Past exam of the quantum field theory course of the University of Oxford / 2025 / 2

For the complex scalar fields, use the conventional particle names: $\varphi$ annihilates a light $\varphi$ particle and creates a light antiparticle $\bar{\varphi }$; $\Phi$ annihilates a heavy $\Phi$ particle and creates $\bar{\Phi }$. These choices make the distinction between particle labels and field symbols explicit throughout ii and iii. The couplings $g,h$ are real as required by the printed Hermitian Lagrangian density.

Independent global phase symmetries would rotate $\varphi \mapsto e^{-i\alpha }\varphi$ and $\Phi \mapsto e^{-i\beta }\Phi$. The cubic interaction vertex preserves only $2\alpha +\beta =0$ modulo $2\pi$. The continuous internal symmetry of a classical field theory is therefore the circle groupNormalize the Noether charge so that $q_{\varphi }=1$. Then $q_{\Phi }=-2$, $q_{\bar{\varphi }}=-1$ and $q_{\bar{\Phi }}=2$. The Noether current iswhere $A\overleftrightarrow{\partial }B=A\partial B-(\partial A)B$. For example, the Euler-Lagrange field equations givewhich cancel with the relative coefficient $-2$. The quartic interaction vertex separately preserves both original phase rotations.

There is also a discrete charge conjugation symmetry $(\varphi ,\Phi )\mapsto ({\varphi }^{†},{\Phi }^{†})$ for these real couplings. Including it, the internal symmetry is $U(1)⋊{\mathbb{Z}}_2\cong O(2)$, with charge conjugation reversing the Noether charge. The connected symmetry and its conserved current are the $U(1)$ above.

The nonzero free scalar propagators join a field to its adjoint:Contractions $⟨T\varphi \varphi ⟩$, $⟨T\Phi \Phi ⟩$ and mixed-species contractions vanish. With all momenta incoming, the Feynman vertices areThe two identical light legs cancel the $1/2!$ in each cubic interaction vertex; there is no remaining factor $1/2$ in its Feynman rule. Multiply each Feynman vertex by $(2\pi {)}^4{\delta }^{(4)}(\sum p)$, and integrate an internal unconstrained four-momentum with $d^{4}p/(2\pi {)}^4$.

The arrows in the figure follow positive Noether charge: a light arrow carries one unit and a heavy arrow two. Consequently a heavy particle $\Phi$ travels opposite its charge arrow, whereas its antiparticle $\bar{\Phi }$ travels along it. At a ${\varphi }^2\Phi$ Feynman vertex, two unit-charge arrows enter and a two-unit arrow leaves; the conjugate Feynman vertex reverses every arrow. The quartic Feynman vertex has one incoming and one outgoing arrow of each weight. These rules prevent attaching a propagator between two unconjugated fields.

Let $p_1,p_2$ be the incoming and $p_3,p_4$ the outgoing four-momenta, and set $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, the Mandelstam variables. The only connected tree-level Feynman diagram is the $s$-channel exchange of $\bar{\Phi }$, with Noether charge $+2$. The two incoming light particles attach to ${\varphi }^2\Phi$, and the two outgoing light particles to its conjugate. There is no four-light contact Feynman vertex, nor a charge-allowed $t$ or $u$ exchange with these external labels.

Define the invariant scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. ThenThe Feynman vertices already contain the identical-leg combinatorics. No extra factor of two multiplies this invariant scattering amplitude; the distinct identical final-state symmetry factor enters a cross section integral.

Keep the particle labels specified under 2. There are two connected tree-level Feynman diagrams: the quartic contact interaction vertex and an $s$-channel light antiparticle exchange. The initial Noether charge is $1-2=-1$, exactly the charge of $\bar{\varphi }$. At the first cubic Feynman vertex, the incoming $\varphi$ and $\Phi$ leave the second $\varphi$ leg internal; the conjugate Feynman vertex supplies both outgoing particles.

ThusHere $s=(p_1+p_2{)}^2$ for incoming $\varphi (p_1),\Phi (p_2)$. Both terms have the same sign in the chosen S-matrix convention.

The distinction from antiparticles matters. If the paper's heavy “$\Phi$ particle” is instead intended to mean the charge-$+2$ state $\bar{\Phi }$, so that its printed decay into $\varphi \varphi$ is allowed, the exchange diagram is crossed. For that alternative physical process,This is a consistent alternative naming convention only if it is also applied to the decay in iii; one cannot use the same heavy particle label for both charge assignments.

With the conventional particle labels, the allowed cubic decay channels areEach has one tree-level Feynman diagram, consisting of one cubic Feynman vertex. The final identical-leg contractions supply $2!$ and cancel the $1/2!$ in the Lagrangian density.

Taken literally with these labels, the printed $\Phi \to \varphi \varphi$ channel is forbidden: its initial Noether charge is $-2$ and its final Noether charge is $+2$. Its decay amplitude is zero, at every order preserving the global phase symmetry. If the source uses “$\Phi$” to name the heavy antiparticle instead, its intended answer is the nonzero decay amplitude $-h$ for $\bar{\Phi }\to \varphi \varphi$. The following decay width calculation concerns either allowed channel, whose widths are equal by charge conjugation.

For an allowed decay channel in iii, $|\mathcal{M}{|}^2=h^2$. Integrating the spatial Dirac delta distribution in the printed Lorentz-invariant phase-space measure sets ${\mathbf{p}}_2=-{\mathbf{p}}_1$. With $p=|{\mathbf{p}}_1|$ and $E_p=\sqrt{p^2+m^2}$, its remaining value isThe positive root is $p_{\ast }=(M/2)\beta$, where $\beta =\sqrt{1-4m^2/M^2}$, and the Dirac delta distribution contributes the Jacobian determinant $E_{\ast }/(2p_{\ast })$. HenceIt follows that the literal printed integral evaluates toThe printed integration ranges count each final configuration twice, once as $({\mathbf{p}}_1,{\mathbf{p}}_2)$ and once in the opposite order. For two identical final bosons, the physical decay width therefore requires an additional identical final-state symmetry factor $1/2!$:There is one allowed decay channel per heavy particle, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the two-body decay threshold $M=2m$, and the limiting value at threshold is zero. For the literally forbidden labels $\Phi \to \varphi \varphi$, the decay width is instead zero because its decay amplitude vanishes, as explained in iii.

For $g=0$, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at $s=M^2$. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,Here $M$ is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor $1/2!$, gives $\sigma =|\mathcal{M}{|}^2/(32\pi s)$. Thus the Breit-Wigner resonance cross section isAt $s\gg M^2$ it reproduces $h^4/(32\pi s^3)$. At the resonance peak,This is also $8\pi /k_{\ast }^2$ for $k_{\ast }=M\beta /2$, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes ${\Gamma }_{\mathrm{phys}}\ll M$ and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized ${\Gamma }_{\mathrm{printed}}=2{\Gamma }_{\mathrm{phys}}$ from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak $8\pi /(M^2{\beta }^2)$. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.