Lets Talk Higgs

Erasmus00 · #1 (**permalink**) 11-15-2007, 07:24 AM

There has been some indication on other threads that perhaps the Higgs mechanism/Higgs boson are rather poorly understood. Hence, this thread. I'm going to attempt a sort of "question and answer" on the Higgs. Unfortunately, before I can really talk about the Higgs I'll need to give a whirlwind tour of some important ideas in particle physics. I'm purposefully brushing past a lot of subtleties to paint the big picture, and there will be some things you may have to trust me on.

First, we'll need to develop some concepts. The first is the idea of a Lagrangian. One of the more interesting developments of the mid 1800s is the idea that classical physics can be described by the statement "particles minimize their action" where the action(S) is defined as

$S = \int \mathcal{L} dV dt$

That funny looking L is the Lagrangian, and it is fundamental in modern physics. It turns out that the way to generalize classical physics concepts (like electricity and magnetism) to the quantum world is to figure out the Lagrangian for the classical theory and work with it. In classical mechanics it is standard to think of the Lagrangian as

$\mathcal{L} = T-V$

where T is kinetic and V potential energy. In classical mechanics the Lagrangian is always minimized, but in quantum mechanics, we can have fluctuations away from the minimum.

The important things in the Lagrangian of particle physics are called fields, and they represent the types of particles we can have. Simple functions (called scalars) represent spin 0 particles, if they are complex that means the particle is charged. Vector represent spin 1 particles, and there are objects called "spinors" that represent spin 1/2 particles that I will do my best to avoid, as they are a bit more complicated.

The next concept we'll need is symmetry. This is the primary tool that helps us build the Lagrangians for our theories. Generally, what we do is we decide on a symmetry our theory should have, and then use for our Lagrangian every possible term that has such symmetry. As a concrete example consider that what we have a theory with a spin 0, charged particle. This is represented by a complex function $\phi(x,t)$ . There will be a term in the Lagrangian that looks like $m^2\phi^*\phi$ . This is called a mass term- any term that is quadratic in a field (i.e. the field squared) generates a mass for the field.

This term has a symmetry: we can multiply $\phi(x,t)$ by $e^{i\theta}$ . If we do this we get

$m\phi^*\phi \to me^{-i\theta}\phi^*e^{i\theta}\phi = m\phi^*\phi$

This type of symmetry is called a U(1) symmetry. Notice that we could also allow $\theta$ to be a function of x and t. In other words, we could use as slightly different transformation at each point in space and time. This is called a "gauge symmetry."

Next, consider the term $\phi^*\frac{\partial}{\partial x^i}\phi$ The i labels which coordinate we are taking a derivative with respect to (x,y,z, or even t) This term wouldn't be in a lagrangian but will be a useful example. What happens if we try to do a gauge symmetry on this guy?

$\phi^*\frac{\partial}{\partial x^i}\phi \to e^{-i\theta}\phi^*\frac{\partial}{\partial x^i}e^{i\theta}\phi = \phi^*\frac{\partial}{\partial x^i}\phi + i\frac{\partial \theta}{\partial x}\phi^*\phi$

Oh no! It doesn't have a gauge symmetry, because of the derivative involved! We will want derivatives in our Lagrangian, so what can we do? The solution is to add a new field, other then $\phi$ . We'll call it A_i. We also need a transformation law for A_i, when we do a gauge transformation of $\phi$ then $A_i \to A_i +\frac{\partial \theta}{\partial x_i}$

So now we could have a term like $\phi^*\left(\frac{\partial}{\partial x} -iA\right)\phi$ This will be gauge invariant.

But notice! In order to "rescue" gauge invariance, we had to add a whole new field, this means we have added a whole new particle type! Further, its not just a function, its a vector! That means it describes a spin 1 particle. This is a "gauge boson" its a particle that we had to add to satisfy the gauge symmetry.

Does it have mass? Well can we add a term to the Lagrangian like $m A_iA_i$ ? The answer is no- it will not satisfy gauge invariance! Try it using the transformation law for A.

