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HQET Lagrangian identity 
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#1
Apr2214, 08:51 AM

P: 328

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
Apr2214, 12:15 PM

P: 1,020

what is ##v_\mu##?



#3
Apr2214, 12:30 PM

P: 328

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



#4
Apr2214, 12:43 PM

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.



#5
Apr2214, 12:46 PM

P: 328

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



#6
Apr2214, 12:55 PM

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.



#7
Apr2214, 12:57 PM

P: 328

Great sounds good! Thanks



#8
Apr2314, 06:01 PM

PF Gold
P: 472

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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