SSB in a Global Gauge Theory

Nambu-Goldstone Model

Consider a scalar field $\varphi (x)$,

$\varphi (x)=\frac{1}{\sqrt{2}}({\varphi }_1(x)+i{\varphi }_2(x))$

with a Lagrangian density

$\mathcal{L}=({\partial }_{\mu }\varphi {)}^{\ast }({\partial }^{\mu }\varphi )-V(\varphi )=({\partial }_{\mu }\varphi {)}^{\ast }({\partial }^{\mu }\varphi )-{\mu }^2{\varphi }^{\ast }\varphi -\lambda ({\varphi }^{\ast }\varphi {)}^2$

$=\frac{1}{2}{\partial }_{\mu }{\varphi }_1{\partial }^{\mu }{\varphi }_1+\frac{1}{2}{\partial }_{\mu }{\varphi }_2{\partial }^{\mu }{\varphi }_2-\frac{1}{2}{\mu }^2({\varphi }_1^2+{\varphi }_2^2)-\frac{1}{4}\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$.

$\mathcal{L}$ is invariant under the global U(1) transformation: $\varphi (x)\to e^{i\alpha }\varphi (x)$. Here, global refers to the fact that $\alpha$ is not a function of the position x. We now consider two cases.

• Case 1: $\lambda >0$ and ${\mu }^2>0$ - The potential has a single minimum at ${\varphi }_1={\varphi }_2=0$. Hence, the ground state is non-degenerate and therefore shares the symmetry of the potential.

• Case 2: $\lambda >0$ and ${\mu }^2<0$ - The extrema can be determined as follows:

  $V(|\varphi |)=\frac{{\mu }^2}{2}|\varphi {|}^2+\frac{\lambda }{4}|\varphi {|}^4$

  $V^{\prime }(|\varphi |)=\frac{\partial V}{\partial |\varphi |}=({\mu }^2+\lambda |\varphi {|}^2)|\varphi |\stackrel{!}{=}0$

  $V^{\prime \prime }(|\varphi |)=\frac{{\partial }^{2}V}{\partial |\varphi {|}^2}={\mu }^2+3\lambda |\varphi {|}^2$

The location of the extrema can be read off V’; they are located at $|\varphi |=0$ and $|\varphi {|}^2=-{\mu }^2/\lambda \equiv v$.

• The former corresponds to a single point with second derivative $V’’(0)={\mu }^2<0$; hence it is a maximum.
• The latter corresponds to a set of points with second derivative $V’’({\mu }^2/\lambda )=-2{\mu }^2>0$. They form a valley of minima at the configurations ${\varphi }_1^2+{\varphi }_2^2=-{\mu }^2/\lambda$, which is a circle in the ${\varphi }_1{\varphi }_2$-plane.

The ground state is degenerate and hence does not share the symmetry of the potential. The value $v$ is called vacuum expectation value (vev) of the field $\varphi (x)$.

A common choice of ground state is ${\varphi }_1=v$ and ${\varphi }_2=0$, leading to the matrix element

$⟨0|\varphi (x)|0⟩=\frac{v}{\sqrt{2}}$.

For this (arbitrary) choice, excitations about the ground state are described by

$\varphi (x)=\frac{1}{\sqrt{2}}(v+\eta (x)+i\zeta (x))$

where both $\eta (x)$ and $\zeta (x)$ are real fields, representing small deviations from the ground state. Then, the potential becomes

$\mathcal{V}=\frac{{\mu }^2}{2}[(v+\eta {)}^2+{\zeta }^2]+\frac{\lambda }{4}[(v+\eta {)}^2+{\zeta }^2{]}^2$

$=-\frac{\lambda v^2}{2}[v^2+2v\eta +{\eta }^2+{\zeta }^2]+\frac{\lambda }{4}[v^4+4v^3\eta +6v^2{\eta }^2+2v^2{\zeta }^2+\text{terms cubic and quartic in }\eta ,\zeta ]$

$=\text{const}-\frac{2\lambda v^2}{2}{\eta }^2+\text{terms cubic/quartic in }\eta ,\zeta$.

Since the terms cubic and quartic in $\eta$ and $\zeta$ correspond to the (self-)interactions of the fields, we will choose to ignore them.

Then,

$\mathcal{L}=\text{const}+\frac{1}{2}({\partial }_{\mu }\eta )({\partial }^{\mu }\eta )-\frac{2\lambda v^2}{2}{\eta }^2+\frac{1}{2}({\partial }_{\mu }\zeta )({\partial }^{\mu }\zeta )$

• The field $\eta (x)$ has acquired a mass of $m_{\eta }=\sqrt{2\lambda }v$. It corresponds to radial oscillations about the ground state. When trying to oscillate radially in the ${\varphi }_1{\varphi }_2$-plane, the field $\varphi (x)$ has to overcome the potential barrier at $|\varphi |=v$. Such a field that acquires its mass through SSB is referred to as a Higgs boson.
• The field $\zeta (x)$ remains massless. It corresponds to angular oscillations about the ground state. When trying to oscillate angularly in the ${\varphi }_1{\varphi }_2$-plane, the field $\varphi (x)$ does not have to overcome any potential barrier; it follows the trajectory of the valley of minima at $|\varphi |=v$. Such a field is called a Nambu-Goldstone boson.