# Representations of Lorentz group

1. Sep 25, 2009

### rntsai

I'm reading the wiki article on Representation theory of the Lorentz group
and they seem to make a distinction between these two reps:

(1/2,1/2) and
(1/2,0) + (0,1/2)

I did some checks and it seems that these two are the same. Am I wrong
or is the wiki article wrong (won't be the first time for either!)

2. Sep 25, 2009

### rntsai

I posted too quickly (couldn't figure out how to delete the post). It does
look like they are different...sorry about that

3. Sep 25, 2009

### RedX

That's a good question that I think I need clarification on too.

If you have multiple particles in quantum mechanics, then this is represented by a direct product, and not a direct sum. However, when you rotate your state, you rotate all particles through the same angle: you do not have a separate angle for each particle. However, the (1/2, 1/2) direct product is the opposite. Here, you rotate through two different independent angles!

A direct sum is always rotated through the same angle. A direct product? It seems sometimes you rotate through the same angle, and sometimes you don't! Anyone know why?

4. Sep 26, 2009

### rntsai

I don't think I have much insight into the geometry of these reps, so I can't address what
you're asking...I am testing my understanding of the 4 dim reps of the group by trying
different combinations :

(1/2,0) + (1/2,0) (not faithful)
(1/2,0) x (1/2,0) = (1,0) + (0,0) (not faithful)
(0,1/2) + (0,1/2) (not faithful)
(0,1/2) x (0,1/2) = (0,1) + (0,0) (not faithful)

(1/2,0) + (0,1/2) (faithful)
(1/2,0) x (0,1/2) = (1/2,1/2) (faithful)

(here + is direct sum and x is tensor product). So it seems there are 2
inequivalent faithful reps of dim=4: (1/2,0)+(0,1/2) and (1/2,1/2) the last
one being equvalent to (1/2,0)x(0,1/2)

Last edited: Sep 26, 2009
5. Sep 26, 2009

### javierR

The representation (1/2,1/2) is the irreducible (spacetime) vector representation of the SL(2,C)xSL(2,C) Lorentz group, while (1/2,0)+(0,1/2) is a reducible representation, each one being a chiral spin-1/2 representation...the sum of them yields a Dirac spinor (since it's still a spin 1/2 object but has both chiralities).

As for rotating through angles, perhaps I misunderstand what you mean, but: in quantum mechanics, identical particles can form an irreducible representation of the rotation group, and for that matter of the Lorentz group. The new entity acts as a new particle of a single new spin, and with the SO(3,1) form of the group you have the usual spatial rotations and boosts. (Distinguishable particles form reducible representations.) If you use the SL(2,C)xSL(2,C) form of the Lorentz group there are independent factors, with independent "rotations", but together they contribute to the usual spatial rotations and boosts of SO(3,1). So there is no contradiction for a particle to transform according a single angle or multiple angles; it depends on which isomorphic group you choose to work with.

6. Sep 26, 2009

### RedX

I think I confused a "direct product group" with a "direct product representation of a group". With a direct product group, you can have multiple rotation angles, one for each group in the direct product. But if you have a direct product representation of a group, then each index on the tensor transforms through the same angle.

If you set the angles of each SL(2,C) to be the same angle, then does this bear any resemblance to direct product representation of just one SL(2,C)?

7. Sep 26, 2009

### rntsai

Thanks for the clear answer...as a followup question (and I have several more of these;-)
does labeling reps of SL(2,C)xSL(2,C) with a single "spin" make sense? what's the "spin"
of an (n,m) rep? what about (n1,m1)+(n2,m2)?...

8. Sep 26, 2009

### javierR

Note on edits: Sorry, for some reason I kept typing SL(2,C)xSL(2,C) as the group isomorphic to the Lorentz group, which should be SL(2,R)xSL(2,R)! (SL(2,C) on its own is isomorphic to the Lorentz group).

A state in the (n,m) representation has spin n+m. The spin-1 vector potentials are (1/2,1/2), the metric tensor is (1,1) and so on. If n is not equal to m, the representation can be assigned a net chirality, like a Weyl spinor (1/2,0) or (0,1/2). There are also 2-form fields that can have spin 1 and carry some chirality, forming the (1,0) and (0,1) representations.

The representation (n1,m1)+(n2,m2) is reducible (due to the + sign) signifies two states, one forming the (n1,m1) rep and the other forming the (n2,m2) rep. Each has a spin of (n1+m1) and (n2+m2). That does not mean that a single particle cannot be represented by an reducible representation of SL(2,R)xSL(2,R) (anyway, the classification of spin representations under the SO(3,1) Lorentz group can be obtained using the little group for either massless or massive states, and you don't see the reducibility manifestly). For example, a dirac spinor can represent a particle but is (1/2,0)+(0,1/2). And there's the example of the Higgs mechanism in which a (0,0) state can join a (1/2,1/2) state to form a single state of a massive vector (2+1 propagating degrees of freedom).

