Can symmetry generators anticommute with the S-matrix? In Coleman and Mandula's proof of the Coleman-Mandula theorem, they define a symmetry transformation as an unitary $U$ which,


*

*turns one-particle states into one-particle states,

*acts on many-particle states as if they were tensor product of one-particle states,

*commutes with $S$, the $S$-matrix.


I think this means that if $T$ is a generator of symmetry, then $$[T,S]=0.$$ If we relaxed the 3rd point so that $T$ can also anti-commute with $S$, what would be the consequences? In particular, would the theorem still hold?
 A: You cannot relax the third point for the following reason.
The $S$-matrix of a theory is defined as
$$
S_{\beta\alpha} \equiv \left( \Psi^-_\beta, \Psi^+_\alpha \right) \equiv \left( \Phi_\beta, S \Phi_\alpha \right) 
$$
The first equality defines the $S$-matrix in terms of 'in' and 'out' states while the second equality defines the $S$-matrix operator which is an operator whose matrix elements with the free particle states of the theory is equal to the $S$-matrix itself. 
A symmetry of the theory implies existence (Wigner's theorem) of a linear unitary operator $U$ (it can also be anti-unitary and anti-linear) that acts on the one-particle states and acts on multi-particle states as if they were tensor product of one-particle states (points 1 and 2). Under the action of this operator
$$
U \Psi^\pm_\alpha = \Psi^\pm_{\alpha'}
$$
Since this operator is unitary, $U^\dagger U = 1$, we must have
$$
S_{\beta\alpha} = \left( \Psi^-_\beta, \Psi^+_\alpha \right)  = \left( \Psi^-_\beta, U^\dagger U \Psi^+_\alpha \right) = \left( U \Psi^-_\beta,   U \Psi^+_\alpha \right) = \left( \Psi^-_{\beta'}, \Psi^+_{\alpha'} \right) = S_{\beta'\alpha'}
$$
In terms of the $S$-matrix operator this statement implies
$$
\left( \Phi_\beta, S \Phi_\alpha \right) = \left( \Phi_{\beta'}, S \Phi_{\alpha'} \right) = \left( U_0 \Phi_\beta, S U_0\Phi_\alpha \right) = \left( \Phi_\beta, U_0^\dagger S U_0 \Phi_\alpha \right) 
$$
where $U_0$ is the operator that generates the symmetry transformation on the free particle states. Now since the equality holds for any set of states $\Phi_\alpha$, we must have
$$
S = U_0^\dagger S U_0  \implies \boxed{ [U_0, S ] = 0 } 
$$
Thus, it is crucial that the generators of the symmetry (acting on the free Hilbert space) commute with the $S$-matrix. The following things were important in this discussion


*

*Existence of a unitary operator $U$ on the Hilbert space. 

*Existence of a unitary operator $U_0$ on the free Hilbert space.

*If the free state $\Phi_\alpha$ corresponds to $\Psi^\pm_\alpha$, then $U_0 \Phi_\alpha$ corresponds to $U \Psi^\pm_\alpha$. The rigorous definition of the phrase 'corresponds to' is the following
$$
\int d\alpha e^{- i E_\alpha \tau} g(\alpha) \Psi_\alpha^{\pm} \stackrel{\tau\to\mp\infty}{\longrightarrow}\int d\alpha e^{- i E_\alpha \tau} g(\alpha) \Phi_\alpha
$$


Side note It is often the case that $U = U_0$ and hence this distinction is not made. For example, in most quantum field theories ${\bf P} = {\bf P}_0$ and ${\bf J} = {\bf J}_0$ where these are the momentum and angular momentum operators respectively. On the other hand, we  have $H = H_0 + V$ where $V$ is some interaction. Another example, where this equality fails to hold is with the boost operator, ${\bf K} \neq {\bf K}_0$. 
