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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?

Qmechanic
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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.

  • 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 – Derek Easte Nov 01 '19 at 18:20
  • 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. – Yuzuriha Inori Nov 01 '19 at 19:26
  • 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. – Derek Easte Nov 01 '19 at 20:03
  • 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. – Yuzuriha Inori Nov 01 '19 at 20:44
  • 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). – Derek Easte Nov 01 '19 at 21:45
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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

mexichat

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.

anna v
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    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. – nox Nov 09 '19 at 19:29
  • https://images.app.goo.gl/HW9Vk4z2KgKxayKK6 – nox Nov 09 '19 at 19:33
  • The mass is generated by the non zero expectation value of the higgs field entering the yukawa interaction term between fermions and the higgs. – nox Nov 09 '19 at 22:26
  • @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 – anna v Nov 10 '19 at 04:58
  • The Higgs bosonand the Higgs field are two different mathematical concepts, – anna v Nov 10 '19 at 05:02
  • The "exchange of energy" with the higgs field happens due to the yukawa interaction term between fermions and higgs in the standard model lagrangian. – nox Nov 10 '19 at 05:10
  • @lux no, you are not correct in assuming this. The Yukawa coupling is the same type of coupling as the electromagnetic one. It is defined for the interaction of particles with each other, exchanging energy momentum. NOT with the Higgs field. The particle and the field are two different things. The electron does not interact with the electron field. Creation and annihilation operators operate on the electron field to define the propagation of the electron( a free particle trajectory for example is described by contiuously creation and annihilation . – anna v Nov 10 '19 at 05:31
  • No momentum and energy are exchanged unless an electron meets another particle when an interaction happens. – anna v Nov 10 '19 at 05:36
  • It is well established that masses are generated via the yukawa coupling to the higgs field. Very loosely with $L \sim \bar \psi \Phi \psi$ the ssb means $\langle \Phi \rangle = v \neq 0$. Expanding about this vev, the first order contribution gives a mass term $v \bar \psi \psi$ which is the mass term. What do you object to here? – nox Nov 10 '19 at 05:42
  • This happens once,during electroweak symmetyr breaking.In our world it happened , the particles acquired there mass, there was energy and momentum exchanged. When an electron goes through in space it does not interact with the Higgs field exchanging energy and momentum. Its mass is fixed. It only follows creation and annihilation operators until it interacts with another particle with the appropriate interactions, including the Higgs boson particle as in your diagrams. – anna v Nov 10 '19 at 06:03
  • Let's be clear and see if we can agree on some things. Certainly electroweak symmetry breaking happened once. Indeed this is when the higgs field relaxed to its non zero vacuum expectation value. I think you agree the mass term arises from the dc component of the higgs field to lowest order in its fluctuations via to yukawa coupling between fermions and higgs field. It is wrong to say that the fermions are no longer interacting with the higgs field (where did it go?). If we turn the higgs field off or kill the yukawa interaction the std model particles would cease to have mass! – nox Nov 10 '19 at 16:17
  • What is happening is the interaction with the higgs field is omnipresent - however the mass generating term involves interactions where the higgs component has zero momentum (it is the space time constant part of the field, just like in the euler heisenberg lagrangian for constant electromagnetic fields). This allows us to treat std model fields as if they have mass without bothering to invoke the higgs field but it is certainly present and is continually maintaining the particle masses, just via exchange of vanishing momentum. – nox Nov 10 '19 at 16:19
  • Finally we can agree that the higgs boson is the fluctuations about the vev, and that these vertices are the ones in the image above. These represent interactions with excitations of the higgs field about the vev and are of yukawa type. But the original yukawa interaction persists and via the vev ensures the particles have mass. This is the only (nearly) way to give a gauge inv masses. – nox Nov 10 '19 at 16:21
  • @lux Look, lets stop here. Zero momentum interactions are not interaction in the way I have been taught, it sounds like sophistry to me. The electron field is not interacting with the electron in my books. The creation and annihilation operators operate on the fields. I am not used to thinking of the electron being a fluctuaton around the VEV of the electron ( which is zero). Let us agree that we have a different vocabulary. – anna v Nov 10 '19 at 17:08
  • A standard reference on the Euler-Heisenberg Lagrangian for constant electromagnetic fields, or any textbook on the Higgs mechanism will show that this point is far from a difference in vocabulary. – nox Nov 10 '19 at 17:30
  • The creation and annihilation operators you refer to generate precisely the localised (in momentum space) fluctuations of the electron field about the vacuum state whose vev is 0. This is how particles are interpreted as excitations of the fields. – nox Nov 10 '19 at 18:08