SSB in Local Gauge Theory

Higgs Mechanism

Consider a complex scalar field $\varphi (x)$, with a Lagrangian density involving the scalar field and a gauge field $A_{\mu }(x)$,

$\mathcal{L}=(D_{\mu }\varphi {)}^{\ast }(D^{\mu }\varphi )-{\mu }^2{\varphi }^{\ast }\varphi -\lambda ({\varphi }^{\ast }\varphi {)}^2-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }$

Here, $D_{\mu }$ is the Covariant Derivative $i{D}_{\mu }=i{\partial }_{\mu }-e{A}_{\mu }$ and $F_{\mu \nu }$ is the field strength tensor of a gauge field $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. The gauge transformations are given by

$\varphi (x)\mapsto e^{i\alpha (x)}\varphi (x)$

$A_{\mu }(x)\mapsto A_{\mu }(x)-\frac{1}{e}{\partial }_{\mu }\alpha (x)$.

We assume that $\lambda >0$ and ${\mu }^2<0$ so that there exists a false vacuum at $\varphi =0$ and a true vacuum at $|\varphi |=v$, where again $v=\sqrt{-{\mu }^2/\lambda }$. Then, the local gauge Lagrangian becomes

${\mathcal{L}}^{\prime }=\frac{1}{2}({\partial }_{\mu }\eta )({\partial }^{\mu }\eta )+\frac{1}{2}({\partial }_{\mu }\zeta )({\partial }^{\mu }\zeta )+\frac{1}{2}{e}^2{v}^2{A}_{\mu }{A}^{\mu }+ev{A}_{\mu }{\partial }^{\mu }\zeta +{\mu }^2{\eta }^2+\text{further terms}$

• The field $\zeta (x)$ is the massless Goldstone boson of the theory.
• The field $\eta (x)$ is the Higgs boson of the theory with mass $m_{\eta }=\sqrt{2\lambda }v$.
• The field $A_{\mu }(x)$ is the massive gauge boson of the theory with mass $m_A=ev$.
• The gauge boson and the Goldstone boson interact through the non-diagonal term $ev{A}_{\mu }{\partial }^{\mu }\zeta$, which is an example of a derivative coupling.

This brief consideration shows a conundrum. Before SSB, we had a total of four real-valued degrees of freedom: two for the complex field $\varphi (x)$ and two for the two physical polarisations of a massless gauge field $A_{\mu }(x)$. After SSB, we find five real-valued degrees of freedom: one for $\eta (x)$, one for $\zeta (x)$, and three for the three physical polarisations of a massive gauge field $A_{\mu }(x)$.

That means, one of the field must be unphysical! We cannot create new degrees of freedom by simply translating the field variables. For small oscillations,

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

i.e., the field $\zeta (x)$ looks awfully similar to a local gauge transformation. Let us identify $\alpha (x)=\zeta (x)/v$. This suggests that the field $\zeta (x)$ is unphysical and can be removed from the theory, achieved by choosing a gauge, e.g., the unitary gauge, in which $\alpha (x)+\zeta (x)=0$. Then,

$\frac{1}{2}{\partial }_{\mu }\zeta {\partial }^{\mu }\zeta +\frac{1}{2}{e}^2{v}^2{A}_{\mu }{A}^{\mu }+ev{A}_{\mu }{\partial }^{\mu }\zeta \mapsto \frac{1}{2}{e}^2{v}^2{A}_{\mu }{A}^{\mu }$.

We are now left with

${\mathcal{L}}^{\prime }=\frac{1}{2}({\partial }_{\mu }\eta )({\partial }^{\mu }\eta )+\frac{1}{2}{e}^2{v}^2{A}_{\mu }{A}^{\mu }+{\mu }^2{\eta }^2-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }+\text{interaction terms}$.

$\zeta$ is a spurious field (false or not what is seems to be). This unwanted Goldstone boson is in fact absorbed by the gauge boson, giving it a third degree of freedom in the form of a longitudinal polarisation state.