In order to understand the Higgs mechanism, we need to first understand how it came to be that the Higgs *field*, unlike all the other fields corresponding to the other 18 elementary particles, came to have a non-zero average value in the vacuum. As I said in the previous post in this series, this is the key fact about the Higgs field that leads to it giving mass to the other particles. So how did that come about?

In general, the *stable* equilibrium values of quantities tend to be those that result in the lowest energy states of the system as a whole, so that once a system enters that state, it stays there. For example, a rolling ball will come to rest such that its position corresponds to a point in space that has the lowest energy for the ball. It will not move from there without an external impetus. The subsequent average value of the position will then correspond to the actual position of the ball.

But there can be states that are in equilibrium that are not stable. For example, a ball exactly on the peak of a mound will be in equilibrium, but it is not stable. As long as it is undisturbed, it will stay on top. But if displaced slightly, it will roll down until it reaches a location that has the lowest energy for the ball and then stay there. It will not, by itself, roll back up to the top of the mound.

There can also be situations where the equilibrium state need not be one in which the *average* value of a quantity gives the lowest energy of the system.

Consider a rod that is hinged at one end to a table, such that it can rotate freely in a vertical plane. If left to itself, the rod will end up in stable equilibrium lying flat on the table, either pointing towards the east (say) or pointing towards the west, because those are the lowest energy states of the rod. So we can say that the lowest energy state of the rod corresponds to the rod’s length *as measured along the east-west axis* having its maximum value, equal to the length of the rod.

But there is another equilibrium state, and that is with the rod vertical. In that position, the distance the rod’s length measures along the east-west axis is zero. If perfectly balanced, the rod could theoretically stay upright but this equilibrium state is at a *higher* energy than the other two and hence, like the ball at the top of the mound, is unstable and the slightest perturbation will cause the rod to go into one of the other two stable equilibrium states.

One way to make the rod vertical is to carefully set it up that way. But there is a more ‘natural’ way of having the *average* orientation of the rod be vertical that does not require careful setting up. If the table to which the rod is hinged is shaken very vigorously and randomly, the rod will flip from one side to the other very rapidly. If so, its *average* position would then be the vertical one. So in the case of high table agitation, the rod’s state will be such that *on average* it has zero length along the east-west axis, even though this is *not* the lowest energy state of the rod.

But as the level of table agitation drops, there will come a point when there is not enough energy to flip the rod from one side to another, and the rod will come to rest in one or other position. The new stable configuration of the rod will be that with the rod’s length along the east-west axis being equal to the full length of the rod.

So to sum up, in a state of high agitation, the rod will be in a state in which the average value of the length of the rod along the east-west axis is zero but as the agitation drops below a certain critical value, there will be a sudden switch to a state in which the average length of the rod along the east-west axis is the full length of the rod, i.e., non-zero.

The vertical state of the rod is symmetric with respect to east-west axis. But when the rod is lying flat on the table in one of the two orientations, we say the ‘symmetry is broken’ because the rod suddenly and unpredictably gets oriented in just one of the states it could have had, and is no longer symmetric about the vertical position.

The Higgs field acquires its non-zero average vacuum value in an analogous way, where we make the identification of the *average* value of the Higgs field as corresponding to the average length of the rod *along the east-west axis*.

In the very, very early stages of the universe immediately after the Big Bang, the temperature was extremely high, corresponding to a state of extremely high agitation, and the Higgs field had an *average* (i.e., ‘vacuum expectation’) value of zero. This was because it was varying rapidly in such a way that the equilibrium state corresponded to a zero average value of the field. But as the universe cooled, the level of agitation dropped and at one particular temperature (which happened about *one-trillionth of a second* (10^{-12}s) after the Big Bang) the wildly varying Higgs field got ‘locked’ into the lower energy stable equilibrium state that corresponded to it having a non-zero value. This process is known as ‘spontaneous symmetry breaking’ corresponding to a symmetric situation about the value zero spontaneously becoming non-symmetric by acquiring a single non-zero value, one that could not be predicted in advance.

This process of a randomly ordered state locking itself into a particular orientation upon cooling is quite common in everyday life. It happens with water molecules with random motion freezing into ice when the temperature drops below the freezing point, and with the little domains of magnetism in ferromagnetic material suddenly becoming all aligned in one direction below the critical temperature, giving rise to large magnetic effects. Such changes are also referred to as ‘phase transitions’. The transition in the case of the Higgs field is referred to as the ‘electroweak phase transition’ because, as we will see in the next post in this series, it is this transition that results in the weak interaction force particles W^{+}, W^{–}, and Z acquiring mass.

Next: The Higgs mechanism

Marshall says

Great analogies, can’t wait for the rest!

ollie says

Mano, you must be a pretty good teacher.

