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HQET Lagrangian identity

by Einj
Tags: hqet, identity, lagrangian
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Einj
#1
Apr22-14, 08:51 AM
P: 305
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
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andrien
#2
Apr22-14, 12:15 PM
P: 1,020
what is ##v_\mu##?
Einj
#3
Apr22-14, 12:30 PM
P: 305
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
#4
Apr22-14, 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.
Einj
#5
Apr22-14, 12:46 PM
P: 305
Yes, it sounds correct. Do you think this is enough to say that it is always zero?
andrien
#6
Apr22-14, 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.
Einj
#7
Apr22-14, 12:57 PM
P: 305
Great sounds good! Thanks
Hepth
#8
Apr23-14, 06:01 PM
PF Gold
Hepth's Avatar
P: 466
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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