Hence, to satisfy gauge invariance, we had to postulate spin 1, massless bosons. BUT, we know the W and Z bosons have masses. How can we understand this? We'll have to understand the idea of spontaneous symmetry breaking. This will have to be next post. For now, feel free to ask questions and I'll try to answer.
-Will

Farsight · #2 (**permalink**) 11-15-2007, 08:30 AM

Thanks for this Erasmus. I'll be asking questions. I have to go now, and it's perhaps a little tangential, but can you tell me what a frame-Higgs is, as per Garret Lisi's paper?

[0711.0770] An Exceptionally Simple Theory of Everything

Qfwfq · #3 (**permalink**) 11-16-2007, 12:06 AM

Godd going, Will!

Quote:

Originally Posted by Farsight

...tell me what a frame-Higgs is, as per Garret Lisi's paper?

First, I think it would be a good idea to let the thread proceed in reasonable order, keeping questions in line with the level of the explanation.

Anyway, I found what the author means in the first section after the introduction and it's the kind of thing that I had imagined, considering what he's up to. 1.1 A connection with everything:

...while the gravitational fields are described by the spin connection, $\omega\in so(3; 1) = Cl^2(3; 1)$ a Clifford bivector valued 1-form, and the frame, $e \in Cl1(3; 1)$ , a Clifford vector valued 1-form. The frame may be combined with a multiplet of Higgs scalar fields, , to interact with the electroweak gauge fields and fermions to give them masses.

The connection, in GR, describes the relation between frames of reference, although it isn't always done with Clifford algebras.

Farsight · #4 (**permalink**) 11-19-2007, 05:31 AM

Quote:

Originally Posted by Erasmus00

One of the more interesting developments of the mid 1800s is the idea that classical physics can be described by the statement "particles minimize their action" where the action(S) is defined as

S = integral L dV dt (sorry, maths error when quoting)

That funny looking L is the Lagrangian, and it is fundamental in modern physics. It turns out that the way to generalize classical physics concepts (like electricity and magnetism) to the quantum world is to figure out the Lagrangian for the classical theory and work with it. In classical mechanics it is standard to think of the Lagrangian as

L = T-V

where T is kinetic and V potential energy. In classical mechanics the Lagrangian is always minimized, but in quantum mechanics, we can have fluctuations away from the minimum.

Erasmus: I have some problems of understanding this, and fear I fall at the first hurdle. I like to think I think in classical terms, but it’s so classical, and so simple, that when I consider a photon, I’m thinking in terms of a transverse wave, where the photon is an action. The kinetic energy is constant, the momentum is hf/c, and whilst people commonly talk about the electric field and the magnetic field, all I can see is something like an S-wave in seismology. There are no “orthogonal earthquake fields”, so I can’t see how it ties in with $S = \int \mathcal{L} dV dt$ and $\mathcal{L} = T-V$ . So, please can you talk a little about the Lagrangian of a photon?

Quote:

The important things in the Lagrangian of particle physics are called fields, and they represent the types of particles we can have. Simple functions (called scalars) represent spin 0 particles, if they are complex that means the particle is charged. Vector represent spin 1 particles, and there are objects called "spinors" that represent spin 1/2 particles that I will do my best to avoid, as they are a bit more complicated.

Please can you talk about spinors? In my simple thinking, the next simplest entity after the photon is the electron. If I skip the electron I feel I’ve got no chance with anything more complicated.

Quote:

The next concept we'll need is symmetry. This is the primary tool that helps us build the Lagrangians for our theories. Generally, what we do is we decide on a symmetry our theory should have, and then use for our Lagrangian every possible term that has such symmetry...

How does symmetry help? I'm afraid I rather got lost on your hypothetical example, I struggle when I try to look this up for myself (eg below), and I don’t understand why the U(1) group describes the electromagnetic force.

Standard Model (mathematical formulation) - Wikipedia, the free encyclopedia
Special unitary group - Wikipedia, the free encyclopedia
Symmetry in physics - Wikipedia, the free encyclopedia

”An important type of physical theory based on local symmetries is called a gauge theory and the symmetries natural to such a theory are called gauge symmetries. Gauge symmetries in the Standard model, used to describe three of the fundamental interactions, are based on the SU(3) × SU(2) × U(1) group. (Roughly speaking, the symmetries of the SU(3) group describe the strong force, the SU(2) group describes the weak interaction and the U(1) group describes the electromagnetic force.)”

