# Concrete example that projective representation of symmetry group occurs in a quantum system except the case of spin half integer?

I for the first time learnt there exist projective representation of symmetry group in Weinberg's QFT vol1. In Weinberg's textbook, the definition of symmetry of a system is that a symmetry is a bijection of Hilbert space such that for any normalized states $|\Psi\rangle$, $|\Phi\rangle$ and the mapped states $|\Psi'\rangle$, $|\Phi'\rangle$ $$|\langle \Psi|\Phi\rangle| = |\langle \Psi'|\Phi'\rangle|.$$

With this definition, we see symmetry group of the system admits a projective representation: $$U(T_2)U(T_1)=e^{i\phi(T_2,T_1)}U(T_2T_1).$$

For example, spin-$1/2$ is a projective representation of $SO(3)$ group. I want to consider whether there are other symmetry except internal symmetry which can occur projective representation in a system, but I can't construct one.

In quantum mechanics, the definition of a symmetry $S$ is a operator which commutes with Hamitonian $H$, i.e $$[S,H]=0.$$ Then we see in the eigenenergy $E_i$ subspace, $$H\psi^i_\mu=E_i \psi^i_\mu$$ with $\mu=1,2,\cdots,f_i$. That is the eigenspace of energy $E_i$ with dimension $f_i$. $$HS\psi^i_\mu=SH\psi^i_\mu=E_i S\psi^i_\mu.$$ So $S\psi^i_\mu$ still have the same energy $E_i$ and it must can be expanded by $\psi^i_\nu$ with $\nu=1,2,\cdots,f_i$. Define the expansion coefficients as $D^i_{\nu\mu}(S)$, $$S\psi^i_\mu= \sum_\nu D^i_{\nu\mu}(S)\psi^i_\nu$$ For any two symmetries $R,S$ with $[R,H]=0=[S,H]$, then $RS\equiv Q$ as a whole must be a symmetry because $[RS,H]=[R,H]S+R[S,H]=0$.

On the one hand, $$RS\psi^i_\nu=R\sum_\mu D^i_{\mu\nu}(S)\psi^i_\mu =\sum_{\mu\gamma}D^i_{\mu\nu}(S) D^i_{\gamma \mu}(R)\psi^i_\gamma= \sum_{\mu\gamma} D^i_{\gamma \mu}(R)D^i_{\mu\nu}(S)\psi^i_\gamma$$ On the other hand, $$RS\psi^i_\nu=Q \psi^i_\nu=\sum_\gamma D^{i}_{\gamma\nu}(Q)\psi^i_\gamma==\sum_\gamma D^{i}_{\gamma\nu}(RS)\psi^i_\gamma.$$ So $$D^i(RS)=D^i(R)D^i(S).$$

That is we can only get the representation of symmetry group other than the projective representation.

My question:

1. Does it mean only representation of symmetry group can occur in quantum system? According to above derivation, if it's true, what's the meaning to discuss the projective representation?

2. If the answer of 1st question is No. What's the loophole in my arguement. And except the case of spin half integer under rotation (or Lorentz group, Poincare group), give me a concrete example projective representation of symmetry group can occur?

• Why is there a Greek index here at all? And why is the representation of the symmetry acting on it and not on the Roman index? – ACuriousMind Sep 9 '17 at 19:29
• @ACuriousMind Roman index $i$ denotes the eigenspace with energy $E_i$. Greek index $\mu\nu$ denote the label of eigenstates in this eigenspace. – user153663 Sep 9 '17 at 20:02
• (Putting aside the issue that the Roman index is entirely superfluous) What exactly is happening in the line after "for any two symmetries"? If $R$ and $S$ are your abstract symmetry operators and $D$ is the representation map, then $RS\psi$ doesn't make any sense - you need to write $D(R)D(S)\psi$ or $D(RS)\psi$ from the start. You've cheated yourself by believing that $RS\psi$ a) makes sense and b) is equal to both of these, where by b) you have implicitly assumed that the representation is not projective. – ACuriousMind Sep 9 '17 at 20:38
• @ACuriousMind $RS\equiv Q$ as a whole must be a symmetry because $[RS,H]=[R,H]S+R[S,H]=0$. – user153663 Sep 9 '17 at 23:09
• @ACuriousMind I didn't directly say $D(R)$ as the representation. I firstly define the expansion coefficients as $D^i_{\nu\mu}(S)$, $$S\psi^i_\mu= \sum_\nu D^i_{\nu\mu}(S)\psi^i_\nu$$ Then I find that the coefficients are the representation of the whole symmetry group. – user153663 Sep 9 '17 at 23:11

The confusion here arises from two different notions of "symmetry":

1. Symmetry in the sense of Wigner's theorem: This is Weinberg's notion of symmetry - a ray transformation that leaves all inner products invariant. Wigner's theorem says that such ray transformations have (anti-)unitary representants determined up to a phase, and when dealing with a whole group of such transformation, this ambiguity in phase is essentially where the notion of projective representations arises from. For a comprehensive review, see also my Q&A on projective representations, central extensions and universal covers.

2. Dynamical symmetry: Physically, the first notion of "symmetry" is too weak: If the (anti-)unitary operators corresponding to the ray transformations do not commute with the Hamiltonian, then applying the symmetry before time evolution yields different results from applying it afterwards. In particular, the expectation values of the "symmetry" operators are not conserved in time, something we want from symmetries. So a dynamical symmetry - what one usually simply calls a symmetry of the system - is an operator (or a set of operators) that commutes with the Hamiltonian.

Now, if you start with the notion of "a symmetry group" as a group of operators on a Hilbert space that commute with the Hamiltonian, then you are correct that there are no projective representations - a group of linear operators is clearly linearly represented on the space the operators are acting on.

The projective representations arise if we know that abstractly, there should be a representation of an abstract symmetry group $G$ (like the rotation group, or the Lorentz group, or any other symmetry group the classical system we are quantizing has) on the space of states. Then we know that:

1. To each $g\in G$ there is a ray transformation $T(g)$ that is a symmetry in the sense of Wigner.

2. For each $T(g)$, the (anti-)unitary representatives commute with the Hamiltonian. If this holds for one representative, it holds for all since phases commute with everything. A projective representation of $G$ is a consistent choice of these representatives.

• Thanks. Could give another example which involves projective representation of other group like $\mathbb{Z}_2\times\mathbb{Z}_2$ other than the spin? – user153663 Sep 11 '17 at 0:53
• @fff123123 A very prominent example is the projective representation of the Witt algebra in conformal field theory, which is usually phrased in terms of the linear representations of its centrally extended version, the Virasoro algebra. – ACuriousMind Sep 11 '17 at 1:08
• But I think any unitary transformation can keep the inner product invariant. So any unitary transformation is a symmetry in first sense? – user153663 Sep 11 '17 at 4:35
• @fff123123 No, but any symmetry in the first sense has (anti-)unitary representants, and all unitary operators induce a ray transformation that is a symmetry of the first kind. It is important to note that the symmetries of the first kind are ray transformations, not operators on the Hilbert space, since this is exactly the origin of the projective representations. – ACuriousMind Sep 11 '17 at 9:17