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Do fermion and scalar running masses run in the same way? Specifically, what are the qualitative differences in the mass beta functions for, say, scalar $\lambda\phi^4$ field theory and the fermion masses in QED?

Next, is there some obvious intuition for whether the masses should get bigger or smaller in the UV? Since particle masses become more and more irrelevant in scattering processes at higher energies I'd think that the running masses would get smaller in the UV, but this might not be the correct intuition.

Finally, can one see that scalar masses are unnatural solely from their mass beta functions? Do they just run very quickly so that they are sensitive to their values in the UV?

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So I am curious about hearing more about this as well. The gist however is that yes, the beta functions are really the point of the hierarchy problem. If you have a heavy particle with mass M_H, a scalar particle coupled to H will have a mass running proportional to M_H^2, whereas the fermion's running is protected by the chiral symmetry and is only proportional to m_f, not M_H. However effective field theory tells you that the M_H contributions only kick in at energies above the mass of the heavy particle. So the hierarchy problem arises as you run down from very high energies. –  Andrew Aug 2 '13 at 2:52
    
Yeah, that's my perception of what's going on, too. The problem being that scalar masses are way more sensitive to higher mass particles running in their own loops than fermions are. However, I'm unclear if this is also a problem just based on the self interaction of a single scalar field or if a scalar is only unnatural if there are other, heavier particles running around. Does the hierarchy problem only exist because it's assumed that there are particles more massive than the top or is it a problem even if the top is the heaviest particle? –  user26866 Aug 2 '13 at 15:20
    
@naturalnessquestions There are no naturalness problems at all in a renormalisable field theory. You just set the renormalised masses/couplings to their physical values when you renormalise and that's it. No further discussion necessary. When naturalness issues come up people are always implicitly embedding the renormalisable theory as an effective field theory with some unknown UV completion. Only then does it make sense to talk about sensitivity to the UV. –  Michael Brown Aug 2 '13 at 15:53
    
Put another way: if you have a UV complete theory what is the point of bothering about sensitivity to the UV? You already have the UV theory in hand. –  Michael Brown Aug 2 '13 at 15:54
    
@naturalnessquestions Well, it's most clear if there's another particle. But I think people rely on the Planck scale as the ultimate "high energy scale" you can rely on existing. You have to imagine, what happens at the planck scale to set the initial condition for the higgs mass running? And basically the point is, even if you don't know how to treat physics above the planck scale in detail, how are you going to get something out that's not the planck scale? –  Andrew Aug 2 '13 at 16:27
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1 Answer 1

Référence : Steven-Weinberg-The-Quantum-Theory-of-Fields-Vol-2-Modern-Applications

p.143 (18.4.15), (18.4.16)(18.4.17), p.144 (18.4.19) (18.4.20)

Scalar $(\phi^4$ theory):

$$\mu \frac{d}{d \mu} m^2_\phi(\mu) = (-2 + \frac{g_\mu}{16 \pi^2} + O(g_\mu^2))~m_\phi^2(\mu) \quad \quad \quad(18.4.17)$$

Electron :

$$\mu \frac{d}{d \mu} m^2_e(\mu) = (-1 - \frac{e_\mu^2}{2 \pi^2} + O(e_\mu^4))~m^2_e(\mu) \quad \quad \quad(18.4.20)$$

Weinberg comment p 144:

"As long as the coupling remain small, the $m(\mu)$ decrease in magnitude. Our previous assumption that masses may be neglected as $\mu \rightarrow +\infty$ is justified if in fact $m(\mu)$ does vanish for $\mu \rightarrow +\infty$. However it is only known to be the case in asymptotically free theories, where the couplings all do remain small for $\mu \rightarrow +\infty$; in all other cases, this assumption is just an educated guess".

[EDIT]

Ref (Ryder, Quantum field theory) pages 314,315, 337-339

From my point of view, The difference of sign relatively to the constant coupling comes, first, from a different expression of the self-energy $\Sigma$, which gives a correction to the inverse propagator (so a correction to $m^2$ for scalars - and a correction to $m$ for fermions), secondly, for the electron, we have to take care of the renormalisation of the field $\psi$ .

For $\Phi^4$ scalars, we have, in dimensional regularisation ($\epsilon = 4 -d$) - You may replace $\frac{1}{\epsilon}$ by $\log (\frac{\Lambda}{\mu})$ for Pauli-Villars regularisation if you want :

$$\Sigma_\phi = \frac{-g m^2}{16 \pi^2 \epsilon} + finite ~ terms$$

For the electron in QED, we have :

$$\Sigma_e = \frac{e^2m}{2 \pi^2 \epsilon} + finite ~ terms$$ For the electron, we have to take care of the the renormalization of the field, but this does not change the global sign (the final result is a $\frac{3}{8}$ term instead of $\frac{1}{2}$).

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Thank you! Do you have any intuition for the relative sign difference in the two beta functions, though? That is, why does the scalar beta function depend on the coupling with a plus sign and the fermionic beta function depends on the coupling with a minus? –  user26866 Aug 2 '13 at 20:26
    
I have made an edit to the answer. –  Trimok Aug 3 '13 at 9:56
    
Yeah, I understand the calculation, but wondered if there was some clear explanation why the self-interactions of the scalar field should cause the scalar mass to run in a way that's qualitatively different than the way the fermion mass runs in QED, due the minus sign. –  user26866 Aug 3 '13 at 20:27
    
What I may say is that the diagrams of self-energy are different, and it involves different fields. For the $\phi^4$ theory, you have a loop of virtual scalar field $\phi$, while, in QED, for the electron, you have one virtual electron and one virtual photon. In QED, the electron current is the source of the electromagnetic field, but in $\phi^4$ theory, in some sense, the field $\phi$ is its "own source". –  Trimok Aug 4 '13 at 9:49
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