What is the difference between the Higgs Boson particle and an electron moving through the Higgs field?

Question

I am watching a lecture by Sean Caroll titled "Particles, Fields, and the Future of Physics". I am not a physicist by any means but enjoy the subject in my spare time hoping to understand it.

This lecture gave rise to a better understanding of quantum field theory for me and how it relates to particle physics.

The first understanding I gained; Carroll mentions that in the view of quantum field theory every particle we know of is basically a perturbation in a field. So an electron is just some "wave" in the electron field. For an electron to "be there" the field itself must be perturbed at that location. (is this correct?)

Another understanding I gained is that these perturbations are energy and that for certain field to be perturbed it takes more or less energy. Specifically that the $W$ boson field takes more energy to be perturbed than the electron field. Even more that if we are given a $W$ boson particle and it decays, its energy can be transferred (by some unknown mechanism?) to both the electron and anti-neutrino fields creating particles of each in that location.

My understanding breaks down when he describes the Higgs field. Specifically that an electron moving through the Higgs Field "gains/is given" mass. Does this mean that the electron is encountering a Higgs boson/particle? Or is it linked to the fact that the Higgs field (forgive me for this butchering of words and physics) is held at a higher energy when not perturbed?

I guess, how is a Higgs particle different from a particle traveling through the Higgs field?

I am pretty sure there could be a good analogue explanation and question between how charged particles experience a force when traveling through an electromagnetic field?

If this question is lack too much understanding, I apologize in advance.

Answer 1

Yes, an electron is just some wave, as you say, in the electron field, as it is for any particle. You can also interpret in a broad sense that a field needs to be perturbed at a particular point in spacetime for you to have a non-zero odd of measuring it a that point, although this simple picture is complicated by quantum phenomenas.

The energy of a decaying particle not only can but needs to end up somewhere. This is conservation of energy! The mechanism are not unknown, they are the possible interactions (read that as forces) between fields, though they are not all clearly understood in their dynamics.

The idea of billard balls particles colliding is really not the best to have in mind when considering QFT. The electron, which is really a wave/excitation in a field, travelling in spacetime in presence of the Higgs field does not need to ''collide'', in a classical view, with a Higgs particle to interact. Keep in mind that these field excitations are not exactly localized, much as a wave is not. What happens is that the electron field interacts with the Higgs field and as seen form the dynamic of the electron field it corresponds to it having a mass. The closest analogy that comes to mind, which is pretty bad: don't give it too much intellectual weight, is of a bullet going through water that acquires a different dynamic behavior by interacting with the surrounding media, but that's as far as it goes.

Your question about the difference between a Higgs particle and another one, is like asking what is the difference between sound and light. They are not excitations of the same medium.

I am sadly not aware of any good and simple analogies for the Higgs mechanism. The closest thing, which is not simple but quite close conceptually speaking, are electrons in crystal having a different effective mass because of their interaction with the crystal lattice. Without using effective field theories, you can model electron wavefunctions moving in the crystal lattice using standard quantum mechanics. From there, you study their dispersion relation which is in essence the equation relating energy and momentum. The dispersion relation, in some cases, will take the a functional form of a free wave from which you can infer an effective mass. You can interpret that as saying that the interaction with the lattice modifies the mass of the free electron.

Answer 2

The Higgs field is actually two complex scalar (spin 0) fields so there are two particle and two anti-particle excitations (quanta). The pair of fields transform as an electroweak doublet which essentially means that the Higgs field quanta interact with the electroweak gauge field quanta (W and B bosons).

In addition, the Higgs field has a peculiar potential energy such that the lowest energy state of the field is a state in which the 'vacuum' is 'filled' with Higgs field quanta. In such a ground state, the Higgs field has condensed.

In effect, the 'vacuum' becomes an electroweak superconductor. And, in a way analogous to photons having effective mass within an electric superconductor, the electroweak bosons have effective within the Higgs condensate.

However, there is a particular combination (mixture) of the $W^0$ and $B$ bosons that can propagate freely in this condensate and this mixture is the physical photon.

The $W^+, W^-$, and a complementary mixture of the $W^0$ and $B$ bosons, the $Z^0$, cannot propagate freely and thus have an effective mass.

Matter fields also couple to the Higgs field via a Yukawa interaction which, in the Higgs condensate, gives the matter field quanta an effective mass too.

Finally, the Higgs boson is the quanta of the remaining degree of freedom of the Higgs field (the other three are taken up by the three massive electroweak bosons). The Higgs boson would be massive even in the absence of a Higgs condensate.

In summary,

• The electroweak bosons become massive in the electroweak-charged Higgs condensate in a way analogous to the photon becoming massive in a superconductor.
• The matter particles, fermions, become massive in the condensate due to a Yukawa interaction.
• The Higgs boson would be massive regardless.

Answer 3

I would like to add a few things to your question "why is the electron passing through the higgs field different from the higgs boson" that the previous answers did not cover?

