Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 59 (Quantum Field Theory)

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

Treat $\varphi ,{\varphi }^{†}$ as independent variables when taking the Legendre transform. Their canonical momenta areThe canonical quantization of a complex scalar field imposes, at equal times,with $[\varphi ,{\varphi }^{†}]=[\varphi ,\varphi ]=[\pi ,{\pi }^{†}]=[\pi ,\pi ]=0$, together with their adjoints, and $[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$. These are canonical commutation relations for two real scalar degrees of freedom, written in a complex basis. Local fields are operator-valued distributions, so these identities are understood after smearing or with a regulator.

The Hamiltonian density is $\pi \dot{\varphi }+{\pi }^{†}{\dot{\varphi }}^{†}-\mathcal{L}$, givingThe Heisenberg equation of motion is $\dot{O}=i[H,O]$. For example, $i[\int {\pi }^{†}\pi ,\varphi ]={\pi }^{†}$, while commuting $\pi$ with the gradient term and integrating its derivative of a Dirac delta distribution gives ${\nabla }^2{\varphi }^{†}$. ThusBoth fields satisfy the Klein-Gordon equation. Their spatial Fourier modes therefore have frequencies $\pm E_p$, $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. A complex scalar field has independent coefficients for these two frequency sectors, so, defining the invariant measure $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, its mode expansion isOne can recover the mode operators at any fixed time using complex scalar mode inversion:Substitution of the expansion verifies these inverses: the spatial integral selects the appropriate momentum, while the two frequency sectors acquire factors $E_p\pm E_q$. The unwanted sector vanishes when the delta function sets $E_p=E_q$. The canonical commutation relations then give, for example,Likewise $[b(p),b^{†}(q)]$ has the same value. The cross commutators vanish; their coefficients contain $E_p-E_q$ on the corresponding momentum delta function. Conversely, these oscillator brackets reproduce the equal-time field brackets, confirming the normalization of the measure.

Insert the mode expansion into $H$ and integrate over space. The terms $a^{†}{b}^{†}$ and $ba$ enforce opposite spatial momenta; their coefficient is $-E_p^2+{\mathbf{p}}^2+m^2=0$. For the diagonal terms the coefficient is $E_p^2+{\mathbf{p}}^2+m^2=2E_p^2$. Combining this with the two invariant measures leavesIn a box, $E_0={\sum }_{\mathbf{p}}{E}_p$, one half-quantum for each of the two real oscillator species. It diverges as the cutoff is removed. Normal ordering removes this vacuum energy, yielding the normal-ordered Hamiltonian of a free complex scalar field:In this nongravitational free theory an additive constant in $H$ changes only an overall state phase, not Heisenberg equations of motion, energy differences or scattering probabilities. It is therefore consistent to choose the Fock vacuum energy as zero. Both creation operators raise the energy: $[H,a^{†}(p)]=E_p{a}^{†}(p)$ and $[H,b^{†}(p)]=E_p{b}^{†}(p)$. Repeated creation builds a bosonic Fock space of spin-zero particles and antiparticles, with the same positive mass and energy. The negative-frequency part of the field does not describe negative-energy physical particles.

The global phase symmetry of a complex scalar field, with orientation $\varphi \mapsto e^{-i\alpha }\varphi$, has the indicated Noether current. Taking its divergence, the two first-derivative terms cancel, givingThe same calculation holds for its normal-ordered version. Integrating its temporal component gives a conserved charge when spatial boundary flux vanishes.

Here the ordering convention is important. Direct substitution into the literally ordered product $i\int ({\varphi }^{†}\dot{\varphi }-{\dot{\varphi }}^{†}\varphi )d^{3}x$ yieldsThe pair terms again vanish, now because their coefficient is a difference of equal energies. Reordering the $b$ term produces a divergent vacuum constant: with a finite-mode regulator $Q_{\mathrm{bare}}=N_a-N_b-N_{\mathrm{modes}}$. Thus the requested finite charge formula uses zero vacuum charge, or equivalently the vacuum subtraction of a complex scalar charge. With that conventional subtraction,The subtraction changes neither current conservation nor the action of $Q$ on fields; without it, the unrenormalized product does not literally equal the displayed normal-ordered operator.

Finally the oscillator commutators giveor $\boxed{Q{a}^{†}=a^{†}(Q+1),Q{b}^{†}=b^{†}(Q-1)}$. Acting on a charge eigenstate, the first creator raises its charge by one and the second lowers it by one. Starting from the neutral Fock vacuum, their one-particle states therefore have charges $+1$ and $-1$ respectively. This is the complex scalar charge operator $N_a-N_b$, whereas the energy counts the sum of the two occupations. A multiparticle state has additive charge equal to the number of particles minus the number of antiparticles.

The real scalar field has canonical momentum $\pi =\dot{\varphi }$. Its Legendre transform givesThe first three terms constitute $H_0$, the free massive Hamiltonian operator. The remaining term is $H_I=\lambda \int {\varphi }^4{d}^{3}x/4!=-\int {\mathcal{L}}_I{d}^{3}x$. This sign is a consequence of having no time derivatives in the interaction: the canonical momentum is unchanged, while the interaction enters the Hamiltonian density with the opposite sign to the Lagrangian density.

In the interaction picture, operators evolve with $H_0$ and states evolve with $H_I(t)$ constructed from the free interaction-picture field. Their evolution operator satisfiesIntegrating once gives $U_I=I-i{\int }_{t_0}^t{H}_I(t_1)U_I(t_1,t_0)d{t}_1$. Repeated substitution orders the integrations as $t\ge t_1\ge \cdots \ge t_n\ge t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a time-ordered product, which proves the Dyson seriesTaking the asymptotic limits with the usual adiabatic or regulated scattering prescription yieldsThe field in this expression is the free field evolving in the interaction picture, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.

Use relativistically normalized two-particle states $|p_1,p_2⟩=a^{†}(p_1)a^{†}(p_2)|0⟩$, with $⟨p|q⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. The connected first-order scattering contribution isHere the connected S-matrix uses a normalized vacuum and physical external one-particle states. At this order, normal ordering the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.

Two fields annihilate the incoming particles and two create the outgoing ones. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=4!$, the external-leg factorial cancellation at a quartic scalar vertex. Each attachment contributes its plane-wave factor, soThe spacetime integral is the four-momentum conservation delta function, givingIn particular, the tree scattering amplitude is independent of angle.

For completeness, derive the relativistic two-body phase space in the centre-of-mass frame. Write $\sqrt{s}=2E_{\ast }$, $p_{\ast }^2=E_{\ast }^2-m^2$. The spatial delta function sets the two final momenta opposite, and the energy delta function is $\delta (\sqrt{s}-2\sqrt{p^2+m^2})$. Its radial Jacobian is $E_{\ast }/(2p_{\ast })$. ThusThe invariant flux factor is $\mathcal{F}=4p_{\ast }\sqrt{s}$ for equal initial masses. Using $d\sigma =|T{|}^{2}d{\Phi }_2/\mathcal{F}$, the labeled angular density is ${\lambda }^2/(64{\pi }^{2}s)$. The outgoing quanta belong to one real scalar field, so they are identical. A full-sphere integration counts each event twice and requires the identical final-state symmetry factor $1/2!$. ConsequentlyEquivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong elastic scattering from a quartic scalar contact interaction cross-section.

Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.