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There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.

A stable symmetry-breaking minimum requires $\lambda >0$ and $m^2<0$. Put $v^2=-m^2/\lambda >0$. The common scalar potential can then be written, up to a constant, asIts minima form the sphere $\Phi \cdot \Phi =v^2$. Choose the vacuum direction ${\Phi }_0=(0,0,v{)}^T$. If instead $m^2>0$ with positive $\lambda$, the minimum is at the origin and the assumed nonzero vacuum expectation value is not a stable vacuum of these models.

In the ungauged theory, the internal symmetry is global $O(3)$; its connected part is $SO(3)$. Constant orthogonal transformations preserve both the kinetic term and the potential. The chosen vacuum is invariant under $O(2)$ acting on its first two components, so the continuous breaking is $SO(3)\to SO(2)$ with two broken generators. The vacuum orientations are genuinely different degenerate global vacua.

The global triplet scalar symmetry breaking is visible directly in the fluctuation masses. Write $\Phi =({\pi }_1,{\pi }_2,v+h{)}^T$. The potential Hessian isThe quadratic Lagrangian is thereforeThere is one radial scalar of mass $m_h^2=2\lambda v^2=-2m^2$ and two massless Goldstone bosons. The two Goldstone modes describe motion tangential to the vacuum sphere. They remain physical, since a global rotation has only constant parameters and cannot remove arbitrary spacetime-dependent fluctuations. All three original real-scalar degrees of freedom remain. Interactions persist: the shifted potential is $V-V_{\min }=\frac{\lambda }{4}(2vh+h^2+{\pi }_1^2+{\pi }_2^2{)}^2$, containing cubic and quartic terms. Masslessness of the angular modes is protected by the exact continuous symmetry through Goldstone's theorem.

In the gauged theory, the three-component real field is the vector representation of local $SO(3)$, equivalently the Adjoint representation of $SU(2)$. The center of $SU(2)$ acts trivially on this field, so both descriptions have the same local field content and perturbative spectrum. Take $(t^a{)}_{bc}=-i{ϵ}^{abc}$, for which $[t^a,t^b]=i{ϵ}^{abc}{t}^c$. The covariant kinetic term and the gauge kinetic term are invariant under local transformations with the associated connection transformation. The scalar inversion $\Phi \mapsto -\Phi$, with the gauge field unchanged, is also a discrete invariance of the displayed action.

The vacuum leaves rotations about its third axis unbroken. The residual gauge group is $SO(2)$, or the corresponding $U(1)$ subgroup in the $SU(2)$ description. Expanding the covariant kinetic term about ${\Phi }_0$ givesThus the gauge-boson mass matrix is $g^2{v}^2\mathrm{diag}(1,1,0)$. The third generator annihilates the vacuum, while the first two do not. The masses are $m_{B^1}=m_{B^2}=gv$, $m_{B^3}=0$, and $m_h=\sqrt{2\lambda }v$.

The quadratic derivative mixing displays what happens to the Goldstone fields. With the generator convention just chosen,Local rotations can set the two angular fields to zero in unitary gauge. Then $\Phi =(0,0,v+h{)}^T$ andThe two Goldstone modes supply the longitudinal polarizations of the two massive vector bosons. They do not survive as additional physical massless scalars. The radial scalar is still physical, and $B^3$ retains only two transverse polarizations. This adjoint triplet Higgs spectrum realizes the Higgs mechanism; the gauge group is only partially broken, so one massless gauge boson remains.

The degree-of-freedom check isIn contrast, the global theory has just the three scalar modes, including its two physical Goldstone bosons. A vacuum expectation value in the local theory is a choice of gauge-fixed description: the direction can be changed by gauge transformations and is not itself an observable. Gauge-independent masses and the physical degree count express the actual effect. Both theories retain Lorentz symmetry in the constant scalar vacuum.

Take real field coordinates ${\varphi }^I$; for complex scalar fields, split them into real and imaginary parts. Assume a smooth scalar potential and positive, canonically normalized kinetic terms in a relativistic theory. At a classical vacuum, $V_I({\varphi }_0)=0$. The Taylor expansion isThus the scalar mass matrix is the Hessian matrix, and its eigenvalues give squared masses of the linearized scalar excitations.

Put $X_a(\varphi )=i{t}^a\varphi$. Infinitesimal invariance of the scalar potential gives the identity $V_I{X}_a^I=0$ at every field value. Differentiate with respect to ${\varphi }^J$ and evaluate at the classical vacuum:Every nonzero infinitesimal symmetry direction therefore lies in the kernel of the scalar mass matrix. This is the classical Goldstone theorem expressed through Goldstone directions in the scalar mass matrix.

