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Use the real Lie-algebra convention of the question, with a gauge covariant derivative $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The gauge field strength is its curvature:Write $a_{\mu }={\delta }_X{A}_{\mu }/ϵ=-{\partial }_{\mu }X+[X,A_{\mu }]$. To first order in $ϵ$,Substitute $a_{\mu }$. The mixed second derivatives of $X$ cancel. The terms containing first derivatives of $X$ cancel in pairs. The remaining terms areThe Jacobi identity combines the last two into $[X,[A_{\mu },A_{\nu }]]$. ConsequentlyThe transformation is homogeneous even though the connection transformation contains an inhomogeneous derivative term.

For a finite-dimensional Lie algebra, define its Adjoint representation by ${\mathrm{ad}}_X(Y)=[X,Y]$ and its Killing form byIt is a symmetric bilinear form by cyclicity of the trace. The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Set $A={\mathrm{ad}}_Z$, $B={\mathrm{ad}}_X$, $C={\mathrm{ad}}_Y$. ThenThis proves the invariant bilinear form on a Lie algebra property, without assuming simplicity or nondegeneracy.

For definiteness use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and take a real compact semisimple Lie algebra as the gauge algebra. Its positive internal metric is $B=-\kappa$. A Killing-form Yang-Mills Lagrangian with the coupling absorbed into the connection isThe spacetime metric is unchanged by internal gauge transformations. HenceThus gauge invariance follows directly from invariance of the Killing form. Other overall conventions are possible, but the energy sign must be checked rather than inferred from a prefactor in isolation.

For physical kinetic terms the internal form must be real, nondegenerate and positive definite after choosing the overall sign. If ${\mathcal{E}}_i=F_{0i}$ and ${\mathcal{B}}_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, the above convention has Lagrangian $[B({\mathcal{E}}_i,{\mathcal{E}}_i)-B({\mathcal{B}}_i,{\mathcal{B}}_i)]/(2g_{\mathrm{YM}}^2)$ and physical energy densityThe canonical Hamiltonian density also contains the nondynamical multiplier and a spatial divergence. With ${\Pi }_i={\mathcal{E}}_i/g_{\mathrm{YM}}^2$ and $D_i{\Pi }_i={\partial }_i{\Pi }_i+[A_i,{\Pi }_i]$, integration by parts using the invariant bilinear form on a Lie algebra givesThe Gauss law constraint in gauge theory sets $D_i{\Pi }_i=0$. If the boundary flux vanishes, or the appropriate boundary contribution is included, the integrated physical Hamiltonian is the positive energy displayed above.

An indefinite internal form would give gauge-field polarizations with opposite kinetic signs. A degenerate form would fail to supply a kinetic term for some directions. The compactness criterion from the Killing form says that negative-definiteness of the Killing form of a real finite-dimensional algebra is equivalent to compact semisimplicity; thus the pure Killing-form construction selects compact semisimple real forms. One cannot use a complex-bilinear Killing form on arbitrary complex field components as if it were a positive Hermitian metric.

This does not prohibit Abelian gauge theories. A compact Abelian factor has zero Killing form, so it needs a separately chosen positive invariant bilinear form on a Lie algebra, rather than the Killing form. More generally an algebra with a positive invariant metric on a Lie algebra is compact reductive, namely a direct sum of a compact semisimple algebra and an Abelian center. A noncompact group can also share the same compact Lie algebra through global covering choices in an Abelian factor; positivity is a statement about the algebra and internal metric, not by itself a classification of global gauge-group topology. Quantum matter anomalies and global restrictions would require additional input; no matter content is specified here.

For the finite matrix transformation, let $g(x)\in SU(N)$ and regard $g$ as a multiplication operator. The identity ${\partial }_{\mu }{g}^{-1}=-g^{-1}({\partial }_{\mu }g)g^{-1}$ givesTaking commutators of these differential operators cancels the adjacent multiplication operators $g^{-1}g$, yieldingThis argument keeps the derivatives acting on test fields and avoids treating $D_{\mu }$ as just a matrix.

