Standard quantum mechanics postulates that, for an isolated system, time evolution is ruled by unitary operators, then one can prove Schrodinger equation (SE), which is not Lorentz invariant.

If we want a quantum relativistic theory we need, for some reasons, a field viewpoint and this leads to QFT.

My question is: can we still say that in QFT time evolution is, by postulate, unitary?


A fundamental postulate of QFT establishes that the theory admits a strongly continuous representation of (orthochronous proper) Poincaré group $\cal P$. A certain one-parameter subgroup of $\cal P$ describes time evolution (with respect to an inertial reference frame) which, as a consequence, turns out to be unitary since it is part of a larger unitary representation.

Alternatively, you can exploit Wigner theorem. Time is homogeneous in inertial systems both in classical and relativistic physics, so time evolution must preserve probability transitions for isolated physical systems. Wigner theorem implies that time evolution is represented by unitary or antiunitary transformations $U_t$. Assuming that $U_t$ is a representation of $\mathbb R$ (the axis of time), i.e. $U_tU_s = U_{t+s}$, one sees that each $U_t$ must be unitary, because $U_t = U_{t/2}U_{t/2}$ is unitary nomatter if $U_{t/2}$ is unitary or antiunitary.

Under some further quite mild requirements (separability of the Hilbert space and the fact that the complex valued maps $\mathbb R \ni t \mapsto \langle \psi | U_t \phi \rangle$ are measurable for every choice of the vectors) a theorem due to von Neumann proves that $\mathbb R \ni t \to U_t$ is strongly continuous and thus admits a unique self-adjoint generator (Stone theorem), that is the Hamiltonian operator, by definition.

  • $\begingroup$ Is this still true in a generally relativistic background metric? $\endgroup$ – CuriousOne Sep 26 '14 at 4:02
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    $\begingroup$ No, it is not in that case. Generally speaking, to have a notion of evolution it is necessary that the spacetime admits a timelike Killing field. In that case, the Killin symmetry induces a continuous one-parameter group of automorphisms on the algebra of observables. If the background quantum state is invariant under that symmetry, when passing to the Hilbert space of the theory by means of the GNS construction, the Killing symmetry representing that time evolution is unitarily implemented. $\endgroup$ – Valter Moretti Sep 26 '14 at 6:24
  • $\begingroup$ Thanks! I have to think trough that for a while... it's not completely clear to me how that construction allows for a "causal" evolution and what e.g. path integrals would integrate over in case of a dynamically changing metric. So it's the Killing symmetry that guarantees unitarity "at all times"? Does that imply that we lose unitarity for non-static spacetimes which do not have a timelike Killing symmetry? $\endgroup$ – CuriousOne Sep 27 '14 at 1:26
  • $\begingroup$ In the absence of timelike Killing symmetries (even only asymptotic) there is no time evolution unitary or not, at least described by a time independent Hamiltonian. $\endgroup$ – Valter Moretti Sep 27 '14 at 8:14
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    $\begingroup$ Actually, in a generally covariant picture, time evolution is not a relevant notion, causality is described by the so called time-slice axiom, which does not require any unitary evolution... $\endgroup$ – Valter Moretti Sep 27 '14 at 8:16

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