From: baez@galaxy.ucr.edu (John Baez) Subject: When can we quantize a symplectic transformation? Date: 31 Aug 2000 18:23:15 GMT Newsgroups: sci.physics.research Summary: [missing] In article <8ojee9$p3o$1@nnrp1.deja.com>, wrote: >In article , > torre@cc.usu.edu (Charles Torre) wrote: >> If you consider the dynamical evolution of a *free* quantum >> field in *Minkowski* spacetime along a suitably generic >> foliation of spacetime then the field evolution is NOT >> represented by a unitary transformation and there is no >> self-adjoint operator generating this evolution. >Wow! Hmm, on second thought, it's not surprising, as there's no time >translation symmetry for an arbitrary foliation. Hmm, on third thought, >it is strange, as the classical physics can be given a Hamiltonian >formulation - can't they?! Yes, there's a Hamiltonian formulation of the corresponding classical time evolution. The problem is that in systems with infinitely many degrees of freedom, not all classical symmetries can be quantized - not even all *linear* classical symmetries. More precisely: not all symplectic linear transformations on the classical phase space H can be implemented as unitary operators on the corresponding Fock space F(H) - only those that are "close to unitary". Remember, the classical phase space H starts out life as a symplectic vector space, but we must make it into a complex Hilbert space before we can define the Fock space F(H) as a Hilbert space. Once H has been made into a complex Hilbert space, it makes sense to ask whether a symplectic linear transformation T: H -> H is unitary or not. We can also ask if it's "close to unitary": this means that it differs from a unitary by an operator X such that tr(X* X) is finite. By results of Irving Segal and his student Shale, only symplectic linear transformations that are close to unitary in this sense can be quantized to give unitary operators on F(H). If I remember it correctly, the essence of Torre's proof was to show that the time evolution corresponding to a wiggly foliation of Minkowski spacetime is not "close to unitary" in this sense. (Except perhaps in 1+1 dimensions? I forget.) Buzzword bonus: instead of "close to unitary", people usually say a symplectic linear transformation "meets the Hilbert-Schmidt condition" if it differs from a unitary operator by an operator X for which tr(X* X) is finite. Why? An operator X with tr(X* X) < infinity is usually called a "Hilbert-Schmidt operator", that's why! All such operators are compact and this was introduced as a nice sufficient condition for compactness by Hilbert and Schmidt back in the early days of functional analysis.