Let me add this to clarify the last sentences: A Dirac spinor appears as a spin-1/2 field under the SO(3,1) Lorentz group (which we technically should understand via the double covering group Spin(3,1)). But under SL(2,R)xSL(2,R), the representation is reducible; i.e. it decomposes as above. As for the Higgs mechanism, the point is that at first you see reducible states (scalar + vector), but that this ends up being equivalent to then writing a massive spin-1 state (in an irreducible representation).

Last edited: Sep 27, 2009
9. Sep 27, 2009

### javierR

Clearly these are typos: SO(3,1) is isomorphic to SL(2,R)xSL(2,R) not SL(2,C)xSL(2,C). (SL(2,C) on it own is isomorphic to SO(3,1)). Sorry.

10. Sep 27, 2009

### rntsai

Actually I didn't catch that before...at any rate I'm actually doing my claculations at the
algebra level (so these are all reps of A1+A1 or D2 over the reals or over the complexes).
this lets me defer complications related to signature, covering groups,...until later.

I think you meant to say :

That does not mean that a single particle cannot be represented by a reducible representation of SL(2,R)xSL(2,R)

I never really got a satissfying definition of what a "particle" is...

11. Sep 27, 2009

### javierR

Yeah, thanks, I should take more time to read what I write. It's fixed now.

I won't presume that that was a prompt for me to go on about what I think. But what the heck. I don't know what a satisfactory definition of particle would be for you, and there are several degrees to which one could go to. You probably already have some notion that the bits that carry energy-momentum in a given (quantum) field, which can be associated with wavefunctions, are particles. The most precise definitions can be found in "axiomatic" or "constructive quantum field theory". I personally feel that a "more physical" notion of a particle should be formed in ones mind after understanding the technical characterizations.

For most purposes, you could call a particle a quantum state created by a oscillator operator (that is, "creation and annihilation" operators), whether or not this is an "elementary particle" or not. The state can be characterized by how it (or the field operator associated with it) transforms under the Lorentz group, gauge groups, global groups like flavor, etc. More specifically, a single particle operator should be in an irreducible representation of those groups (that's why I mentioned the caveat: if you work with isomorphic groups, the irreducible reps may become reducible reps). What one calls a particle *is* dependent upon the energy scale, since going from larger to smaller distance scales may show the particle to be composite, which means it becomes more appropriate to refine the original irreducible representation into a reducible representation of the various groups (in particular the Lorentz group). That corresponds to replacing the single (composite) field operator with a group of field operators that correspond to more elementary particles. This is not to say the picture of composite vs. constituents is always clear (even when we can identify the more elementary operators) due to the fact that this is a *quantum* theory, and a wavefunction is a functional of the various fields involved. If we follow the notion of "particle" above, then we can still say that there are quarks and gluons that are more fundamental (when we write down the QCD Lagrangian e.g.), but we just can't say that they're free or unentangled.

12. Sep 27, 2009

### rntsai

To be honest (and going out on a limb a little), I'm not sure a definition is necessary!
(I'm fairly conservative by nature so I won't go much beyond this).

The bit on the isomorphic groups doesn't seem right. Reducability can change if the base field is changed (complexes vs. reals for example), but for isomorphic groups it won't.
If you go to smaller group howerver (as in a proper subgroup for example), then an irrep of
the bigger group usually (but not always) splits up into several irreps of the smaller one.

13. Sep 28, 2009

### javierR

I was speaking pretty loosely. The reference to "isomorphic group" was a sloppy attempt to make a short remark about going from a general Dirac spinor of the Lorentz group to Weyl spinors of SL(2)xSL(2).

The thing is, for most purposes we are technically talking about the proper orthochronous Lorentz group, which is a subgroup. If we talk about the (full) Lorentz group (we can cover this by taking the proper orthochronous group with time and space inversions), then the irreducible spinor rep is [(1/2,0)+(0,1/2)], the Dirac spinor representation that appears naturally when we write down the 4x4 gamma matrices of SO(3,1). Then under the restriction to the proper orthochronous Lorentz group, the irreps become (1/2,0) and (0,1/2) separately, which is clear when we use SL(2,R)xSL(2,R). Hence in going between the ideas of (1) generally having to use the Dirac spinor of the Lorentz group and (2) having separate Weyl spinors, I referred to "going to the isomorphic group" (the groups are locally isomorphic).

Note: The notation (a,b) for reps was also used kind of loosely. In case you're wondering, it arises as follows. The algebra so(3,1) has J and K elements, being 3D rotations and boosts. The subalgebra so(3) of 3D rotations is the classifying algebra for particle states forming representations labeled by j= 0, 1/2, 1, .... indices (i.e. spin values). We can rewrite the algebra by defining A=1/2(J+iK) and B=1/2(J-iK). Then the A's and B's each form their own algebras that are disjoint from each other: sl(2,R)xsl(2,R). The notation for spins then arises by noting that the original spins were labels for J reps, and that now J=A+B. So we write j-->(a,b) to *label* the reps of the lorentz group, even though they are really labels from the 3D rotation group. E.G. j=1 --> (1/2,1/2).