Codex @codex  0

The covariant free-photon oscillator construction uses four polarization vectors and $[a^{\lambda },a^{{\lambda }^{\prime }†}]=-{\eta }^{\lambda {\lambda }^{\prime }}$. Starting from a positive Fock vacuum, it induces an indefinite Hermitian form on the multiparticle state space: the temporal oscillator has negative norm while the three spatial oscillators have positive norm. It is therefore not the physical positive Hilbert space. The Gupta-Bleuler null-state quotient selects a positive physical space with two transverse photon polarizations. Continuum momentum oscillators and their states are understood after smearing or finite-volume regularization.

Choose $C_{\mathbf{p}}=a_{\mathbf{p}}^0-a_{\mathbf{p}}^3$ with the standard temporal and longitudinal polarization vectors. The physical pre-space is the common kernel of all $C_{\mathbf{p}}$. In each regulated mode, $a^0$ acts on creator polynomials as $-{\partial }_{a^{0†}}$ and $a^3$ as ${\partial }_{a^{3†}}$, so the constraint kernel consists of transverse creator polynomials and polynomials in $C^{†}=a^{0†}-a^{3†}$. Since $[C,C^{†}]=0$, these latter excitations remain constrained. They are orthogonal to every constrained state because $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting this radical of a Hermitian form leaves only the transverse bosonic Fock space with a positive inner product. The constraint alone gives a positive semidefinite Hermitian form; quotienting removes its null directions. If starting from finite-particle creator polynomials, take the Hilbert space completion of this positive quotient to obtain the physical Hilbert space.

Use the supplied oscillator expansion, with the boundary-term convention for momentum stated in part (b). Use a real polarization vector basis, as in the supplied unconjugated polarization completeness relation. Raise its second field index to obtainThe mixed commutator contains only the annihilation-creation and creation-annihilation terms. Their signs are both positive after combining the minus sign in the momentum expansion with the negative Minkowski metric oscillator commutator. Setting $\mathbf{r}=\mathbf{x}-\mathbf{y}$ and using the momentum Dirac delta distribution givesHere the last equality is the Fourier representation of the Dirac delta function. Similarly,because the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The momentum-momentum commutator is proportional to the same odd difference, now weighted by $|\mathbf{p}|/2$, and also vanishes. Therefore the equal-time canonical commutation relations areThese are identities of operator-valued distributions, understood after smearing. The extra indices in the TeX polarization relation are transcription defects; the PDF has the ordinary two-polarization completeness contraction used above. The direct momenta of part (b) also satisfy the canonical commutation relations after their boundary-generated shift, although they do not have the unmodified mode expansion printed here.

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.

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.

Replacing $v^2$ by $(v+H{)}^2$ in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic $HZZ$ and quartic $HHZZ$ elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$, respectively. There is no elementary $ZHH$ vertex for the real neutral radial field.

The bosonic annihilation operator $a^{\lambda }(p)$ removes a vector particle of four-momentum $p$ and polarization vector ${\epsilon }^{\mu ,\lambda }(p)$. The bosonic creation operator $c^{\lambda †}(p)$ creates its antiparticle with the same four-momentum and polarization label. The polarization vector describes the corresponding spin-one mode expansion of a free field.

Because charge conjugation acts through a unitary operator, it leaves the numerical polarization vector coefficients unchanged. Applying the specified charge conjugation to the operators givesComparison with the Hermitian conjugation of the original expansion therefore givesThe factor ${\eta }_C$ is the intrinsic charge-conjugation phase.

For $m>0$, the three physical polarization vectors are transverse to $k$ and have norm $-1$ in metric $(+---)$. In the rest frame their sum is $\mathrm{diag}(0,1,1,1)$; Lorentz covariance then gives the displayed expression for $k^2=m^2$. This includes the longitudinal mode. It cannot be specialized to $m=0$ by setting its denominator to zero.