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Polarization vectors of spin-1 particles

  1. Apr 22, 2015 #1
    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. jcsd
  3. Apr 22, 2015 #2

    ChrisVer

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    Gold Member

    I think this uses the spherical coordinates?
     
  4. Apr 22, 2015 #3
    And why that? I thought you in the fourms here have an experience about these calculations..
     
  5. Apr 22, 2015 #4

    Hepth

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    Gold Member

    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.
     
  6. Apr 23, 2015 #5
    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
     
  7. Apr 25, 2015 #6

    vanhees71

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    2016 Award

    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".
     
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