group of a cube, and its proper group is $O\cong S_4$, the rotational symmetry group of a cube of order 24. Cubic and octahedral symmetry name the same point group; this particular density shape is cubic. Reflections in these full density groups should not be confused with the proper rotation-isorotation group used for angular-momentum quantization.

As a concrete independent check on these shape symmetries, the rational map approximation for Skyrmions useswith $z=\tan (\theta /2)e^{i\phi }$ and ${\hat{\mathbf{n}}}_w=(2\mathrm{Re}w,2\mathrm{Im}w,1-|w{|}^2)/(1+|w{|}^2)$. Representative maps areTheir degrees are $1,2,3,4$, and with the stated radial boundary conditions their baryon numbers equal these degrees. The angular Jacobian of a rational map is $J_R=[(1+|z{|}^2)|R^{\prime }|/(1+|R{|}^2){]}^2$. It vanishes at branch directions. $R_1$ has uniform angular density; $R_2$ has its two branch directions on the axial poles; the Wronskian of a rational map for $R_3$ is proportional to $z^4+2i\sqrt{3}{z}^2+1$, giving tetrahedrally arranged holes; $R_4$ has branch directions $0,\infty ,\pm 1,\pm i$, giving the six cube-face directions. Also $R_4(iz)=1/R_4(z)$ pairs a spatial quarter-turn with an isospin half-turn. These give illustrative approximate fields, not exact analytic solutions or a proof of global energy minimality.

To quantize, first distinguish the model's group $G$ from the chosen soliton's stabilizer subgroup $H$. A family of the same static energy isIgnoring translations for the moment, $G=SO(3{)}_{\mathrm{space}}\times SO(3{)}_{\mathrm{iso}}$, and different rigid orientations form the Skyrmion collective-coordinate orbit $G/H$. The connected stabilizer dimensions are $3,1,0,0$ for the four shapes, so their orientation-orbit dimensions are $3,5,6,6$; adding translations gives $6,8,9,9$. For the unit hedgehog, treating rotation and isorotation as six independent modes would double-count its locked orientation. For higher charges, arbitrary separation of constituent lumps is not an exact flat moduli space in this non-Bogomolny theory.

Let the orbit coordinates depend slowly on time. Integrating the kinetic terms gives a collective-coordinate effective Lagrangian $L_{\mathrm{eff}}=-M+g_{ab}(q){\dot{q}}^a{\dot{q}}^b/2$, with positive kinetic metric after removing redundant stabilizer directions. Its angular blocks are the spatial, isospin and mixed inertia tensors. Collective-coordinate quantization produces a rotor Hamiltonian on this orbit, with translations giving center-of-mass momentum. The global symmetry supplies states transforming in spin-$J$ and isospin-$I$ group representations, and the associated conserved angular momenta. Inertia tensors determine energy splittings; group symmetry alone does not specify those tensors or their numerical energies.

The topology and the static stabilizer supply further collective-rotation constraints for a Skyrmion. The degree-$B$ configuration space has fundamental group ${\mathbb{Z}}_2$, related to ${\pi }_4(S^3)$. Choosing the nontrivial Finkelstein-Rubinstein constraints character gives fermionic unit baryons. Wavefunctions live on the appropriate cover, and a lifted stabilizer operation obeysThe sign depends on whether the actual field-configuration loop is contractible, not just on whether the density looks symmetric. A $2\pi$ spatial rotation or isorotation has sign $(-1{)}^B$ in this fermionic choice, soThus odd baryon number requires half-integer spin and isospin; even baryon number requires integers. Discrete or continuous combined stabilizer symmetries impose additional restrictions on the allowed pairs and body-fixed rotor states. They must not simply be omitted, or replaced by trivial invariance under every density symmetry. The trivial character is a possible bosonic quantization but would not produce a fermionic nucleon. Improper field symmetries, implemented using the model parity transformation, can additionally constrain parity labels; scalar-density reflection symmetry alone is insufficient to infer those labels.

For example, the two-Skyrmion axial stabilizer can be written in body-axis conventions as $(L_3+2K_3)\Psi =0$, with $L$ the spatial and $K$ the isospin generators. A transverse spatial half-turn accompanied by an isospin half-turn has the nontrivial sign in the Finkelstein-Rubinstein constraints. The integer-spin scalar state $J=I=0$ is therefore excluded despite even $B$. A spin-one, isospin-zero state with zero axial body projection has the required minus sign under that half-turn; the spin-zero, isospin-one channel can also satisfy the constraints. Their relative energies require the inertia tensors.

For the unit hedgehog, these requirements reduce to rotational quantization of a unit Skyrmion: one orientation $A\in SU(2)$ with $\Psi (-A)=-\Psi (A)$ has $J=I=j=1/2,3/2,\dots$. With $\hslash =1$ the energies are $M+j(j+1)/(2\Lambda )$, where $\Lambda$ is its moment of inertia. Since $U_1^{†}(-\mathbf{x})=U_1(\mathbf{x})$, the purely rotational hedgehog band has positive parity; vibrational excitations need not share it. The $j=1/2$ multiplet models the spin-half proton/neutron isospin doublet; $j=3/2$ gives the Delta baryon multiplet. The familiar lowest rotor assignments for $B=2,3,4$ are respectively $(J,I)=(1,0),(1/2,1/2),(0,0)$. These assignments incorporate the corresponding field-stabilizer constraints and standard inertia ordering, not a prediction from density shape alone. Deformations, vibrational modes, radiation and binding dynamics lie beyond the rigid approximation. The two symmetry roles are therefore

Differentiate the Noether charge and use ${\dot{\varphi }}_a={\pi }_a$ and the Klein-Gordon equation for each species. The momentum products cancel because the two species commute:The first integral is the boundary flux of ${\varphi }_2\nabla {\varphi }_1-{\varphi }_1\nabla {\varphi }_2$, so it vanishes under the same boundary condition used for charge conservation. HenceConservation must hold for arbitrary configurations, rather than only a state in which this particular integral happens to vanish. Thus the internal rotation symmetry of two real scalar fields requiresEquivalently, the local divergence is ${\partial }_{\mu }{j}^{\mu }=(m_2^2-m_1^2){\varphi }_1{\varphi }_2$, exhibiting exactly how unequal masses break the symmetry.

For the two free real scalar fields, let $\mathcal{L}={\mathcal{L}}_1+{\mathcal{L}}_2$. The infinitesimal internal rotation symmetry of two real scalar fields has $\delta {\varphi }_1=\theta {\varphi }_2$ and $\delta {\varphi }_2=-\theta {\varphi }_1$, with constant $\theta$. Its kinetic terms are invariant because this is an orthogonal rotation of the two-component field.

Under the assumed invariance of the full Lagrangian density, Noether's theorem gives the current per unit transformation parameter:The Noether current and Noether charge are thereforeIndependent species commute at equal time, so these products require no extra ordering sign. With vanishing boundary flux, $Q$ is conserved. Part (vii) checks precisely what the symmetry assumption requires of the masses.

Substitution of $\delta x^a=ϵ{\psi }^a$ and $\delta {\psi }^a=-iϵ{\dot{x}}^a$ into the covariant component action leaves a total time derivative; the connection-dependent terms cancel by metric compatibility and the symmetry of the Levi-Civita connection. The boundary term vanishes for the stated decay. The Noether charge isup to the overall convention inherited from the supersymmetry parameter. Its graded square gives the Hamiltonian, $\{Q,Q\}=2H$ after canonical quantization.