HQET Lagrangian identity

by Einj
Tags: hqet, identity, lagrangian
 P: 321 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 $Q_v=P_+Q_v$, where $P_\pm=(1\pm \displaystyle{\not} v)/2$ are projection operators. Thanks a lot
 P: 1,020 what is ##v_\mu##?
 P: 321 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
 P: 1,020 HQET Lagrangian identity 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.
 P: 321 Yes, it sounds correct. Do you think this is enough to say that it is always zero?
 P: 1,020 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.
 P: 321 Great sounds good! Thanks
 PF Gold P: 470 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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