As I see it the Higgs boson is the mediating particle of the Higgs field and get its own mass from the Higgs field. Isn´t this circular? I mean, for instance, an electron creates a radial electric field but in no way it can interact with the field it created. So my question is: how can a particle get its mass from the field it mediates?

  • 1
    $\begingroup$ The Higgs is no less an effective (mean) field theory of a so far undiscovered microscopic theory than the other fields in the standard model. In that sense there is really nothing special about it. As for the total contribution of the Higgs to the mass of the universe, it's somewhere in the vicinity of a fraction of a percent. $\endgroup$
    – CuriousOne
    May 6, 2015 at 13:51

3 Answers 3


I suspect you have misunderstood what is meant by the term field in relation to the Higgs boson. You say:

an electron creates a radial electric field but in no way it can interact with the field it created

and you are quite correct that the electron creates an electrostatic field around it. However in this context the term field refers to a quantum field, and that is something completely different.

Staying with electrons for the moment, in quantum field theory we assume there is an electron quantum field that occupies all of spacetime. The particles we see as electrons are excitations of this field i.e. if you add a quantum of energy to the electron field it appears as an electron. Likewise, remove a quantum of energy from the electron field and an electron will disappear. All the interactions of electrons are a result of interactions of the electron quantum field with other quantum fields. For example, the electrostatic force that we started out with is the result of the electron quantum field interacting with the photon quantum field.

So when you say:

As I see it the Higgs boson is the mediating particle of the Higgs field

this is completely wrong. There is a Higgs quantum field that occupies all of spacetime, and the Higgs boson recently discovered at the LHC is an excitation of this quantum field. The properties of the Higgs boson, including its mass, derive from the interactions of the Higgs field.

So there is no sense in which the Higgs boson mediates the Higgs field. However, and at the risk of confusing matters still further, there is a Higgs analogue to electrostatic field. Just as exchange of virtual photons creates the electrostatic field, exchange of virtual Higgs bosons does (in principle) create a Higgs-o-static field. If you're interested to know more have a look at the question Why isn't Higgs coupling considered a fifth fundamental force?.

  • $\begingroup$ Guess we must find the ziggs boson then. Otherwise the Higgs field would be the only one without a mediator. $\endgroup$ May 6, 2015 at 20:31
  • $\begingroup$ @PedroMalafayaBaptisa: No. "Mediators" are the quanta of gauge fields (i.e. photons, gluons and W/Z for the electromagnetic, strong and weak field, respectively). The Higgs field is not a gauge field, hence its quantum should not be thought of as a mediator. $\endgroup$
    – ACuriousMind
    May 7, 2015 at 10:53

Take an Abelian example. The $U(1)$ gauge invariant kinetic term of the photon is given by $$\mathscr{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$ Where $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$$

That is $\mathscr{L}$ is invariant under the transformation: $A_{\mu}(x) \rightarrow A_{\mu}(x)-\partial_{\mu} \eta(x)$ for any $\eta$ and $x$. If we naively add a mass term for the photon to the Lagrangian $$\mathscr{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}+\frac{1}{2}m^{2}A_{\mu}A^{\mu}$$.

It will become clear that the mass term violates the local gauge symmetry. The $U(1)$ gauge symmetry thus requires the photon to be massless. Now extend the model by introducing a complex scalar field with charge $-e$ that couples both to itself and to the photon: $$\mathscr{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}+(D_{\mu} \phi)^{\dagger}(D^{\mu}\phi)-V(\phi)$$. Where we have defined \begin{eqnarray} D_{\mu} &=& \partial_{\mu}-ieA_{\mu} \\ V(\phi) &=& -\mu^{2} \phi^{\dagger} \phi +\lambda (\phi^{\dagger} \phi)^{2} \end{eqnarray} it is clear that this Lagrangian is invariant under the gauge transformations \begin{eqnarray} A_{\mu}(x) &\rightarrow & A_{\mu}(x)-\partial_{\mu}\eta(x) \\ \phi(x) &\rightarrow & e^{ie \eta(x)} \phi(x) \end{eqnarray} It is convenient to parameterise $\phi$ as $$\phi = \frac{v+h}{\sqrt{2}}e^{i \frac{\chi}{v}}$$ where $h$ and $\chi$, which are referred to as the Higgs boson and the Goldstone boson, respectively are real scalar fields. Substituting these relations back into the Lagrangian, we find \begin{eqnarray} \mathscr{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu} - evA_{\mu}\partial^{\mu} \chi +\frac{e^{2}v^{2}}{2}A_{\mu}A^{\mu} +\frac{1}{2}\left( \partial_{mu}h\partial^{\mu}h -2 \mu^{2} h^{2}\right) +\frac{1}{2}\partial_{\mu}\chi \partial^{\mu}\chi +(h, \chi \quad \text{interactions}) \end{eqnarray} This Lagrangian now describes a theory with a photon of mass $m_{A} = ev$, a Higgs boson $h$ with $m_{h}=\sqrt{2}\mu=\sqrt{2 \lambda} v$ and a massless Goldstone $\chi$. The strange $\chi - A_{\mu}$ mixing can be removed by making the following gauge transformation $$A_{\mu} \rightarrow A^{'}_{\mu} = A_{\mu}-\frac{1}{ev}\partial_{\mu}\chi$$. The gauge choice with the transformation above is called the unitary gauge. The Goldstone $\chi$ will then completely disappear from the theory and one says that the Goldstone has been eaten to give the photon mass.


Mass comes from things other than the Higgs field--the latter is just the main contributor. What gives the Higgs boson its mass is still up for debate--for a more detailed discussion, see the following post:

How does the Higgs Boson gain mass itself?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.