Graduate What is the relationship between QFT and QM?

[PeterDonis]

Strictly speaking, QFT is about variable number of particles.

No, it isn't, because there are quantum field states that do not even have a particle interpretation. The fundamental objects in quantum field theory are fields, not particles.

 

[A. Neumaier]

Strictly speaking, QFT is about variable number of particles.

No. A quantum field theory is primarily about fields, as the name says. Particles appear in general only as approximate asymptotic concepts.

Starting with them is just a didactical tradition.

 

[A. Neumaier]

As far as I understand, the "perturbative/nonperturbative" refers to the method chosen to calculate the S-matrix (including the lifetime of positronium). If you are lucky, you can do that without using the perturbation series, but in most cases it is easier to use the perturbative approach. The chosen calculation method should not have any effect on the (Fock) structure of the Hilbert space. The Fock space structure simply tells you that numbers of particles are good quantum-mechanical observables that commute with each other. So, each Fock sector is a common eigenspace of all particle number operators in the theory.

The particle number operator cannot be renormalized in the standard perturbation scheme. To get finite results one needs to renormalize, and to get approximate lifetimes, one needs to sum infinite number of renormalized diagrams - which is neither fixed order perturbation theory nor doable in Fock space.

In your personal approach to QED, you get a renormalized particle number operator by using a nonstandard formulation with a nonstandard perturbation theory, but then you can no longer determine what to sum to get the lifetimes. At least you haven't shown that you can...

 

[meopemuk]

The particle number operator cannot be renormalized in the standard perturbation scheme. To get finite results one needs to renormalize, and to get approximate lifetimes, one needs to sum infinite number of renormalized diagrams - which is neither fixed order perturbation theory nor doable in Fock space.

I don't quite understand... In the standard renormalization procedure we add counterterms to the Hamiltonian to make sure that the S-matrix is finite and satisfies a few renormalization conditions. Why would we want to renormalize the particle number operator? Are you talking about the Greenberg-Schweber transition to "dressed particles"? Or something else?

Besides, in QED we are dealing with stable particles - photons, electrons, protons. What the "lifetimes" have to do with them?

Eugene.

 

[A. Neumaier]

Why would we want to renormalize the particle number operator?

Because unrenormalized operators have no meaning in the renormalized theory - all expectations containing it are ill-defined.

Besides, in QED we are dealing with stable particles - photons, electrons, protons. What the "lifetimes" have to do with them?

Positronium is an unstable composite particle which has a finite lifetime computable by QED.