What is the reason behind the HQET Lagrangian identity?

[Einj]

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

 

[andrien]

what is $v_\mu$?

 

[Einj]

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

 

[andrien]

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.

 

[Einj]

Yes, it sounds correct. Do you think this is enough to say that it is always zero?

 

[andrien]

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.

 

[Einj]

Great sounds good! Thanks

 

[Hepth]

I thought it was because : (bear with me i don't 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
$$ \bar{Q} \left( \gamma^{\nu} - \gamma^{\nu} \right) Q $$