A No-Nonsense Explanation of How the Higgs Gives Particles Their Masses

A No-Nonsense Explanation of How the Higgs Gives Particles Their Masses

Particle physicists have developed, mostly in the second part of the 20th century, an elegant theoretical model that constitutes a framework for our understanding of the fundamental particles and forces in our universe.

Particle physicists have developed, mostly in the second part of the 20th century, an elegant theoretical model that constitutes a framework for our understanding of the fundamental particles and forces in our universe. This robust framework is the so-called Standard Model, and it has as one of its main ingredients the Higgs field, a ubiquitous quantum field that is responsible for giving masses to particles. Due to the wave-particle duality in quantum mechanics, the Higgs field has a fundamental particle associated with it, the so-called Higgs boson.

The goal of the present article is to provide a more mathematical explanation of the Higgs boson and its properties.

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Fig. 1: This picture shows the event in which the Higgs boson decays into two photons where the green lines are their energy deposits. This experiment allowed scientists to narrow down the mass of the Higgs (source).

Spontaneous Symmetry Breaking

As observed by Zee, though the fundamental laws of Nature have several symmetries, the world we experience is far from symmetric. In physics jargon, our Lagrangians are symmetric, but the world they describe is not. In fact, the study of how such symmetries are broken is a central topic in physics.

To start, let us consider an example of symmetry breaking where the system consists of interacting N-component scalar fields.

Scalar Fields

Consider a Lagrangian of an N-dimensional scalar field with components:

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Eq. 1: N-dimensional scalar field.

with Lagrangian density given by:

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Eq. 2: Lagrangian density with rotational symmetry (in N dimensions).

Notice that the sign of the mass term is negative (on purpose). The Lagrangian is:

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Eq. 3: Lagrangian corresponding to the Lagrangian density Eq. 2.

Case N=1

If N=1, the Lagrangian has the extrema

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Eq. 4: Extrema of the Lagrangian L. Here, the first ϕ is local maximum

where ϕ=0 is a maximum. Any of the other two ϕs, namely, v or -v, are equivalent ground states and we can choose either one to be our ground state. The ground state is referred to as VEV (vacuum expectation value).

Now, note that the tunneling barrier between the two extrema is infinite:

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Eq. 5: The barrier between the extrema of the Lagrangian L in QFT is ∞.

This occurs because the spacetime integral in Eq. 5 is infinite. Since the potential barrier is infinite, the ground state wave function (or functional) cannot tunnel between the extrema. It must stay at the value v determined by the positive version of Eq. 4. Hence, the reflection symmetry ϕ → -ϕ, whichis present in ordinary quantum mechanics is broken. But, as noted by Zee, the symmetry breaking is not caused by the (artificial) introduction of new terms in L. The ϕ → -_ϕ _symmetry is broken “by itself”.

Case N=2

The potential, in this case, is the famous Mexican hat potential. We now have an infinite number of physically equivalent minima (or vacua). All these vacua have the same (squared) value, namely:

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Eq. 6: The squared value of the all minima for N=2.

Now note that each minimum points in a different direction. But the results cannot depend on the choice of ϕ-direction so I will follow Zee and choose:

  • _ϕ _to be in the 1-direction (such that ϕ₂=0).

  • The value of ϕ₁=+v given by Eq. 6 (positive sign).

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Fig 2: The “Mexican hat” potential function V(ϕ) and its infinite minima.

We now study fluctuations around ϕ₁ and ϕ₂ by writing the minima as:

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Eq. 7: Fluctuations around the minima. The 1-component fluctuates in the radial direction and the 2-component fluctuates around the circle at the minimum of this potential.

The 1-component fluctuates in the radial direction and the 2-component fluctuates around the circle at the minimum of the potential (thus, its absolute value is unchanged).

Substituting Eq. 7 in the Lagrangian we obtain:

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Eq. 8: The second component of ϕ is massless!

This equation tells us that the 2-component of _ϕ _is massless.

physics quantum-physics science data science

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