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

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The discussion explores the representation of particles in quantum field theory (QFT), specifically contrasting the photon as a 4-vector and the electron as a 2-component spinor. It examines whether the photon could be represented as a three-component object transforming under a different Lorentz group representation, emphasizing that 3-vectors lose their geometric meaning under boosts. The conversation highlights the distinction between the symmetry groups of nonrelativistic quantum mechanics and relativistic QFT, noting that the photon must be treated within the full 4D spacetime framework. It also touches on the concept of projective representations in quantum physics and how they relate to Lie algebras and universal covering groups. Overall, the discussion underscores the complexity of particle representations in the context of different symmetry groups.
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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?
 
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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.
 
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).
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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