I have a question about mass and energy and the Higgs field.

My understanding is that fermions 'gain' rest mass by interacting with the Higgs field. But mass is a form of energy, so is the rest mass in fact an interaction?

Likewise when two fermions interact (via say, Coulomb potential) they exchange a virtual photon, so is the virtual photon the interaction? Or am I mixing up the meaning of interaction and energy?


It's more of a mathematical tool rather than some physical interaction. To see what the maths is, we try and use the Higgs mechanism on a very simple case, which will be an abelian $U(1)$ gauge theory, and you will in the end see where the mass comes from.

The $U(1)$ invariant kinetic term of the photon is: $$\mathcal{L}_{kin}=-\frac14F_{\mu\nu}F^{\mu\nu}$$ where $$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\ .$$ That is, $\mathcal{L}_{kin}$ is invariant under the transformation $A_{\mu}(x)\to A_{\mu}(x)-\delta_{\mu}\eta(x)$ for any $\eta$ and $x$. Now, if we try to naively add a mass term for the photon: $$\mathcal{L}=-\frac14F_{\mu\nu}F^{\mu\nu}+\frac12m^2A_{\mu}A^{\mu}$$ we soon find out that the mass terms violates local gauge symmetry, and hence the $U(1)$ gauge symmetry thus requires the photon to be massless.

But what happens if we can break the symmetry? We try to do this by introducing a complex scalar field with charge $-e$ that couples to the photon as well as with itself: $$\mathcal{L}=-\frac14F_{\mu\nu}F^{\mu\nu}+(D_{\mu}\phi)^{\dagger}(D^{\mu}\phi)-V(\phi)$$ where $D_{\mu}=\partial_{\mu}-ieA_{\mu}$ and $V(\phi)=-\mu^2\phi^{\dagger}\phi+\lambda(\phi^{\dagger}\phi)^2$. We can see that the Lagrangian is invariant under the gauge transformations: $$A_{\mu}(x)\to A_{\mu}(x)-\partial_{\mu}\eta(x)$$ $$\phi(x)\to e^{ie\eta(x)}\phi(x)\ .$$ If $\mu^2<0$, the state of minimum energy will be that with $\phi=0$ and the potential will preserve the symmetries of the Lagrangian. Then the theory is simply normal QED with an extra charged scalar field $\phi$ with mass $\mu$.

However, if $\mu^2<0$, the field $\phi$ will acquire a vacuum expectation value: $$\langle \phi \rangle =\sqrt{\frac{\mu^2}{2\lambda}}\equiv \frac{v}{\sqrt{2}}$$ and the global $U(1)$ symmetry will be spontaneously broken!

We can parametrize $\phi$ as: $$\phi=\frac{v+h}{\sqrt{2}}e^{i\frac{\chi}{v}}$$ where $h$ and $\chi$ are referred to as the Higgs boson and the Goldstone boson respectively. They are real scalar fields with no vacuum expectation values. Substituting, we find: $$\begin{align*}\mathcal{L}=&-\frac14F_{\mu\nu}F^{\mu\nu}-evA_{\mu}\partial^{\mu}\chi\\&+\frac{e^2v^2}{2}A_{\mu}A^{\mu}+\frac12(\partial_{\mu}h\partial^{\mu}h-2\mu^2h^2)\\&+\frac12\partial_{\mu}\chi\partial^{\mu}\chi+\dots\end{align*}$$ This now describes a theory with a massive photon with mass $m_A=ev$, a Higgs boson $h$ with $m_h=\sqrt2\mu=\sqrt{2\lambda}v$ and a massless Goldstone $\chi$. We can remove the Goldstone boson from the theory with a transformation called the unitary gauge, but that's beside the point.

Thus we have successfully incorporated mass into our gauge boson with the help of symmetry breaking using the Higgs mechanism.

Although this does not happen in our universe, what (probably) does happen is that the gauge symmetry of the electroweak force $SU(2)\times U(1)$ is spontaneously broken to give the gauge bosons of the weak force their mass (the photons remains massless due to $SU(2)_L\times U(1)_Y\to U(1)_Q$, i.e. electromagnetism is unbroken by the scalar vacuum expectation value). The fermions in an analogous (but non-trivial) fashion gets their mass from the mechanism.

You can see that nowhere above have we mentioned 'interaction' because the Higgs mechanism is not an interaction (although the public eats such words easily). The correct interpretation of an 'interaction' is as Anna has mentioned in her answer, so I won't elaborate on that.

