# Intuition behind mass corrections to massless fermions

+ 8 like - 0 dislike
788 views

I'm trying to understand the intuition behind the mass correction to massless fermions. To be concrete lets consider a theory with a massless Weyl fermion ($\psi$), as well as two massive particles, a complex scalar ($\phi$) and another Weyl fermion ($\psi '$), \begin{equation} {\cal L} = {\cal L} _{ kin} - \frac{1}{2} m ^2 \left| \phi \right| ^2 - M \left(\psi '\psi ' + h.c. \right) - \frac{ \lambda }{ 4!} \left| \phi \right| ^4 - g \phi \psi \psi ' + h.c. \end{equation} This Lagrangian would be valid if for example we impose a $U(1)$ symmetry such that, \begin{align} & \psi \rightarrow e ^{ i \alpha } \psi \\ & \phi \rightarrow e ^{ - i\alpha } \psi \\ & \psi ' \rightarrow \psi ' \end{align} Now consider the lowest order mass correction the for massless fermion,

$\hspace{4cm}$ The mass correction can be calculated by setting external momenta to zero: \begin{align} i {\cal M} & \sim g ^2 \int \frac{ d^4 \ell }{ (2\pi)^4 } \frac{ 1 }{ \ell ^2 - m ^2 } \frac{ M }{ \ell ^2 - M ^2 } \\ & \sim g ^2M \left( \frac{1}{ \epsilon } + \log \left( \frac{ \mu ^2 }{ \Delta ( m , M ) } \right) \right) \end{align} where $\Delta$ is a function of $m$ and $M$. In $\overline{MS}$ we toss the infinity and the first order correction is \begin{equation} m _{ \psi } \sim g ^2 M \log \frac{ \mu ^2 }{ \Delta } \end{equation} This has three interesting features which I am trying to understand:

1. It seems that the fermion doesn't decouple from the theory in the limit that its mass goes to infinity. Why are we not justified in integrating it out in this case?
2. The calculation is very insensitive to the boson mass. Is this an accident or has a deeper explanation?
3. The calculation has large logs unless $\mu \sim m, M$. It seems if you renormalize at this scale the log goes away and the physical mass goes to zero! So if I am understanding correctly the particle is massless when doing experiments at the scale of the other particles in the theory (at least at lowest order in perturbation theory) but heavy when its much below this scale. This seems very strange!
This post imported from StackExchange Physics at 2014-06-13 12:35 (UCT), posted by SE-user JeffDror
retagged Jun 13, 2014
+1 for the question. You say that $\Delta$ is a function of $m$ and $M$, so $m_\psi$, function of $\Delta$, should be a function of the boson mass $m$ too. Or am I missing something ?

This post imported from StackExchange Physics at 2014-06-13 12:35 (UCT), posted by SE-user Trimok
@Trimok: Yes, I agree that its dependent on $m$ - but only logarithmically. In other words its a very weak dependence when compared to the linear dependence on the fermion mass.

This post imported from StackExchange Physics at 2014-06-13 12:35 (UCT), posted by SE-user JeffDror

+ 3 like - 0 dislike

One first important point is that what is called "mass"in the question is not the physical on-shell mass. The physical on-shell mass is defined as a pole in the two-point function of the field, this is a non-perturbative defintion which makes clear that the physical on-shell mass does not depend on an energy scale. As indicated in the question, the mass considered is the $\overline{MS}$ mass, computed in a $\overline{MS}$ renormalization scheme, which depends on an energy scale and whose physical interpretation is the difference between the bare mass at the given energy scale and the on-shell mass.

1. Indeed, the fermion of mass $M$ does not decouple from the $\overline{MS}$-mass of $\psi$ when $M$ goes to infinity. This means that the $\overline{MS}$-mass of $\psi$ is "UV-sensitive", it depends on the physics at very high energies.

2. I think the explanation is a variant of the chiral symmetry (it would be exactly the most standard chiral symmetry for $\psi' = \psi$).

If $M=0$, the Lagrangian has the following symmetry, variant of the chiral symmetry: $\psi_L \mapsto e^{i \alpha} \psi_L$, $\psi_R \mapsto e^{-i \alpha} \psi_R$, $\psi^{'}_L \mapsto e^{-i \alpha} \psi^{'}_L$, $\psi^{'}_R \mapsto e^{i \alpha} \psi^{'}_R$. This symmetry forbids a mass term for $\psi$ or $\psi'$. This means that at the perturbative level, there will be no $\overline{MS}$ mass correction.

Now if $M \neq 0$, there will be some $\overline{MS}$-mass correction for $\psi$ but this correction should vanish in the limit $M$ goes to zero. This explains proportionality to $M$ and the absence of other terms proportional for example to the boson mass.

3. It is not so strange because it is not the physical on-shell mass which varies in this way but the $\overline{MS}$-mass.

Remark: the story of the $\overline{MS}$-mass corrections for massless fermions is of great phenomenological importance. The UV-sensitiveness of the correction (see 1.) means that it is "unnatural" to have massless fermions in some theory whithout having some symmetry explaining it: as the $\overline{MS}$-mass is the difference between some bare mass defined at some UV cut-off and the on-shell mass, having a on-shell mass equal to zero means having the bare mass being fine-tuned to compensate the $\overline{MS}$-corrections. One symmetry making massless fermions "natural" is chirality (see 2.). In the Standard Model, there are chiral gauge couplings making massless fermions "natural". This "explains" why there exists massless fermions or rather fermions light with respect to some "natural" scales (GUT, Planck scales), with masses coming only from some spontaneous breaking symmetry phenomenon (Higgs mechanism).

answered Jun 13, 2014 by (5,140 points)

Thank you for the very insightful and interesting answer! It is very helpful. Quick follow up: Since ${\overline MS}$ just gives the difference between the bare and physical mass, is it still possible to use it to estimate the mass corrections to massless particles or are you forced to use a physical regularization scheme (if you just want a rough estimate)?

What do you mean by the "mass corrections to massless particles" ? If by mass, you mean "physical on-shell mass" then there are no such corrections. If by mass corrections, you mean the renormalization of the mass coefficient in the Lagrangian then you can use any renormalization scheme you want (as you did in your question with $\overline{MS}$).

I was thinking about inducing a mass in the context of neutrinos where you can give neutrinos Majorana masses through loop corrections (which is what I'm actually working on). If I'm not mistaken in these cases the chiral symmetry isn't there and so the neutrino can get a mass through loop corrections (after SSB). Am I still able to use $\overline{MS}$ in this case to estimate these mass corrections?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.