Saturday, August 20, 2016

quantum mechanics - Conserved charges and generators


For the Klein Gordon field, the conserved charge for translation in space is given by: P=12d3kk{akak+akak}


If we were to find the generators for a space translation, we would find that, Pj=ij, where j=1,2,3.


If we act both of the above operators on the field ϕ, the result matches! My question is whether both of these, the generators and the conserved charges of a symmetry are always the same thing? What would be a simple way to see this connection?



Answer




OP is wondering whether the conserved charge associated to a continuous symmetry always generates the symmetry itself. We can say, in full generality, that the answer is


Yes.


Let us see how this works.


Classical mechanics.


We use a notation adapted to classical field theory rather than point-particle mechanics, but the former includes the latter as a special sub-case so we are losing no generality.


Consider a classical system which may or may not include gauge fields and/or Grassmann odd variables. For simplicity, we consider a flat space-time. Assume the system is invariant under the infinitesimal transformation ϕϕ+δϕ. According to Noether's theorem, there is a current jμ jμL˙ϕ,μδϕ which is conserved on-shell, μjμOS=0


This in turns implies that the associated Noether charge Q Qdef=Rd1j0dx is conserved, ˙QOS=0


In Ref.1 it is proved that the charge Q generates the transformation δϕ,



δϕ=(Q,ϕ)




where (,) is the DeWitt-Peierls bracket. This is precisely our claim. The reader will find the proof of the theorem in the quoted reference, as well as a nice discussion about the significance of the result.


Furthermore, a similar statement holds when space-time is curved, but this requires the existence of a suitable Killing field (cf. this PSE post).


Moreover, for standard canonical systems, Ref.1 also proves that (,) agrees with the Poisson bracket {,}.


Quantum mechanics.


This is in fact a corollary of the previous case. Ref.1 proves that, up to the usual ordering ambiguities inherent to the quantisation procedure, the DeWitt-Peierls bracket of two fundamental fields agrees with the commutator [,] of the corresponding operators.


If we assume that the classical conservation law μjμ0 is not violated by the regulator (i.e., if the symmetry is not anomalous), then we automatically obtain the quantum analogue of our previous result, to wit



δϕ=i[Q,ϕ]




as required.


References



  1. Bryce DeWitt, The Global Approach to Quantum Field Theory.


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