  • $\begingroup$ Thanks for the replies. However I feel a bit more confused about the 'mechanism' that gives fermions their rest mass. In some articles this is called the Yukawa interaction. This interaction between fermion fields and the Higgs field is ongoing, as I interpret what the authors of such articles are saying $\endgroup$ – Derek Easte Nov 1 '19 at 18:20
  • $\begingroup$ The Yukawa interaction ( or we should rather say the Higgs mechanism applied to the Yukawa Lagrangian ) gives mass to the fermions, and technically speaking, it's not an interaction. $\endgroup$ – Yuzuriha Inori Nov 1 '19 at 19:26
  • $\begingroup$ Ok. Well, one other idea you see offered to the public, is that fermions in a Higgs field are like swimmers in water. Or that the Higgs field is like a kind of fluid, so the concept of mass is supposed to be like the concept of drag, or viscosity. The example of photons being massive in superconductors is also offered, but that involves more understanding of say, voltages and currents than most people have. $\endgroup$ – Derek Easte Nov 1 '19 at 20:03
  • $\begingroup$ Definitely not a very correct description. It's just an (kind of incorrect) analogy to make the mathematics known to the public. There is simply no correct analogy (that I know of) of the Higgs mechanism. $\endgroup$ – Yuzuriha Inori Nov 1 '19 at 20:44
  • $\begingroup$ I have a copy of t'Hooft's article in SciAm (1980). He says that the mechanism suggested by Higgs et al. was a way to "endow some of the Yang-Mills fields with mass, while retaining exact gauge symmetry". The gauge symmetry in the article is isotopic-spin symmetry. The Higgs field then represents a way to "gauge" the difference between protons and neutrons (his example). $\endgroup$ – Derek Easte Nov 1 '19 at 21:45

The Higgs mechanism is not an interaction. It is a mathematical method of giving mass to the gauge bosons of electroweak theory, because in the laboratory, in contrast to the photon, they are massive.

To understand how this works beyond the popularized narrative, one has to study quantum field theory. The standard model of particle physics uses the mathematics of Quantum Field Theory (QFT) to describe existing data and (important) predict future data.

For every particle in the table , QFT posits that there exists a field that covers the whole space, from -infinity to +infinity. These fields are a context on which creation and annihilation operators act. This formalism is behind the Feynman diagrams that so successfully calculate the interactions of the elementary particles in the table. So an interaction means a Feynman diagram. The existence of the fields provide a context, like a coordinate system, on which the interactions of elementary particles with the SU(3)xSU(2)xU(1) form of the standard model take place.Example of Feynman diagrams:

enter image description here

These are a prescription one to one to integrals which calculated will give measurable quantities, as decay probabilitie and crossections. Interactions happen at the vertices , and as you can see there is no mention of a higgs field anywhere, or an electron or a neutrino or ... at that. The fields exist as a context to the diagram.

There is what is called the VEV of a field, the vacuum expectation value.

In quantum field theory the vacuum expectation value (also called condensate or simply VEV) of an operator is its average, expected value in the vacuum.

AFAIK the VEV of all the fields given by the particles in the table is zero, except for the Higgs which due to symmetry breaking is given as 246 GeV. Here is the mexican hat


Note that the value has nothing to do with the experimentally measured mass of the particle Higgs. At the once off symmetry breaking during the cosmological time of the standard Big Bang, electroweak symmetry breaking happens once, and since then the gauge bosons are what they are in the table, and all the rest of the particles acquire a unique mass at the time. It is not by interaction, but as a consequence of symmetry breaking once .

You have to separate the concept of the Higgs mechanism from the concept of interaction.

  • $\begingroup$ The higgs boson is definitely coupled to the particles of the standard model via a Yukawa coupling. Thus there is, for example, a quark-higgs-quark vertex that can be used in Feynman diagrams. See for example self energy calculations involving a higgs loop. Moreover the feynman diagrams given in the answer are mapped to algebraic expressions with no momentum integrals because they are tree level diagrams. Momentum integrals enter the map only when virtual particle loops are present. $\endgroup$ – lux Nov 9 '19 at 19:29
  • $\begingroup$ images.app.goo.gl/HW9Vk4z2KgKxayKK6 $\endgroup$ – lux Nov 9 '19 at 19:33
  • $\begingroup$ The mass is generated by the non zero expectation value of the higgs field entering the yukawa interaction term between fermions and the higgs. $\endgroup$ – lux Nov 9 '19 at 22:26
  • $\begingroup$ @lux the off mass shell Higgs are generated the same way the electron off mass shell are generated. We do not say that the electron iinteracts with the electron field. It is two different concepts, that is the problem. We keep the "interacts" for the exchange of momentum and energy. The exchange of energy and momentum with the Higgs field to give the mass happens only at electroweak symmetry breaking energy, which is not our present world $\endgroup$ – anna v Nov 10 '19 at 4:58
  • $\begingroup$ The Higgs bosonand the Higgs field are two different mathematical concepts, $\endgroup$ – anna v Nov 10 '19 at 5:02

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