Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 48 / 3

The Maxwell field has the gauge symmetry $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$ for an arbitrary smooth scalar function $\alpha$. Commuting the two derivatives shows that $F_{\mu \nu }$ is unchanged, so the Maxwell Lagrangian is invariant. Vary the action, use antisymmetry of $F$ and integrate by parts with variations vanishing on the boundary:The Euler-Lagrange field equations are therefore ${\partial }_{\mu }{F}^{\mu \nu }=0$. Expanding $F$ givesIn Lorenz gauge the second term vanishes, leavingThe PDF calls this “Lorentz gauge”; the standard name is Lorenz gauge. A residual gauge symmetry preserves this condition when $\square \alpha =0$.

Put $D={\partial }_{\rho }{A}^{\rho }$. The gauge-fixing variation isAfter integration by parts its contribution to the Euler-Lagrange field equations is $+{\partial }^{\nu }D$. It cancels the $-{\partial }^{\nu }D$ from the Maxwell Lagrangian, giving $\square A^{\nu }=0$ for all four components. This is the Feynman gauge equation; the physical Lorenz gauge condition must still be imposed on states rather than inferred as an additional classical equation from this gauge-fixed density.

Treating the covariant components $A_{\mu }$ as coordinates, the derivative of the actual displayed density with respect to ${\partial }_0{A}_{\mu }$ gives the canonical momentaIn terms of lower-index spatial coordinates,The gauge fixing has supplied the nonzero momentum of $A_0$, absent from the ungauge-fixed Maxwell Lagrangian.

There is a boundary-term convention needed for the next part's printed momentum expansion. Direct expansion gives the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction:To verify it, the difference of the densities is $\frac{1}{2}[({\partial }_{\rho }{A}_{\sigma })({\partial }^{\sigma }{A}^{\rho })-D^2]$; differentiating $K^{\rho }$ gives these two terms, while the remaining second-derivative terms cancel after renaming indices. The subtracted density has ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$, which is exactly the momentum used by the mode expansion in part (c). The literal momenta asked for here are the boxed expression, not $-{\dot{A}}^{\mu }$. Both densities have the same field equations, and their canonical commutation relations agree through the boundary-induced canonical transformation in Feynman gauge.

Denote the momentum in the printed mode expansion by ${\tilde{\pi }}^{\nu }$, as it equals $-{\dot{A}}^{\nu }$ for the boundary-subtracted density discussed in part (b). Take a real complete basis of polarization vectors. Its completeness relation with one index raised isHere raising or lowering the polarization labels uses the same diagonal matrix $\eta$, so this is just the completeness relation in the PDF. Write $\mathbf{r}=\mathbf{x}-\mathbf{y}$. In the mixed field-momentum commutator, the $a{a}^{†}$ term gives the product of the minus sign in $\tilde{\pi }$'s creation part and the minus sign in the oscillator commutator. The $a^{†}a$ term gives the same sign. The two mode normalization factors multiply to $1/2$, and the momentum delta function removes one integral. ConsequentlyFor the field-field commutator, the polarization sum gives insteadbecause the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. Similarly,These are the requested canonical commutation relations, derived by photon oscillator completeness and canonical brackets.

They also hold for the literal momenta of part (b). The shifts areSince the fields commute at equal times, their mixed brackets with the momenta do not change. The only potentially new momentum bracket isAll other momentum brackets vanish separately. Thus both canonical conventions have $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ and vanishing coordinate-coordinate and momentum-momentum brackets, even though the momentum operators differ by spatial derivatives. This resolves the density/expansion mismatch without changing the printed oscillator algebra.

The oscillator commutator givesThe temporal mode $\lambda =0$ has negative norm. More precisely, the nonzero wave packet $|f,0⟩=\int d^{3}pf(\mathbf{p})a_{\mathbf{p}}^{0†}|0⟩/(2\pi {)}^3$ has norm $-\int d^{3}p|f(\mathbf{p}){|}^2/(2\pi {)}^3<0$. The difficulty is an indefinite Hermitian form, not the divergent normalization of an unsmeared momentum eigenstate. The covariant photon Fock space is therefore not a positive physical Hilbert space.

Gupta-Bleuler quantization imposes only the annihilation, or positive-frequency part of a quantum field, of the divergence:A strong operator identity $\partial \cdot A=0$ is incompatible with the four unconstrained canonical commutation relations; in the literal momentum convention it would set ${\pi }^0=0$. The weaker condition instead implies $⟨\Phi |\partial \cdot A|\Psi ⟩=0$ between physical states: the positive-frequency part annihilates the ket, and its adjoint annihilates the bra.

For a nonzero momentum choose upper-index polarization vectorswhere ${\mathbf{e}}_1,{\mathbf{e}}_2$ are an orthonormal pair perpendicular to $\mathbf{p}$. Labels 1 and 2 have transverse polarization, label 3 has spatial longitudinal polarization, and label 0 has timelike photon polarization. With $p^0=|\mathbf{p}|=\omega$, the divergence of the annihilation part is proportional to $-i\omega (a_{\mathbf{p}}^0-a_{\mathbf{p}}^3)$. The physical condition is thereforeFor a one-mode state $|\chi ⟩={\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, the condition reads $-c_0-c_3=0$. ThusThe last direction has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to ${ϵ}^{\mu 0}+{ϵ}^{\mu 3}=p^{\mu }/\omega$, hence it is a pure gauge symmetry direction. A purely longitudinal creator $a^{3†}$ alone is not physical, despite having positive norm.

For the multiparticle sketch, put $C=a^0-a^3$ and $C^{†}=a^{0†}-a^{3†}$. The commutator $[C,C^{†}]=0$ allows polynomials in $C^{†}$ times transverse creator states to obey the constraint. Conversely, on creator polynomials the constraint is $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, so these are all the constrained polynomials. Every term containing $C^{†}$ is orthogonal to every constrained bra, since $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting those null vectors and completing the remaining positive space is the Gupta-Bleuler null-state quotient. It givesThe one-particle instance is the transverse one-photon physical quotient.