A Transforming lepton basis to diagonlise charged lepton mass

venus_in_furs
Messages
19
Reaction score
1
To determine the mass of charged leptons, we rotate such that the matrix of yukawa couplings (which gives the mass matrix after EWSB) is diagonal.
We also call this flavour basis for neutrinos, because the flavoured neutrinos couple directly to the correspondong flavoured lepton in weak charged current interactions.

we often see the notation in text that we rotate from ##e'_L \rightarrow e_L ## and ## \nu'_{e L} \rightarrow \nu_{eL} ##
Where the UN-primed is the mass basis of charged leptons.
What is the primed basis?for neutrinos
##\nu_{eL}## is flavour/interaction basis (e, mu tau)
##\nu_1## is mass basis (1,2,3)
what is ##\nu'_{eL}## ?? (e, mu, tau)

for leptons
## e_L ## is mass/flavour/interaction basis
what is ## e'_L ## ?##\nu'_{eL}## and ##\nu_{eL}## are both LH and both electron flavour.. so I am not sure what the difference is?
Is it just that we rotate nu' simply because we rotated the charged leptons ?
Even if this is so, I am still not sure what the dashed basis ##e'_L## was in the first place

Is this to do with interactions before and after EWSB?Thanks in advance
 
Last edited:
Physics news on Phys.org
The primed basis is any general basis. Unless you are working in some extended theory where this basis has a special meaning, it is not relevant what it is. You can just as well choose to work in the mass basis of the charged leptons from the beginning.
 
  • Like
Likes venus_in_furs
Oh, ok, well I guess that's simple enough then!

I was reading the 'Fundamentals of neutrino physics and astrophysics' book, seesaw chapter, and always they start everything with primed basis then convert to un-primed so I thought it must have some meaning, then I looked in earlier chapters and saw they do it everywhere. I guess they are just being as general and thorough as possible and to highlight how things transform together.

OK great, thanks for the reply.
 
Sorry I thought about it some more and I'm still a bit confused.
If the prime and non-prime are different basis, why do they keep the flavour index in both?

##\nu'_{e L} = A\nu_{eL} + B\nu_{\mu L} + C\nu_{\tau L} ##
-> which looks weird... e is a flavour state on the right ... why would you also use e as the label on the left?
-> why does the primed keep the flavour label ?##\nu_{eL}## is a flavour state and couples to the charged lepton.
Then if ##\nu'_{eL}## is not a flavour state, and does not couple directly to corresponding charged lepton, what significance does the 'e' index have?

Do you see what I'm confused about?

Thanks again
 
venus_in_furs said:
which looks weird... e is a flavour state on the right ... why would you also use e as the label on the left?
This is also done also in the quark sector, with the difference that it is the down states that are usually denoted by a prime and that the prime then refers to the down-type quarks in the up-type quark mass basis, while the unprimed down-type quarks usually refer to the down-type quark mass basis. You still need three indices to denote the three different states, regardless of basis.

venus_in_furs said:
Then if ν′eLνeL′\nu'_{eL} is not a flavour state, and does not couple directly to corresponding charged lepton, what significance does the 'e' index have?
None, it is pure convention.
 
  • Like
Likes venus_in_furs
hmm, yes ok, I guess this makes sense then if you put it like that. Ok thank you.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top