Eye_in_the_Sky said:
... the first description tells us that, in the nonrelativistic domain, we have no particular need for QFT, while the latter suggests to us that, nevertheless, in that domain, QFT can still be applied.
The above remark was intended to dispel the not uncommon misconception in which the notion of a "quantized field" is thought to apply strictly to a relativistic domain.
So, what is an example of a nonrelativistic domain in which QFT can be applied, yet for which it may be said that QFT satisfies no particular need?
Well, quite simply ... take the single-particle Schrödinger field and quantize it.
Upon doing that ... what do we get?
(note: x is a spin-space index; i.e. x=(s,
x), δ(x-x')=δ
3(
x-
x')δ
ss')
case (1): [Ψ(x,t),Π(x',t)]=ih
barδ(x-x') → Fock space for identical Bosons
case (2): {Ψ(x,t),Π(x',t)}=ih
barδ(x-x') → Fock space for identical Fermions
In QM, on the other hand, the Fock space is constructed by taking a direct sum of a one-dimensional Hilbert space plus a series (n = 1,2, ...) of appropriately symmetrized n-particle Hilbert spaces. Thus, QFT gives us the
exact same thing back (except (with the "bonus") that the symmetrization postulates for multiparticle states in QM have been replaced by the commutator-anticommutator rules of QFT (which themselves are on par with (i.e. no stronger than) QM's rule [Q
j,P
j'] = ih
barδ
jj')).