If a symmetry operator S in a QFT annihilates the vacuum, why does S preserve the space of 1-particle states? In the paper "Supersymmetry and Morse Theory", on the third page (p. 663 in the journal version), Witten says:

"Now in any quantum field theory if a symmetry operator (an operator
  which commutes with the Hamiltonian) annihilates the vacuum state,
  then the one particle states furnish a representation of the
  symmetry."

Why is this true? Is there a simple explanation or computation that doesn't go too far afield of Witten's relatively informal discussion in the introduction of this paper, or is it more complicated?
 A: This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization.
A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the momentum operator, they can have further quantum numbers. These states should exist even in an interactiong field theory, describing a single particle away from any interaction.
In a local quantum field theory, these states are associated with local field operators:
 $$| p, \sigma \rangle = \int e^{ipx} \psi_{\sigma}^{\dagger}(x) |0\rangle d^4x$$
Where $\psi $ is the field corresponding to the particle and $\sigma$ describes the set of other quantum numbers additional to the momentum.
A symmetry generator $Q$ being the integral of a charge density according to the Noether's theorem
$$Q = \int j_0(x') d^3x'$$
should generate a local field when it acts on a local field:
$[Q, \psi_1(x)] = \psi_2(x)$
(In the case of internal symmetries $\psi_2$ depends linearly on the components of $\psi_1(x)$, in the case of space time symmetries it depends on the derivatives of the components of $\psi_1(x)$)
Thus in general:
$$[Q, \psi_{\sigma}(x)] =  \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x)])$$
Where the dependence of the coefficients $ C_{\sigma\sigma'}$ on the momentum operator $\nabla$ is due to the possibility that $Q$ contains a space-time symmetry.
Thus for an operator $Q$ satisfying $Q|0\rangle = 0$, we have
$$ Q | p, \sigma \rangle = \int e^{ipx} Q \psi_{\sigma}^{\dagger}(x) |0\rangle d^4x = \int e^{ipx} [Q , \psi_{\sigma}^{\dagger}(x)] |0\rangle d^4x = \int e^{ipx} \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x) |0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) \int e^{ipx} \psi_{\sigma'}^{\dagger}(x) |0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) | p, \sigma' \rangle $$
Thus the action of the operator $Q$ is a representation in the one particle states. 
The fact that $Q$ commutes with the Hamiltonian is responsible for the energy degeneracy of its action, i.e., the states $| p, \sigma \rangle$ and  $Q| p, \sigma \rangle$ have the same energy.
