# Polarization vectors of spin-1 particles

1. Apr 22, 2015

### Safinaz

Hi there,

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.

2. Apr 22, 2015

### ChrisVer

I think this uses the spherical coordinates?

3. Apr 22, 2015

### Safinaz

And why that? I thought you in the fourms here have an experience about these calculations..

4. Apr 22, 2015

### Hepth

Its just a convention? One points along the Z, two are circular polarizations (those with 1+-i) around the Z axis. Allows you to define handedness in terms of these easily.

5. Apr 23, 2015

### Safinaz

So why for the W boson helicity states, the time component is mentioned $\bar{\epsilon} (t)$ while not for the D meson. the spin of D meson is 0 , and for W is 1, while it's often the polarization four vectors of a massive spin-1 particle described by transverse $\bar{\epsilon} (\pm)$ and longitudinal $\bar{\epsilon} (0)$ vectors.

Also as B meson the decaying particle, has not it helicity states?

Thanx

6. Apr 25, 2015

### vanhees71

I don't understand, where you need polarization vectors here. The internal vector-boson lines stand for propagators, for which you can choose any gauge you like. The most convenient one is the Feynman gauge, where the Propgator is simply
$$D_{\mu \nu}(k)=-\frac{g_{\mu \nu}}{k^2-M^2+\mathrm{i} 0^+}.$$
Everything else is given by the Feynman rules and some "Diracology".