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Following up on this question: Weinberg says

In general, it may be possible by using suitable linear combinations of the $\psi_{p,\sigma}$ to choose the $\sigma$ labels in such a way that $C_{\sigma'\sigma}(\Lambda, p)$ is block-diagonal; in other words, so that the $\psi_{p,\sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogenous Lorentz group.

But why inhomogeneous Lorentz group if, in the first place, we performed a homogeneous Lorentz transformation on the states, via $U(\Lambda)$? I also want to be clear what is meant by the states "furnishing" a representation.

Regarding the above confusion, the same scenario again shows up during the discussion on the little group. Here's a little background: $k$ is a "standard" 4-momentum, so that we can express any arbitrary 4-momentum $p$ as $p^{\mu} = L^{\mu}_{\nu}(p) k^{\nu}$, where $L$ is a Lorentz transformation dependent on $p$. We consider the subgroup of Lorentz transformations $W$ that leave $k$ invariant (little group), and find that:

$U(W)\psi_{k \sigma} = \sum_{\sigma'} D_{\sigma' \sigma}(W)\psi_{k \sigma'}$. Then he says:

The coefficients $D(W)$ furnish a representation of the little group; i.e., for any elements $W$ and $W'$ , we get $D_{\sigma' \sigma}(W'W) = \sum_{\sigma''}D_{\sigma' \sigma''}(W)D_{\sigma''\sigma}(W')$.

So is it that even in the first part about the Lorentz group, $C$ matrices furnish the representation and not $\psi$?

Also, for the very simplified case if $C_{\sigma'\sigma}(\Lambda, p)$ is completely diagonal, would I be correct in saying the following in such a case, for any $\sigma$?

$$U(\Lambda)\psi_{p,\sigma} = k_{\sigma}(\Lambda, p)\psi_{\Lambda p, \sigma}$$

Only in this case it is clear to me that $U(\Lambda)$ forms a representation of Lorentz group, since $\psi_{p,\sigma}$ are mapped to $\psi_{\Lambda p, \sigma}$.

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Just a tip: no need to mark the questions as follow-up questions so boldly. I've made some cosmetic edits to this and your last question, but if I changed the meaning anywhere from what you wanted to ask, please do fix it. :-) –  David Z May 25 '13 at 6:33
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A representation is a vector space with a group action attached to it. In linear algebra, it's a bunch of vectors $v^i$ which move around under the group action, i.e. when $g$ is a group element then there is a matrix $D(g)^i{}_j$ which acts on them as $D(g)^i{}_j v^j$. In QM/QFT, the vector space is spanned by states $|\psi(p,\sigma)\rangle$ (or whatever notation Weinberg uses). That what he means by "furnishing" a rep. –  Vibert May 25 '13 at 6:51
    
@Vibert: That's what I thought. But $U$ matrices map the states $\psi$ to Lorentz-transformed states, so then it should be the $C_{\sigma \sigma'}$ matrices that furnish a representation of the Lorentz group. I'm confused why Weinberg says that the states $\psi$ are the ones furnishing such a representation. (See the edited question) –  1989189198 May 26 '13 at 5:14
    
OK, I see, it's a matter of language. Formally a representation is a linear map from the group to your vector space, so in this case the map $W \mapsto D(W).$ But when you use this construction, of course it depends on both the cases $\psi$ and the Lorentz matrices acting on them. That's why in physics we use the term representation in a sloppier way than our mathematician friends. –  Vibert May 26 '13 at 7:02

2 Answers 2

In the inhomogenous Lorentz group $ISO(1,3)$, you have the space-time translation group, and the Lorentz group $SO(1,3)$.

You begin to find a representation of the space-time translation group, by choosing a momentum $p$. So your representation must have a $p$ indice :

$$\psi_p$$

After this, you will have to get the full representation, by finding a representation of the Lorentz group compatible with the momentum $p$, this will add an other indice $\sigma$ which corresponds to the polarization, so you will have a representation :

$$\psi_{p, \sigma}$$

which is the representation of the inhomogenous Lorentz group.

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Re the meaning of representation, here is a definition from Peter Woit's "Quantum Mechanics for Mathematicians" lecture notes (available on-line), section 1.3:

Definition (Representation). A (complex) representation ($\pi, V$) of a group $G$ is a homomorphism $$ \pi: g \in G \rightarrow \pi(g) \in GL(V) $$ where $GL(V)$ is the group of invertible linear maps $V \rightarrow V$, with $V$ a complex vector space.

Saying a map is a homomorphism means $$ \pi(g_1) \pi(g_2) = \pi(g_1g_2) $$ When $V$ is finite dimensional and we have chosen a basis of $V$, then we have an identification of linear maps and matrices $$ GL(V) \simeq GL(n,\boldsymbol{C}) $$ where $GL(n,\boldsymbol{C})$ is the group of invertible $n$ by $n$ complex matrices.

So the representation is the homomorphism (the operation-preserving map) from the group $U(\Lambda)$ to the transformation matrices (Weinberg's C's and D's), but these matrices require a vector space (the $\psi$s), on which to act.


For the rest, here's my answer (caveat emptor, I'm just a slow student):

This section 2.5 is titled "One Particle States". If $C$ turns out to be reducible (block-diagonalizable), the different blocks are independent of one another (no mixing between blocks) and are interpreted as different particles species. So, for a single particle state a single irreducible block is assumed.

In this argument it's OK to generalize from homogeneous to inhomogeneous transformations, because translations don't mix $\sigma$'s and hence don't affect the block structure of $C$:

$$U(1,a) \Psi_{p,\sigma} = e^{-ip\cdot a} \Psi_{p,\sigma} $$

Finally, in the case you posit of a completely diagonal $C$, I think you are left with a bunch of particle species with no $\sigma$-mixing at all, i.e. scalars, each with a trivial little group ($k=1$).

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