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.

  • 3
    $\begingroup$ 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. $\endgroup$ Mar 1, 2016 at 4:31
  • $\begingroup$ Related (possible duplicate?): physics.stackexchange.com/q/243029 $\endgroup$
    – Art Brown
    Mar 13, 2016 at 4:31

1 Answer 1


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.


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