# Spin-1 rep of su(2) vs. vector transformations

1. Aug 3, 2013

### copernicus1

Hi,

From perusing books on QFT, I've gathered that the photon is written as a 4-vector in field theory and transforms under the standard Lorentz group operators, while an electron for instance is a 2-component spinor and transforms under a special representation of the Lorentz group as part of a bispinor with its positron partner.

But I remember as an undergrad working out the spin-1 representation of SU(2), which gives basically a set of 3x3 "Pauli matrices". Could the photon be represented as a three-component object that transforms under some other representation of the Lorentz group? Is the reason we write the photon as a 4-vector fundamental or does this just make it easy to work with the 4-component bispinor? Thanks!

I guess a more general follow-up question might be: is there any fundamental connection between su(2) and spin-1/2? Or is it just convenient to use su(2) for spin-1/2 particles and so(3) for spin-1?

2. Aug 3, 2013

### rubi

The Photon transforms under the $(\frac{1}{2},\frac{1}{2})$ representation of $SL(2,\mathbb C)$. If you want a 3-component spinor, you would need something like $(1,0)$ or $(0,1)$, but then you can't have a representation of the discrete symmetries, so the remaining option for spin 1 is $(1,0) \oplus (0,1)$, which is the representation in which the field strength tensor transforms.

3. Aug 3, 2013

### king vitamin

First of all, the reason the representations look different between nonrelativistic QM and relativistic QFT is that the symmetry groups are different. Specifically, the symmetry group of rotations in Galilean relativity is $SO(3)$ while the full symmetry group of all Lorentz transformations is $SO(3,1)$. Clearly the former is a subgroup of the latter, but when we go to Minkowski space we need to treat objects on the full 4D spacetime. You cannot write a photon as a 3-component vector because 3-vectors are no longer a good definition of a geometric object: their length gets contracted under boosts. We define our particle to be in an irrep of the Lorentz group so that different observers don't have different descriptions of the same particles.

Now, in quantum physics, demanding that a state be in an irreducible representation is actually too strong, a state can actually transform up to a phase factor under the symmetry. So for a representation $R(g)$ of a group element $g$, we can have $R(g)| \psi \rangle = e^{i \phi}| \psi \rangle$ for some $\phi \in \mathbb R$. Such a phase factor cannot be measured. These are called "projective representations" of the group. It turns out that one can obtain these through studying Lie algebras. In general, one can consider many Lie groups with the same Lie algebra, but the group obtained by exponentiating the Lie algebra is the "universal covering group" of the other Lie groups. It turns out that the representations of the universal covering group are precisely the projective representations of the other Lie groups with the same Lie algebra.

This can all be found in chapter 2 of Weinberg's QFT text.

Since you mentioned working with Sakurai in another thread, consider how he builds representation theory of angular momentum in chapter 3. First, he introduces the Lie group $SO(3)$ (the symmetry group of rotations) in the standard way. Then, he computes its Lie algebra. He considers the 2D representation of this algebra (spin-1/2) and he exponentiates it. He finds that this representation actually takes the quantum state $R(2 \pi) | \psi \rangle = -| \psi \rangle$, that is, it picks up a sign under $2 \pi$ rotations, so this is a projective representation of $SO(3)$. Additionally, he shows that an arbitrary rotation built up this way is really an element of $SU(2)$. So the groups $SO(3)$ and $SU(2)$ have an identical Lie algebra (usually denoted $su(2)$), and under exponentiation the Lie algebra goes to $SU(2)$, the universal covering group. If he started with the 3D rep of the Lie algebra and exponentiated it, he would have obtained the standard representation of $SO(3)$ (up to isomorphisms), so the covering group contains the normal representations of the symmetry as well as the projective ones.

As a final note, when one considers relativistic QFT, $SL(2,\mathbb C)$ is the universal cover of $SO(3,1)$.