Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 3

Because $({\gamma }_5{)}^2=1$, its even and odd powers are ${\gamma }_5^{2r}=1$ and ${\gamma }_5^{2r+1}={\gamma }_5$. Separate the matrix exponential into these two power series:No representation of the gamma matrices is needed. Hermiticity of ${\gamma }_5$ also makes this exponential unitary for real $\alpha$, while the projectors $(1\pm {\gamma }_5)/2$ acquire opposite phases.

Put $U=e^{-i\alpha {\gamma }_5/2}$. The transformed Dirac adjoint is ${\bar{\psi }}^{\prime }={\psi }^{†}{U}^{†}{\gamma }^0=\bar{\psi }U$, since ${\gamma }^0$ anticommutes with ${\gamma }_5$. It is therefore incorrect to transform the two sides of a Yukawa bilinear with mutually inverse matrices.

Write $M(\varphi )={\varphi }_1+i{\gamma }_5{\varphi }_2$. The transformed Yukawa interaction is invariant exactly whenSince $M$ commutes with ${\gamma }_5$, this is $M({\varphi }^{\prime })=e^{i\alpha {\gamma }_5}M(\varphi )$. Use i and equate the coefficients of $1$ and $i{\gamma }_5$:This is the global SO(2) rotation of a chiral scalar-doublet Yukawa model. Equivalently, $\Phi ={\varphi }_1+i{\varphi }_2$ transforms as ${\Phi }^{\prime }=e^{i\alpha }\Phi$, while the two chiral spinor components transform with opposite half-angle phases. The sign of the scalar rotation follows from the negative sign in the specified fermion transformation.

The scalar kinetic and potential terms depend only on the rotation-invariant combinations ${\partial }_{\mu }{\varphi }_1{\partial }^{\mu }{\varphi }_1+{\partial }_{\mu }{\varphi }_2{\partial }^{\mu }{\varphi }_2$ and ${\varphi }_1^2+{\varphi }_2^2$. They are unchanged by the constant SO(2) rotation. The Yukawa term is invariant by ii. Finally, anticommutation gives $U{\gamma }^{\mu }={\gamma }^{\mu }{U}^{-1}$, soNo derivative of $U$ appears because the parameter is global. Thus the entire classical Lagrangian density has the combined chiral symmetry.

An explicit Dirac mass term instead changes asIt is not invariant for arbitrary $\alpha$ unless $m=0$. For the concrete choice $\alpha =\pi$, the original mass term reverses sign. The invariant Yukawa term can nevertheless generate a mass after a vacuum is selected, because the scalar field transforms along with the fermions; this is spontaneous rather than explicit breaking.

The scalar potential isSince ${\mu }^2,\lambda >0$, all minima form the circle ${\varphi }_1^2+{\varphi }_2^2=v^2$, with $v=\mu /\sqrt{\lambda }>0$. Choose the representative vacuum $⟨{\varphi }_1⟩=v$, $⟨{\varphi }_2⟩=0$, and write ${\varphi }_1=v+h$, ${\varphi }_2=\pi$. A nontrivial scalar rotation moves this vacuum to another minimum, so the continuous symmetry undergoes spontaneous symmetry breaking. The invariant action and the selected ground state have different stabilizers; the circle of equivalent minima is not evidence that the action lost its symmetry.

The Yukawa terms becomeComparison with the free Dirac equation givesFor the usual choice $g>0$, this is simply $gv$. The absolute value matters if the real coupling is negative: the dispersion depends on $m_{\psi }^2$, and a field redefinition can change the mass coefficient's sign. There is no explicit symmetry-breaking fermion mass parameter in the original theory.

The scalar symmetry is broken completely. Because the spinor transformation uses a half angle, $\alpha =2\pi$ leaves the scalars fixed while sending $\psi \to -\psi$; this discrete fermion parity is unbroken in the full combined transformation group. It does not remove the broken continuous generator or the associated Goldstone boson.

Using ${\mu }^2=\lambda v^2$, expand the potential around the chosen vacuum:The linear term vanishes at the minimum. With canonical real-field kinetic terms, comparison with $m_h^2{h}^2/2+m_{\pi }^2{\pi }^2/2$ givesThe massive radial scalar changes the magnitude of the vacuum field; the massless Goldstone boson changes its orientation. The symmetry is global here, so no gauge field eats that mode.

The genuinely new scalar cubic interactions are $-\lambda v{h}^3-\lambda vh{\pi }^2$. Including identical-field factorials, their Feynman vertex factors areThe quartic interactions remain $-\lambda h^4/4-\lambda h^2{\pi }^2/2-\lambda {\pi }^4/4$, with respective vertex factors $-6i\lambda ,-2i\lambda ,-6i\lambda$. The shifted Yukawa terms from iv give $h\bar{\psi }\psi$ vertex $-ig$ and $\pi \bar{\psi }\psi$ vertex $g{\gamma }_5$. These Yukawa structures were present before the shift; the new fermion bilinear is the mass term, not an additional independent interaction coupling. There are no derivative interactions from the linear Cartesian field shift.

The quadratic part of the covariant-derivative term contains $g^2{v}^2{A}_{\mu }{A}^{\mu }/2-gv{A}_{\mu }{\partial }^{\mu }b$. Its mixing term cancels the $+gv{A}_{\mu }{\partial }^{\mu }b$ in the stated gauge fixing term. Thus the vector mass and diagonal quadratic action areFor nonzero $p^2$ introduce the transverse and longitudinal transverse and longitudinal momentum projectors $P_T^{\mu \nu }=g^{\mu \nu }-p^{\mu }{p}^{\nu }/p^2$ and $P_L^{\mu \nu }=p^{\mu }{p}^{\nu }/p^2$. After integration by parts, the inverse quadratic kernel isTheir orthogonality makes the inversion immediate. The gauge-boson propagator isEquivalently,The projector formula explains the physical massive pole and the gauge-dependent longitudinal pole. The apparent $1/p^2$ in the projectors cancels in the equivalent expression; it is not an extra massless physical state. At $\xi =1$ the numerator reduces to $g^{\mu \nu }$. At fixed momentum the $\xi \to \infty$ limit gives the unitary gauge propagator $-i(g^{\mu \nu }-p^{\mu }{p}^{\nu }/M_A^2)/(p^2-M_A^2+i0)$. The relation between this limit and ultraviolet power counting is not uniform in momentum.^[ref]

At a stationary symmetry-breaking vacuum of a gauge-invariant scalar potential, the angular field has zero potential curvature. Its free kinetic term is $(\partial b{)}^2/2$, while the displayed gauge-fixing term adds $-\xi g^2{v}^2{b}^2/2$. HenceThis is the would-be Goldstone boson, whose gauge-dependent pole accompanies the longitudinal pole in i; it is not an additional physical scalar particle of that mass. For $M_A\ne 0$, fixed finite $p$ and $\xi \to +\infty$,The angular field becomes infinitely heavy in this gauge and is absent from the unitary gauge propagating description. The qualification about fixed momentum matters: loop integrations probe arbitrarily large $p$, so one cannot automatically interchange the $\xi \to \infty$ limit with ultraviolet integration. If $gv=0$, there is no massive Higgs phase and the asserted heavy-field limit does not follow. A separate radial Higgs boson may remain; its mass depends on the unspecified potential and is not determined by this gauge-fixing term.