Let $H_0$ be the full stabilizer subgroup of ${\varphi }_0$. The linear map from the Lie algebra of $G$ to field space, $T\mapsto iT{\varphi }_0$, has kernel equal to the Lie algebra of $H_0$. The rank-nullity theorem givesThere are consequently at least $\dim G-\dim H_0$ massless scalar directions, namely the tangent directions to the symmetry orbit through the classical vacuum. If the intended $H$ is $H_0$, these are the $\dim G-\dim H$ symmetry-required Goldstone bosons. Exactly that many massless modes occur if the scalar mass matrix is positive definite on a complement of those tangent directions. A nonsingular positive field-space kinetic metric changes normalization, but not the number of zero masses.

Two qualifications are needed for the literal assumptions. A subgroup fixing the classical vacuum need not be the full stabilizer subgroup. For example, take $G=SO(3)$ acting on a real triplet and $V=\lambda (|\varphi {|}^2-v^2{)}^2/8$ with $\lambda ,v>0$. At ${\varphi }_0=v{e}_3$, the scalar mass matrix is $\lambda v^2\mathrm{diag}(0,0,1)$: there are two massless modes. Choosing $H=\{1\}$ satisfies the printed invariance condition but would incorrectly predict three. The full stabilizer subgroup is $SO(2)$.

Even with the full stabilizer subgroup, symmetry does not exclude an accidental massless scalar. Take $G=SO(2)$ rotating $(x,y)$, with an invariant singlet $z$, andAt $(v,0,0)$ the full continuous stabilizer is trivial, but the scalar mass matrix is $\mathrm{diag}(2\lambda v^2,0,0)$. One zero direction is the Goldstone boson; the other is an accidental massless scalar at quadratic order. The proof establishes the symmetry-required count, not unconditional equality with the total number of massless fields. The statement concerns global internal symmetry; gauging it changes the physical interpretation through the Higgs mechanism.

Assume $\lambda >0$ and $v>0$. The vacuum manifold is the sphere $|\varphi |=v$. Choose ${\varphi }_0=v{e}_3$. For the given generators, $t^3{\varphi }_0=0$ whereas $t^1{\varphi }_0$ and $t^2{\varphi }_0$ are nonzero and independent. The Adjoint representation of SU(2) rotates this sphere, and the full connected stabilizer subgroup of $v{e}_3$ is generated by $t^3$. Hence the classical symmetry pattern isThe central element of SU(2) acts trivially on the triplet and belongs to this stabilizer; it does not add a separate broken direction. The gauge-field-free representative has zero gauge field strength and constant scalar magnitude. Locally near this nonzero classical vacuum, unitary gauge aligns the triplet along the third axis.

First make the gauge normalization explicit. Write the gauge kinetic term as $-\mathcal{N}{F}_{\mu \nu }^a{F}^{a\mu \nu }/4$, where $\mathrm{Tr}(t^a{t}^b)=\mathcal{N}{\delta }^{ab}$ if $F=F^a{t}^a$ is the matrix used inside the printed trace. Canonical normalization of a gauge kinetic term givesFor a conventionally normalized component trace, $\mathcal{N}=1$ and $g_c=g$. If the trace is the ordinary matrix trace in the displayed three-dimensional representation, $\mathrm{Tr}(t^a{t}^b)=2{\delta }^{ab}$ and $\mathcal{N}=2$. The PDF does not specify which trace convention is intended; both are covered by this formula.

In unitary gauge, the gauge covariant derivative isExpanding this kinetic term and the scalar potential gives the complete physical-field LagrangianHere a square of a vector means its contraction with the Minkowski metric, in signature $(+---)$. The labels 1 and 2 represent massive vectors; 3 represents the surviving Abelian vector.

To display their charge and all interactions more clearly, put $a_{\mu }={\mathcal{A}}_{\mu }^3$, $W_{\mu }^{\pm }=({\mathcal{A}}_{\mu }^1\mp i{\mathcal{A}}_{\mu }^2)/\sqrt{2}$, and defineWith the sign of gauge field strength printed in the paper, ${\mathcal{F}}^3=f+i{g}_{c}C$ and $G^{\pm }=({\mathcal{F}}^1\mp i{\mathcal{F}}^2)/\sqrt{2}$. Thus the same Lagrangian becomesThe physical masses areIn particular, literal adjoint matrix trace gives $m_W=gv/\sqrt{2}$; the standard canonical component convention gives $m_W=gv$. These are descriptions with differently normalized couplings, not different physical spectra.

The complex vector pair carries opposite charges under the unbroken U(1) gauge symmetry. The gauge field strength terms contain $a{W}^{+}{W}^{-}$ cubic interactions, $aa{W}^{+}{W}^{-}$ quartic interactions, and four-vector interactions involving the charged fields. There is no pure Abelian cubic or quartic self-interaction. The Higgs mode $\eta$ has cubic and quartic scalar interactions and couples through $2g_c^{2}v\eta W^{+}{W}^{-}+g_c^2{\eta }^2{W}^{+}{W}^{-}$. It is neutral and has no tree-level $\eta aa$ interaction. This is the physical charged-vector Lagrangian for an adjoint SU2 Higgs model.

