# 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?

• The case $\hat{H}\propto\hat{n}$ is trivial. $\hat{H}\hat{S}|n\rangle = \hat{S}\hat{H}|n\rangle = n\hat{S}|n\rangle$, so $\hat{S}|n\rangle$ has the same number of particles as $|n\rangle$. In other words, $\hat{S}$ preserves particle number. A "free" field theory is precisely one in which $\hat{H}$ is a sum of $\hat{n}$ operators for every possible mode, so we've shown that the statement is true for a free theory. For an interacting theory I don't know. May 14, 2015 at 4:57
• More answers (among them mine) together with an extensive discussion can be found at physicsoverflow.org/30822 Jun 16, 2015 at 8:38

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.

• 1) Why does $[Q,\psi]$ have to produce a strictly local field? I guess that the symmetry of the Hamiltonian may lead to the requirement that it does but I do not see it. 2) Do indeed the statements hold in interacting field theory? Could you maybe point to a reference where it is shown that the fullblown self-interacting one-particle states also furnish a representation of the symmetry?
– Void
May 20, 2015 at 19:56
• @Void 1) Please think for a moment of $Q$ as the electric charge operator, you can write it uzing the Gauss' law as a surface integral of the electric field over a very large sphere at infinity. Due to the large distance these fields do not produce singularities when multiplied by other fields, thus the only singualrities coming from the commutator are those due to the local field $\psi$, thus the commutator itself is also a local field at $x$. May 21, 2015 at 13:50
• @Void 2) Assuming Lorentz symmetry, the renormalization factors Z in ψ R =Z(ψ)ψ are Lorentz scalars, thus nothing essential changes in the analysis when the true renormalized field is used: Its transformation properties remain the same. May 21, 2015 at 14:39