# Representations of SO(3)

1. May 16, 2013

### silverwhale

Hi Everybody!

I am working on QFT and learning representation theory from Coleman's lecture notes. Just the necessary stuff to go to the Dirac equation.
To my question:

From the generators of SO(3) I get through exponentiation an element of SO(3), this holds naturally for any Lie group.

Now, the generators themselves are n x n dimensional.

Then, if I take as my generators the pauli sigma matrices which satisfy the commutation relations of the Lie group SO(3) then the resulting representations of the elements of SO(3) should be rotations which act on two component spinors.

BUT how do I get an element of SO(3) which acts on vectors? Or better say which acts on a multiplett such that it transforms as a vector.
A 3 x 3 representation of the Lie algebra should give me an element of SO(3) which acts on 3 compenent spinors I guess, but not vectors.

Let me repeat, when constructing the represention of SO(3) with a 3 x 3 representation of the Lie algebra, I should get a rep that transforms 3 spinors, but how do I get a rep that transforms 3 vectors?
Does it depend on the choice of the representation of the Lie algebra?

I saw the words vector representation and spinor representation, are these in any way related to my problem?

Thanks for any help in advance!!

2. May 16, 2013

### DrDu

E.g. take the vector $S=(\sigma_x,\sigma_y,\sigma_z)^T$. Then use $[\sigma_i, \sigma_j]=i\epsilon_{ijk}\sigma_k$ to define the 3x3-matrix $A^i=i\epsilon_{ijk}$ corresponding to the group action $[\sigma_i,\cdot]$.
I think this is called the adjoint representation.

3. May 16, 2013

### Bill_K

You may be confusing SO(3) with the 8-parameter group SU(3). SO(3) does not have a 3-spinor representation.

4. May 16, 2013

### DEvens

Your lecture notes should be telling you some of this. If not, you probably want to get an appropriate text, maybe from your university library. It's a big topic.

SO(3) stands for "special, orthogonal, 3 dimensional." A related group is SU(2), the "special, unitary, 2 dimensional" group. Each is a Lie group and has three generators in a related Lie algebra. The Lie algebra is basically the tangent space to the manifold the group represents.

SO(3) is essentially rotations in 3 dimensions.

Basically, SO(3) has representations in terms of matrices that are 1x1, 2x2, 3x3, etc.

The 1x1 is the "scalar" rep. It basically does not do anything to the things that transform that way. In other words, it's just the number 1. It acts on things that don't change under rotation, in other words, scalars.

The 2x2 rep is the spinor rep. That's just the Pauli spin matrices as generators. That operates on 2 component spinors. But recall that we usually don't try to squeeze electrons into two component spinors. We usually use 4 component Dirac spinors.

The 3x3 rep is just rotations. You can build that a few ways. One way is as generated by rotations around x, y, or z axis.

One 4x4 rep is the usual Dirac rep. That's the one that operates on Dirac spinors as generated by two copies of the Pauli spin matrices, with the appropriate signs and i's and such. Those are the gamma matrices.

You need to distinguish between reducible and non-reducible reps. Think about the 2x2 Pauli sigma matrix rep. You could make a new rep by combining two 2x2 reps.

[sigma 0 ]
[ 0 sigma]

That is, you just make a 4x4 matrix by sticking in two copies of the 2x2 rep. Transforming all elements of a rep to this block-diagonal form (using similarity transforms) is called reducing it. You can see this is just two copies of a smaller rep. If you can't transform all elements of a rep this way at the same time it is irreducible. The 4x4 rep for Dirac spinors has to be irreducible. Otherwise, you could rotate the two parts into separate objects.

The usual 3x3 rep of rotations is irreducible. (Though it isn't instantly obvious.) The generators look like so. First the x one.

[0 1 0]
[-1 0 0]
[0 0 0]

Then the y one.

[0 0 1]
[0 0 0]
[-1 0 0]

Then the z one.

[0 0 0]
[0 0 1]
[0 -1 0]

You can't use a similarity transform to make all three block diagonal at the same time.

However, sticking a scalar and a spinor rep together is reducible.

[1 0]
[ 0 sigma]

A reducible rep can't be elementary, because you can rotate things around so that the (in the example) scalar and spinor parts are separate. An irreducible rep acts on a sub-space that does not transform into another sub-space. So a (3 dimensional for example) vector gets transformed onto another vector, not into scalar and spinor parts.

Now the killer thing: Scalars are spin 0. Vectors are spin 1. Spinors are half integral spin. What that means for the generators and the reps is this. Scalars are invariant under rotations. So they don't change. Vectors are spin 1, so you rotate them 360 degrees to get back to where they started.

But half integral spin things are anti-symmetric. You need to rotate them 720 degrees to get them back where they started. And this is shown in the reps. Rotating by 360 degrees in a spinor rep gets you a -1. And you need another 360 to get back to 1. They are, in effect, a double covering of SO(3). This is related to (insert really a lot of Lie group theory, Lie algebra theory, manifold theory, etc., and some caveats) SU(2).

The important reps for quantum mechanics are the scalar spin-0 rep, the vector spin-1 rep (photons, gluons), and the Dirac spin-1/2 rep (leptons). If you start into quantum gravity you may see a spin 2 rep, sometimes called a tensor rep, for gravitons. If you get into more complicated particle physics you might see a spin 3/2 rep for super-spin particles. Ordinarily we don't deal in higher than spin 2, since there are reasons to think that higher than spin 2 can't be elementary. Compound particles, like molecules or nuclei, can sit in just about any rep if they have high enough spin.

