# Explicit expression for the vector polarisation vectors

The polarisation vectors of particles of arbitrary spin $j$ are typically defined selecting a standard "representative" momentum $p_\star$ and then boosting into a general frame. A simple example is the case of spin $j=1/2$ massive particles, where $p_\star=m(1,\boldsymbol 0)$, and $$u(p_\star)=\sqrt{2m}\begin{pmatrix}1\\0\\1\\0\end{pmatrix}$$

One can show that the boosted polarisation vector $u(p)=L(\Lambda)u(p_\star)$ is given by $$u(p)=\begin{pmatrix}\sqrt{p\cdot\sigma\vphantom{\int}}\begin{pmatrix}1\\0\end{pmatrix}\\ \sqrt{p\cdot\bar\sigma\vphantom{\int}}\begin{pmatrix}1\\0\end{pmatrix} \end{pmatrix}$$ and a similar expression for the $s=-1/2$ case.

In the case of spin $j=1$ massive particle, I've never seen any explicit expression for $\varepsilon^\mu(p)$, not even a formal one (besides $\varepsilon^\mu(p)=\Lambda \varepsilon^\mu(p_\star)$). I expect that there should exist an expression similar to the $j=1/2$ case because, for one thing, $\varepsilon^\mu\sim u^\dagger\sigma^\mu u$ (in a formal sense; more precisely, $\frac12\otimes\frac12=1$ in the sense of representations of the Lorentz group). My question is : what is the explicit form of $\varepsilon^\mu$ for arbitrary $p^\mu$? I'm pretty sure that the result is well-known and can be found in many books, but I have failed to find it, so here I am.

For definiteness, let us consider $p_\star=m(1,\boldsymbol 0)$ and $$\varepsilon_+(p_\star)=\frac{1}{\sqrt{2}}\begin{pmatrix}0\\+1\\-i\\0\end{pmatrix}\qquad\quad \varepsilon_0(p_\star)=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\qquad\quad \varepsilon_-(p_\star)=\frac{1}{\sqrt{2}}\begin{pmatrix}0\\+1\\+i\\0\end{pmatrix}$$ and the standard boost is chosen such that $\Lambda p_\star=p$.

Question: what is $\varepsilon_\pm(p),\varepsilon_0(p)$? Thanks in advance!

D$\vphantom{asd}$ear OP,
I hope you are doing ok. As usual, Weinberg's got you covered. See in particular equation 2.5.24: the standard boost that takes you from $p_\star$ to $p$ is \begin{aligned} L^i{}_k&=\delta^i_k+(\gamma-1)\hat p_i\hat p_k\\ L^i{}_0&=\hat p_i\sqrt{\gamma^2-1}\\ L^0{}_0&=\gamma \end{aligned} where $\hat p_i=p_i/|\boldsymbol p|$ and $\gamma=\sqrt{1+\boldsymbol p^2/m^2}$. This matrix satisfies $$A(p)=LA(p_\star)$$ for any vector $A$, such as $p$ itself or the polarisation vectors $\varepsilon_\sigma$. I guess you can take it from here.