# A What is the relationship between QFT and QM?

#### meopemuk

But there are many examples of QFTs (e.g., conformal field theories in 1+1 dimension) whose Hilbert space does not have such a Fock space structure. The latter is specific to the nonrelativistic case and to free relativistic theories.
When talking about QFT, I have in mind mainly QED, which is a very successful example of a physical theory. QED is both relativistic and interacting. It can be formulated as a theory with a fixed Fock space structure.

Eugene.

#### A. Neumaier

It can be formulated as a theory with a fixed Fock space structure.
... but only by neglecting nonperturbative aspects. The latter are already needed for determining the lifetime of positronium.

#### meopemuk

But in a quantum field theory, symmetry principles give no position operator. They only give total momentum, and angular-momentum operators, not even (the nonexisting) momentum operators for single particles!
The global representation of the symmetry (=Poincare) group is acting in the entire Fock space. The generators of this representation are identified with "total" observabes: total momentum, total angular momentum and total energy (Hamiltonian). Using these operators and the generator of boost one can build other useful operators: total spin, center-of-energy position, total mass, etc.

But there are many more useful operators/observables in the Fock space. For example, each sector is build as as a tensor product of one-particle spaces, so each sector inherits all one-particle observables, such as position, spin, momentum etc. So, in the entire Fock space, there is no problem in defining, for example, a one-electron position and tracing its time evolution during the collision/interaction process.

Eugene.

#### meopemuk

... but only by neglecting nonperturbative aspects. The latter are already needed for determining the lifetime of positronium.
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.

Moreover, the (Fock) structure of the Hilbert space has to be independent on the presence/absence of interactions. According to Dirac, relativistic interaction is defined through specific representation of the Poincare group in the Hilbert space of the system. This definition does not require you to change the Hilbert/Fock space structure depending on the interaction.

Eugene.

#### meopemuk

Strictly speaking, QFT is about an infinite number of degrees of freedom.
Strictly speaking, QFT is about variable number of particles.

Eugene.

#### PeterDonis

Mentor
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 someting 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.

"What is the relationship between QFT and QM?"

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