# A question about a computation in a paper of Effective Lagrangian

1. May 9, 2015

### 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

Last edited: May 9, 2015
2. May 9, 2015

### Orodruin

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

3. May 10, 2015

### 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

4. May 10, 2015

### Orodruin

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

5. May 10, 2015

### 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

6. May 10, 2015

### ChrisVer

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

7. May 10, 2015

### Orodruin

Staff Emeritus
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.

8. May 10, 2015

### Breo

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?

Last edited: May 10, 2015
9. May 10, 2015

### Orodruin

Staff Emeritus
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.

10. May 10, 2015

### Breo

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?

11. May 10, 2015

### Orodruin

Staff Emeritus
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.

12. May 15, 2015

### Breo

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

13. May 15, 2015

### Breo

The Weyl equations?

14. May 16, 2015

### Orodruin

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

15. May 16, 2015

### 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?:

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

16. May 16, 2015

### Orodruin

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

17. May 16, 2015

### 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.

18. May 16, 2015

### Orodruin

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

19. May 16, 2015

### Breo

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?

20. May 16, 2015

### Orodruin

Staff Emeritus
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.