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In the decay of ## B \to D^* l \nu ##, I found that the polarization vectors are described as following:

In the B rest frame the helicity basis

## \bar{\epsilon}(0)= \frac{1}{\sqrt{q^2}} (p_{D^*},0,0,-q_0), \\

\bar{\epsilon}(\pm)=\pm \frac{1}{\sqrt{2}} (0,\pm 1,- i,0), \\

\bar{\epsilon}(t)= \frac{1}{\sqrt{q^2}} (q_0,0,0,-p_{D^*}). ##

and the polarization vectors for ## D^*##

## \epsilon(0)= \frac{1}{m_{D^*}} (p_{D^*},0,0,E_{D^*}),\\

\epsilon(\pm)= \mp \frac{1}{\sqrt{2}} (0,1,\pm i,0). ##

While for leptons the polarization vectors of W boson into its rest frame:

## \bar{\epsilon}(0)= (0,0,0,-1), \\

\bar{\epsilon}(\pm)= \frac{1}{\sqrt{q^2}} (0,\pm 1,- i,0), \\

\bar{\epsilon}(t)= \frac{1}{\sqrt{q^2}} (1,0 0,0). ##

Have any one an explanation for this convention ?

I know from Ryder's book for example that the polarization vectors of a massive spin-1 particle are described by 3 components.

Also what is meant by the helicity basis ## \bar{\epsilon} ##, are they different than the polarization vectors of D, B mesons or W ?

Bests.

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# Polarization vectors of spin-1 particles

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