I've been referring to Chapter 2 of Introduction to Quantum Field Theory by Weinberg where he talks about symmetries and how they go about. Now, there are two points that he mentions. A ray, which by his definition in Section 2.2 is an equivalence class of vectors on a Hilbert space with norm 1 such that there is a $\theta$ and $\Psi_1 = e^{i\theta}\Psi'_1$

First, when he starts talking about symmetries on Pg. 76, he says. If $T_1$ is a transformation on a ray , $\mathcal{R}_1$ which takes it to $\mathcal{R}_2$ by the following action $T_1:\mathcal{R}_1\mapsto \mathcal{R}_2$ then by winger's theorem, this induces a unitary transformation on vectors on the Hilbert space $\Psi_1 \in R_1$ such that $U(T_1)\mathcal{\Psi}_1 = \mathcal{\Psi}_2$ where $\Psi_2 \in \mathcal{R}_2$.

But, when we look at Section 2.3, he considers a symmetry tranformation on "spacetime" $$x^\mu \mapsto \Lambda^\mu_\nu x^\nu + a^\mu$$ (the standard Lorentz transformation) and infers on Pg.57 that by Wigner's theorem this induces a unitary transformation $U(\Lambda,a)$ such that $\Psi \mapsto U(\Lambda,a)\Psi$. As per the first point, doesn't the Wigner's theorem just say for transformation on Rays and not spacetime? Arn't these two completely different spaces, respectively the Hilbert Space $\mathcal{H}$ and 4-D Real space $\mathcal{R}^4$

My Doubt:

The above statements has put me in a fix! As per my understanding of the Hilbert space, it a infinite dimensional vector space that facilitates the existence of physical state vectors such as the position,momentum,spin, etc. as $|x,p,s,..\rangle$.

Do they mean that the space time transformation induces a transformation on the rays $|x^\mu,p,s,..\rangle$ of Hilbert space as $$T_1|x^\mu,p,s,..\rangle \mapsto |\Lambda^\mu_\nu x^\nu + a^\mu,p,s,..\rangle$$ and therefore, my wigners theorem, there exists a unitary matrix such that the above two are related as $$U(\Lambda,a)|x^\mu,p,s,..\rangle \mapsto |\Lambda^\mu_\nu x^\nu + a^\mu,p,s,..\rangle$$

If the above is the case, my understanding would be right. Could someone heck if this is what it means? On a side note, when we say $U(\Lambda,a)$ is a transformation relating two Hilbert spaces, aren't both the Hilbert spaces the same? Just the basis in a sense changes right? If not, can you give examples/ understanding how 2 Hilbert spaces can be different?

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    $\begingroup$ "As per my understanding of the Hilbert space, it a finite dimensional vector space" [emph. mine] No. The Hilbert spaces of quantum mechanics are typically infinite-dimensional. And yes, the rest of what you say is mostly correct. Please don't ask question that merely ask for checking your understanding, because "That's right" is too short to even post as an answer. $\endgroup$ – ACuriousMind Mar 8 '16 at 14:25
  • $\begingroup$ Hello, It was a typo, I meant infinite dimesional. Well, I wanted to see if that's what is means. But the last part of the answer. What about that? I dont see a difference between 2 hilbert spaces but I'm sure I've come across it in textbooks where they look at maps between 2 hilbert spaces.in the context of QM. $\endgroup$ – Sai krishna Deep Mar 8 '16 at 14:30
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    $\begingroup$ All separable Hilbert spaces are abstractly isomorphic. From that point of view, there'd only ever be one Hilbert space. Being isomorphic as Hilbert space doesn't mean you cannot write down different objects, which may be isomorphic as Hilbert spaces, but are surely not the same object. $\endgroup$ – ACuriousMind Mar 8 '16 at 14:35

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