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?

  • $\begingroup$ I doubt the unitary equivalence if you include non-separable Hilbert spaces. $\endgroup$ – Urgje Oct 22 '15 at 9:44
  • $\begingroup$ I should have mentioned separable. You are right, the equivalence only holds for H.S with countable bases. $\endgroup$ – Nirmalya Kajuri Oct 22 '15 at 13:20

A subrepresentation is not the same as a subspace on which a representation exists (which exists on any subspace): it has to be the same representation (group action), restricted to that space.

  • $\begingroup$ 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? $\endgroup$ – Nirmalya Kajuri Oct 21 '15 at 23:24
  • $\begingroup$ Sorry, that should read $ U_1 \otimes \mathbb{1} \oplus \mathbb{1} \otimes U_2 $ $\endgroup$ – Nirmalya Kajuri Oct 21 '15 at 23:30

Each separable infinite-dimensional Hilbert space carries both irreducible and reducible representations of any noncompact Lie groups you can name. But this information in itself is of little use.

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.

The physics is in the representation, not in the Hilbert space itself.


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