Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 301 / 1

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

Write $D={\partial }_{\mu }{A}^{\mu }$. The additional gauge fixing variation is $\delta (-D^2/2)=-D{\partial }_{\mu }\delta A^{\mu }$. Integration by parts gives the Euler-Lagrange field equation ${\partial }_{\mu }{F}^{\mu \nu }+{\partial }^{\nu }D=0$, hence $\square A^{\nu }=0$. This equation follows from the gauge-fixed density without imposing $D=0$ as a separate operator identity.

Direct differentiation of the density as printed gives the canonical momentum components conjugate to the lower-index fields:There is a boundary-term convention to reconcile with part (c). The Feynman-gauge Maxwell kinetic density after a boundary-term subtraction isIt has the same Euler-Lagrange field equations, but its canonical momentum components are ${\pi }^{\prime \nu }=-{\dot{A}}^{\nu }$. The mode expansion supplied in part (c) is the expansion of these latter momenta. Thus it is valid after the stated boundary-term subtraction; it is not the direct derivative of the original density. For example, a time-independent spatially varying $A_0$ with $A_i=0$ gives ${\pi }^i=-{\partial }_i{A}_0$ but ${\pi }^{\prime i}=0$.

The boundary-term shift of canonical field momenta is a canonical transformation. If spatial boundary terms vanish, $B=\int d^{3}x{K}^0=\int d^{3}x{A}_0{\partial }_i{A}_i$, and ${\pi }^{\nu }={\pi }^{\prime \nu }+\delta B/\delta A_{\nu }$. This explains why the two conventions have the same dynamics while their momentum formulas differ.

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.