If the angular fields ${\pi }_1$ and ${\pi }_2$ were retained, their vanishing potential masses would identify the two Goldstone bosons of the ungauged triplet. In the gauge theory they mix with the broken-direction gauge fields and can be removed by gauge fixing; the Higgs mechanism uses them as the longitudinal polarizations of the two massive vectors. They are not additional physical massless scalars. The physical degrees of freedom are conserved: $3+3\times 2=9$ before rearrangement, and $1+2\times 3+2=9$ afterwards.

This is not the Standard Model electroweak interaction. It has three original gauge bosons, leaving two massive charged vectors and one massless neutral vector, with no massive neutral Z boson. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y$, a complex Higgs doublet, and three massive vectors plus the photon. Its charge is $Q=T_3+Y$, not just the surviving $T_3$. Adding fermions cannot supply the missing gauge generator or turn the surviving neutral vector into both a photon and a Z boson. In particular, the usual right-handed fermions in the Standard Model are SU(2) singlets; they would have zero charge if only $T_3$ were available, instead of the charges produced by hypercharge.

A homogeneous static saddle minimizes $f(\rho )=-\mu \rho +g{\rho }^2/2$ for $\rho =|\psi {|}^2\ge 0$. With $g>0$ and $\mu >0$,The mean-field Bose-Einstein condensate has a macroscopic coherent occupation of the uniform mode. Its continuous particle-number phase symmetry is $\psi \mapsto e^{i\chi }\psi$, the circle group $U(1)$ generated by particle number. Choosing one phase breaks it, and the saddle manifold is a circle. For $\mu \le 0$, the constrained minimum is the vacuum ${\rho }_0=0$, so the condensed saddle requires the stated positive-density regime.

Spontaneous symmetry breaking is understood through a phase-selected thermodynamic description: an exact finite-volume number eigenstate has vanishing field expectation. The resulting Goldstone boson is a gapless phase/sound mode, while phase gradients carry the superfluid velocity $\mathbf{v}=\nabla \varphi /m$. This is the saddle-point conclusion requested, not a proof of true condensate order in every dimension or temperature. Long-wavelength phase fluctuations can invalidate the assumed order, as quantified below.

Take the Minkowski metric with signature $(+---)$ and a coupling $e\ne 0$. For a charged scalar field use the gauge covariant derivativeA scalar electrodynamics Lagrangian isAny real potential bounded below and depending only on the modulus gives gauge invariance. Indeed,The derivative of the phase cancels the shifted gauge field in the first identity; commuting partial derivatives proves the second. These identities verify invariance of every term, including the interaction hidden in the kinetic term.

For an unbroken example choose $V=m_{\varphi }^2|\varphi {|}^2+\lambda |\varphi {|}^4$ with $m_{\varphi }^2>0$ and $\lambda >0$. The minimum is at zero. There is no vector mass term in the quadratic expansion and no Higgs mechanism. Pure electromagnetic theory has a neutral massless spin-one photon with two physical transverse polarizations, equivalently helicities $+1$ and $-1$; longitudinal and time-component polarizations are gauge redundancies. In the unbroken scalar theory, the same massless photon is accompanied by a spin-zero charged particle and its oppositely charged antiparticle, both of mass $m_{\varphi }$. A complex scalar has two real physical degrees of freedom, rather than two unrelated charged species.

For a Higgs example chooseThe vacuum modulus is nonzero. Around one vacuum representative, unitary gauge removes the phase and writes $\varphi =(v+h)/\sqrt{2}$. ThenThus the quadratic spectrum isThe gauge boson is a massive spin-one particle with three polarizations, and $h$ is a neutral massive spin-zero Higgs boson. The scalar phase supplies the longitudinal vector polarization; it is not an extra physical massless Goldstone boson. The degree count is unchanged: two massless-vector polarizations plus two scalar degrees become three massive-vector polarizations plus one radial scalar degree. The underlying gauge invariance remains a redundancy of the description.

For the two-charge theory, useand retain the same gauge transformation of $a_{\mu }$. Both derivatives transform with the phase of their own field. A manifestly stable two-charge scalar gauge potential is, for positive $m_{\varphi }^2,M^2,{\lambda }_{\varphi },{\lambda }_{\psi }$, $g\ge 0$ and a nonzero complex constant $b$,Since $\psi -b{\varphi }^2$ has charge $2e$, its modulus is gauge-invariant. Expanding its square exhibits the direct couplingIn particular the charge in ${\psi }^{\ast }{\varphi }^2$ is $-2e+2e=0$, which uses the stated charge ratio. The potential is real, bounded below and has the zero-field minimum, while the kinetic terms have the standard positive signs. A complete example is thereforeThere is genuine direct interaction even with $g=0$ because $b\ne 0$. In four spacetime dimensions $b$ has mass dimension $-1$, but the expanded potential contains only quadratic, cubic and quartic field monomials; the cubic coefficient $M^{2}b$ has dimension one. Thus the example is also power-counting renormalizable.

