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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.

The oscillator-generated covariant photon Fock space has an indefinite Hermitian form, not a positive Hilbert space inner product. For a normalizable one-photon wave packet of polarization $\lambda$, its squared norm is proportional to $-{\eta }^{\lambda \lambda }\int d^{3}p|f(\mathbf{p}){|}^2$. Thus $\lambda =0$ gives a negative-norm photon state. The divergent ${\delta }^3(0)$ of an unsmeared momentum eigenstate is a separate normalization issue, avoided by the wave packet.

Choose contravariant polarization vectors ${ϵ}^{0\mu }=(1,0)$, ${ϵ}^{3\mu }=(0,\hat{\mathbf{p}})$, and two ${ϵ}^{r\mu }=(0,{\mathbf{e}}_r)$ with ${\mathbf{e}}_r\cdot \mathbf{p}=0$ for $r=1,2$. The third spatial polarization is longitudinal polarization; the first two are transverse polarization. The timelike photon polarization is distinct from the longitudinal one.

The Gupta-Bleuler quantization condition sets the divergence of the positive-frequency part of a quantum field to zero on physical states:Restore the factors $e^{-i|\mathbf{p}|t}$ to the annihilation terms. Since $p_{\mu }{ϵ}^{0\mu }=|\mathbf{p}|$ and $p_{\mu }{ϵ}^{3\mu }=-|\mathbf{p}|$, Fourier transform gives the equivalent conditionIts sign depends on the chosen sign of the longitudinal polarization vector; the covariant condition does not.

To see its content for a general Fock state, temporarily discretize momentum and decompose one unphysical oscillator sector as $|\Psi ⟩={\sum }_{n,m\ge 0}|n,m{⟩}_{0,3}\otimes |{\chi }_{nm}{⟩}_{\perp }$, where $|n,m⟩=(a^{0†}{)}^n(a^{3†}{)}^m|0⟩/\sqrt{n!m!}$. The temporal oscillator obeys $a^0|n,m⟩=-\sqrt{n}|n-1,m⟩$, while $a^3|n,m⟩=\sqrt{m}|n,m-1⟩$. The condition therefore becomesEquivalently, the allowed finite-particle states use the transverse creation operators and only $b^{†}=a^{0†}-a^{3†}$ in the unphysical sector. Indeed the constraint acts on a polynomial of the two unphysical creation operators as $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, whose kernel consists of polynomials in their difference. Also $[b,b^{†}]=-1+1=0$, and a state containing $b^{†}$ is orthogonal to every constrained state because $b$ annihilates every such state. This argument applies mode by mode and extends by smearing to continuum momentum.

For one photon, $c_0{a}^{0†}|0⟩+c_3{a}^{3†}|0⟩$ is constrained only when $c_3=-c_0$, producing a null state. The Gupta-Bleuler null-state quotient removes these null directions. The condition excludes negative-norm physical states; quotienting its null states leaves the two positive-norm transverse photon polarizations. The condition alone gives a positive semidefinite Hermitian form, not yet a positive definite Hilbert space.

In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to $k$; taking the positive-frequency branch gives $\omega =c|k|$.

For $E_0=E(1,i,0)$, $k=k\hat{z}$, and $\phi =kz-\omega t$, the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is