# Lorentz Group = SU(2) x SU(2)?

1. Jun 10, 2014

### Xenosum

In Ryder's Quantum Field Theory it is shown that the Lie Algebra associated with the Lorentz group may be written as
\begin{eqnarray} \begin{aligned}\left[ A_x , A_y \right] = iA_z \text{ and cyclic perms,} \\ \left[ B_x , B_y \right] = iB_z \text{ and cyclic perms,} \\ \left[ A_i ,B_j \right] = 0 (i,j = x,y,z).\end{aligned} \end{eqnarray}
by performing a change of basis in the original algebra. This suggests that the Lorentz algebra consists of two SU(2) subalgebras which commute with each other, because the SU(2) algebra is given by $\left[ J_i , J_j \right] = i \epsilon_{ijk} J_k$. Now what I don't understand is why this implies that the Lorentz group is "essentially SU(2) x SU(2)".
This is admittedly independent study so I'm honestly not too sure if I know what SU(2) x SU(2) means in the first place. Of course I understand what a tensor product is with respect to vector spaces, but it seems rather elusive with respect to algebras or groups. If someone could explain this and elucidate why the above commutation relations suggest an SU(2) x SU(2) algebra I would very much appreciate it.

Thanks for any help~

2. Jun 10, 2014

### Xenosum

Er, nevermind I think I get it. But please correct me if I'm wrong.

The point is that every algebra has a corresponding representation, which is given by a vector space equipped with a set of operators whose commutation relations (or whatever operations over which the algebra is defined) are the same as the parent algebra. The tensor product of two algebras is then easily understood as a tensor product of their corresponding representations (the tensor product of operators, or matrices). In this case, the tensor product of two orthogonal representations of SU(2) will give you a set of operators (matrices) which, when worked out, is found to satisfy the commutation relations given above. Thus the vector space given by SU(2) x SU(2) forms a viable representation of the Lorentz algebra; and ultimately the group itself.

One can work this out by noting that the defining representation of SU(2) is given by the 1/2 the Pauli matrices. Then one may take the tensor products of these matrices (with themselves) and determine that the results satisfy the above relations.

3. Jun 10, 2014

### dauto

You're essentially correct. Note that this product is called the direct product of the algebras, not the tensor product of the algebras.

4. Jun 12, 2014

### vanhees71

A bit more accurately the Lorentz group is homomorphic to $\mathrm{SL}(2,\mathbb{C})$, the special linear group in 2D complex vector space, consisting of all complex $2\times 2$ matrices with determinant 1. This group is the covering group of the orthochronous proper Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$.

On the level of the Lie algebras this covring group is isomorphic to $\mathrm{SU}(2) \otimes \mathrm{SU}(2)$.

5. Jun 12, 2014

### Xenosum

Thanks a bunch.

As a follow-up question I should like to ask whether this immediately implies the existence of a spin-1/2 particle. By this I mean to ask, since SU(2) acts on (spin-1/2) spinors, and the generators of SU(2) are the quantum operators corresponding to spin, what does this imply about the Lorentz group in regards to quantum mechanics?

It seems that the Dirac equation for massive spin-1/2 particles can be derived simply by making an appeal to this isomorphism. So is it safe to say that whenever SU(2) appears in the reduction of some group, this group (or, more accurately, its representation) can be taken to act on a quantum particle (i.e., a spinor)? If so, it's cool to think that relativity actually forces spin-1/2 particles to exist.

If not, this is the distinction between pure math and physics that really confuses me. Sorry.

Last edited: Jun 12, 2014
6. Jun 15, 2014

### king vitamin

In my opinion, it's actually slightly less immediate (or at least just as immediate) than the non-relativistic case. If you began only insisting on rotational (SO(3)) invariance, you would be led to the su(2) Lie algebra, so your irreducible SO(3) invariant (up to a phase) states transform under irreducible representations of SU(2) - the first non-trivial such state is the spin-1/2 state. Of course, you also get spin 1,3/2,2,5/2,...

In the relativistic case, you've expanded the symmetry group to all Lorentz transformations SO(1,3) - so rotations AND boosts. You actually take linear combinations of rotations and boosts to decompose the Lie Algebra into su(2)xsu(2). To check whether the states you get through these representations are really spin-1/2, you need to go back and reconstruct the rotation operators from your A and B operators, and check how your states transform under rotations. It turns out that the two inequivalent two-dimensional representations are spin-1/2 (which one could probably guess), but the more complex symmetry group leads to the addition of the notion of chirality - you now have "right-handed" and "left-handed" spin-1/2 states which go to each other under parity (the Dirac fermion is a direct sum of these two representations, so it is parity invariant).

So in other words, you do get spin (including fermions) through this approach, but the notion that you need relativity for spin is a misconception (a rather common one as well).

7. Jun 15, 2014

### vanhees71

Just to add to king vitamin's very important statement.

From a group-theoretical point of view (and to really understand quantum theory one should take this point of view!) the meaning of spin is pretty much the same in relativistic and non-relativistic physics, as long as you consider only massive particles (massless particles make no sense in the non-relativistic case anyway, but that's another very interesting group-theory story). It simply determines how the momentum-eigenvectors of a particle at rest behave under rotations. For a spin-1/2 particle they transform in the SU(2) representation, i.e., the fundamental representation of the covering group of the rotation group SO(3).

In the relativistic case, there are two irreducible representations of the proper orthochronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$ (i.e., the group built by all Lorentz-transformation matrices with ${\Lambda^0}_0 \geq 1$ and $\mathrm{det} \Lambda=1$). They are denoted by $(1/2,0)$ and $(0,1/2)$. The corresponding spinors are called Weyl spinors.

The extension to a representation to the somewhat larger group $\mathrm{O}(1,3)^{\uparrow}$, the orthochronous Lorentz group, which in addition to the proper orthochronous Lorentz transformations (which are all composed of boosts and rotations) also contains the space-reflection operation (parity), requires to take the direct sum of the two Weyl-spinor representations, making up a Dirac spinor. In this representation the space reflection interchanges the two Weyl spinors. That's why one calls them left and right-handed parts and introduces chirality, which is a very important concept in the physics of the standard model (be it to understand the electroweak part of the standard model (quantum flavor dynamics) or the approximate chiral symmetry in the light-quark sector of QCD, which enables us to build effective models of hadrons).