Take $D_{\mu }={\partial }_{\mu }+i{K}_{\mu }$, with the Pauli matrices acting on the Higgs doublet. For electroweak hypercharge $Y=1/2$,This specifies every component of the gauge covariant derivative for the electroweak interaction. Expanding the gauge-covariant kinetic term makes its derivative, trilinear and quartic interactions explicit:where the Pauli matrix multiplication law givesThe antisymmetric Pauli contribution vanishes because $W_{\mu }^a{W}^{b\mu }$ is symmetric in $a,b$. Reversing the sign convention for $D_{\mu }$ reverses the linear gauge interactions consistently, without changing the masses.

A nonzero vacuum expectation value requires ${\mu }^2<0$. Minimizing the Higgs potential gives $v^2=-{\mu }^2/\lambda$. By an gauge transformation choosein unitary gauge. The electroweak doublet gauge-boson mass matrix follows by inserting the vacuum expectation value in the gauge-covariant kinetic term:Define the charged electroweak gauge bosons and the neutral rotation through the Weinberg angle bywith inverse $W_{\mu }^3=c_W{Z}_{\mu }+s_W{A}_{\mu }$ and $B_{\mu }=-s_W{Z}_{\mu }+c_W{A}_{\mu }$. ThenThe factors differ because $W^{+}$ and $W^{-}$ are conjugate fields whereas $Z$ is real. The massless photon corresponds to the unbroken Lie algebra generator $Q=T_3+Y$, which annihilates $⟨\varphi ⟩$. Thus three of the four real gauge bosons acquire mass, with $m_W=m_Z{c}_W$. The three would-be Goldstone bosons provide their longitudinal polarizations; the remaining scalar $h$ is the Higgs boson.

Introduce the left-handed lepton doublet $L=({\nu }_{eL},e_L{)}^T$, with $Y_L=-1/2$, and the right-handed singlet $e_R$, with $Y_R=-1$. Use the chiral projectors $P_L=(1-{\gamma }^5)/2$ and $P_R=(1+{\gamma }^5)/2$ in the course's convention. The minimal Standard Model has no right-handed neutrino. The gauge-invariant fermion terms areThe doublet contraction in the Yukawa interaction is a singlet, and its total hypercharge is $+1/2+1/2-1=0$. A bare term ${\bar{e}}_L{e}_R$ would fail electroweak gauge invariance. The gauge covariant derivatives above already give all the requested fermion-gauge couplings. In terms of mass eigenstates, $e=g{s}_W=g^{\prime }{c}_W$ and they becomeThus the weak charged current is chiral and the neutrino has zero electric charge. The gauge-invariant electron Yukawa mass follows from the Yukawa interaction after electroweak symmetry breaking:Rephase $e_R$ to make $y_e$ real and positive. It givesThe electron Dirac mass is therefore compatible with the original gauge symmetry through the Higgs mechanism. The neutrino remains massless in this minimal renormalizable lepton sector.

For a differentiable curve $R(t)$ in the SO(3) group with $R(0)=I$, differentiating $R(t{)}^{T}R(t)=I$ gives $X^T+X=0$, where $X=R^{\prime }(0)$. Conversely, if $X$ is a real skew-symmetric matrix, $e^{tX}$ is an orthogonal matrix and has determinant $e^{t\mathrm{tr}X}=1$. HenceThe displayed generators of the SO(3) Lie algebra areThey are linearly independent and span the three independent entries of a real skew-symmetric matrix. Equivalently, $T_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The vector triple-product identity then givesThus the structure constants of a Lie algebra in this basis are

Use a real, antisymmetric convention throughout the SO(3) vector Higgs model. With gauge coupling $g$, the gauge covariant derivative and gauge field strength areThere is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local non-Abelian gauge transformation ${\Phi }^{\prime }=R\Phi$, with $R(x)$ in the SO(3) group, the compatible transformation of the gauge potential isIt gives $D_{\mu }^{\prime }{\Phi }^{\prime }=R{D}_{\mu }\Phi$ and $F_{\mu \nu }^{\prime }=R{F}_{\mu \nu }{R}^{-1}$. These covariance laws fix the relative signs in the gauge covariant derivative and gauge field strength.

An SO(3) invariant quartic scalar potential isSince $(R\Phi {)}^T(R\Phi )={\Phi }^T\Phi$, this scalar potential is gauge-invariant. Its minimum is the sphere ${\Phi }^T\Phi =v^2$, rather than a preferred direction in the internal vector space.