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HQET Lagrangian identity

  1. Apr 22, 2014 #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. Apr 22, 2014 #2
    what is ##v_\mu##?
  4. Apr 22, 2014 #3
    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. Apr 22, 2014 #4
    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. Apr 22, 2014 #5
    Yes, it sounds correct. Do you think this is enough to say that it is always zero?
  7. Apr 22, 2014 #6
    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.
  8. Apr 22, 2014 #7
    Great sounds good! Thanks
  9. Apr 23, 2014 #8


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

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

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