Quantum Field Theory and Perturbation Approximation

[rogerl]

Pls. answer in the simplest and the most intuitive way.

1. What is the reason our quantum field theory needs perturbative approach. Is it because in the concept of fields, there is an infinite number of freedom in the oscillations of the virtual particles, or is it because the field is infinitely close to one another?

2. Classical field theory needs perturbative approach too. How does it differ to quantum field theory?

3. Some said we use perturbative approximation because we haven't encountered the right theory yet. Why. Would the final right QFT version no longer use perturbation? How would this occur?

 

[tom.stoer]

I wouldn't say that we always need the perturbative approach.

PT has - w/o doubt - some nice features:
1) PT allows for a rather quick start with a rather good approximation b/c tree level approx. is often well known from other (e.g. classical) considerations
2) PT allows for a systematic calculation of higher order
3) PT allows for a simple classification of states (Fock space, asymptotic plane wave states)

But there are situations where PT failes:
1) it cannot be used in a strong coupling regime, i.e. when the expansion parameter is not small (whatever this means exactly)
1') in principle PT assumes that g=0 is a reasonable expansion; this need not be the case
2) there are non-perturbative contributions, e.g. instanton effects in QCD, scaling with 1/g
3) in QCD it is known (and can be "derived" from PT) that PT itself breakes down in the IR at and below the QCD scala lambda

 

[dextercioby]

The main point about PT is that sometimes, if the expansion parameter is small enough, it's the only tool to extract physical information from non-linear and non-homogenous differential equations.

 

[tom.stoer]

I would like to add one comment: even if the expansion parameter is small and even if the series can be defined order by order using regularization/renormalization in order to achieve finiteness, it is not guarantueed that the whole series will converge; usually it will diverge / it will be an asymptotic expansion only. Therfeore I would say that PT is a tool to extract physics, but not a tool to define QFT.

 

[friend]

Question: can the quantum fields of QFT be made to resemble the 1/(r^2) electric force?

 

[dm4b]

Pls. answer in the simplest and the most intuitive way.

1. What is the reason our quantum field theory needs perturbative approach. Is it because in the concept of fields, there is an infinite number of freedom in the oscillations of the virtual particles, or is it because the field is infinitely close to one another?

2. Classical field theory needs perturbative approach too. How does it differ to quantum field theory?

In reply to (1) and (2) the reason is because there are no exactly solvable interacting field theories (in the dimensionality required for spacetime)

Perturbative techniques are one way to get an answer w/o solving the equations exactly.

3. Some said we use perturbative approximation because we haven't encountered the right theory yet. Why. Would the final right QFT version no longer use perturbation? How would this occur?

I think the general consensus is that QFT is on the right track. It's just that it's only valid within its domain, like all the other theories we have. QFT has had remarkable success as a theory.

With that said, if they don't find the Higgs (or find something else unexpected) one can't help but wonder what kind of reformulation of the Standard Model might be required ;-)

 

[dm4b]

Question: can the quantum fields of QFT be made to resemble the 1/(r^2) electric force?

If you calculate the magnitude for the leading-order contribution to the basic QED scattering process, it can be shown that in the nonrelativistic limit it does indeed produce a repulsive Coulomb Potential, which would give rise to the 1/(r^2) force you mention.

Seems like the early chapters in most QFT texts do something along these lines for various fields theories, to demonstrate how QFT produces the expected repulsive or attractive potentials, as required.

So, to answer your question: sort of ;-)

 

[tom.stoer]

Question: can the quantum fields of QFT be made to resemble the 1/(r^2) electric force?

QED in Coulomb gauge produces this result exactly, not only in a perturbation expansion, as an interaction operator.

\hat{V} \sim e^2 \int d^3x \int d^3y \, \frac{\hat{\rho}(\vec{x})\,\hat{\rho}(\vec{y})}{|\vec{x}-\vec{y}|}

where rho is the fermionic charge density. This result is not widely known as in QED one usually uses (or teaches) a covariant gauge. In QCD one can do something similar but there the operator 1/|x-y| is replaced by an integral operator with gauge-field dependent kernel which can only be written down symbolically.

 

[friend]

QED in Coulomb gauge produces this result exactly, not only in a perturbation expansion, as an interaction operator.

I would be interested in seeing the QFT perturbation expansion for the Coulomb force on an electron. Thanks.

 

[tom.stoer]

What would you like to calculate? Energy levels? You would instead use different approximations like "heavy baryons" for the nucleus, mean field ore something like that.

