Past exam of the theoretical physics major option course of the University of Oxford / 2017 / 4

The continuous internal global symmetry in field theory isBoth $|\phi {|}^2$ and ${\partial }_{\mu }{\phi }^{\ast }{\partial }^{\mu }\phi$ are unchanged. The group is the circle group, acting by phase rotations. The ordinary derivatives do not make it a local gauge symmetry: spacetime-dependent $\alpha$ would add derivative terms. There is also the discrete symmetry of complex conjugation; in the two-real-field description the full internal invariance is $O(2)$, whose connected component is this $U(1)$.

The scalar potential is $V=-m^2|\phi {|}^2+\lambda |\phi {|}^4/4$. Completing the square givesThe Hamiltonian density also contains the nonnegative terms $|\dot{\phi }{|}^2+|\nabla \phi {|}^2$. Its lowest-energy configurations therefore have no time or spatial variation and minimize the displayed square. ThusThe minima form a circle of phases. Selecting the real positive representative $\phi =v$ chooses one vacuum and spontaneously breaks the continuous global symmetry in field theory. The stationarity at $\phi =0$ is a local maximum of the radial potential, not the ground state.

Put $r=v+\sigma /\sqrt{2}$. Direct differentiation of $\phi =r{e}^{i\pi /f}$ cancels the mixed radial-angular terms, leavingUse $\lambda =2m^2/v^2$ from part b to expand the potential. The complete local expression isThe radial field is canonically normalized, giving $m_{\sigma }^2=2m^2$. For arbitrary $f$, the canonically normalized angular field is ${\pi }_c=\sqrt{2}v\pi /f$ and has zero mass:The interactions comprise radial cubic and quartic self-couplings, a derivative $\sigma \pi \pi$ coupling and a derivative $\sigma \sigma \pi \pi$ coupling. In canonical variables the latter coefficients are $1/(\sqrt{2}v)$ and $1/(4v^2)$ respectively. The additive constant shifts the vacuum energy and does not affect these masses or interactions.

The original phase symmetry becomes the exact shift $\pi \mapsto \pi +f\alpha$, with $\sigma$ unchanged. Any undifferentiated polynomial ${\pi }^3$ or ${\pi }^4$ would violate this shift, so such terms are absent. This argument allows derivative interactions; it forbids an angular potential. The massless angular excitation is the Goldstone boson for the broken global $U(1)$.

Choose the current convention corresponding to $\delta \phi =i\alpha \phi$. Noether's theorem givesFor the requested terms with no $\sigma$ and $f=\sqrt{2}v$, the Lagrangian is $\frac{1}{2}(\partial \pi {)}^2$ and the current becomes $J^{\mu }=f{\partial }^{\mu }\pi$. The supplied covariant oscillator normalization cancels the invariant measure when the annihilation operator contracts with $a^{†}(p)$, soThe momentum is on the massless shell, $p^2=0$. Differentiating once more therefore givesThis Goldstone current matrix element in a phase-only scalar theory shows that a conserved current can have a nonzero matrix element proportional to a null momentum. Thus current conservation is fully consistent with the current creating a massless one-particle state from the vacuum; the nonzero coefficient $f$ is its Goldstone-current coupling. It is the divergence that vanishes, not the current matrix element. Reversing the current convention reverses the nonzero matrix element but leaves this zero divergence unchanged.