Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 3

Use metric signature $(+,-,-,-)$ and the model's rescaled units $c=\hslash =1$. Its exponential implicitly uses dimensionless rescaled field and coupling combinations; the numerical coefficients of the potential are those supplied. Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables in taking canonical derivatives:The Hamiltonian density isThe kinetic and gradient terms are nonnegative, so a static homogeneous minimum is obtained by minimizing the potential. Put $s=|\varphi {|}^2\ge 0$. For $\lambda =0$, $V=1-s$ is unbounded below. For $\lambda <0$, the exponential decays while $-s$ tends to minus infinity. Thus a stable vacuum requires $\boxed{\lambda >0}$.

For positive $\lambda$,The exponential complex-scalar potential consequently hasFor $0<\lambda <1$, the minima form a circle $\varphi =v{e}^{i\theta }$, with $v^2=-\ln \lambda /\lambda$ and $V_{\min }=(1+\ln \lambda )/\lambda$. For $\lambda \ge 1$ the minimum has $V_{\min }=1$. The symmetry remains global phase rotation in either phase, but a chosen nonzero vacuum breaks it spontaneously. At $\lambda =1$ the origin is a unique minimum with vanishing quadratic curvature and a positive leading quartic term; this critical masslessness should not be confused with a Goldstone mode.

Let $v=\sqrt{-\ln \lambda /\lambda }>0$ and write $\varphi =v+\chi$. ThenSince $V^{\prime }(v^2)=0$ and $V^{\prime \prime }(v^2)=\lambda$, the quadratic potential isThereforeThe constant can be dropped from the fluctuation equations. For canonical real fields, use $\chi =(h+i\pi )/\sqrt{2}$:These are the radial and angular masses of a phase-invariant scalar potential. The radial Higgs mode changes the field magnitude and costs energy to second order. The angular Goldstone boson changes the phase along the vacuum circle without a quadratic restoring force. Its linear dispersion is ${\omega }^2=|\mathbf{k}{|}^2$, whereas the radial mode has ${\omega }^2=|\mathbf{k}{|}^2+m_h^2$. The zero angular mass follows the continuous broken global symmetry, not from an accidental cancellation in a particular coordinate choice.

Write $\varphi =(v+h/\sqrt{2})e^{i\vartheta }$. The covariant kinetic term containsA local U(1) gauge symmetry can remove $\vartheta$ in unitary gauge. The angular fluctuation no longer appears as an independent physical massless Goldstone boson; its degree of freedom supplies the longitudinal polarization of the massive vector field. At quadratic order,The factor of two follows from the convention that $v$ is the complex field's vacuum value. This Higgs mechanism changes a two-polarization massless electromagnetic field into a three-polarization massive Proca field; the remaining radial scalar keeps $m_h^2=2\lambda v^2$. Static fields in the homogeneous broken phase are screened on a scale of order $m_A^{-1}$ in these units. The gauge symmetry is a redundancy of the description; the physical statement is the changed spectrum and disappearance of an independent massless angular mode.

Write $\varphi =x+iy$. The added term is $-(x^2-y^2)$, so the full potential simplifies toThe phase-rotation subgroup is reduced from U(1) to the two-element symmetry $\varphi \mapsto -\varphi$. More fully, the kinetic term and potential are invariant under independent reflections $x\mapsto -x$ and $y\mapsto -y$, a ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$ internal symmetry generated by $\varphi \mapsto -\varphi$ and complex conjugation. There is no continuous internal rotation symmetry for the explicit twofold anisotropy of a complex scalar field.

For fixed $x$, positive $\lambda$ makes the exponential grow with $y^2$, so every global minimum has $y=0$. On the real axis,ThusFor $0<\lambda <2$, the nonzero stationary points are the two global minima; the origin is unstable along the real direction. For $\lambda \ge 2$ the origin is the unique minimum, with zero real-direction quadratic curvature at the critical value $2$. Selecting $+v$ breaks the sign reversal that exchanges the two vacua, while complex conjugation remains unbroken.

Set $\varphi =v+\chi$, $\chi =u+iw$. Since $e^{\lambda v^2}=2/\lambda$, expansion givesEquivalently, in canonical fields $h=\sqrt{2}u$, $\pi =\sqrt{2}w$,Both masses are positive throughout the broken phase $0<\lambda <2$. A broken discrete symmetry does not demand a massless Goldstone boson. At $\lambda =2$ the radial mass tends to zero at a phase boundary, so a claim of strictly positive mass there would be incorrect. For completeness, expanding around the symmetric origin gives squared masses $\lambda -2$ and $\lambda$ for the canonical real and imaginary fields.