There is a normalization issue in the printed last paragraph. The standard Killing form defined through the adjoint trace on $\mathfrak{su}(N)$ isnot simply $\mathrm{tr}(XY)$. For example, with $N=2$ and $X=Y=-i{\sigma }_3/2$, the defining trace is $-1/2$, whereas the adjoint trace is $-2$. The printed trace formula can be used as a rescaled invariant form, with the constant absorbed into the gauge coupling; it has exactly the invariance needed here. For either normalization,by cyclicity. This proves finite Yang-Mills gauge transformation invariance, with the healthy sign chosen for anti-Hermitian gauge fields. The normalization discrepancy is not a failure of gauge invariance.

Finally let $g(x)=e^{ϵX(x)}=I+ϵX(x)+O({ϵ}^2)$. Since $X\in \mathfrak{su}(N)$ is traceless and skew-Hermitian, this exponential lies in $SU(N)$. Expanding the finite formula givesThus the stated infinitesimal transformation is the derivative of the finite transformation at the identity. It describes transformations in the identity component; arbitrary global or large gauge transformations need not be generated by one globally defined infinitesimal parameter.

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

Use a dimensionless Hamiltonian in the functional weight $e^{-H}$, assume $\alpha >0$, and let $\varphi (\mathbf{x})={\int }_{|\mathbf{p}|\le \Lambda }{d}^{D}p(2\pi {)}^{-D}{e}^{i\mathbf{p}\cdot \mathbf{x}}\tilde{\varphi }(\mathbf{p})$. Reality means $\tilde{\varphi }(-\mathbf{p})=\tilde{\varphi }(\mathbf{p}{)}^{\ast }$. At zero field, the quadratic form isIn the momentum-shell renormalization group, split into ${\varphi }_{<}+{\varphi }_{>}$ below and above $\Lambda /b$. Disjoint Fourier supports make the quadratic form split into $H_0[{\varphi }_{<}]+H_0[{\varphi }_{>}]$. Integration over the shell gives a source-independent Gaussian functional integral multiplying the partition function, or an additive constant in the effective free energy; it leaves the slow-mode quadratic coefficients unchanged before rescaling. A uniform field has support only at zero momentum, so it does not change this shell integration.

Restore the ultraviolet cutoff by ${\mathbf{p}}^{\prime }=b\mathbf{p}$, ${\mathbf{x}}^{\prime }=\mathbf{x}/b$, and chooseThe measure, two gradients and two fields have scale factors $b^D$, $b^{-2}$ and $b^{-(D-2)}$, whose product is one. The mass term has factor $b^D{b}^{-(D-2)}=b^2$, while the uniform source term has factor $b^D{b}^{-(D-2)/2}=b^{(D+2)/2}$. Hence the Gaussian momentum-shell scaling isFor a slowly varying nonuniform source the corresponding formula is $h^{\prime }({\mathbf{x}}^{\prime })=b^{(D+2)/2}h(b{\mathbf{x}}^{\prime })$. Here the nonuniform source on the right is its projection onto the retained Fourier modes; an eliminated source component contributes only to the field-independent Gaussian normalization. The field engineering dimension is $(D-2)/2$ and its Gaussian anomalous dimension is zero. The mass and uniform source are relevant perturbations of the Gaussian fixed point. Positivity of $\alpha$ is necessary for a stable kinetic term near this point. One can use a finite volume and positive mass as infrared regulators and then take the critical limit; at exactly zero mass the integral over the zero Fourier mode alone is not a normalized finite-volume Gaussian measure.