Erasmus00 · #5 (**permalink**) 11-19-2007, 07:13 AM

Quote:

Originally Posted by Farsight

So, please can you talk a little about the Lagrangian of a photon?

My first goal is to describe the basics behind the Higgs mechanism. Maybe after I finish I'll open up the thread a bit and push into some other areas.

Quote:

Please can you talk about spinors?

Spinors are probably going to (mathematically) stay beyond the scope of this thread. Mathematically, they require a bit more group theory to really get a handle on.

Quote:

How does symmetry help?

Symmetry is perhaps the only guiding principle we have when writing down a lagrangian from first principles. This is why its a helpful concept. For now, just think of U(1) symmetries as a symmetry where we multiply by $e^{i\theta}$ where $\theta$ is an arbitrary parameter. The 1 is for one dimensional. The U means unitary, which is the property $U^{\dagger}U = 1$ .

Quote:

I'm afraid I rather got lost on your hypothetical example

Specific questions would be helpful to resolve any problems you are having.
-Will

Erasmus00 · #6 (**permalink**) 11-19-2007, 08:06 AM

Onward and upward. In the first post in this thread, I tried to touch as gently as I could on the following points
1. Lagrangian's determine the physics of a system.
2. Insisting on a gauge symmetry requires adding new particle content to the theory, massless "gauge bosons."

BUT, experimentally, we know some gauge bosons have mass(the W and Z bosons)? How can this possibly happen?

The answer is the Higgs mechanism. The idea of the Higgs mechanism is that while the "laws of physics" (in this case the Lagrangian) DO have a symmetry, the ground state of the system itself does not. Lets work through this via a specific example. Lets start with the following "toy" Lagrangian.

$\mathcal{L} = \mu \phi\phi^* + \lambda \left(\phi\phi^*\right)^2$

We will call this a "symmetry breaking" lagrangian.

Notice that this has the same sort of symmetry in the examples in the first post. But what about its ground state? If we assume the solutions to $\phi$ are constants, then we just have to minimize the Lagrangian the same way we minimize anything- take a derivative and set it to 0. Treating $\phi$ and $\phi^*$ as independant, we see that

$\frac{\partial \mathcal{L}}{\partial \phi} = 0 = \mu \phi^*+2\lambda\left|\phi\right|^2\phi^*$

We notice a funny thing, if we pick $\mu$ negative and $\lambda$ positive, we have a non-zero minimum with a value

$\left|\phi\right|^2 = \frac{\mu^2}{2\lambda} = \nu^2$

Here I'm defining $\nu$ . This isn't actually one minimum, but infinitely many equivalent minima, as we can expect from the symmetry. $\phi = \nu e^{i\theta}$ describes our "family" of minima.

But the system can only be in ONE of those states, and so it has to "pick" one. This means that the system will fall into one specific choice of the "angle" $\theta$ describing our family of minima. This is why we call this "spontaneous" symmetry breaking. There is nothing that rules out one of the minima as special, so the system "spontaneously" picks one.

This can have profound consequences. Consider a term in the lagrangian of the form

$\mathcal{L} = \frac{\partial \phi^*}{\partial x^i}\frac{\partial \phi}{\partial x}$

To make this gauge invariant, we have to add terms to the derivatives as we did above.

$\mathcal{L} = \left(\left(\frac{\partial}{\partial x^i} - iA_i\right)\phi\right)^*\left(\frac{\partial}{\partial x^i} - iA_i\right)\phi$

Now, if we add to this Lagrangian our "symmetry breaking" Lagrangian considered above, our solution for $\phi$ isn't changed, for two reasons. First, the derivative terms with $\phi$ don't matter if $\phi$ is a constant. Second, the terms that have BOTH $\phi$ and $A_i$ are considered "couplings" between the gauge bosons and the field, these won't change the background solution.

Notice in the Lagrangian directly above, there are "coupling terms" that are quadratic in the A fields.

