# Does every Hilbert Space carry a representation of Poincare group?

We know all infinite dimensional Hilbert Spaces are unitarily equivalent. It should follow therefore that if I have an unitary representation of say Lorentz or Poincare group on one infinite dimensional Hilbert Space A I should be able to induce a representation on any other infinite dim Hilbert Space B using the unitary map between A and B. So for instance the Simple Harmonic Oscillator Hilbert Space will carry a representation of the Poincare group. Is this statement correct?

If so, suppose I have Hilbert Space which carries an irreducible representation of Poincare Group (like the space of positive frequency solutions to Klein Gordon equation in Minkowski space). Suppose also that it admits a tensor decomposition into some factor Hilbert Spaces. Then by the argument above, each factor Hilbert Space carries a representation of the Poincare Group (with possibly no geometric interpretation). But this would contradict the statement that the original representation was irreducible. Then what gives?

• I doubt the unitary equivalence if you include non-separable Hilbert spaces. Oct 22, 2015 at 9:44
• I should have mentioned separable. You are right, the equivalence only holds for H.S with countable bases. Oct 22, 2015 at 13:20

The Hilbert spaces in quantum mechanics always come with distinguished representations that give certain operators an interpretation as distinguished observables. Without these distinctions, there would not be any difference between the Hilbert space of a harmonic oscillator, of a hydrogen atom, of a $N$-particle system, of a bosonic quantum field, or of a fermionic quantum field.
• Thanks for the answer! So you're saying that the original representation is irreducible, as that has nothing to do with the induced representations that one may construct on the factor Hilbert Spaces via unitary maps. I can also construct a reducible representation on the same space -$U_1 \otimes \mathbb{1} \oplus \mathbb{1} \oplus U_2$ where $U_1, U_2$ are the representations on the factor subspaces, but this is an entirely different representation. Is that correct? Oct 21, 2015 at 23:24
• Sorry, that should read $U_1 \otimes \mathbb{1} \oplus \mathbb{1} \otimes U_2$ Oct 21, 2015 at 23:30