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In this question let me call unbroken Lagrangian simply the form of the SM Lagrangian $\mathcal{L}$ before the redefinition of fields involved after performing the shift $h \rightarrow h+v$, where $v$ is the Higgs vev. By contrast, I call broken Lagrangian the form of $\mathcal{L}$ after the 'change of basis'.

Q.1 Does one obtain different beta functions (at 2 loops for concreteness) and running couplings depending on whether one uses the 'broken' or 'unbroken' forms of $\mathcal{L}$? I understand of course that there are different couplings in the two cases (e.g. yukawas in the unbroken $\mathcal{L}$, fermion mass terms in broken $\mathcal{L}$) but how does, for example, the running of the strong coupling $g_s$ (present in both forms) compare?

Q.2 Are there situations (for example scales below or above the EW scale) when one should use running couplings computed specifically from one form or the other?

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No, the beta functions do not depend on how you write your Lagrangian.

However, they do depend on what particles can be produced at a given energy scale and this is where the "broken symmetry" makes a difference.

The $W$ and $Z$ bosons are massive ($m_W \approx 80$ GeV and $m_Z \approx 90$ GeV). This means that at energy scales significantly below 100 GeV these particles do not have a big effect on the computations, because the energy is too low to "produce" them. Therefore, we integrate them out and this changes the beta functions.

Concretely this means that at energies much above the electroweak scale $m_Z \approx 90$ GeV, we need to take the $W$, and $Z$ bosons into account when we compute the beta functions. At energies significantly below the electroweak scale, we integrate them out and thus get different beta functions.

Take also note that the symmetry does not become "unbroken" above the electroweak scale, because this would mean that the $W$ and $Z$ would be suddenly massless. This is certainly not what we observe at colliders. Instead, the thing is that at energy scales much larger than their masses, i.e. $\gg 100 $ GeV their masses do not really play a role (because they are so small compared to the momentum in the diagram) and only in this sense they become "massless".

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