$A_i^2\phi\phi^*$

If we expand $\phi$ around its minimum, the first term would yield

$\nu^2A_i^2$

And we have produced, from a gauge invariant theory, a mass term for our gauge bosons. This is the essence of the Higgs mechanism- a non-zero ground state for one of our fields allows for the generation of new terms in our Lagrangian.

This was probably pretty muddled, and full of typos, as I'm writing this in an airport. Please ask questions so I can clean this up and hopefully add some clarity.
-Will

Edit: slight rewordings, hopefully for clarity

Qfwfq · #7 (**permalink**) 11-20-2007, 01:44 AM

Perhaps Will the only problem could be a bit too much brevity, on what is pretty much the heart of the matter.

Maybe I can help if I point out that, where Will says: "But the system can only be in ONE of those states, and so it has to 'pick' one.", this is what he meant by spontaneous symmetry breaking at the end of the first post. Choosing one, out of the whole "family" of equivalent minima, means taking a value of the angle $\theta$ in $\phi = \nu e^{i\theta}$ , which is called Weinberg's angle. This ought to make the words "if we add to this Lagrangian our 'symmetry breaking' Lagrangian considered above" a bit more clear.

I'll try to address two questions that were specific enough:

Quote:

Originally Posted by Farsight

So, please can you talk a little about the Lagrangian of a photon?

Essentially, it would be the non-zero side of the Klein-Gordon equation, left multiplied by the conjugate field, but this is only one term in the whole thing and may be called the boson's free term. In order to treat QED for photon and one kind of charged fermion (along with its anti, an introductory textbook treatment usually does it for electron/positron) one must also have the fermion's free term, which is the non-zero side of the Dirac equation in my avatar (also left multiplied by the conjugate field), and the interaction term. Note that the funny symbol $\cancel{\partial}$ in my avatar is a shorthand for a contraction in which there are the $\frac{\partial}{\partial x_i}$ derivations; to go into more exact detail would require a digression on spinors that we'll avoid. Summing these two terms, the fermion's free term and its interaction term with the boson, basically amounts to the trick Will already did when he "added the $A_i$ field" in order to have a gauge invariant term with a derivative, it gives a term:

$\psi^*\left(\cancel{\partial} -i\cancel{A}\right)\psi$

Now if one momentarily considers only this (without the left $\psi^*$ ) positing it equal to zero is the Dirac equation for the fermion in an electromagnetic field $A_i(\vec{x})$ regarded as "given" (IOW not determined by the equation). At this point we maybe can look at:

Quote:

Originally Posted by Farsight

I don’t understand why the U(1) group describes the electromagnetic force.

OK consider an infinitesimal displacement $dx^i$ and contract it with the EM field: $A_i(\vec{x})dx^i$ . This of course is an infinitesimal Lorentz scalar, but what is it? You might guess it's an infinitesimal increment of "something" along the infinitesimal displacement, and it is exactly a change $d\theta$ in the angle $\theta$ of the U(1) symmetry that Will introduced in the OP.

Farsight · #8 (**permalink**) 11-20-2007, 08:11 AM

Thanks for the replies Erasmus and Qfwfq. I'll try to do some reading to get up to speed.

Rade · #9 (**permalink**) 11-27-2007, 03:58 PM

A question. Suppose I have two "particles". The first, I call [A-2^] where ^ indicates the particle is pure anti-matter having 2 mass units. The second, I call [B-3] which is made of pure matter having 3 mass units. Now, suppose the [A-2^] is a spin 1 vector and [B-3] is a spin 1/2 spinor. What is the Lagrangian that would allow these two particles to form stable union of the form {[A-2^]~[B-3]} where ~ is a strong binding force? If not possible, why not ? Thanks for any help.

modest · #10 (**permalink**) 12-06-2007, 04:47 AM

Erasmus00,

Wonderful topic and excellent gap-filler to what is usually not found in a usual introduction.

For anyone,

I'm thinking about the high-range in expected mass of the higgs. Is there an easy way to show how its mass is not protected by a symmetry? I honestly am not trying to start a discussion on fine-tuning or hierarchy problems - just looking for how the higgs' mass is derived differently, or rather, how it cannot be derived similarly (as say - a fermion).

thank you

-modest