# Irreducible representations of translations

1. Dec 10, 2009

### RedX

I read somewhere that the irreducible representations of Lie groups were countable. But what about translations? Isn't each momentum value its own irreducible singlet, and there are a continuum of momentum values?

For example take $$e^{ipx}$$. If you translate it, it doesn't mix with anything else, it remains the same state. There are infinite number of momentum p, therefore infinite number of irreducible represetations.

2. Dec 11, 2009

### hamster143

You were probably reading about compact Lie groups. Compact Lie groups are sufficiently well-behaved that there may be such a theorem (I'd have to check the liturature, though). The translation group is non-compact, as is the Lorenz group.

3. Dec 11, 2009

### dextercioby

The group of translations is non-compact. Its linear, unitary representations (required by the Wigner theorem to describe this symmetry in the quantum mechanical formalism) are infinite-dimensional. The basis of the space of representation is not countable. The operators of the representation don't have a discrete spectrum. One needs to go beyond the standard Hilbert space <environment> to account in order to fully describe such representations.

4. Dec 11, 2009

### RedX

The Lorentz group is SO(4), and I don't know if it's compact (I don't know what compact means), but its irreducible representations are countable. The way that's done is to show that SO(4)=SU(2)xSU(2), and so the irreducible representations are (0,1/2), (1/2, 0), (1/2, 1/2), ... which are the right spinor, left spinor, and vector representations, respectively.

I thought all groups, Lie or not Lie, have unitary representations, and that all representations are linear (linearity of matrices). If the representations are infinite-dimensional (an infinity by infinity matrix), then could there still not be countable irreducible representations, where each irreducible representation is of size infinity?

What's confusing me a little is SO(3). The irreducible representations are countable: they are the spin 0, 1/2, spin 1, etc. Now take spin 1: this is 3x3. But when changing from the spin 1 basis to the position basis, you go from 3x3 to infinityxinfinity (azimuthal and polar angle basis). So how can the two representations, one 3x3 and the other infinityxinfinity, be equivalent, when they are not even the same size?

5. Dec 12, 2009

### Fredrik

Staff Emeritus
Actually, it's SO(3,1), or equivalently $SL(2,\mathbb C)/Z_2$. But in QM, we take the symmetry group to be $SL(2,\mathbb C)$ instead, because it simplifies the math without really changing the physics.

For subsets of $\mathbb R^n$ you can take the definition to be "closed and bounded". (But the actual definition is that every open cover has a finite subcover). SU(2) is compact because it's homeomorphic to a 3-sphere. (There's a continuous bijection from SU(2) to the unit 3-sphere, and it has a continuous inverse).

Representations are linear by definition. I don't remember exactly what Wigner's theorem says, but it implies that we sometimes (when the group isn't simply connected?) have to consider operators that are antlinear and antiunitary and instead of linear and unitary. (I think the time-reversal operator is the only relevant operator in QM that has this property).

An irreducible representation of SO(3)=SU(2)/Z2, or SU(2), which is what we actually use, is just a group homomorphism from SU(2) into $GL(\mathbb R^{2S+1})$. The unit rays of $GL(\mathbb R^{2S+1})$ represent the possible spin states, and those have nothing at all to do with position. To get the full quantum theory of a single particle of spin S, you have to take the tensor product of $\mathbb R^{2S+1}$ and $L^2(\mathbb R^3)$, which is the Hilbert space whose unit rays represent all possible states of a spin-0 particle.

Last edited: Dec 12, 2009
6. Dec 12, 2009

### Fredrik

Staff Emeritus
I think this is only true if you want to describe them in terms of eigenvectors of the generators. I'd be surprised if you can't just work around the fact that these generators don't have eigenvectors, and describe these things in other terms.

7. Dec 12, 2009

### hamster143

As Fredrik says, the Lorentz group is SO(3,1). SO(4) is compact, SO(3,1) is not compact.

Representations you describe do exist, but they are not unitary. Non-unitarity limits their usefulness by making it impossible to define Lorentz-invariant norms and wavefunctions. Instead, we look at the bigger picture (the Poincare group: Lorentz + translations). For example, for any positive m, the space of all possible definite-momentum scalar particles of mass m is infinite-dimensional, and Poincare group induces a transformation on this space. This space and this transformation can be formulated in such a way that the norm is preserved, giving us a unitary, infinite-dimensional representation. Similarly, there's another representation that acts on Dirac spinors. We can't write down a 4-d rep that preserves spinor norm ($|\psi_0|^2+|\psi_1|^2+|\psi_2|^2+|\psi_3|^2$), but we can write an infinite-dimensional rep that boosts spinors into different spinors.

Last edited: Dec 12, 2009
8. Dec 12, 2009

### RedX

Let me see if I got this: SO(3,1) is not compact because boosts parameters (in group parameter space) form a set that maps to velocities: (-c,c), where c is the speed of light, and (-c,c) is not compact as it is not a closed interval. Is it possible to find a closed set that maps to (-c,c)? If you could, then by reparameterization of your group, you could make the group compact in group parameter space?

Also, the parameter space for rotations is $$[0, 2\pi)$$. But this space is not compact, because it is not a closed set. $$[0, 2\pi ]$$ would be compact though?

Also, the parameter space of translations is infinite, from (-infinity,infinity)? Is this the reason translations are not compact, because infinitely-sized spaces can't be compact?

thanks

9. Dec 12, 2009

### Fredrik

Staff Emeritus
It doesn't seem to work out really well to consider the parameter space instead of the actual group. You're getting the wrong result for rotations in a plane. Every circle is compact in the topology inherited from $\mathbb R^2$, so SO(2) is too.

10. Dec 12, 2009

### dextercioby

The (restricted) Lorentz group is not compact as a Lie group, because it's not compact as a topological space. Its manifold contains the submanifold of Lorentzian boosts which is known to be homeomorphic to the R^3 endowed with the usual metric topology.

As for the manifold formed by the spatial rotations, this is known to be homemorphic to a 3-sphere with any two antipodal points identified. This latter manifold is compact, again in the usual metric topology.

A nice discussion of these topics can be found in the famous 2nd chapter of the 1st volume of Weinberg's QFT treatise.

11. Dec 12, 2009

### hamster143

The set of boosts in one direction is really "infinitely long". There's no "end", it goes to infinity. (Consider what happens to the velocity subset [0,1] after successive boosts). That gives another distinction between SO(4) and SO(3,1). In both cases, you can define a local metric and calculate the volume of the entire group. SO(4) is a 4-sphere, its volume is finite. The integral over SO(3,1) diverges.

[0, 2\pi] is compact. We can have ends of the interval identified with each other, it would still be compact, and it would be homeomorphic to SO(2) (the subspace of SO(4) that corresponds to rotations about any one axis).

Not necessarily.

Last edited: Dec 12, 2009
12. Dec 12, 2009

### RedX

This is probably what's messing me up. What does the metric between two different group elements look like? I assume the metric is written in terms of the group parameters of the two elements, call them 'x' and 'y'. Can you say the metric is just d(x,y) and d(x,y) is any function that satisfies the axioms for metrics? How is distance defined between two group elements?

Does this have anything to do with the Haar measure?