Question on observables of QFT

[wangyi]

Hi,
In Weinberg's QFT book(section 2.2), he said after proved the generator of the symmetry group is Hermitian and can be a candidate for an observable:

Indeed, most(and perhaps all) of the observables of physics, such as angular momentum or momentum, arise in this way from symmetry transformations.

My questions:
1. Does the observable here mean at least in principle, we can measure it in experment?
2. Does generator of any symmetry group stand for an observable? (I know a exception, SUSY grnerator seems not to be an observable. But are there other exceptions or what is the most common case?)
3. Does every observable come up this way? Or what are observables in QFT? Does it the same with observables in QM? In Dirac's book of QM he said a Hermitian operator with a complete set of eigenstates can in principle be observed in experiments. If QM and QFT are the same in this sense, then a Hermitian scalar field or a Majorana Fermion field is an observable in every space-time point. But does it true?

Thank you very much and happy new year!

 

[dextercioby]

Hi,
In Weinberg's QFT book(section 2.2), he said after proved the generator of the symmetry group is Hermitian and can be a candidate for an observable:

Depends on the observable, really. It can be unbounded in an a priori chosen Hilbert space and so, by virtue of Stone's theorem it's actually at least essentially self-adjoint.

My questions:
1. Does the observable here mean at least in principle, we can measure it in experment?

Yup.

2. Does generator of any symmetry group stand for an observable?

If the symmetry is well implemented at quantum level (i.e. using the theorem of Wigner), then the generators, according to Stone's theorem are (essentially) self-adjoint and, by means of the second postulate can describe observables.

3. Does every observable come up this way?

Almost all of them do.

Or what are observables in QFT? Does it the same with observables in QM?

Yes.

In Dirac's book of QM he said a Hermitian operator with a complete set of eigenstates can in principle be observed in experiments.

By virtue of Gelfand-Maurin spectral theorem, only a self-adjoint operator has a complete set of generalized eigenvectors in a rigged Hilbert space.

If QM and QFT are the same in this sense, then a Hermitian scalar field or a Majorana Fermion field is an observable in every space-time point. But does it true?

"Hermitean" scalar field means

\varphi =\varphi^{*} ,

where "star" is the involution operation on the Grassmann algebra in which the classical field is valued. It can mean at classical level simply "complex conjugation", while at qauntum level it means "charge conjugation".

Majorana Fermion field means

\psi =\psi ^{*} ,

where "star" is the operator of charge conjugation.

None of the above 2 operators (when acting in the context of algebras of qauntum operator) coincide with the operation of taking the adjoint of an operator acting in a (rigged) Hilbert space.

Daniel.