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I'm not a particle physicist, but I did manage to get through the Feynman lectures without getting too lost. Is there a way to explain how the Higgs field works, in a way that people like me might have a hope of understanding?

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Related: physics.stackexchange.com/q/6450/2451 –  Qmechanic Dec 7 '11 at 2:59
Thanks for the answers. It will take me a few days to dig in and comprehend them. –  Mike Dunlavey Dec 7 '11 at 14:02
See also this popular description by L. Susskind, or this other by 't Hooft including pictures of the LHC and some nice diagrams. –  Eduardo Guerras Valera Mar 9 '13 at 23:25

3 Answers 3

up vote 18 down vote accepted

The Higgs mechanism is no different from superconductivity, except the condensate responsible for superconductivity is a relativistically invariant scalar field.

If you have a bosonic field, its particles can be in a Bose-Einstein condensate. When this condensate is charged, you call it a superconductor. A photon in a superconductor gets a mass, and this is the Higgs mechanism. For a relativistic boson described by a scalar field, you give the field a constant nonzero value to make a condensate. When the field has charge, this makes a superconducting condensate which gives the gauge boson a mass.

The whole effect is described in detail on the Wikipedia page on the Higgs mechanism, starting from a nonrelativistic superconductivity model of bosonic particles, and continuing analogously to relativistic condensates.

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I hate to choose between these answers, but this one helps me more, given my extremely limited understanding. –  Mike Dunlavey Feb 21 '12 at 14:33

Explaining the Higgs mechanism properly is a fair bit beyond the level of the Feynman lectures, but here's an attempt.

Spontaneous symmetry breaking

In order to understand the Higgs mechanism in detail, you need to know about two concepts that are involved in quantum field theory. The first is spontaneous symmetry breaking. This is actually a pretty simple idea: even if the physical laws that govern a system are symmetric in some way, the actual behavior of the system will not necessarily be symmetric in the same way.

A common example is balancing a pencil on its tip. If you get the pencil perfectly balanced, the system is rotationally symmetric about the axis of the pencil, and the physical laws that govern the behavior of the pencil (gravity, Newton's laws, etc.) are also symmetric, in the sense that they don't prefer any one direction over another. Despite this, the pencil is going to fall in some direction, and at that point the system is no longer rotationally symmetric. Once the pencil is rolling around on the table, you don't see the effects of that rotational symmetry in the physical laws anymore.

In its application to particle physics, spontaneous symmetry breaking occurs when there is some potential which has a local maximum that lies on the symmetry axis.

Mexican hat potential

This is the famous "Mexican hat potential" shown in this site's favicon. Mathematically, it's represented by an expression of the form

$$V(\phi,\phi^\dagger) = -\frac{1}{2}\mu^2|\phi|^2 + \frac{\lambda^2}{4}|\phi|^4$$

In this expression, $\mu$ and $\lambda$ are constants and $\phi$ is a quantum field. Note the unusual usage of $\lambda^2$ for the quartic coupling which simplifies subsequent algebra (conventionally, the coefficient is taken to be $\lambda$). $\phi$ can be thought of as a complex-valued variable which is related to the quantum state at a point in spacetime. Now, for our purposes, the universe will "try" to "seek out" whatever state minimizes its potential, just like the balanced pencil will "seek out" the state that minimizes its potential. The only difference is that whereas the pencil changes its state by adjusting its position, the universe changes its state by adjusting the value of the field $\phi$. Just as the pencil has to fall in some direction, the universe will have to experience a transition from its initial state on the peak in the center to some other state for which $\phi$ lies at or near the bottom of the valley, namely $|\phi| = \frac{\mu}{\lambda}$.

Local gauge invariance

The other concept involved is local gauge invariance. This one is a little harder to explain, because there's no simple physical analogy, and it takes a fair amount of math.

Basically, gauge invariance means that you should be able to make local gauge transformations to the fields involved in the theory without the physical laws changing as a result. An example of a gauge transformation is the phase rotation $\phi(x)\to e^{i\alpha}\phi(x)$. If the parameter $\alpha$ depends on spacetime position, $\alpha(x)$, then we call it "local."

In quantum field theory, there are basically two ways the fields themselves can show up: either directly, or as a derivative. They always show up in pairs or larger groupings, usually of a conjugate and a normal field, like $\phi^\dagger\phi$. ($\phi^\dagger$ is the Hermitian conjugate, which is the complex conjugate and transpose of a matrix-valued field.) Terms like that which involve only the fields themselves are automatically gauge invariant, because when you make the gauge transformation, you get $\phi^\dagger e^{-i\alpha(x)}e^{i\alpha(x)}\phi = \phi^\dagger\phi$; the phase factor just cancels out.

But when you have a term that involves derivatives, like $\partial\phi^\dagger\partial\phi$ (called the kinetic term), you run into problems:

$$\partial[e^{i\alpha(x)}\phi(x)] = i\alpha'(x) e^{i\alpha(x)}\phi(x) + e^{i\alpha(x)}\partial\phi(x)$$

If you plug this into the gauge transformed kinetic term $\partial[e^{-i\alpha(x)}\phi^\dagger(x)]\partial[e^{i\alpha(x)}\phi(x)]$, you'll see that it will not simplify back to the original $\partial\phi^\dagger(x)\partial\phi(x)$, because of those extra terms introduced by taking the derivative of $e^{i\alpha(x)}$.