Or are you interested in scattering solutions? I think Bjorken-Drell present Coulomb gauge quantization and scattering theory.

 

[friend]

What would you like to calculate?

Let me tell you where I'm at so maybe you can help me better. I seem to have a derivation from principles alone of the path integral for a free particle. (PM me if you want to see it) My lagrangian only cantains the m*v^2. Now I'm thinking about how to come up with the potential energy term in the lagrangian. And I'm wondering if I can do this with what I have.

I consider that the potential energy in a system changes a particle's momentum. And I'm thinking that a change in momentum can also be changed by an interaction with another free particle, which I already have. So perhaps a potential energy term can be constructed as an accumulation of interactions. The question then becomes: how can a potential energy field be constructed as a series of interactions; how can a field be dissolved into interactions?

So I ask about quantum field theory, virtual particles, and perturbative expansions. Do I actually need a perturbative expansion to achieve the virtual particles of a static field? Or can I get a potential energy field non-perturbatively? Maybe this should have been put in a separate thread. But I'm stuck at this point; I'm not really sure what I should be trying to calculate in the QFT formulism. Any insight you could give would be appreciated.

 

[tom.stoer]

Let me tell you where I'm at so maybe you can help me better. I seem to have a derivation from principles alone of the path integral for a free particle. (PM me if you want to see it) My lagrangian only cantains the m*v^2. Now I'm thinking about how to come up with the potential energy term in the lagrangian. And I'm wondering if I can do this with what I have.

How shall this work? There are numerous possible potential terms, so somehow you must plug in at least a construction principle for a specific potential.

So perhaps a potential energy term can be constructed as an accumulation of interactions. The question then becomes: how can a potential energy field be constructed as a series of interactions; how can a field be dissolved into interactions?

The answer is "mean field approximation". What you do is to transform a system of N (N large) interacting particles into a system of one particle interacting with a "mean field" emerging from integrating out the other particles. This is standard quantum mechanics for many body systems.

Do I actually need a perturbative expansion to achieve the virtual particles of a static field? Or can I get a potential energy field non-perturbatively?

I think the question is wrongly put: you do not construct the potential from perturbation theory, but you apply perturbation theory (as a mathematical tool) to find approximate solutions. The potential (or more generally - the interaction term) is more fundamental than the perturbation series and the virtual particles. This is no limited to my specific example.

 

[friend]

The answer is "mean field approximation". What you do is to transform a system of N (N large) interacting particles into a system of one particle interacting with a "mean field" emerging from integrating out the other particles. This is standard quantum mechanics for many body systems.

Yes, I think that's what I want to do. I'm wondering if any potential can be described as a mean field approximation of a multiparticle system.

I think the question is wrongly put: you do not construct the potential from perturbation theory, but you apply perturbation theory (as a mathematical tool) to find approximate solutions. The potential (or more generally - the interaction term) is more fundamental than the perturbation series and the virtual particles. This is no limited to my specific example.

I think this is the opposite of what you just said above. What I want to do is actually prove that any legitimate potential term can be constructed as a series of interactions. Then I can construct the potential term out of the kinetic term that I already have.

So what does the lagrangian look like for the interaction between the collision of two otherwise free particles? Thank you.

 

[tom.stoer]

I think this is the opposite of what you just said above. What I want to do is actually prove that any legitimate potential term can be constructed as a series of interactions. Then I can construct the potential term out of the kinetic term that I already have.

We are talking about two different potential terms. The interaction (potential) term I presented in post #8 is the full Coulomb potential term in QED. What you are talking about is an effective potential for one single (effective) particle. What I still do not understand is how you want to construct a potential from nothing else but a kinetic energy term

 

[friend]

We are talking about two different potential terms. The interaction (potential) term I presented in post #8 is the full Coulomb potential term in QED. What you are talking about is an effective potential for one single (effective) particle. What I still do not understand is how you want to construct a potential from nothing else but a kinetic energy term

What I meant is using the type of information in the kinetic energy term, namely the momentum, to construct the potential energy term. Since the force of that potential changes the momentum of a particle, and since an interaction with another particle also changes the momentum of a particle, I was asking if it is possible to construct the potential energy term from an aggregate formulation of a series of interactions.

So I suppose the place to start is with the lagrangian of simpliest form of interaction, two otherwise free particles bouncing of each other. I wonder what that looks like. Is that just two free particles with a cross term --- p1^2/m1 +p2^2/m2 +/-p1p2/m1m2?