Question about mass of particles given by the Higgs

Question

The higgs basically gives mass, or adds an interaction that looks like a mass, by introducing a term in the hamiltonian of the kind $$L_{int}=gH\psi^2,$$ where $gH$ looks like the mass of the quanta of the field $\psi$ (normally a mass term would look like $m^2\psi^2$.

the mass $m$ is always a constant, and so I guess is the mass of a quanta in a lagrangian containing such mass term for a field. However, in the Higgs casem, $H$ doesn't necessarily have a constant value, for instance, it could be oscillating about the minimum of $V(H)$, depending of the dynamics and energies of that particular situation.

So my question is, should we expect some kind of oscillation, or time change, in the mass of a particle? or is the higgs field assumed to be basically frozen at the minimum of the potential on any experiments that we could currently make or measure from cosmological data?

Answer 1

I think you are mixing up the Higgs field before spontaneous symmetry breaking (SSB), the vacuum expectation value (vev) and the Higgs field after SSB.

Before SSB, you could have a term in your Lagrangian such as $\mathcal{L} = g\bar{\psi}H\chi$ where $\psi$ and $\chi$ are fermionic fields, $H$ is the Higgs field before SSB and $g$ is the coupling strenght of this interaction.

Suppose now that $\psi$ is an $SU(2)$ doublet meaning that $\psi$ has two components $\eta$ and $\chi$ which are Dirac fermions. The Higgs field is also an $SU(2)$ doublet and therefore has two complex scalar fields $H_1$ and $H_2$ as components:

$$ H= \begin{pmatrix} H_1 \\ H_2 \end{pmatrix},\hspace{3mm} \psi = \begin{pmatrix} \eta \\ \chi \end{pmatrix} $$

In the standard model (SM), the Higgs field attains it's vev after the $SU(2)_L \times U(1)_Y$ symmetry of the SM is broken.

After that, the Higgs field $H$ changes into

$$ H= \begin{pmatrix} H_1 \\ H_2 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \\ v+h \end{pmatrix} $$

where $v$ is the Higgs vev and $h$ is a scalar field. $v$ is a constant and describes for which value of the potential the Higgs field is in it's ground state. $h$ is a scalar field and it describes the dynamics of the Higgs field.

You are correct when you say that the Higgs field can flucuate and does not need to have a constant value, the Higgs field can indeed oscillate near the minimum of it's potential, but this motion is described by the field $h$, the vev $v$ on the other hand is always a constant. And because it is the vev $v$ which will lead to a mass term, the mass of a particle will also be constant, unless you can somehow change the vev of the Higgs field.

To see why only the vev will lead to a mass term, we can look at the contraction of the $SU(2)$ doublets in $\mathcal{L}$:

$$ g\bar{\psi}H\chi = g\begin{pmatrix} \bar{\eta} & \bar{\chi} \end{pmatrix} \begin{pmatrix} 0 \\ v + h \end{pmatrix}\chi = gv\bar{\chi}\chi + g\bar{\chi} h \chi $$

We actually get two terms! The first one, $gv\bar{\chi}\chi$ contains a mass which is given by $gv$ and because $v$ has mass dimension one, we really can interpret it as a mass. The second term describes an interaction between the field $\chi$ and $h$ and contains only a coupling strength, $g$, but not a mass since $g$ alone has mass dimension zero.

So to answer your question, no, we do not expect the mass of the field to change in time because the mass given to the field $\chi$ by the Higgs mechanism depends only on the vev, wheras the dynamics of the Higgs field after SSB is described by the scalar field $h$ which does not affect the mass.