Abstract

It is known that a massless free field on a (1 + 1)-dimensional de Sitter space-time does not allow a de Sitter invariant Fock vacuum on a Hilbert space. We show that this result no longer holds on an indefinite inner product space. To this end we use the Gupta-Bleuler formalism to construct a causal, de Sitter and conformally covariant massless free quantum field on the (1 + 1)-dimensional de Sitter space-time admitting a de Sitter invariant vacuum in an indefinite inner product space. The field is defined rigorously as an operator-valued distribution and is covariant in the usual strong sense: Vg−1ϑ(x)Vg = ϑ(g · x) for any g in the de Sitter group, where V is a unitary representation of the de Sitter group on the space of states. The formalism of Gupta-Bleuler triplets also allows for an explicit description of the gauge degree of freedom. We show that, although the field itself is not an observable (it is gauge dependent), the stress tensor is. The component t00(x) is positive in all physical states, and vanishes in the vacuum. In addition, since the field is conformally covariant, the model does not exhibit a conformal anomaly in the trace of the energy-momentum tensor.

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