# SU(N) Vector vs. Spinor Representations

1. Feb 25, 2005

### fliptomato

I'm a little bit confused about the difference between the spinor and vector representations of SU(N)--I guess I could start with asking how a spinor and a vector differ: is this only a matter of how they transform under Lorentz transformations?

Following up, the covariant derivative for a spinor of SU(2) is (i.e. for a scalar field $$\phi$$ that transforms as a spinor of SU(2)):

$$D_\mu \phi = (\partial_\mu - i g A^a_\mu \tau ^a )\phi$$

While the covariant derivative for the vector representation of a scalar $$\phi$$ is:

$$D_\mu \phi = \partial_\mu_a + g \epsilon_{abc} A^a_\mu \phi_c$$

(these are from Peskin and Schroeder p. 694-5, eq. (20.22) and (20.27) resp.)

My understanding is that this means we have a scalar field $$\phi$$ that has a nonabelian gauge symmetry in some abstract (internal) SU(2) space.

The spinor covariant derivative seems to make sense from the general definition of the covariant derivative:

$$D_\mu = \partial_\mu - igA^a_\mu t^a$$

where $$t^a$$ is a generator of the gauge group. does this mean that the generator of the vector representation is something like $$\epsilon_{abc}$$? Where does this $$\epsilon_{abc}$$ come from?

Thanks,
Flip

2. Feb 25, 2005

### dextercioby

There is something really puzzling here.On normal basis,the matter field (in this case a scalar) would transform under the fundamental representation of the gauge symmetry group,which in this case is SU(2).The generators of the fundamental reperesentation of the group SU(2) are indeed $T_{a}=\frac{1}{2}\sigma_{a}$...So they should be there...I don't know which kind (what spin bears) this fundamental representation is.

Daniel.

3. Mar 17, 2005

### junebuglaura

I actually just had this problem myself (on an exam) and was stuck! Luckily I figured it out!

First, the covariant derivative is representation dependent:

$$D_\mu = \partial_\mu - igA^a_\mu t_R^a$$

Where $$t_R^a$$ are the generaters for a particular representation R. Now when you have a field that transforms like a vector, you can use the fact that SU(2) is really the same as SO(3) (the infinitesimal transformations are measured by a vector).

The Fundamental representation of SO(3) are the angular momentum generators J, that satisfy the commutation relations:

$$[J_i ,J_j ] = i\epsilon_i_j^k J_k$$

So the $$\epsilon^i^j^k$$ are the structure constants of SO(3). The Adjoint representation for any group has generators given by:

$$(t_A^a)^b^c=-i f^a^b^c$$

Where $$f^a^b^c$$ are the structure coeffiecients. So in particular for SO(3):

$$(t_A^a)^b^c=-i \epsilon^a^b^c$$

So then for the adjoint reprensentation of the "spin-one" (SO(3)) version of SU(2) we have:

$$D_\mu = \partial_\mu + g\epsilon^a^b^c A^a_\mu$$

I hope this helps!

-Laura

4. Mar 17, 2005

### marlon

Let me add that a spinor is a special kind of vector. I mean, it has the property that if you rotate it 360° you get the exact opposite (A ---> -A) of what you originally rotated. Rotate another 360° and you get where you started off in the first of the two rotations (A---> -A--->A).

Now let us look at the rotationgroup SO(3) or even any other group, it don't matter :

An object v transforms as a vector if you can write v' = Uv where U is a representation for the group in question, U represents a rotation. Another way to say this is if you transform an object under a certain group, the 'image' of this transformation will be a linear combination of the object that you transformed. So transforming like a vector really means that the object you transform will be written out as a linear combination of it's components after the transformation.

An object transforms as a tensor if you can write v'=UU'U''v
So this means that v transforms 'as a product of vectors' because of the multiple U-matrices.

Now, transforming like a spinor really means that the object tranforms like a vector (you know what that means) but not just any vector. This is a special case, where the U-matrix does not represent just any transformation but a transformation that gives you the opposite of the initial object after a rotation of 360°.

One can recognize a spinor by the way it transforms under a group. If the generator is a Pauli-matrix you are done...Just like in the case of SU(3), if you now the generator is a GellMann matrix, you know you are working with anobject in the adjoint representation and these objects are GLUONS

regards

marlon

5. Mar 20, 2005

### BlackBaron

Since a long time ago, I've been looking for a simple explanation of what a spinor is, without much luck.
So far, I had simple patched up together many definitions and different points of view until I finally had an acceptable idea on what a spinor was.
Your explanation, Marlon, is, so far, the simplest and clearest I've seen, thanks a lot for that.

Now, my question is, does ANY Lie group have a spinor representation?
If not, wich ones do?

6. Mar 20, 2005

### marlon

No, for example : under addition, the real line is a Lie group.

It is important to realize the the spinor representation is indeed the representation of a Lie Group because it is a rotation and any rotation can be put into a Lie Group like SO(3) and so on...

Perhaps we should exclude the rotations (in the complex plane) that do not put an object onto itself after 360° of turning...but i am not sure of that anymore, it has been too long for me...

regards
marlon