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 Poincare 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

In the weak, non-crossing, leading-eccentricity model the exterior resonant coefficient has positive sign: write its disturbing function as $R_{\mathrm{res}}=B_q{e}_c^q\cos {\varphi }_{bc}$, with $B_q>0$ for the orders considered. The physical feedback can be seen for first order. Before an outer conjunction near apoapsis, the faster inner planet pulls the outer one backwards; after passing it pulls forwards. On the outward side, the nearer, stronger earlier pull wins, reducing the outer semi-major axis and increasing its mean motion. Thus ${\varphi }_{bc}<\pi$ is driven upward. On the inward side, the later forward pull wins and drives ${\varphi }_{bc}>\pi$ downward. The paired impulses cancel at the symmetric centre.

For higher order, sum the feedback over all $q$ conjunction phases. The first surviving angular Fourier harmonic is $\sin {\varphi }_{bc}$; the lower Fourier harmonics cancel over the equally spaced directions. A local Hamiltonian reduction with canonical momentum $J$ hasThe negative curvature of the Kepler orbit Hamiltonian gives the displayed sign. Linear stability analysis at $\pi$ gives a restoring acceleration. The symmetric libration centres of an eccentricity resonance are thereforeThe corresponding conjunction phases relative to periapsis are $\{\pi \}$, $\{\pi /2,3\pi /2\}$ and $\{\pi /3,\pi ,5\pi /3\}$. They suppress repeatedly close passages and provide resonance protection.

These are expected centres of the specified leading-harmonic model, not centres fixed solely by the integer order for arbitrary orbital eccentricity and masses. Asymmetric resonant-argument libration can occur when higher Fourier harmonics matter. For example, if $R=A\cos \varphi +B\cos 2\varphi$, $A>0$, $B>A/4$, the symmetric $\pi$ equilibrium loses stability and stable centres satisfy $\cos \varphi =-A/(4B)$. Thus further dynamical information is needed beyond the PDF's unrestricted eccentric orbit to assert unique centres. Exterior asymmetric branches are documented by Winter and Murray, “Resonance and chaos. II. Exterior resonances and asymmetric libration”.

Now the eccentric planet $c$ is inside the circular perturber $d$. The interior eccentricity-type mean-motion resonance hasIn the mean-conjunction approximation, ${\varphi }_{cd}=n(\Lambda -{\varpi }_c)$, so ${\Lambda }_j={\varpi }_c+({\varphi }_{cd}+2\pi j)/n$. Successive conjunctions advance by $2\pi m/n$; coprime $m,n$ visit $n$ distinct phases.

The near-conjunction tangential pulls are reversed for an eccentric body inside its perturber. Just before conjunction the inner body is pulled forwards and afterwards backwards. Near periapsis, an inward-moving inner body experiences the stronger earlier forward pull; its semi-major axis grows, its mean motion falls, and ${\varphi }_{cd}$ increases because its coefficient of ${\lambda }_c$ is negative. On the outward side the feedback reverses, restoring the first-order centre zero. Higher-order feedback is strongest on the side closest to the external perturber, near apoapsis; the leading coefficient alternates in sign with order.

Writing $R_{\mathrm{res}}=B_n{e}_c^n\cos {\varphi }_{cd}$ gives $B_n<0$ for odd $n$ and $B_n>0$ for even $n$ in the leading non-crossing model. The same negative-curvature Hamiltonian argument therefore givesConjunction phases relative to periapsis are $\{0\}$, $\{\pi /2,3\pi /2\}$ and $\{0,2\pi /3,4\pi /3\}$, respectively, as shown in the lower row of the figure. As before, these are small-eccentricity symmetric predictions; additional resonant Fourier harmonics and coupled secular dynamics can change the stable branches.

For actual alignment of the eccentric body at true longitude $\Lambda$, the exact phase relation is instead ${\varphi }_{cd}=(m+n)x-my$, where $x=\Lambda -{\varpi }_c$ and $y=M_c(x)$. The simple equally spaced angles replace $y$ by $x$ at leading order.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

The $-\mu \psi$ term means that the generator is the grand potential, rather than just the physical energy. With boundary variations vanishing, the Hamiltonian functional isIts functional derivative isso the Gross–Pitaevskii equation is $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$. Treating $\psi$ and ${\psi }^{\ast }$ as independent variables in this variation gives the factor $g$, not $2g$.

