2) why the Higgs field is on the average non-zero / SurprizingFacts

How the Higgs field works:

1. The main idea
2. Why is the Higgs field on average nonzero

How is it that the Higgs field in nature has an average value that is not zero, and in others (apparently elementary) fields of nature known to us, it is zero? [Очень мелкий шрифт: другие поля, за исключением гравитационного поля самого нижнего уровня, зовутся метрическими, это позволяет определить существование пространства и времени].

First, fermion fields can not have a large constant non-zero value in nature. This is due to the difference between fermions and bosons. Bosons can be on the average non-zero, but fermions can not. So you can forget about the electrons (and their cousins ​​muons and tau), about neutrinos and quarks. Fine: Fermions can form pairs with each other or with antifermions and make composite bosons, which can be on average non-zero. This is so for upper and lower quarks and their antiquarks, and for electrons in a superconductor. But this is a long story, and it does not concern us directly.

What about the fields of photons, gluons, W and Z? All these bosons. In principle, these fields could have a constant nonzero average over the universe. But experiments, not theory, say that this is not so. A sufficiently large nonzero magnitude of the electric field would lead to the appearance of various effects, which we do not observe. The most important of these would be a violation of rotational invariance on a large scale. An electric field is a vector (spin-1), it points in a certain direction, so if it is non-zero, then the direction in which this value is indicated must be different from all the others. (Figure 1, bottom left).

Fig. 1

Higgs field is also scalar (spin-0), it does not indicate anywhere. Among the other scalar fields (not elementary and not relativistic) are examples of the air density field, the pressure field inside the Earth, the temperature of the ocean. At each point of space and time, the density or pressure or temperature is simply a number, and the electric field is a number and a direction. So if the Higgs field has a nonzero value, there is no preferred direction (Fig. 1, lower right. What is more strange (since it is relativistic), the Higgs field does not generate any preferred frame of reference. For air density, there is a preferred frame of reference, since you are either at rest with respect to air, or moving through it. But for the Higgs field this is not so; All observers rest on it. Therefore, the success of Einstein's SRT describing all phenomena does not contradict the presence of a relativistic scalar field of non-zero value, such as the Higgs field. In short, in the presence of a nonzero Higgs field, the vacuum behaves exactly as it would behave when H = 0; Its presence can be detected only through the impact on the particle masses (or through something more cardinal, for example, using the LHC to create Higgs particles).

The simplest thing is that the Higgs field would maintain a nonzero value throughout the universe, If it had a non-zero equilibrium value H 0 participating in its equation of motion of class 1:

 $ d ^ 2H / dt ^ 2 - c ^ 2 d ^ 2H / dx ^ 2 = - (2  pi  nu_ {min}) ^ 2 (H-H_0) $ "data-tex =" display

(It should be class 1, not class 0, for reasons that will become clear to us after discussing the Higgs particle). In fact, the situation is a bit more complicated. The correct equation will look like this:

 $ d ^ 2H / dt ^ 2 - c ^ 2 d ^ 2H / dx ^ 2 = a ^ 2 H - b ^ 2 H ^ 3 $ "data- Tex = "display

Where a and b are constants (their squares are positive! Notice the plus sign before a 2 H, and compare it with the minus in the previous equation), which we will learn about a little later. This can be rewritten as:

 $ d ^ 2H / dt ^ 2 - c ^ 2 d ^ 2H / dx ^ 2 = - b ^ 2 H (H ^ 2 - [a/b] ^ 2) $ "Data-tex =" display

If H (x, t) is a constant in space and time, then dH / dt = dH / dx = 0, so

 $ 0 = - b2 H (H2- [a/b] ^ 2) $ "Data-tex =" display

(When H (x, t) is a constant with respect to x and t), and it has solutions (for the time being we will simplify everything):

1. H = 0
2. H = + a / b
3. H = -a / b

In other words, there are three equilibrium positions, and not one. Fine: I simplify this here, but without detriment.

This is not immediately clear, but the solution H = 0 is unstable. The situation is similar to the equation of motion of a ball in a bowl of the form shown in Fig. 2 – similar to the bottom of the wine bottle. He also has three equilibrium positions, one at 0 and two at ± x 0 . But obviously the position in 0 is unstable – any push will cause the red ball to go far away from x = 0, a cardinal change in the situation. Conversely, the equilibrium at x = x 0 is stable, since any push will cause the green ball to oscillate with a small amplitude around the point x = x 0 – not such a drastic change. The same will be true for a light green ball at x = -x 0 . Similarly, although H = 0 is the solution of the equation for the Higgs field, the history of our universe has turned out to be quite complex in order to guarantee that the Higgs field is properly padded, so it could not remain in that position. Instead, the Higgs field turned out to be in a solution with a nonzero value in a stable situation.


For decades, thanks to a combination of experiments and theory, we knew that the value of the Higgs field (Which is traditionally called "v") is 246 GeV. This gives us the notion of those constants, a and b:

a = vb = (246 GeV) b

Thus, we can define a through b, and we can rewrite the equation of the Higgs motion:

 $ d ^ 2H / dt ^ 2 - c ^ 2 d ^ 2H / dx ^ 2 = - b ^ 2 H (H ^ 2 - v ^ 2) $ "data-tex = "Display

But this does not give us an idea of ​​the b itself. In the next article, we learn more about it.

Now, although I've prepared everything so that H can be equal to v or -v, it does not matter whether the value of the Higgs field is positive or negative (in fact, the possibilities are even greater, below); The world will be exactly the same, with the same physics, since nothing depends on the sign of H. This does not become obvious at once, but it is so; One hint – wherever you find H in the equations I describe or in the description of how the Higgs field works, H 2 appears everywhere, and not just H-aH 2 Depends on whether H = v or H = -v. [Fineprint:infactHisacomplexfieldwitharealandimaginarypartsoHcanbeequaltovmultipliedbyanycomplexnumberzforwhich|z|=1;AndinfactH*H=|H| 2 always appears in the equations, and it does not depend on z. And even that's not all! But for now, that's enough for now.]

If you find a way (let's say, by colliding protons with the Large Hadron Collider) somehow push or make an indignation in the Higgs field, it will fluctuate there and there – then There are waves in the form

 $ H = v + A cos [2 pi (nu t – x / lambda)] $ "data-tex =" display

Where A is the wave amplitude, ν and λ are the frequency and wavelength, and the relationship between ν and λ depends on the exact shape of the equation of motion, in particular, on b and v. Since the Higgs field is a quantum field, these waves will have an amplitude quantized, and we call the quantum of these waves the Higgs particle. Next time, we will look at the properties of these particles.

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