A question about a computation in a paper of Effective Lagrangian

[Breo]

I was hitting against a wall for the last hours.

I am not able to obtain the 1/2 terms in the eq. 5 of this paper and left all in terms of only $N_i$ and $\overline{N_i}$,neither. Anyone could give me a tip?

http://arxiv.org/pdf/hep-ph/0210271v2.pdf

Thank you in advance.

 

[Orodruin]

I assume you mean from eq 3? Remember that $N$ and $\bar N$ contain both $N_R$ and $N_R^c$.

 

[Breo]

I took that in consideration:

$N_{R_i} = \sqrt{\eta^*}N_i - \eta^*N_{R_i}^c$ and $N_{R_i}^c = \sqrt{\eta}N_i - \eta N_{R_i}$

And the conjugate ones which change $\eta^* \longrightarrow \eta$ and viceversa.

Right?

But still do not reach the eq (5) :s

 

[Orodruin]

Can you show your attempt? It is difficult to see what you are doing otherwise.

 

[Breo]

Starting with the derivative term:

$i(\sqrt{\eta}\overline{N_i} - \eta\overline{N_{R_i}^c}) \not\partial(\sqrt{\eta^*}N_i - \eta^*N_{R_i}^c) = i(\overline{N_i}\not\partial N_i - \sqrt{\eta^*}\overline{N_i} \not\partial N_{R_i}^c - \sqrt{\eta}\overline{N_{R_i}^c}\not\partial N_i + \overline{N_{R_i}^c}\not\partial N_{R_i}^c )$

Equalities: $\sqrt{\eta}\sqrt{\eta^*} = \eta\eta^* = 1$ and $\sqrt{\eta}\eta^* = \sqrt{\eta^*}$ and $\eta\sqrt{\eta^*} = \sqrt{\eta}$

But I can not reach a 1/2 and also if I try to go further the whole term vanishes! I did something wrong

 

[ChrisVer]

Why don't you try it the other way around?

 

[Orodruin]

I took that in consideration:

$N_{R_i} = \sqrt{\eta^*}N_i - \eta^*N_{R_i}^c$ and $N_{R_i}^c = \sqrt{\eta}N_i - \eta N_{R_i}$

You are here essentially writing $z = w - z^*$. This is not a way to remove $z$ from your equations in favour of $w$. Remember that $N_{R_i}$ and $N_{R_i}^c$ contain the same degrees of freedom! Your final expression should contain neither $N_{R_i}$ nor its conjugate.

 

[Breo]

Why don't you try it the other way around?

I can see what do you mean. That would be very "direct" but I should be able to compute it using the way I tried, right?

 

[Orodruin]

I can see what do you mean. That would be very "direct" but I should be able to compute it using the way I tried, right?

No, you first need to write the kinetic term in a way that is symmetric wrt $N_R$ and $N_R^c$, or you will have to use some identities at some point.

 

[Breo]

No, you first need to write the kinetic term in a way that is symmetric wrt $N_R$ and $N_R^c$, or you will have to use some identities at some point.

Sorry, I was replying to Chris :P

You mean to consider Majorana Fermions so play with $N_{R_i} = N_{R_i}^c$ ? in the most convenient way?

 

[Orodruin]

You mean to consider Majorana Fermions so play with $N_{R_i} = N_{R_i}^c$ ? in the most convenient way?

No you cannot do that. They are not the same in the notation used in the paper. If I remember correctly, the paper uses Dirac notation for Majorana particles, i.e., $P_L N_R = 0$ and $P_R N_R^c = 0$. You need to use other ways to rewrite the kinetic term in two.

 

[Breo]

No you cannot do that. They are not the same in the notation used in the paper. If I remember correctly, the paper uses Dirac notation for Majorana particles, i.e., $P_L N_R = 0$ and $P_R N_R^c = 0$. You need to use other ways to rewrite the kinetic term in two.

Can I compute in momentum space and then comeback to position space again freely? Or there is a way to split the ordinary derivative once for right-handed and once for left-handed interactions? if I understood correctly

 

[Breo]

The Weyl equations?

 

[Orodruin]

Try transposing (half of) the kinetic term. It is a scalar and should remain the same. Then use partial integration.

 

[Breo]

I have doubts if I can do the following as the mass terms would have free $N_R$ and $\overline{N_R}$ while integrating:

$\frac{1}{2} i (\overline{N_R}\overset{\leftarrow}{\not{\partial}} +\overset{\rightarrow}{\not{\partial}}N_R)$

So, since $N_R$ and $N_R^C$ have the same degrees of freedom, could I decompose the kinetic term as follows?:

$\frac{1}{2} i (\overline{N_R^c}\not{\partial}N_R^c + \overline{N_R}\not{\partial}N_R)$

 

[Orodruin]

could I decompose the kinetic term as follows?

Yes. It simplifies to
$\frac 12 \overline N i \not\partial N$

 

[Breo]

Thank you very much!

Now the last doubt I have about the computations is what allows me to put the propagator in the action (eq.11) and what to remove the kinetic part.

 

[Orodruin]

Are you familiar with how to integrate out heavy degrees of freedom?

 

[Breo]

Are you familiar with how to integrate out heavy degrees of freedom?

I have read something about heavy particles in effective lagrangians. Briefly, looks like this but not at all as it uses fluctuation operators(=propagators?) which formula is as follows:

$\frac{\delta^2 S_{heavy}}{\delta h(x) \delta h (x')} |_{h=h_o}$ you mean smoething about that?

 

[Orodruin]

At tree level, which is essentially all you need here, it is basically inserting the classical equations of motion for the field into the Lagrangian to remove the heavy degrees of freedom and obtain an effective Lagrangian. This is what they do if I remember correctly.