Does the location of the Hilbert space of momentum eigenstates in QFT change under time translations and boosts?

Question

I have two questions concerning Wigner's transformation law for irreps of the Poincare group: \begin{equation} U[\Lambda,\vec{a}]\vert p,\sigma\rangle=e^{ip\cdot a}D_{\sigma'\sigma}[\Lambda;p]\vert \Lambda p,\sigma'\rangle \end{equation}

Question 1: Should this transformation law be thought of as taking place on some Cauchy surface $\Sigma$ in Minkowski space? For example the Cauchy surface $t=constant$? i.e. Is the Hilbert space of states $\vert p,\sigma\rangle$ located on a three dimensional surface in Minkowski space? In which case would it not be more accurate to index the states with a reference to which hypersurface they are associated to?

\begin{equation} \vert p,\sigma\rangle\rightarrow \vert p,\sigma\rangle_{\Sigma}? \end{equation}

Question 2: If the answer to question $1$ is "yes, the Hilbert space of the $\vert p,\sigma\rangle$ is located on a co-dimension 1 hypersurface $\Sigma$", then should one take into account that the hypersurface is not invariant under all Poincare transformations?

For example if we take the Cauchy surface to be a constant time slice $t=c_1$, then under a time translation by an amount $t_2$ the hypersurface will also change, so a more accurate transformation law would be

\begin{equation} e^{-i \hat{H} t_2}\vert p,\sigma\rangle_{t=c_1}=e^{-iE t_2}\vert p,\sigma\rangle_{t=c_1+t_2} \end{equation} where the subscript on the states indicate the changing Cauchy surface under a time translation.

Answer 1

This just the representation theory pf the group. There no time involved.

In applications the representation states are in the Heisenberg picture, in which neither they nor the Hilbert space depend on time.

Answer 2

1. The Hamiltonian formulation of (quantum) field theory generally breaks Lorentz covariance - we have to single out one direction as time, and the Hamiltonian generates translations in that particular direction. People often say that the Hamiltonian formulation is "not manifestly covariant" because in principle you need to double-check all of your "final" results for proper covariance/invariance to demonstrate that the choice of time direction for the intermediate Hamiltonian steps did not matter.

   In addition, standard QFT happens on Minkowski space and we just make a time split $\mathbb{R}^{1,3}\cong\mathbb{R}\times \mathbb{R}^3$, i.e. all the Cauchy surfaces are just $\mathbb{R}^3$ and so all the Hilbert spaces you might attach to those are isomorphic. But it is indeed the case that you may think about a QFT as a machinery that lives on some sort of "cylinder" spacetime $[-\infty,\infty]\times \Sigma$ that assigns Hilbert spaces to the asymptotic past $-\infty\times\Sigma$ and the asymptotic future $\infty\times\Sigma$ and has an "S-matrix" operator that connects the two. This is the fundamental idea behind the attempt of formalizing QFTs as functors on bordisms and is mostly (but not exclusively) useful in the context of topological field theories.

2. The representation theory of the Lorentz/Poincaré group really doesn't care about any of this - we're asking "what are the irreducible representations of this group?", and we're getting an answer. A priori there is no physical meaning at all attached to such a representation.

   In the Heisenberg picture, the states in a Hilbert space do not evolve in time, and so it makes no sense to "attach" the Hilbert space to any particular surface - the states are unchanging, and so should be transforming under the full Poincaré group. This justifies looking for the irreducible representations - the Heisenberg picture is the "default" viewpoint in QFT. You might be worrying about this again when you start encountering "derivations" in the interaction picture, but no worries: The interaction picture is rigorously nonsense anyway due to Haag's theorem, so this is just another inconsistency.