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Vector Mesons decay constant

  1. Jun 16, 2013 #1
    I'm studying the [itex] \tau \rightarrow \rho \; \nu_{\tau} [/itex] decay. I'm asked to calculate the decay width, using a parameterization of the matrix element of the hadronic current. I actually find a matrix element of the form:
    [tex] \left<\rho |\; \bar{u} \gamma^{\mu} \left( 1-\gamma^{5} \right ) d \; | 0\right> [/tex]
    in which I have both the vector and the axial current (u and d are up and down quarks). The [itex] \rho [/itex] meson is a spin 1 vector meson, so I expect that only the term from the vector current survives. I've infact verified this statement in many articles which report:
    [tex] \left<\rho |\; \bar{u} \gamma^{\mu} d \; | 0\right> = f_{\rho} m_{\rho} \epsilon ^{\mu} [/tex]
    with [itex]\epsilon ^{\mu} [/itex] the polarization vector of the meson.

    The problem is that I'm not able to demonstrate it. How can I formally demonstrate it? Is there any parity argument which allows me to exclude the axial term?
     
  2. jcsd
  3. Jun 16, 2013 #2
    Last edited by a moderator: May 6, 2017
  4. Jun 21, 2013 #3
    I know understand! The vector current is even under a G-parity transformation. I recall that G-parity is a multiplicative quantum number used for hadrons classification. It is defined as a rotation about the 2-axis, followed by charge conjugation, i.e. [tex] C \mathrm{e}^{i\pi I_{2}} [/tex] where [tex] I_{2}[/tex] is an isospin generator. The [tex] \rho [/tex] is even to. The axial current is odd under such a transformation.
    For more details, I suggest Quarks and Leptons by Lev Okun.
     
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