## Saturday, November 17, 2007

### Modest Understanding of Lie Groups Part 0: U(1)

This semester I had the pleasure to take a very little nice course on mathematics, mathematics for physicists that is. What this means is that half the course we dealt with lie groups and the remaining month or so we studied path integrals. Now, why is this interesting? It just happens to be the sexiest mathematics available to me at this point.

In case you didn't know, finding symmetries in physics leads to a deeper understanding of the phenomena at hand. This is obvious to any undergraduate student facing for the first time electromagnetism. The most basic problem of this course is finding the electrical field a distance d above an very long line of uniform density charge. Needless to say, you want to know how much the line would pull (or repel) a charge, should you feel like putting one a distance d above it. Of course you don't need to understand much about physics to eventually see that it doesn't matter where you place it, as long as it is a distance d perpendicular to the cable. Clearly this is because the line of charge is very long and this places are practically the same to the line. From this information you then can guess that the electric field must only depend on the coordinate perpendicular to the line, a trivial conclusion, but proves the point just fine I guess.

Again, why am I talking about this? Turns out our most precious tool (for the moment) allowing us understanding the world, the Standard Model, is based on symmetry groups. Namely it is usually represented by SU(3)xSU(2)xU(1). Let's start by understanding the simplest part of this: U(1). Imagine a circle, or rather, its points:
In this figure, A and B are two points on the circle. All the points on this circle are characterized by some properties. For example, if a point on the circle is represented by the vector
$X = \begin{bmatrix}x\\y\end{bmatrix}$
the point
$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$
is also on the circle! Having seen this, it's easy to see that the points on a circle with the operation of addition (since each point is characterized by an angle, we can understand it as the sum of their angles) form a group. If we see this circle on the complex plane, a point on the circle can be represented by a complex number
$e^{ia}$

and a rotation about the center of the circle will be given by multiplying this number by the following phase factor
$e^{i\theta}$
This will take us
$e^{ia}\rightarrow e^{i\theta}e^{ia}=e^{i(a+\theta)}$
that is, another point on the circle (closure). If this is new to you, try and find the identity and inverse elements.
We can see this phase factor as a 1X1 matrix, and call it U. It's clear then that in this case
$UU^*=1$
But in general for bigger matrices
$UU^\dagger=1$
I hope then to have explained how this implies that the complex numbers of norm 1 form a group under the operation of multiplication. This group is the most simple I can think of for now, the U(1) group (unitary matrices of rank 1, which satisfy the last equation).
Tune in next time for a brief explanation on all the other classic groups.

vannern said...

And what about SU(2) and SU(2)xU(1)? Do they play an individual role?

Eric Pulido said...

I'll get to it in time. There are more important things in sight right now.

Luis Sanchez said...

Just as a sneak avance, I can tell you right now that SU(2)xU(1) describes the electroweak force. And yes, post about it, Eric!

Eric Pulido said...

It's been a rough couple of weeks, so a week may pass before I make a proper post. Meanwhile maybe Luis could tell us a bit about he trip he might be making next year...

Cheers

Luis Sanchez said...

Well, about that trip. Just for the record on the rumor mill: I have been selected by the SMF for a summer internship at DESY.

Nonetheless, my application still needs to be approved by DESY itself, so it is still not sure. I will keep my fingers crossed 'til then.

Charis said...
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