# What is a 'representation' in Field Theory

1. Oct 1, 2005

### robousy

I see this term a lot. I know its something to do with a group but I'm not too sure what it is.

Also - how does a representation (mathematical concept) translate into particle physics concept.

2. Oct 1, 2005

Staff Emeritus
Okay, you seem to know what a group is. Well a representation of a group is a homomorphism of that group into a bunch of matrices acting on some vector space, such that the group multiplication goes into the ordinary matrix multiplication as it exists in this set of matrices. A group will have many representations, some more useful than others, for example you always have the Trivial Representtion - every element of the group maps into the zero matrix (0), so a*b goes into (0)(0) = (0) and the product is preserved, as desired.

Suppose all the matrices in a representation are in some block diagonal form:
$$\left( \begin{array}{cc} (A) & 0 \\ 0 & (B) \end{array} \right)$$

Then the under lying vector space the matrices act on can be written as a cartesian product $$\mathcal{V} = \mathcal{V_A} \oplus \mathcal{V_B}$$ and the matrices (A) and (B) act on the factors independently, so each factor provides a representation of the group too. This a said to be reducing the original representation down to the two smaller ones. Obviously we are intoroduced in representations that cannot be so reduced. These irreducible representations are the 'atoms' out of which all the representations can be formed.

When the group is a Lie group of motions on some space, its representations form invariant subsets of motions, like rotations in a plane among all rotations. How particles behave under such representations defines what kind of particles they are. For example the non-relativistic electrons transform under the representations of SU(2), the unitary 2X2 complex matrices with determinant +1. The vector space these matrices act on is the space of two-component spinors. The counterintuitive property of spinors, that you have to turn them through 720 degrees rather than 360 degrees to return them to their original state, is due to their being controlled by these representations, rather than representations of the three dimensional rotations SO(3), the 3X3 orthogonal real matrices with determinant +1. Fermions go by the unitary representations, bosons by the rotation ones.

3. Oct 1, 2005

### robousy

Thanks so much for the in depth response. I have a couple of points I want to clarify:

If we take the $$SO(2)$$ rotation group
$$\left(\begin{array}{cc}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{array}\right)$$

then is a representation essentially ANY value of theta? I am guessing not. How would one obtain a represention of $$SO(2)$$ for example.

Are representations the same as generators? I know that the pauli matrices are generators of $$SU(2)$$ but does that make them a representation also?

I have some more questions but they will probably build on these questions so I'll save them.

4. Oct 1, 2005

Staff Emeritus
I don't think there are any good nontrivial representations of SO(2). No the different thetas correspond to different matrices; it's the whole set of matrices that constitutes the representation. In this case think of the rotations as a physical group of turnings, with the product of two turns being doing one turn after another and the inverse of a turn being the same turn in the opposite direction. Then the SO(2) matrices are a representation of this abstract group; in this case, a faithful representation because the mapping is one-to-one, given a prechosen basis for the plane you are acting on. Theta is a parametrization of the representation.

5. Oct 1, 2005

### robousy

you say that the whole set of matrices constitutes the representation - but this reads to me that the entire SO(2) group constitutes a representation. Is that right?

Also - how closely related are generators to representations?

6. Oct 2, 2005

### Blackforest

Does some one knows something about a) a structure of the set of $2^{\nu}$ components spinors ? b) the existence of "fields" of spinors ? and c) the representation of these fields if they exist ?
By the magie of a theorem of Cartan, any matrix of degree $2^{\nu}$ can be regard in one and in only one way as the sum of a scalar, a vector, a bi-vector , ... a p-vector in the space of p = 2.${\nu}$ dimensions. But such a matrix is a $(2^{\nu}, 2^{\nu})$square matrix and a set of $2^{\nu}. 2^{\nu}$ components; thus could be seen as a set of $2^{\nu}$ spinors ? (definition: a spinor is a set of $2^{\nu}$ components ) or do I forget something important: e.g. the set of these $2^{\nu}$ components is not any one because it must behave in a precise way relatively rotations and reversals ? Thanks for more explaination.

7. Nov 17, 2010

### Gambla

What you wrote above describes an element of the group for each value of theta. So really in that case there are an infinite number of group elements (also a property of Lie groups). SU(2) for example has only three generators, BUT, an infinite number of group elements.

Hope this Helps! Georgi ("group theory for particle physicists") is an excellent book for this. He clears things up in the first chapter

8. Nov 18, 2010

### tom.stoer

There is a very simply example. The SO(2) matrices and the U(1) complex rotation are so to speak two different representations of te same "thing". You can map every rotation in R² described by an angle and the 2*2 matrix into a rotation in the complex plane. So the underlying "thing" is the same but you can construct two different representations on top of it.

The same works for SU(2) which can be described by 2*2 unitary matrices. You can construct SO(3) matrices where for each rotation in SU(2) you get an associated matrix in SO(3). The underlying "thing" which is described by abstract relations like a multiplication table exists even w/o constructing matrix representations.

So the SU(2) group has a 2-dim. spinor rep. described by 2*2 unitary matrices, it has a 3-dim. vector rep. described by 3*3 orthogonal matrices ... and it has infinitly many higher dimensional representations. Note that SU(2) has the reps 0, 1/2, 1, 3/2, 2, ... whereas in SO(3) you only have 0,1,2,3,... so SU(2) is "larger" than SO(3).

There is one additional stepin quantum mechanics

In qm the representations do not act on ordinary vector spaces but on a Hilbert space. For every rep. the dimensionality agrees with the dim. of the vector space. For the 2-dim. spinor rep. you have 2 basis vectors |+1/2>, |-1/2>, for the 3-dim. vector rep. you have 3 basis vectors |-1>, |0>, |+1>; the dim. D of a rep. with spin S is directly calculated by D = 2S+1.

Now you can construct operators acting on the Hilbert space and generating rotations. These are not matrices but operators! But of course you construct wave functions living in 3-space and you can construct rotation matrices generating rotations of 3-vectors. You find that for any rotation operator in the Hilbert space acting on a ket |...> you find an associated rotation of the qm wave function <r|...> which is described by a rotation matrix acting on r. So there is a relation of rotations in the Hilbert space and in 3-space - but they are not the same thing.

Then you can look at an infinite dimensional Hilbert space, e.g. the Hilbert space for the hydrogen atom (forget about the r-part of the wave functions, for rotations only the angles are relevant). Represented in 3-space this Hilbert space is spanned by the spherical harmonics Ylm. l labels the representation l=0, l=1, l=2, ... and m labels the basis vector for each l with m=-l, -l+1, ..., 0, .., +l; the dimensionality is just Dl = dim rep(l) = 2l+1

That means that the infinite dimensional Hilbert space is spanned by infinitly many finite-dimensional representations of SO(3).

The relation between the "rotation acting on the Hilbert space" and the "rotation acting in 3-space" is a relation like

Ylm(RΩ) = Dmn Yln(Ω)

R rotates the unit vector described by the two angles in Ω, whereas D "rotates" the spherical harmonic. D can be related directly to a qm operator rotating the abstract state U|lm>

<Ω|lm> = Ylm(Ω)
<Ω|U|lm> = <Ω|U(lm)> = Dmn Yln(Ω)
<Ω|U|lm> = <U-1Ω|lm> = Ylm(RΩ)

Last edited: Nov 18, 2010