# HQET Lagrangian identity

1. Apr 22, 2014

### 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

2. Apr 22, 2014

### andrien

what is $v_\mu$?

3. Apr 22, 2014

### 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

4. Apr 22, 2014

### 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.

5. Apr 22, 2014

### Einj

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

6. Apr 22, 2014

### 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.

7. Apr 22, 2014

### Einj

Great sounds good! Thanks

8. Apr 23, 2014

### Hepth

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