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,σ⟩=∫eipxψ†σ(x)|0⟩d4x
Where
ψ is the field corresponding to the particle and
σ 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=∫j0(x′)d3x′
should generate a local field when it acts on a local field:
[Q,ψ1(x)]=ψ2(x)
(In the case of internal symmetries
ψ2 depends linearly on the components of
ψ1(x), in the case of space time symmetries it depends on the derivatives of the components of
ψ1(x))
Thus in general:
[Q,ψσ(x)]=∑σ′Cσσ′(i∇)ψσ′(x)])
Where the dependence of the coefficients Cσσ′ on the momentum operator ∇ is due to the possibility that Q contains a space-time symmetry.
Thus for an operator Q satisfying Q|0⟩=0, we have
Q|p,σ⟩=∫eipxQψ†σ(x)|0⟩d4x=∫eipx[Q,ψ†σ(x)]|0⟩d4x=∫eipx∑σ′Cσσ′(i∇)ψσ′(x)|0⟩d4x=∑σ′Cσσ′(p)∫eipxψ†σ′(x)|0⟩d4x=∑σ′Cσσ′(p)|p,σ′⟩
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,σ⟩ and
Q|p,σ⟩ have the same energy.
This post imported from StackExchange Physics at 2015-06-16 14:50 (UTC), posted by SE-user David Bar Moshe