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

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.