Confusion about the quantum field Lorentz transformation

[HomogenousCow]

On page 59 of Peskin & Schroeder, there's a section on the lorentz transformation of field operators which I've attached. I'm confused about the part towards the end where he does a change of variable on the integration measure; it seems like he's only rewriting the lorentz-invariant integration measure in terms of the boosted momentum, and not the rest of the integrand. Why is this allowed? Surely the integral has to be boosted together.

EDIT: Nevermind, just wasn't thinking about it the right way

 

[vanhees71]

Although Peskin&Schroeder are of course correct, I find the definition of $\hat{U}$ also a bit ad hoc. There are two ways to treat the problem. My favorite is straight forward using the "canonical quantization prescription", i.e., you use the Hamiltonian formulation for the classical field and then quantize it using commutator or anticommutator relations for the canonical field operators and their canonically conjugated momenta (depending on whether you have bosons or fermions, but you are forced from the analysis of the free fields that you have to quantize (half-)integeger spin fields as bosons (fermions) in order to get a microcausal theory with a Hamiltonian bounded from below). Then the canonical formalism tells you the generators of the symmetries via Noether's theorem (up to normal ordering to make them well-defined and finite of course).

The other way, going much along the lines as PS is to demand that there's a unitary operator operating such that you get the usual local transformations for the field operators as for the classical field analogues. Then you get the transformation of the creation and annihilation operators wrt. the usual momentum-spin basis, PS start with.

It's of course independent of which normalization condition for the $\hat{a}$ and $\hat{a}^{\dagger}$ you choose. The quantum transformation is always unitary by definition. Of course, there are no finite-dimensional unitary representations of the Lorentz group (except the trivial one), because it's not a compact group, but this has nothing to do with the unitarity of the transformations in QT. By definition a symmetry must always be represented by a unitary ray representation. In the case of the proper orthochronous Poincare group you can always lift any unitary ray representation to a proper unitary representation of its covering group. The ray representations of the Poincare group has no non-trivial central charges (as the Galileo group fortunately has, because otherwise there'd be no useful non-relativistic QM to begin with :-)).

 

[HomogenousCow]

Although Peskin&Schroeder are of course correct, I find the definition of ^UU^\hat{U} also a bit ad hoc. There are two ways to treat the problem. My favorite is straight forward using the "canonical quantization prescription", i.e., you use the Hamiltonian formulation for the classical field and then quantize it using commutator or anticommutator relations for the canonical field operators and their canonically conjugated momenta (depending on whether you have bosons or fermions, but you are forced from the analysis of the free fields that you have to quantize (half-)integeger spin fields as bosons (fermions) in order to get a microcausal theory with a Hamiltonian bounded from below). Then the canonical formalism tells you the generators of the symmetries via Noether's theorem (up to normal ordering to make them well-defined and finite of course).

This gives the same transformation rules on the creation/annihilation operators though right? Once you exponentiate the generators.

 

[vanhees71]

Indeed. You can derive P&S's definition of the $\hat{U}$'s in this way. P&S just go the other way and define how the $\hat{U}$s act on the creation and annihilation operators wrt. to the standard momentum-spin single-particle basis and then prove that this definition leads to the construction of a local unitary representation of the Poincare group.