DonDueed says

Is there any connection between this “non-zero average value” for the Higgs field, and the situation of Hawking radiation from a black hole?

In that case, as I understand it, the radiation arises when one of a pair of virtual particles near an event horizon falls into the black hole, preventing recombination and thereby allowing the remaining particle to escape. The black hole thereby loses mass. But that seems odd, since the infalling particle also has mass, so how is there a net loss?

The explanation I heard involved zero-point energy, which (in some way I didn’t quite grok) ensures that the infalling particle has less than zero energy (and therefore negative mass). I have to admit that I didn’t find this explanation very convincing, but maybe it involves a mechanism similar to this symmetry-breaking one for the Higgs field.

M, Supreme Anarch of the Queer Illuminati says

This series has been great for me. As a post-undergrad trying to make up the difference between my physics minor and a genuine start at real physics, I’ve been working on the math and the tight-focus version of some of this; it’s a huge help to see enough of the big picture now and then to orient myself.

Rob Grigjanis says

Don, Hawking radiation has nothing to do with symmetry-breaking.

The virtual particle scenario was always a kind of simplified picture that Hawking came up with to make it more accessible, and I’m not sure how (or

ifit corresponds to the actual calculation he did. Still, you can look at it like this:One of the virtual pair escapes the black hole, carrying energy with it. Because it is “no longer virtual”, energy must be conserved, therefore the energy of the black hole has been reduced. Worrying about negative energy particles might be taking the picture too seriously.

There’s some discussion here.

Mano Singham says

So far, no student has cursed at me or thrown things at me so I think I am getting by.

Rob Grigjanis says

Matt Strassler is always worth reading, and here he talks about virtual particles. If you scroll down, one of his comments also discusses Hawking radiation.

Mano Singham says

To add to what Rob said, you can have energy violation for a short time limited by the uncertainty principle. Also, the particle-antiparticle produced are ‘virtual’ which means that the normal relationship that connects it energy to is mass and momentum no longer applies (the technical term being that it is ‘off the mass shell’. What Hawking showed was that in certain situations, if a pair is produced just outside the black hole, one virtual particle could fall back into the black hole while the other ‘escapes’ and goes away. But is can only do that if it becomes ‘real’, i.e., it regains the normal relationship of energy to momentum and mass. This implies that its energy must be positive. But since the total energy of the pair must be zero (since they were produced out of ‘nothing’), that implies that the particle that fell into the black hole must have negative energy. Hence the energy of the black hole decreases.

As Rob says, this is different from the spontaneous symmetry breaking process.

Eugene Hill says

i have been following particle physics in scientific american for 20+ years. your posts connect all the dots for me and i can actually visualize/conceptualize how it all works. thank you.

Jean says

Does that mean that the Higgs field could have ended up in a different stable state? And if so, what would have been the consequences?

Mano Singham says

All the options for the Higgs field to end up in had the same value for the magnitude of the vacuum expectation value, just like the rod’s length along the east-west axis is the same for both final states.

Eric Riley says

That’s my question too – could there be a ‘-m’ state where gravity is a repulsive, rather than attractive force? Or are there multiple states, lest we read a little too much into the table analogy?

I can see that the magnitude would be the same, but does the ‘directionality’ play a role?

Rob Grigjanis says

The Standard Model says nothing about gravity. It’s basically a theory established on a ‘flat’ background spacetime. The content of the model is just the electromagnetic, weak, and strong interactions, plus the Higgs field. And all choices (directions) for a stable vacuum are physically equivalent.

Rob Grigjanis says

Sorry, that was meant as a response to Eric.

DonDueed says

Thanks, Mano and Rob.

I guess what made me connect the two concepts was the idea of a non-zero value for the field (in the Higgs case) and the vacuum energy (in the Hawking radiation case, as I apparently misunderstood it).

dmcclean says

dmcclean says

Oops, blockquote fail.

Paul Jarc says

As the universe continues to cool, could other fields settle into nonzero vacuum expectation values?

Rob Grigjanis says

The only possible future phase transition I’ve heard about would be due to the observed mass of the Higgs particle. The mass is just about the value which suggests that our vacuum

mightnot be the lowest energy vacuum state. In other words, our vacuum might be metastable, and could decay (by quantum tunneling) to a different, lower energy vacuum with very different properties from ours.This wouldn’t be due to cooling, though. Just probability of decay. The Ultimate “Playing Dice”!

Mano Singham says

It is theoretically possible, I suppose, but unlikely. For one thing, at 2.7K, the universe has not much more to cool. Another is that a new property would have to emerge for the other particles that would require postulating a new effect. Right now, only a non-zero Higgs field is needed.

Eric Riley says

I, perhaps, wasn’t being very clear with my question – as we find out later in this series, there are four states into which the Higgs field may have settled. Are there any gross effects we might observe if it had collapsed into one of the other states?