Equivalence Theorem

We require the description of a massive gauge boson V that carries momentum k. It also carries a polarisation state, which we denote as the four-vector $ϵ$. They are

$ϵ(k,\lambda =+1{)}^{\mu }=\frac{1}{\sqrt{2}}(0,1,+i,0),ϵ(k,\lambda =-1{)}^{\mu }=\frac{1}{\sqrt{2}}(0,1,-i,0)$

$ϵ(k,\lambda =0{)}^{\mu }=\frac{1}{\sqrt{2}}(0,0,0,1)$.

The polarisation states are ${\lambda }_i=\pm 1,0$, corresponding to two transverse polarisations and one longitudinal polarisation.

Now, let us assume that our boson $V$ is highly-boosted, along the z-direction. This leads to a momentum $k=(E,0,0,|k|)$, where $E=\sqrt{|k{|}^2+m_V^2}$. For a highly-relativistic gauge boson, we have $E\approx |k|\gg M_V$ and therefore, $k^{\mu }\to (|k|,0,0,|k|)$. The boost along the z-direction leaves the two transverse polarisation vectors unchanged, since they are orthogonal to the boost direction. However, for the longitudinal polarisation state, we find

$ϵ(k,\lambda =0{)}^{\mu }=\frac{1}{M_V}(|k|,0,0,E)\stackrel{E\gg M_V}{\to }\frac{1}{M_V}(|k|,0,0,|k|)=\frac{1}{M_V}{k}^{\mu }$.

Hence, we can identify the longitudinal polarisation vector of the gauge bosons as the momentum $k$ in the limit of highly-boosted gauge bosons. This relation between the Nambu-Goldstone bosons and the gauge bosons at high energies is called the equivalence theorem.