For the Klein Gordon field, the conserved charge for translation in space is given by: →P=12∫d3k→k{a†kak+aka†k}
If we were to find the generators for a space translation, we would find that, Pj=i∂j, 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 ϕ→ϕ+δϕ. According to Noether's theorem, there is a current jμ jμ∼∂L∂˙ϕ,μδϕ which is conserved on-shell, ∂μjμOS=0
This in turns implies that the associated Noether charge Q Qdef=∫Rd−1j0dx 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 ∂μjμ≡0 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
- Bryce DeWitt, The Global Approach to Quantum Field Theory.
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