Nambu-Goldstone Bosons

When a continuous symmetry of the Lagrangian density is spontaneously broken, the Nambu-Goldstone theory that one or more massless scalar particles emerge. The number of such Nambu-Goldstone bosons is equal to the number of broken generators of the symmetry group.

In SM, the symmetry group $SU(2{)}_L\times U(1{)}_Y$ breaks down to $U(1{)}_{em}$. The three emerging Nambu-Goldstone bosons ${\varphi }_1$, ${\varphi }_2$ and ${\varphi }_4$ are absorbed by the $W^{\pm }$ and $Z$ bosons. As a consequence, each of the $W^{\pm }$ and $Z$ bosons acquire mass and a third degree of freedom in the form of a longitudinal polarisation of the field. Additionally, these Goldstone bosons are unphysical; they are gauged away.

In a general $R_{\xi }$ gauge, we break or fix a gauge by adding the term

$\mathcal{L}\to \mathcal{L}-\frac{({\partial }_{\mu }{A}^{\mu }{)}^2}{2\xi }$, where $\xi$ is an arbitrary real-valued constant and $A$ represents any of the gauge fields present in $\mathcal{L}$.

The propagator of a gauge boson $V\in W,Z$ is then given by

${\Delta }_{\mu \nu }^V(q)=\frac{-i}{q^2-m_V^2+iϵ}\left[g_{\mu \nu }+(\xi -1)\frac{q_{\mu }{q}_{\nu }}{q^2-\xi m_V^2}\right]$

where the $\xi$ can be the Unitary gauge ($\xi \to \infty$), Feynman gauge ($\xi \to 1$) or the Landau gauge ($\xi \to 0$).

The propagator for the Nambu-Goldstone boson absorbed by $V\in W,Z$ is given by

${\Delta }^V(q^2)=\frac{-i}{q^2-\xi m_V^2+iϵ}$.

As we can see, the mass of the scalar boson depends on the gauge fixing parameter $\xi$.

• In the Unitary gauge, the mass of the Nambu-Goldstone boson tends toward infinity and the propagator vanishes. Hence, we do not need to account for it in our calculations.
• In the Feynman or Landau gauge, we need to include the Nambu-Goldstone bosons as degrees of freedom with mass $m_V$ or $0$ respectively.