In order to keep local phase rotations from changing the physical laws, we need to add a connection $A(x)$ to the derivative, turning it into the gauge covariant derivative

$$D\phi(x) = \partial\phi(x) - iqA(x)\phi(x)$$

We stipulate that when you make a local phase rotation, you also alter the connection as $A(x) \to A(x) + \frac{1}{q}\partial\alpha(x)$. Essentially, making a gauge transformation not only alters the field, but also alters the operation of differentiation so that the combination of $D\phi(x)$ doesn't have any extra terms, $D\phi(x) \to e^{i\alpha(x)}D\phi(x)$. (It's not all that hard to do the calculation to prove to yourself that this is true.) That means that the phase factor cancels out of $D\phi^\dagger D\phi$, just as it did with the normal field terms.

Putting it all together

Where spontaneous symmetry breaking and local gauge invariance come together is in the Lagrangian. If you're not familiar with what a Lagrangian is, you can kind of think of it as an energy density, but it's not really that important. Just know that there are particular terms in it with particular meanings that I'll identify as they come.

In real-world mechanical systems, the Lagrangian is formed by subtracting the potential energy from the kinetic energy. We can do the same thing in quantum field theory: form a Lagrangian by subtracting the potential $V(\phi)$ from the kinetic term $\frac{1}{2}D\phi^\dagger D\phi$.

$$\mathcal{L} = \frac{1}{2}\bigl(\partial + iqA\bigr)\phi^\dagger\bigl(\partial - iqA\bigr)\phi +\frac{1}{2}\mu^2\phi^\dagger\phi - \frac{\lambda^2}{4}(\phi^\dagger\phi)^2$$

Most of quantum field theory is done perturbatively. This means we need the magnitude of the field (in this case $|\phi|$) to be very small so that we can calculate interesting things as perturbation series, e.g. $F(\phi) = F_0 + \phi F_1 + \cdots$. That would be all well and good if $\phi$ were close to zero. But as I said in the section on spontaneous symmetry breaking, that's typically not going to be the case. The universe will "seek out" the lowest potential, which means $|\phi|$ is going have a value closer to $\frac{\mu}{\lambda}$, the minimum.

The solution? Just redefine the field. Instead of using $\phi$, we'll use $\eta = \phi - \frac{\mu}{\lambda}$, which will be close to zero. So spontaneous symmetry breaking causes us to rewrite the Lagrangian as

$$\begin{multline} \mathcal{L} = \frac{1}{2}\bigl(\partial + iqA\bigr)\eta^\dagger \bigl(\partial - iqA\bigr)\eta + \frac{2q^2\mu}{\lambda}A^2\eta + \frac{q^2\mu^2}{\lambda^2}A^2 \\ - \frac{\lambda^2}{4}\eta^4 + \lambda\mu\eta^3 - \mu^2\eta^2 + \frac{\mu^4}{4\lambda^2}\end{multline}$$

I would definitely encourage you to plug the definition of $\eta = \phi + \frac{\mu}{\lambda}$ into the original Lagrangian, expand out the terms, and see that it really does work out to the last expression. It's just algebra, nothing too complicated, but there are a couple of neat cancellations that take place.

Now, as I said, I'm going to explain the meanings of some of the terms. In a QFT Lagrangian, there are three basic types:

  • Kinetic terms involve derivatives of the fields. They are of the form $D\eta^\dagger D\eta$, where $\eta$ is a field. Loosely speaking, they represent the energy of motion of the field.
  • Mass terms involve a product of the field and its conjugate, with some numerical coefficient. They are of the form $M\eta^\dagger\eta$. The constant $M$ is the mass of the particle that we associate with the field (although it takes a fair bit of calculation to show it).
  • Interaction terms involve a product of three or more fields. They are of the form $cA^2\eta$, where $c$ is the coupling constant for the interaction. These terms represent fundamental interactions which are shown as individual vertices in Feynman diagrams. The term $cA^2\eta$, for example, would represent an interaction between two $A$ particles and one $\eta$ particle.

Since we're talking about the Higgs mechanism, the mass terms are of particular interest. Take a look at the two forms of the Lagrangian above. You'll notice that both of them have a mass term for the field $\eta$ (or $\phi$). But in the first expression, there is no mass term for the field $A$, whereas in the second one, the mass term $\frac{q^2\mu^2}{\lambda^2}A^2$ has "magically" appeared! Just because of what amounts to a change of coordinates, a massless particle has acquired a mass! That is the Higgs mechanism in a nutshell, and $\eta$ is the Higgs field.

Of course the real thing is a little more complicated because there are multiple fields instead of just one $A$, plus they have Lorentz indices, plus there are more complicated gauge transformations involved. But the basic procedure is the same.

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In the part of "local gauge invariance", the "derivative terms" cannot be gauge invariance if we have $q=0$? Does it mean we do not have a gauge invariant Lagrangian for superfluid? –  buzhidao Jun 19 at 10:12
@buzhidao Right, it's an implicit requirement of the procedure that $q\neq 0$. Having $q=0$ corresponds to a particle with the quantum number of the vacuum, i.e. with no charge. But that would have to happen for all quantum numbers at once in order to make LGI invalid, and there are no known fundamental particles with vacuum quantum numbers. (There is the Pomeron, but that is believed to be a composite of gluons.) –  David Z Jun 20 at 13:05
  • simplest explanation without complex mathematical equation for a person don't have a wide knowledge about physics. Higgs Field, think of it as a field of people(audiences) waiting for a celebrity. Then think the celebrity as a particles and this celebrity(particle) walks toward the audience(higgs field). The more popular the celebrity is, the more people will mob him/her or let say the more interaction between the celebrity and the people, and less popular celebs less mob/interaction from the people. More interaction, more mass gains.
  • everytime I discuss Higgs Boson/ Higgs Field/ Higgs Mechanisms, I'll start with this story so that the people I discuss it with already have a vision before I introduce to them heavy mathematics required to further explain the Higgs Boson/field/mechanisms.
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