HQET Lagrangian identity

  1. Hi everyone. I'm studying Heavy Quark Effective Theory and I have some problems in proving an equality. I'm am basically following Wise's book "Heavy Quark Physics" where, in section 4.1, he claims the following identity:

    $$
    \bar Q_v\sigma^{\mu\nu}v_\mu Q_v=0
    $$

    Does any of you have an idea why this is true??

    I think that an important identity to use in order to prove that should be [itex]Q_v=P_+Q_v[/itex], where [itex]P_\pm=(1\pm \displaystyle{\not} v)/2[/itex] are projection operators.

    Thanks a lot
     
  2. jcsd
  3. what is ##v_\mu##?
     
  4. Is the four velocity of the heavy quark. However, I don't think it really matters. The only important thing is that the P's are projectors. I think I solved it, it's just an extremely boring algebra of gamma matrices
     
  5. Yes, that is what it seems. But if you go in the rest frame of the particle, the term you will be having is like ##σ^{4\nu}##,which is a off diagonal matrix in the representation of Mandl and Shaw ( or may be Sakurai).Those projection operators are however diagonal in this representation and hence it's zero.
     
  6. Yes, it sounds correct. Do you think this is enough to say that it is always zero?
     
  7. Of course, you can always go to the rest frame of a heavy quark. That is how we evaluated the matrix elements in qft in old days.
     
    1 person likes this.
  8. Great sounds good! Thanks
     
  9. Hepth

    Hepth 516
    Gold Member

    I thought it was because : (bear with me i dont remember the slash command for the forums right now) the equation of motion:

    $$ v^{\mu}\gamma_{\mu} Q_v = Q_v$$
    $$ \bar{Q}_v v^{\mu}\gamma_{\mu} = \bar{Q}$$

    so
    $$ v_{\mu} \bar{Q} \left( \gamma^{\mu} \gamma^{\nu} - \gamma^{\nu} \gamma^{\mu}\right) Q $$
    becomes
    $$ \bar{Q} \left( \gamma^{\nu} - \gamma^{\nu} \right) Q $$
     
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