1
$\begingroup$

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 \begin{equation} \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} \end{equation} and the standard boost is chosen such that $\Lambda p_\star=p$.

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

$\endgroup$
0
$\begingroup$

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{equation} \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} \end{equation} where $\hat p_i=p_i/|\boldsymbol p|$ and $\gamma=\sqrt{1+\boldsymbol p^2/m^2}$. This matrix satisfies \begin{equation} A(p)=LA(p_\star) \end{equation} for any vector $A$, such as $p$ itself or the polarisation vectors $\varepsilon_\sigma$. I guess you can take it from here.

Sincerely, OP.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.