For repulsive interactions $g>0$ and $\mu >0$, complete the square in the local energy:The gradient contribution is nonnegative, so the homogeneous ground state has arbitrary constant complex argument andThe infinite-volume constant may be subtracted to define a relative grand potential; it does not affect the functional derivative. At fixed particle number one instead minimizes the physical energy $H+\mu N$, with $\mu$ as its Lagrange multiplier. For $g>0$, $\mu \le 0$ the grand-potential minimum over $n\ge 0$ is the vacuum $n=0$. For $g<0$ the displayed unconstrained functional is unbounded below; the positive-density stable uniform state assumed in the following parts requires repulsion.

Let $v_1\cdots v_r$ be a longest path in the finite simple graph. Every neighbour of either endpoint lies on it. Among indices $1\le i<r$, consider those with edges $v_1{v}_{i+1}$ and those with edges $v_i{v}_r$. Their cardinalities are the endpoint degrees, whose sum is at least $n\ge r$, exceeding the $r-1$ available indices. They intersect, producing the cycleEvery connected component has at least $\delta (G)+1>n/2$ vertices, so the graph is connected. If $r<n$, an edge joins a vertex outside this cycle to it, and breaking the cycle there gives a longer path, a contradiction. Hence $r=n$ and this is a Hamilton cycle, proving Dirac theorem.

The complete graph $K_4$ is planar, has chromatic number four and a Hamilton cycle. Adding one leaf attached to one of its vertices preserves planarity and chromatic number four but prevents a Hamilton cycle, since a leaf has degree one.

For the last claim, the given plane embedding has all vertices on the outer cycle. Add noncrossing diagonals until its polygonal interior is triangulated. A triangulated polygon has a vertex incident with only its two boundary neighbours: for example the dual graph of its triangles is a tree and a leaf triangle supplies an ear. Remove this vertex. Inductively three-colour the smaller triangulated polygon, starting with a triangle. The removed vertex's neighbours are adjacent and have different colours, so it can receive the third. Restriction to the original graph givesThis uses the specified embedding, not the false assertion that every planar Hamiltonian graph is three-colourable; $K_4$ is already a counterexample to that broader assertion.

Invert the operator definitions to obtain $a=\sqrt{m\omega /(2\hslash )}\hat{x}+i\hat{p}/\sqrt{2\hslash m\omega }$ and the adjoint formula for $a^{†}$. The canonical commutation relation yields $[a,a^{†}]=1$, with $[a,a]=[a^{†},a^{†}]=0$. Substitution givesWrite $N=a^{†}a\ge 0$. Since $[N,a]=-a$ and $[N,a^{†}]=a^{†}$, these are lowering and raising operators. Applying $a$ to a ground state would lower its energy unless it vanishes, so $a|0⟩=0$. From the normalized unique ground state constructThe commutator gives the factorial normalization and $a|n⟩=\sqrt{n}|n-1⟩$. For any energy eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered norm forces its $N$ eigenvalue to be a nonnegative integer and lowering eventually reaches the ground state. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its Hilbert space.

For the perturbation $\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$, nondegenerate time-independent perturbation theory gives $E_n^{\prime }=E_n+\lambda W_{nn}+{\lambda }^2{\sum }_{m\ne n}|W_{mn}{|}^2/(E_n-E_m)+O({\lambda }^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared matrix elements $(\hslash \omega {)}^{2}n(n-1)$ and $(\hslash \omega {)}^2(n+1)(n+2)$ respectively. ThusThe absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also givesFor $|\lambda |<1/2$ this is another oscillator with mass $m^{\prime }=m/(1-2\lambda )$ and frequency ${\omega }^{\prime }=\omega \sqrt{1-4{\lambda }^2}$. ThereforeIts expansion is $E_n[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic Hamiltonian is not bounded below.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.