There really is a great deal more to this stuff. You should get a good text. Get something with a title like "group theory and quantum mechanics" or something like that.

5. May 16, 2013

### The_Duck

The 3-dimensional representation of SO(3) is the vector representation. One possible choice of generators is

$\sigma^{(3)}_x = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i\\ 0 & i & 0 \end{array} \right)$

$\sigma^{(3)}_y = \left( \begin{array}{ccc} 0 & 0 & i \\ 0 & 0 & 0\\ -i & 0 & 0 \end{array} \right)$

$\sigma^{(3)}_z = \left( \begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0\\ 0 & 0 & 0 \end{array} \right)$

You can check that these have the right commutation relations to be generators of SO(3) (well, you'll probably have to fix whatever sign mistakes I've made). Then you can check that when you exponentiate these matrices you get the usual rotation matrices for 3-vectors.

For this reason, spin-1 particles, which live in the 3-dimensional representation of the rotation group, are often called "vector" particles. For example we speak of "vector mesons" (like the rho meson) or "vector bosons" (usually referring to the W and Z particles).

6. May 16, 2013

### kof9595995

I think OP might have confused the 3-dimensional rep of SO(3) that describes spin-1 in QM and 3-dimensional real vector rep of SO(3). If I remember correctly, the only difference is that the former is defined on complex vector space and the latter is on real vector space, besides this, these two reps are unitarily equivalent. (my memory might be inaccurate though, so someone please correct me if something is wrong)

7. May 17, 2013

### dextercioby

No, they are 3 dimensional. You cannot exponentiate an nxn matrix and get a matrix from SO(3) in the vecinity of the unit matrix, unless n=3.

They are not, since you problem is about SO(3) and <vector and spinor> representations are in the context of SU(2).

8. May 17, 2013

### silverwhale

First thanks to everybody for their comments!
I will go through them one by one.

Let me start with the last one! :)

To dextercioby, I do not agree with your first answer. Simply because we do define generators of SO(3) which obey the commutation relations of the Lie algebra i.e. [J_i, J_j] = i e_ijk J_k, which are the 2 x 2 pauli matrices!

This is called the spin 1/2 representation.
But correct me if I'm wrong!

9. May 17, 2013

### Bill_K

I don't understand this, since SO(3) has spinor representations and SU(2) does not.

Last edited: May 17, 2013
10. May 17, 2013

### dextercioby

You're making a confusion, the Pauli matrices form a 2-dimensional representation of so(3), by exponetiating a linear combination of them you don't get an element of SO(3), but of SU(2) or a unitary operator representing the group SU(2).

Last edited: May 17, 2013
11. May 17, 2013

### DrDu

As far as I understand, the algebras SO(3) and SU(2) are isomorphic while the corresponding groups are not: SU(2) is the double cover of SO(3).
People also tend to mix up the algebra and it's representations, which is understandable as these algebras get their name from some special representations (namely the one of orthogonal 3x3 matrices and unitary 2x2 matrices).

12. May 17, 2013

### DrDu

Alternatively you can say that you get a projective representation of SO(3). I think this is the more physical interpretation why we speak of vectorial and spinorial representations, the latter ones being projective reps.

13. May 17, 2013

### dextercioby

Literally speaking, both groups have only vectorial representations, since the representation space is a vector(!) space. BUT, basis vectors from a representation of SU(2) are called <spinors>, and for integer weights the spinors of dimension 2n+1 they are called <vectors>, due to the fact that the representations of SU(2) with integer weights can be 1-1 traced back to representations of SO(3).

14. May 17, 2013

### Bill_K

The difference is that the spinorial representations are double-valued.

15. May 17, 2013

### DrDu

Yes, that's the characteristic of a projective representation.

16. May 17, 2013

### dextercioby

Correct, but projective representations are hard to work with, since the projective Hilbert space fails to be linear. We need to get somehow from projective reps to ordinary reps and that's where the work of Bargmann and later Mackey kicks in.

17. May 17, 2013

### dextercioby

No, of course not. Projective representations of SU(2) are not double-valued, using this faulty usage of the term.

There's no really such thing as a double-valued rep., simply because the definition will not allow it. See the definition from wiki (handy resource, don't feel like looking it up in a serious book).

A representation of a Lie group G on a complex Hilbert space V is a group homomorphism Ψ:G → B(V) from G to B(V), the group of bounded linear operators of V which have a bounded inverse, such that the map G×V → V given by (g,v) → Ψ(g)v is continuous.

Unfortunately guys working in physics have propagated this term.

Last edited: May 17, 2013
18. May 17, 2013

### DrDu

Who claimed that SU(2) has projective reps?

19. May 17, 2013

### dextercioby

It does, but it's not the point. I was addressing a seemingly generalization of yours: that the so=called <double valued> character is typical of projective reps shifted back to regular vector space, which is wrong, of course, for groups that are simply connected (i.e. their universal covering group is isomorphic to the group itself).

20. May 17, 2013

### silverwhale

Ok.

But then I don't understand why the pauli matrices obey the commutation relations of the lie algebra of SO(3)? If exponentiating them produces a representation of SU(2).
I do know that SU(2) is isomorphic to SO(3) but that doesn't really help me.

And also how then do we get spinors from SO(3)? If not by looking at all the different representations of SO(3)?

I feel a little bit confused now (well I was already but now I am more..)

I am very new to this stuff. And I really want to learn it! So any good and short literature advice for physicists is highly welcomed.