Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook