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]

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.

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]

Do the basic axioms really need to be modified for QFT?

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]

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?

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]

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.

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]

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).

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]

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?

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

 

[snoopies622]

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

. . .to get an operator you must multiply by a nice function and integrate over x.

A nice function? What nice function?

 

[A. Neumaier]

A nice function? What nice function?

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

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]

...multiply by an arbitrary nice function...

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

 

[snoopies622]

The fields themself are not observables, but auxiliary operators from which the observables (like the Hamiltonian) can be constructed.

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

 

[A. Neumaier]

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

No. Second quantization means that you quantize a classical field theory that itself can be considered as the quantization of a classical particle. - Thus the ''second''.

 

[lugita15]

No. Second quantization means that you quantize a classical field theory that itself can be considered as the quantization of a classical particle. - Thus the ''second''.

How can a classical field theory be considered as a quantization of a classical particle?

 

[DarMM]

In a formal sense. Many classical field theories are mathematically no different from the
quantum mechanical theory of a single particle. For instance the equations for a free
massive scalar classical field and a single relativistic quantum mechanical boson are identical as they are both the Klein-Gordon equation.

However Second Quantisation is an out dated piece of terminology. It original arose because you would quantised a single particle and get an PDE for its quantum mechanical version. The wavefunction appearing in this PDE behaved as a classical field, which itself could then be quantised. However it's better nowadays to think of simply quantising some given classical field, as it is just a coincidence that the equations of some classical fields are identical to the quantisation of single particles.

 

[martinbn]

How can a classical field theory be considered as a quantization of a classical particle?

Take for example the Schrodinger equation. It describes the dynamics of the quantized systems, say a classical particle, at the same time it is a PDE so it can be viewed as describing a classical field.

 

[A. Neumaier]

However Second Quantisation is an out dated piece of terminology.

Only in some quarters. Entering ''second quantization'' at http://scholar.google.com with the option ''since 2003'' still gives about 7,510 hits to papers using the term.

In fact, I find it a meaningful concept. There is even the established concept of the second quantization operator that takes an operator A acting on classical fields and turns it into a corresponding 1-particle operator $$\Gamma(A)$$ acting on the corresponding Fock space.

It is just a coincidence that the equations of some classical fields are identical to the quantisation of single particles.

This is not true. Maxwell's equations are perfectly legitimate classical field equations, and their second quantization gives precisely the photon quantum field. This is not a coincidence, as my lecture http://arnold-neumaier.at/ms/optslides.pdf shows.

 

[martinbn]

This is not true. Maxwell's equations are perfectly legitimate classical field equations, and their second quantization gives precisely the photon quantum field. This is not a coincidence, as my lecture http://arnold-neumaier.at/ms/optslides.pdf shows.

But in this case there isn't first quantization, is there? So second quantization is not the best terminology. In fact in all cases there is just one quantization.

 

[DarMM]

Only in some quarters. Entering ''second quantization'' at http://scholar.google.com with the option ''since 2003'' still gives about 7,510 hits to papers using the term.

Oh, certainly people use it. I simply meant that it is not used anywhere near as much as it was a few decades ago and it is becoming less popular as a concept. People now prefer to just think of themselves as quantising a classical field, not second quantising a classical particle.

For instance, how does one get Yang-Mills via second quantisation?

This is not true. Maxwell's equations are perfectly legitimate classical field equations, and their second quantization gives precisely the photon quantum field. This is not a coincidence, as my lecture http://arnold-neumaier.at/ms/optslides.pdf shows.

What classical particle system is the Maxwell equation a quantisation of? In other words what is the first quantisation in this case?

 

[snoopies622]

Just a basic math question here - perhaps it belongs in another place:

I read here and there that the Klein-Gordon equation is the same as the classical equation of motion for a free scalar field.

If we represent it as
$$\frac {1}{c^2} \frac {\partial ^2 \psi}{\partial t^2} = \nabla^2 \psi - \frac {m^2 c^2}{\hbar ^2} \psi$$

I recognize the first two terms as a generic wave equation. The third term must place limits on solutions.

What do these limits look like?

 

[A. Neumaier]

But in this case there isn't first quantization, is there? So second quantization is not the best terminology. In fact in all cases there is just one quantization.

In this case, there is also first quantization, as the Maxwell equations are the equations for the wave function of a single photon, when the latter is coded in terms of the Riemann--Silberstein vector. See e.g. quant-ph/0511011.

 

[A. Neumaier]

For instance, how does one get Yang-Mills via second quantisation?

What classical particle system is the Maxwell equation a quantisation of? In other words what is the first quantisation in this case?

The quantization of a classical massless particle with spin (in an associated Poisson manifold) gives Maxwell for spin 1, and Yang-Mills for spin 1 plus inner symmetries.

 

[A. Neumaier]

Just a basic math question here - perhaps it belongs in another place:

I read here and there that the Klein-Gordon equation is the same as the classical equation of motion for a free scalar field.

If we represent it as
$$\frac {1}{c^2} \frac {\partial ^2 \psi}{\partial t^2} = \nabla^2 \psi - \frac {m^2 c^2}{\hbar ^2} \psi$$

I recognize the first two terms as a generic wave equation. The third term must place limits on solutions.

What do these limits look like?

The massless Klein-Gordon equation _is_ the wave equation. The massive case appears classically as an inflation field in general relativity.

 

[snoopies622]

The massive case appears classically as an inflation field in general relativity.

Ah. By "classical" I was thinking "before the 20th century". Does it not appear then? I don't mean with Planck's constant of course, just an equation of the same form but with different physical constants. I thought it might also represent some kind of wave.

 

[A. Neumaier]

Ah. By "classical" I was thinking "before the 20th century". Does it not appear then? I don't mean with Planck's constant of course, just an equation of the same form but with different physical constants. I thought it might also represent some kind of wave.

I don't know of any use of the Klein-Gordon equation before Klein & Gordon. (If one existed, the equation would probably carry a different name).

On the other hand, classical in physics means ''not quantum''.
For example, general relativity is a classical field theory, although it wasn't considered before the 20th century.