# Confusion regarding Poincare group representations in Weinberg's Quantum Field Theory, Volume 1

I am a beginner to quantum field theory and my course is based on Weinberg's QFT (vol. 1; chapter 2) - I have quite a few confusions.

I have had an introductory group theory course before, so, I know that there are mathematical structures called "groups" (sets of elements and an operator) satisfying some basic axioms and that we can construct representations which are linear maps (homomorphisms) from the group to a space of linear operators (which act on different linear vector spaces).

First, the text says that the Lorentz transformation matrices ($$\Lambda$$) form a "group" - this is easy enough (satisfies group axioms), but I have a question. When we talk about these forming a group, do we mean an actual "group" or some representation of an abstract group ("Poincare" group) that acts on the 4D spacetime - the $$\Lambda$$'s themselves act on the spacetime 4-vectors, $$a^{\mu}$$ (like operators - which are exactly what representations map group elements to)? For example, in case of SU(2), the matrices $$e^{i \theta \sigma\cdot \hat{n}}$$ are a representation of SU(2) that act on the space of complex 2D vectors while the group element is characterised by the continuous parameter, $$\theta$$. However, in this case, the unitary representations acting on the quantum states, $$U(\Lambda)$$, are characterised by $$\Lambda$$ as well - I am confused what these actually are; parameters characterising the group, the Poincare group elements themselves or a representation that acts on spacetime but is also used to characterise a different representation?

Second, the text talks about irreducible representations of the Poincare group - we define the action of the unitary representation on a state $$|\Psi(p,\sigma)\rangle$$ as, $$U(\Lambda)|\Psi(p,\sigma)\rangle = \sum_{\sigma'} C_{\sigma', \sigma} (\Lambda, p) |\Psi(\Lambda p,\sigma')\rangle$$

It says,

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

What exactly are we doing here? $$U$$ is the representation of the Poincare group and I understand the mathematical steps that follow this - choosing the standard 4-momentum, $$k^{\mu}$$, and looking at how $$D_{σ′,σ}$$ (which comes from the transformations that leave $$k^{\mu}$$ invariant) are related to $$C_{σ′,σ}(Λ,p)$$ and so on - but I do not understand why we are doing this. Why do we want to find out $$C_{σ′,σ}(Λ,p)$$ and why is it being called an irreducible representation of the group (if I understand correctly) and how does $$D_{\sigma',\sigma}$$ help in this?

In case of rotations, we say, $$\mathcal{D}(R) |j,m\rangle = \sum_{m'} \mathcal{D}^{(j)}_{m',m}|j,m'\rangle$$ with $$\mathcal{D}^{(j)}_{m',m}$$ being an irreducible representation. If we try to compare the two equations, we should expect a superposition over $$p$$ rather than $$\sigma$$ ($$\mathcal{D}(R)$$ has elements $$\mathcal{D}^{(j)}_{m',m}$$, but why does $$U(\Lambda)$$ have elements like $$C_{σ′,σ}(Λ,p)$$?) - however, it is different here. It is not even a sum over both $$(p, \sigma)$$ but only $$\sigma$$.

I am sorry if these questions are dumb, but I am new to this and unable to relate what I learnt in my group theory classes to what is being done here and am really confused. I will be glad for any clarification, thanks!

I also did go through the various questions on this forum on the same paragraph I quoted - but I didn't really find anything that answers my questions much.

Edit:

Following the equation in Weinberg (given above), we take the cases of the massless and massive particles and choose a proper 4-momentum, $$k^{\mu}$$ and define the state above as, $$|\Psi(p, \sigma)\rangle = N(p)U(L(p))|\Psi(k^{\mu}, \sigma)\rangle$$ $$N$$ is a normalisation and $$p^{\mu} = L^{\mu}_{\phantom{-} \nu}(p) k^{\nu}$$. This is clearly contrary to the first equation - why is there no sum over $$\sigma'$$ here?

I understand that it may be a lot to answer, so I would be happy if someone just points me to a more accessible reference as well. Thanks again!

To answer your first question $$\Lambda$$ is a element of the Lorentz group (or the Poincare group) and $$\Lambda^\mu_\nu$$ is its matrix representation on Minkowski space. Consider the Minkowski space $$\mathbb{R}^{1,3}$$ with the bilinear form $$(x,y) = \eta_{\mu\nu}x^\mu y^\nu$$. A Lorentz transformation L is a linear transformation on $$\mathbb{R}^{1,3}$$ which preserves the bilinear form ie., $$(L(x),L(y)) = (x,y)$$. It can be easily seen that the L's with composition of maps form a group . This is the Lorentz group. Now the matrix representation of L on $$\mathbb{R}^{1,3}$$ is denoted by $$L^\mu_\nu$$. These $$L^\mu_\nu$$'s also form a group of $$4 \times 4$$ matrices under matrix multiplication.

