In Weinberg's treatment of symmetries, is he actually considering one underlying abstract group?

Question

I'm studying Weinberg's QFT books, and regarding symmetries I'm quite confused about the distinction between a group and its representations in Weinberg's presentation.

1. First of all, Weinberg states that symmetry transformations are ray transformations preserving the probabilities $P(\mathscr{R}\to \mathscr{R}_n)$ defined for a ray $\mathscr{R}$ and a family of mutualy orthogonal rays $\mathscr{R}_n$ to be

   $$P(\mathscr{R}\to \mathscr{R}_n)=|(\Psi,\Psi_n)|^2,\quad \forall\Psi\in \mathscr{R},\Psi_n\in \mathscr{R}_n.\tag{2.1.7}$$

   He then points out (p. 52) that:

   The set of symmetry transformations has certain properties that defines it as a group. If $T_1$ is a transformation that takes rays $\mathscr{R}_n$ into $\mathscr{R}_n'$ and $T_2$ is another transformation that takes $\mathscr{R}_n'$ into $\mathscr{R}_n''$, then the result of performing both transformations is another symmetry transformation, which we write $T_2T_1$, that takes $\mathscr{R}_n$ into $\mathscr{R}_n''$. Also, a symmmetry transofrmation $T$ which takes rays $\mathscr{R}_n$ into $\mathscr{R}_n'$ has an inverse written $T^{-1}$ which takes $\mathscr{R}_n'$ into $\mathscr{R}_n$ and there is an identity transformation $T =1$ which leaves rays unchanged.

2. He also states Wigner's theorem that states that symmetry transformations defined as above can be realized either by unitary linear or antiunitary and antiliear operators on the Hilbert space $\mathscr{H}$. In his notation, for every symmetry transformation $T$ one gets a unitary operator $U(T)$. Then Weinberg proves that

   $$U(T_2)U(T_1)=e^{i\phi(T_2,T_1)}U(T_2T_1), \tag{2.2.14}$$

saying that $U(T)$ is a projective representation of the symmetry transformations.

After this recap, here are my questions:

So by (1) above, to deal with symmetryes, Weinberg is actualy implictly considering that there is a group $G$ such that we have one homomorphism $T$ mapping $G$ into the group of ray transformations?

In other words, for every $g \in G$ we have $T(g)$ a ray transformation. Upon imposing the symmetry requirement by (2) above, we have that all $T(g)$ descends to one $U(T(g))$ and these $U(T(g))$ form a projective representation of $G$ .

Because of that in the end he forgets $T$ altogether and directly works with projective representations of $G$ on the Hilbert space of states. Is that it?

The main difference between what I'm writing and Weinberg is that I'm trying to abstract a group from its representations.

So I'm guessing there is one underlying group $G$ of symmetries, which give rise to the ray transformations and then to the projective representations, while Weinberg seems to identify $G$ with the ray transformations themselves.

Is my point of view correct of considering there is one abstract group behind all of this? Or there is actually no group behind the ray transformations, and Weinberg is actually defining a group with the ray transformations themselves, instead?

Answer 1

You are essentially right. Weinberg's philosophy is basically this: Suppose you know (due to experimental results, say) that the physical system you are studying has symmetries described by a group $G$. There are many examples of this, e.g. rotational symmetry in 3 dimensions which is described by the group $SO(3)$. How can you construct a Hilbert space $\mathcal H$ which describes this quantum mechanical system?

Any symmetry transformation must satisfy two properties:

• It should take physical states to physical states.
• It should preserve probabilities when the transformation is applied to both the initial and final states.

These define a ray transformation. That is to say, by the first condition, we have a map from $G$ to the space of self-maps on the projective Hilbert space $P\mathcal H$ (the space whose points correspond to physical states). The second condition that probabilities are conserved means that any self-map $T:P\mathcal H \rightarrow P \mathcal H$ which is in the image of $G$ must descend from a unitary and linear or antiunitary and antilinear operator $U(T):\mathcal H \rightarrow \mathcal H$. This is the content of Wigner's theorem. With a bit of extra work, you can prove that this indeed defines a homomorphism. You can think of this as a 2-step homomorphism, the first map being from $G$ to the isometry group of $P\mathcal H$ with the Fubini-Study metric (this is the "group of ray transformations" in your language), and the second from there to the group of unitary-and-linear or antiunitary-and-antilinear operators on $\mathcal H$. The first step is physical, while the second is purely mathematical.

The end result is that the Hilbert space must be a projective representation of $G$. This is not an immediately obvious fact; one might have thought that a symmetry could be realized nonlinearly on the Hilbert space, but Wigner's theorem tells us that can't happen. Because ray transformations are not so easy to work with in practice we generally prefer to work with the operators.

You also raise the question of whether we can construct the group $G$ from its (projective) representation. This is not really a totally sensible question as the group $G$ is part of the data of the representation. With that said, under certain conditions if you know all the representations of $G$ it can be possible to reconstruct $G$, though this is a different direction from what Weinberg is talking about. More precisely, given the set of representations (without any of their data) together with information about how they are related (see Category of Representations), it is possible to reconstruct $G$ if it is a compact topological group (up to a central extension if we include projective representations). This is known as Tannaka-Krein duality.