Alternative quantization of quantum electrodynamics?

Question

A Quantum field theory is determined, if a Hilbert space Basis with Operators acting on it (such that one element of an Hilbert space is also an element of the same Hilbert space if an Operator acting on it) and a Hamiltonian and commutator relations exists. Now I want to know whether Quantum electrodynamics can be reformulated in Terms of cobordisms instead of field distributions in spacetime.

Quantum electrodynamics has the Hamiltonian:

$$H = \int d^3x [\psi^\dagger \gamma_0(\gamma_i (\partial^i+ieA^i) +m_0) \psi + \frac{1}{2} (E^2+(rot A)^2)].$$

The commutator Relations for vector potential $A^i$ and electric fields $E^i$ are

$$[E^j(x),A^i(x')] = -i\delta_{ij} \delta(x-x')$$

for Fermion fields $\psi$ anticommutators are given by

$$\{\bar{\psi}(x),\psi(x')\} = -i\delta(x-x').$$

All other commutator Relations are zero. In ordinary Approach to Quantum electrodynamics one has a vacuum state $|0>$, where the Action of a field operator on it gives the corresponding field function and a creation or Annihilation of a field on the vacuum.

Is it possible to assume the vacuum state on a fixed time $t=t_0$ as a collection of 3-dimensional boundary, where the Hamiltonian Operator $H$ defines how this 3-dimensional boundary evolves to another 3-dimensional boundary at $t=t_0+dt$? Can I assume field Operators as cobordisms on such states? Maybe I can create a theory with same hamiltonian and commutator relation, but where I use boundaries as objects and cobordisms as morphisms?

I would say yes, because this can be achieved by a functor mapping form Hilbert space category to cobordism category.

Another question: Will I obtain different scattering amplitudes if I treat Quantum electrodynamics with cobordisms instead of ordinary well-known Hilbert space Framework?

Maybe also yes, because quantization is different.

Answer 1

You're basically asking for QED to be formulated as a functorial quantum field theory (FQFT). FQFT is not a method of quantization, it is a method of recasting a quantized theory in a rigorous axiomatic form, but doing to for QED seems out of reach for now:

The problem is that QED is neither topological nor conformal - the partition function, which would be the functor mapping a cobordism to an amplitude in this approach, has no well-defined value on a morphism in the category of cobordisms, since the objects of this category are customarily diffeomorphism (or homeomorphism) classes of manifolds and a morphisms $\Sigma$ between two such classes $M_1,M_2$ is the equivalence class of a cobordism $\sigma$ with $\partial\sigma = M_1 \sqcup M_2$. The partition function only gives a well-defined value $Z(\sigma)$ if $Z(\sigma) = Z(\sigma')$ for all other $\sigma'$ in this equivalence class, but this is essentially the definition of a topological quantum field theory, which QED is not.

However, what you can do is weaken the equivalence classes we're taking - for a relativistic QFT, you should take isometry classes of Riemannian manifolds as the objects and isometry classes of Lorentzian manifolds whose boundaries are the objects as the morphisms. Then, the path integral of QED, if it were well-defined, would yield amplitudes for these cobordisms and the assignment of states to the objects (which we should think of as spatial slices) would be the space of state "at that time".

But, in the end, you haven't gained anything here - we don't know how to make the path-integral so well-behaved that we could really define it as a functor on this category of "Lorentzian cobordisms", and anything you might compute here in the end reduces to "ordinary" computations in QED with the path integral formalism. It is not evident at all how you might obtain different predictions from the standard QFT, given that the only way to define the functor is from the standard path integral QFT. The reason this formalism succeeds on topological and conformal theories is that the path integral there "degenerates" into computing some topological invariant of the cobordism, which may be done by rigorous mathematical means other than the physicists' path integral.