# A Seesaw formula convensions

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1. Sep 9, 2016

### venus_in_furs

ok so this is a bit of a boring question, so sorry in advance, but for some reason im struggling with this.

I am deriving the seesaw formula.

Now I have gone through the derivivate and I get A : $m_{\nu} = - m_D^T M^{-1} m_D$

Now I have seen other derivations where they get B : $m_{\nu} = - m_D M^{-1} m_D^T$
Note( transpose on LH or RH side )

So I think the reason for this is that in the langrangian you always have + h.c.
And so depending on which term you write, and which term you shove in h.c. I think you get these two different versions.
But essentially it must be describing the same physics, obviously.

Now, in a paper I am reading they have version B and they have the diagonalisation

$-Dm = U^{\dagger} m_{\nu} U^* = U^{\dagger} m_D M^{-1} m_D^T U^*$

They have used B. I need some somehow reconcile A with B and I am a bit confused.

$D_m$ should be the same whatever convension, its real and diagonal, so it is just mass eingenvalues on the diagonal of the mass matrix.

but it doesnt seem obvious that $U^{\dagger} m_D M^{-1} m_D^T U^* = U^{\dagger} m_D^T M^{-1} m_D U^*$ ??? these dont look equal ..
but it must do, if both give a diagonal matrix of mass values?
is this right? Or have I missed something?

Basically I have written deriviation A in my report. but now I realise some work I did used deriviation B. So I need some smooth transition between the two.

I hope this makes sense, and apologies again for such a boring convension based question, but I guess it means I'm lacking some fundamental understanding if I am struggling with this.

Thanks

2. Sep 9, 2016

### Staff: Mentor

I'm not into the subject, but have you considered the simple possibility that one author denotes vectors as columns (A) and the other as rows (B)?

3. Sep 9, 2016

### venus_in_furs

All the terms in A and B are matrices with real values.

the right hand side of A is the hermitian conjugate of B

so I think it comes from in the lagranian and the hermitian conjugate L = mass_term + h.c.
So if you work with 'mass_term' and rearrange to find mass matrix you get A and if you work with the hermitian conjugate you get B
But either way should bring you to the same final answer at the end, which is the bit I cant seem to get my head around.

4. Sep 10, 2016

### Orodruin

Staff Emeritus
They are not. Yukawa couplings may very well be complex, there is nothing preventing this.

The only thing is that $m_D$ means different things in different conventions so you should not be surprised to find the formulas differing. Obviously, if you have a parametrisation with $m_D$ and change to one where it is replaced by $m_D^T$, this change will occur also in all derived expressions.

When you refer to a particular paper, please give the reference so we can check it.

5. Sep 10, 2016

### venus_in_furs

So when I say they have real values, I meant because you chose to do a rephasing of the fields such that you decide $m_D$ and $D_M$ are real. (and you let $m_L$ stay complex but you set this to zero anyway ).
But yes, sorry, I should be careful as this is a convention which may not be taken by everyone.

OK so I understand that $m_D$ in on convention is different to $m_D$ in the other, and that if you place $m_D$ with $m_D^T$ you must do it everywhere.
But when you have a formula, where in both conventions the result is a diagonal matrix of real values, $D_m$, then won't this matrix look the same in all conventions? both convensions take the first 3 to be the light neutrino mass, and the remaining N to be for the heavy neutrinos.
I would expect that I can assume $D_m$ amd $D_M$ to be the same in all conventions
( $D_M$ is diagonal heavy mass matrix and $D_m$ is diagonal light mass matrix )

In the paper I am looking ( https://arxiv.org/abs/hep-ph/0502082 )

They have $D_m = U^{\dagger} m_D (D_M)^{-1} m_D^T U^*$ ( convention B)
where U is PMNS mixing maxtrix,

so if I want to write to write this in convention A , can I just swap $m_D$ and $m_D^T$ to get to convention A ?

$D_m = U^{\dagger} m_D^T (D_M)^{-1} m_D U^*$ ( convention B)

Or do I need to reconsider how U diagonalises it ?

Edit: Note: I just realised what I wrote about setting the two conventions equal was obviously a load of rubbish in the first post, sorry, it was very late! Now I get why you said what you said

Thanks for the help

Last edited: Sep 10, 2016
6. Sep 10, 2016

### Orodruin

Staff Emeritus
You should be aware that a very common convention is to work in the basis where M is diagonal and real. In this basis, it is not always possible to make $m_D$ real or diagonal.

It will have the same values in all parametrisations. This explicitly means that the expression you get cannot be expressed in the same way - it must be expressed taking the difference between the parametrisations into account.

It is the same matrix that diagonalises the dimension five operator in both cases.

7. Sep 10, 2016

### venus_in_furs

From the paper
convention B: $\, \, m_{\nu} = - m_D \, (D_M)^{-1 } \, M_D^T$
$-D_m = U^{\dagger}\, m_{\nu} \, U^*$
$D_m = U^{\dagger} \, m_D (D_M)^{-1} M_D^T \, U^*$

The convention I use
convention A: $\,\, m_{\nu} = - m_D^T \, (D_M)^{-1} \, M_D$

since $D_m$ is the same in both conventions, and U is the same in both conventions, it is clear I cannot use the same formula to diagonalise

thinking about it, seeing as the two conventions come from working with the two different h.c. parts of the lagrangian, it would make sense to me if I did

$-D_m = U^* \, m_{\nu}^T \, U^{\dagger}$
$D_m = U^* \, (\,\, m_D^T (D_M)^{-1} m_D \,\,)^T \, U^{\dagger} = U^* \, m_D^T \, (D_M)^{-1} \, m_D \, U^{\dagger}$

8. Sep 10, 2016

### Orodruin

Staff Emeritus
You can, you just have to replace $m_D$ by $m_D^T$ everywhere.

9. Sep 10, 2016

### venus_in_furs

ahhhh, ok yes, so when i swap the round the $m_D$ and $m_D^T$ I then get $m_{\nu}$ which is the same in the two conventions, so i just diagonlise the same and carry on from there. Ok great. Thanks for the help, sorry this was more drawn out than it probably needed to be!