what are the postulates of QFT?

snoopies622

I like the way quantum mechanics can be expressed as a set of five or six axioms, like in Daniel T. Gillespie's A Quantum Mechanics Primer or David McMahon's Quantum Mechanics Demystified.

Is there a similar set of axioms for quantum field theory?

dextercioby

An example would be Wightman's axioms which are pretty well-known. See the books by Bogolubov et.al. 1975 or Lopuszanski. There's also an axiomatic formulation due to Haag in terms of (nets of) operator algebras, but this harder to grasp, unless you're also reformulating normal quantum mechanics (the one mentioned in the OP) through operator algebras and their representations.

snoopies622

Thank you.

A. Neumaier

Unfdrtunately, while various axiom systems for QM indeed cover all quantum mechanics, the Wightman axioms are known not to cover any of the fundamental quantum fields theories (QED, QCD, and the standard models), as these are gauge theories and the Wightman axioms are not applicable to these. There have been attempts to repair this (notably by Strocchi), but with partial success only.

The sad truth therefore is that we currently have no adequate system of axioms for QFT, only a number of ideas how it could possibly look like.

Closest to the standard model are in fact not the nonperturbative Wightman axioms but the perturbative Epstein-Glaser approach to quantum field theory; see
http://en.wikipedia.org/wiki/Causal_perturbation_theory (and much more by querying scholar.google.com).

snoopies622

How strange!

kith

When we talk about the axioms of QM, I think of the basic axioms (states in Hilbert space, observables as self-adjoint operators, von Neumann measurements, Schrödinger equation) and not the specific axioms of non-relativistic single particle QM.

Do the basic axioms really need to be modified for QFT? I thought the problems would occur in in constructing the operators and calculating transition amplitudes.

snoopies622

That's what I've been wondering for a long time. According to McMahon, in QFT "fields are promoted to operators". I don't know what this means.

I understand the idea of a physical state being represented by a vector and an observable by an operator and how all that works, but if the field is an operator, upon what does it operate? What kind of mathematical object, and what does this object represent? This seems like a completely different formulation to me.

kith

Also on the physical state.

Let's look at the electron field for example (this is probably flawed but will show the basic features). A simple physical state is a number state |n>, where n is the number of electrons you have. The operators of the electron field are ψ(x) and ψ⁺(x). The first one destroys an electron at space time point x, the second one creates one. The fields themself are not observables, but auxiliary operators from which the observables (like the Hamiltonian) can be constructed. So in general, the field equations (like the Dirac equation) are neither Schrödinger equations (dynamics of the states) nor Heisenberg equations (dynamics of the observables), but dynamical equations for these auxiliary quantities which have no direct physical significance.

A. Neumaier

It means that insterad of the classical electromagnetic field $E(x)$ you have operator-valued fields. Thus each component E_k(x) is an operator. More precisely, it is an operator-valued distribution, whch means that to get an operator you must multiply by a nice function and integrate over x. This operator operates as usual on a Hilbert space of wave functions in infinitely many variables.

atyy

Why aren't the Wightman axioms applicable to gauge theory? I know they haven't been shown to be applicable, but have they been shown to be inapplicable?

A. Neumaier

This was discussed at length in
http://physicsforums.com/showthread.php?t=468492
and threadss citedthere.

snoopies622

Thank you kith and A. Neumaier, I think I understand a tiny bit now.

A nice function? What nice function?

A. Neumaier

In order that the product of a function and a distribution is integrable, the function must be sufficiently well-behaved. An infinitely often differentiable function with compact support will always do, and hence is nice enough. But for many distributions, much less is sufficient. Saying nice allows me to hide all these technical details.

snoopies622

I'm sorry, I meant - what functions are you referring to?

A. Neumaier

I had answered that already: To get an operator rather than a distiribution, multiply by an arbitrary nice function (e.g., a C^inf function with compact support) and integrate over x.. (i.e., rather than integrating over a bounded domain of interest, one typically integrates over a C^inf approximation of its charactersistic function.)

There isn't any more to the informal notion of ''nice''.

snoopies622

Oh ok, I figured there was a specific one that depended on the physical state, or something.

snoopies622

Is this the reason for the phrase "second quantization"?