Momentum eigenstate definition in Eq (2.5.5) of Weinberg Vol. 1 clarification

This question is related to one asked here: Questions concerning some parts of the section on one-particle states in Weinberg's first volume on QFT.

In Eq (2.5.5) of Weinberg's "The Quantum Theory of Fields" Vol. 1, he defines momentum eigenstates $$\Psi_{p,\sigma}=N(p)U(L(p))\Psi_{k,\sigma}$$ where $$k$$ is some reference momentum, $$L(p)$$ is a Lorentz transformation such that $$L(p)k=p$$, $$U(L(p))$$ is the unitary operator associated to $$L(p)$$ that acts on the Hilbert space of states, and $$\sigma$$ denotes a separate discrete set of eigenvalues that further label the state.

I'm wondering how it's justified that the same $$\sigma$$ appears on both sides of the equation. A general Lorentz transformation will mix $$\sigma$$'s, for example when $$\sigma$$ labels spins and the Lorentz transformation is a rotation.

• It's just a convenient definition. Note as you read on that when general Lorentz transformations act on these states that the $\sigma$ eigenstates do get mixed up. The big picture is that we need a way to connect the finite-dimensional representations of the rotation subgroup to the infinite-dimensional reps of the full group. This is one intuitive way to do it. – Luke Pritchett Mar 1 '16 at 4:31
• Related (possible duplicate?): physics.stackexchange.com/q/243029 – Art Brown Mar 13 '16 at 4:31

1 Answer

If you are describing a quantum system with relativistic symmetry the Hilbert space $${\cal H}$$ of this system must carry one unitary representation of the universal cover of the Poincaré group, say $$U(\Lambda,a)$$. Focusing on just the translations $$U(1,a)$$ these can be written in terms of their generators, the four-momentum components $$P^\mu$$, as $$U(1,a)=e^{-i a_\mu P^\mu}\tag{1}.$$

Since the $$P^\mu$$ are four mutually commuting Hermitian operators we can find a basis of simultaneous eigenvectors for them. Suppose $$\Psi_{p,\sigma}$$ is such a basis of eigenvectors of the $$P^\mu$$,

$$P^\mu \Psi_{p,\sigma}=p^\mu \Psi_{p,\sigma}\tag{2},$$

where $$\sigma$$ labels the degeneracies. Now the coice of basis (2) diagonalizes $$P^\mu$$ and hence diagonalizes $$U(1,a)$$ so that the translations are understood. We therefore need to understand the Lorentz transformations $$U(\Lambda):=U(\Lambda,0)$$. One notices that $$U(\Lambda)\Psi_{p,\sigma}$$ is an eigenvector of $$P^\mu$$ with eigenvector $$\Lambda^\mu_{\phantom{\mu}\nu}p^\nu$$: $$P^\mu U(\Lambda)\Psi_{p,\sigma}=\Lambda^\mu_{\phantom{\mu}\nu}p^\nu U(\Lambda) \Psi_{p,\sigma}\tag{3}.$$

Therefore $$U(\Lambda)\Psi_{p,\sigma}$$ must lie in the eigenspace of $$P^\mu$$ with that eigenvalue and that is, by definition of $$\sigma$$, the space spanned by $$\Psi_{\Lambda p,\sigma}$$ for the various $$\sigma$$: $$U(\Lambda)\Psi_{p,\sigma}=\sum_{\bar \sigma}C_{\bar \sigma\sigma}(\Lambda,p)\Psi_{\Lambda p,\bar{\sigma}}\tag{4}.$$

Now the idea is this: you have a space spanned by the various $$\Psi_{p,\sigma}$$, that is your base. A Lorentz transformation mixes eigenspaces associated to $$p$$ and $$\Lambda p$$. Now look for sets of momenta $$\Omega\subset \mathbb{R}^4$$ with the property that if $$p\in \Omega$$ then $$\Lambda p \in \Omega$$. These are invariant subspaces of momenta with respect to the Lorentz group action on four-vectors. They are characterized by the fact that all $$p$$ in such set have the same $$p^2$$ and the same sign of $$p^0$$. Therefore we name these as $$\Omega_m^\pm$$ where $$p^2=-m^2$$ and $$\operatorname{sgn}p^0 = \pm$$. Inside each $$\Omega_m^\pm$$ we can pick one $$k\in \Omega_m^\pm$$ such that all other $$p\in \Omega_m^\pm$$ can be written as $$p = L(p)k$$ for a Lorentz transformation $$L(p)$$.

We now suppose that all $$\Psi_{p,\sigma}$$ are such that $$p\in \Omega_m^+$$ for some $$m$$. The reason is that if it were not the case we could slice the Hilbert in terms of subspaces like that. The reason for doing that is that $$U(\Lambda)$$ takes $$\Psi_{p,\sigma}$$ to $$\Psi_{\Lambda p,\sigma}$$ and therefore does not mix each of hese subspaces with one another.

The catch is that we now change basis. Pick the vector $$\Psi_{k,\sigma}$$ that lies in the subspace we are concerned with. Now define $$\widetilde{\Psi}_{p,\sigma}:= N(p) U(L(p))\Psi_{k,\sigma}\tag{5}.$$

In particular we demand $$N(k)=1$$ so that $$\widetilde{\Psi}_{k,\sigma} = \Psi_{k,\sigma}$$. In that case notice that for the basis $$\widetilde{\Psi}_{p,\sigma}$$ the relation $$\widetilde{\Psi}_{p,\sigma}=N(p)U(L(p))\widetilde{\Psi}_{k,\sigma}\tag{6},$$

does hold with the same $$\sigma$$ in both sides by construction.

The reason to choose such a basis is because it is convenient. In fact in this basis Weinberg shows that $$U(\Lambda)\widetilde{\Psi}_{p,\sigma}=\dfrac{N(p)}{N(\Lambda p)}\sum_{\bar \sigma} D_{\bar \sigma \sigma}(W(\Lambda,p))\widetilde{\Psi}_{\Lambda p,\bar \sigma} \tag{7}.$$

The point is that in this basis we see vividly that $$U(\Lambda)$$ is fully determined by one unitary representation $$D(W)$$ of the subgroup of the Lorentz group characterized by $$\Lambda k = k$$. This is known as the little group of $$k$$ or stabilizer of $$k$$ and this procedure turns the problem of finding unitary representations of the Poincare group to the problem of finding unitary representations of the stabilizer of each standard momentum $$k$$ characterizing each possible mass shell. For example, when $$m> 0$$ this ends up turning to the problem of classifying the possible angular momenta in non-relativistic QM which is very well understood.