1. the electron is an excitation of the EM field, which exist everywhere in space, that is how the electron can propagate, since it propagates as an excitation in the EM field.

2. the electron is a lepton, and must obey the Pauli exclusion principle, but the higgs bozon is a bozon and does not obey the Pauli exclusion principle. Why is it important? Because Higgs bozon number must not be conserver, we can put infinite number of Higgs bozons into any point in space (just like we can put infinite number of photons into any point in space). But very simplified, only one electron can occupy the same quantum state within a quantum system simultaneously.

3. So we can create a 'sea' of higgs bozons just like a 'sea' of photons. That 'sea' could then be infinitely (since they do not obey the Pauli exclusion pr.) filled with higgs bozons. Does it have a 'higgs bozon density'(it does not matter, because they do not obey the Pauli exclusion pr.)? But the 'density' will be the minimal value of the field at every point. The higgs field is like a mexican hat. It's lowest value is not in the middle, but at every single position in the field, it will take it's lowest vacuum expectation value. Why is that important? Because this way this Higgs field will have energy by default(without excitation). This is why it differs from all the other fields (except the gravity field, but the graviton is not yet confirmed, but that field must take a vector value everywhere too, where the higgs field is coupling with another field thus giving rest-mass to a particle, and so pointing towards the center of mass).

4. So the higgs field takes on a minimal scalar value at every point, and that will be equal to the mass it gives to the particles it couples with.

5. So the electron in your question couples (the EM field couples) with the Higgs field, and so the EM field takes the energy from the higgs field, and 'gives' it to it's own excitation, the electron. That energy is a scalar and will be the electron's rest-mass.

6. How does this affect the electron? It will slow down in the space dimensions, and will speed up in the time dimension.

7. How does this differ from the Higgs bozon? Well in this context, the Higgs bozon is affected similarly. It gains rest-mass from the Higgs field.

8. The higgs bozon is the excitation of the Higgs field, just like the electron is an excitation of the EM field.

9. The right question is, how does in QM gain the electron the rest-mass from the Higgs-field? It does by bumping into the Higgs-bozons 'sea' and thus slowing down? No. It gets it's rest-mass from the Higgs-fields unique character that it has a minimal non-zero scalar vacuum expectation value (that will be the mass of the 'passing-through' electron) at every point, even if the Higgs-field is not excited. This value is energy, the Higgs field is like a charged field by default. When the electron (which is the excitation of the EM field) is passing through the Higgs-field, it couples with it and takes it's energy, converting it into rest-mass.

So to answer your question, the electron does not gain it's rest-mass by bumping into the Higgs-bozons. the Higgs Bozons are the excitation of the Higgs field itself. When the Higgs bozon comes to existence (the Higgs field is excited), the Higgs bozon gains it's rest-mass the same way as the electron, by getting the v.e.v, the default energy of the Higgs-field converted into it's rest-mass.

Another interesting question would be, the photon. Why does it not couple (since it's and excitation of the EM field, like the electron) with the Higgs-field? The answer is because the photon is a combination of the W and Z bozons in a way that they cancel this rest-mass affect. That is why the electroweak theory is proven, and the photon and the W and Z bozons are somewhat the same thing, at our lower energy levels they combine, and 3 out of those combinations does not cancel out the Higgs-mechanism, so the W and Z bozons gain rest-mass. But the photon does not because it's a combination that cancels out that affect. The gluon on the other hand does not couple with the Higgs field, and I have no information on why, would be interesting to know.

Answer 4

To justify giving mass to a would-be massless particle, scientists were forced to do something out of the ordinary. They assumed that vacuums (empty space) actually had energy, and that way, if a particle that we think of as massless were to enter it, the energy from the vacuum would be transferred into that particle, giving it mass. A mathematician named Jeffrey Goldstone proved that if you violate a symmetry, (for example, a symmetry with a Rubik's cube would be if you state that the corners must always be rotated 0 or 3 times to be solvable (it works)), a reaction will occur. In the case of the Rubik's cube, the cube will become unsolvable if violated. In the case of the Higgs field, something named after Jeffrey (and another scientist who worked with him named Yoichiro Nambu) is produced, a Nambu-Goldstone Boson. Wikipedia

If we are to talk about the classical picture of the higgs field we must assume that particles are like balls, with a given dimensions. You must forget the quantum explanation of particles that they are excitations of a field. In tha same we must think about classical the higgs field that is some kind of liquid that fills the empty space (aka vacuum).

Mass itself is not generated by the Higgs field- the creation of matter or energy would conflict with the laws of conservation. However, mass is "imparted" to particles from the Higgs field, which contains the relative mass in the form of energy. Once the field has endowed a formerly massless particle the particle slows down because it has become heavier

We have not yet fully understand the classical picture of the higgs field. This is due to the fact we have not yet a solution of the navier stokes equations about the higgs field.