For your second question I will first explain how we find the irreducible unitary representaions of the Poincare group and what $$C_{\sigma'\sigma}$$ is and then I will explain why we are interested in irreps of the Poincare group.

1. Consider a particle of mass $$m \geq 0$$. In general the state of the particle is not completely specified by its momentum as the particle can have internal degrees of freedom (like spin). So the state is represented by $$\Psi_{p,\sigma}$$ which should be thought of as $$|p\rangle \otimes |\sigma\rangle$$ where $$|\sigma\rangle \in \mathscr{V}$$ is an eigenvector of some hermitian operator with eigenvalue $$\sigma$$ in the Hilbert space $$\mathscr{V}$$ (assumed to be finite dimensional) representing the internal degrees of freedom of the particle.

2. Now we can choose a standard momentum k as explained in Weinberg's book. The set of all Lorentz transformations W such that $$W^\mu_\nu k^\nu = k^\mu$$ form the little group $$G_k$$ of k which is a subgroup of the Lorentz group. The space spanned by $$\Psi_{k,\sigma}$$'s is a copy of $$\mathscr{V}$$ (since k is fixed). Then $$$$U(W)\Psi_{k,\sigma} = \sum_{\sigma'} D_{\sigma' \sigma}(W) \Psi_{k, \sigma'}$$$$ (thought as $$|p\rangle \otimes |D(W)\sigma\rangle$$) provides a representation of the little group on this space provided D(W) is a representation of the little group on the Hilbert space $$\mathscr{V}$$ ie., $$$$D(W')D(W)=D(W'W) \Rightarrow U(W')U(W)=U(W'W)$$$$ $$D_{\sigma' \sigma}(W)$$ is the matrix representation of $$D(W)$$ in the $$|\sigma\rangle$$ basis.

3. If we know a unitary representation of the little group, we can find a unitary representation of the Lorentz group as follows. We define (I do not understand this definition) the states of momentum p as $$$$\Psi_{p,\sigma}=N(p)U(L(p))\Psi_{k,\sigma}$$$$ where L(p) is some standard Lorentz transformation which takes k to p. Following Weinberg we get $$$$U(\Lambda)\Psi_{p,\sigma} =\left(\frac{N(p)}{N(\Lambda p)}\right)\sum_{\sigma'} D_{\sigma' \sigma}(W(\Lambda,p)) \Psi_{\Lambda p, \sigma'} = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,p) \Psi_{\Lambda p, \sigma'}$$$$ where $$W(\Lambda,p) = L^{-1}(\Lambda p ) \Lambda L(p)$$ and $$C_{\sigma' \sigma}(\Lambda,p) = \left(\frac{N(p)}{N(\Lambda p)}\right) D_{\sigma' \sigma}(W(\Lambda,p))$$. It can be easily shown that C forms a representation of the Lorentz group (said to be induced by D) using the fact that D forms a representation of the little group ie., $$$$D(W')D(W)=D(W'W) \Rightarrow U(\Lambda')U(\Lambda)=U(\Lambda'\Lambda)$$$$ There is no integral over p because Weinberg has shown that $$U(\Lambda)$$ takes a state of momentum p to a state with momentum $$\Lambda p$$ ( which is intuitive as well).

4. In general the representation C is reducible and can be written as a direct sum of irreps. This is what Weinberg means when he says that " we can choose the σ labels in such a way that $$C_\sigma',\sigma(\Lambda,p)$$ is block-diagonal". It can also be shown that the representation induced by D(W) is unitary and irreducible whenever D(W) is unitary and irreducible. Wigner has proved that every irreducible unitary representation of $$SL(2,\mathbb{C})$$ (the double cover of the Lorentz group) arises in this way. So to findout the unitary irreps of the Poincare group it is sufficient to find the unitary irreps of the little group.

Now coming to why are we interested in unitary irreps of the Poincare group. It is believed that laws of physics should be invariant under space-time translation and Lorentz transformations as they are symmetries of spacetime. This is the Poincare group. Another fundamental assumption is that every quantum system comes equipped with a projective representation of the Poincaré group. Since elementary particles cannot be broken into smaller pieces we assume that they correspond to minimal representations ie., irreps. We assume that to each elementary particle one may associate an irreducible representation of the Poincaré group.