Why I am REALLY disappointed about string theory

[suprised]

For decades, trying to unify the gravity with other forces. But how unify, if the thing does not exist? I suggest philosophers of science to think about it.

What can philosophers possibly add to this? The few times I discusssed with philosophers, I spent hours explaining them time dilatation, geometry of curved space-time, principles of GR but they understood nothing. But they kept having ideas about the "very nature" of space and time, etc.

 

[Fra]

For decades, trying to unify the gravity with other forces. But how unify, if the thing does not exist? I suggest philosophers of science to think about it.

One possible viewpoint is that - if we see the laws of nature, not as statements of what nature IS and WHY, but rather as statements of what we (or an observer genereally) can rationally EXPECT from "nature", and WHY - then all laws of nature (seen from human observers) are already unified by the scientific method.

Ie. all our knowledge of nature, including it's "laws" are the result of the inference we call the scientific method. All laws are inferred from feedback from nature (the observers environment).

But what's lacking is a consistent inferential picture that applies to ANY observer, at any scale. For example how is the inferential system, scaled from human perspective to atomic scale?

We know exactly how we humans have inferred the laws of nature. The question is how a proton does the same? Ie how does the proton "know" what laws to follow? Maybe it doesn't? What it's just doing some constrained random walk? Could the constraint be the correspondence to the laws?

In particular can one ask; is there an observer (at some scale or complexity) that are inabled to distinguish gravity from the other forces, by available inference? Then, there you have the unificiation.

So I'd say unification is the starting point. The question isn't how to unify the forces, the question from inference is how to separate them in a way that reproduces the broken symmetries we see. That's how I see it.

/Fredrik

 

[marcus]

What can philosophers possibly add to this? The few times I discusssed with philosophers, I spent hours explaining them time dilatation, geometry of curved space-time, principles of GR but they understood nothing. But they kept having ideas about the "very nature" of space and time, etc.

:rofl:
They sound charming. I'll bet the discussion was in German. Please tell us how to say "very nature" in whatever was the original language. I want to add this to my vocabulary.

Perhaps you should tell them that the very nature of space and time is nothing else than the measurements possible to make of the geometric relationships between events.

Or else you could hurl them into the river. For this to be possible there must be a river near where you live.

 

[tom.stoer]

Do you know Feynman's discusssion with philosophers regarding "the true nature of an electron"? As a warm-up and in order to understand the expectations of the philosophers he first asked a simple question regarding the "true nature of a brick". He never got to the point to discuss the electron as the philosophers were not able to finish the brick discussion ...

 

[marcus]

Do you know Feynman's discusssion with philosophers regarding "the true nature of an electron"? As a warm-up and in order to understand the expectations of the philosophers he first asked a simple question regarding the "true nature of a brick". He never got to the point to discuss the electron as the philosophers were not able to finish the brick discussion ...

Wonderful story! No, I hadn't heard it.

 

[jaquecusto]

Do you know Feynman's discusssion with philosophers regarding "the true nature of an electron"? As a warm-up and in order to understand the expectations of the philosophers he first asked a simple question regarding the "true nature of a brick". He never got to the point to discuss the electron as the philosophers were not able to finish the brick discussion ...

I also know a little story about Feynman. After a conference, he angrily returned home and told his wife to remind him never to participate in discussions about gravity.:uhh:

 

[tom.stoer]

Wonderful story! No, I hadn't heard it.

Just read his book "surely you're joking, ..."

 

[Fra]

Just a brief comment. I'm not sure if I was the only one making the distinction, but to me "philosophy of physics" is not really same thing as "philosophy of science".

Stuff like "true nature of electron" is not something I can imagine would be discussed within philosophy of science. An electron is not a fundamental concept there.

In PoS, one MIGHT raise questions such as "nature of knowledge" or "information", the problem of induction (that Popper so famously wrote a whole book about and still failed to solve satisfactory) etc.

/Fredrik

 

[tom.stoer]

So in ads4/cft3 you have the vector model on one side and Vasiliev's higher spin theory on the other. ... It's pretty complicated, doesn't have a Lagrangian formulation yet etc.

Perhaps this is a stupid question: what does ít exactly mean to have a theory w/o Lagrangian formulation? What is the definition" of such a theory?

 

[mitchell porter]

http://www.ictp.it/media/101047/schwarzictp.pdf" regarding the M5-brane worldvolume theory.

 

[atyy]

http://www.ictp.it/media/101047/schwarzictp.pdf" regarding the M5-brane worldvolume theory.

Why is weak coupling the classical limit? Is that the same as hbar->0 ?

Are notions of classical limits for the SCFT and the bulk different? I thought these SCFTs were supposed to have classical bulk gravity at large N. How is it that it wasn't/isn't apparent that the SCFTs have classical limits?

 

[mitchell porter]

Why is weak coupling the classical limit? Is that the same as hbar->0 ?

Are notions of classical limits for the SCFT and the bulk different? I thought these SCFTs were supposed to have classical bulk gravity at large N. How is it that it wasn't/isn't apparent that the SCFTs have classical limits?

There are some fundamental issues of quantum field theory here for which I have only an intuitive understanding. However:

I believe any classical limit corresponds to some version of hbar->0. But there can be inequivalent ways to do this, so a single quantum theory can have several, inequivalent classical limits. In fact, this is one way, or even the best way, to understand the meaning of the dualities in string theory - one quantum theory, several classical limits - but the phenomenon of multiple classical limits already exists in QFT.

Suppose we work with path-integral quantization. Under certain circumstances, the path integral will be dominated by the classical solutions to the equations of motion (i.e. the gradient of the amplitude in the space of histories will be flat in the vicinity of the classical solutions, so amplitudes for histories which are "almost classical" and clustered around a single classical solution will additively reinforce each other and make a large contribution to the total amplitude), so you can approximate the quantum theory as the classical theory plus fluctuations. I might have supposed that a theory which is "always strongly coupled" is one without such flat regions in the space of histories, so you can't apply perturbation theory, but then Schwarz later comments that for the 6D SCFT, people are "focus[ing] on the equations of motion", so I'm not sure. But even if I can't rigorously see why strong coupling implies no formulation of the quantum theory in terms of an action functional, it seems sensible that weak coupling does imply perturbative tractability and the existence of a classical limit.

The original example of an "always strongly coupled" non-Lagrangian QFT would be the "E6" superconformal field theory constructed by Minahan and Nemaschansky in the late 1990s. This actually shows up as the worldvolume theory of a D3-brane near an E6 singularity in F-theory; Heckman and Vafa have a paper about the phenomenological implications. The "T_n" theories mentioned by Schwarz were discovered by Gaiotto in his paper "N=2 dualities"; there's a review http://arxiv.org/abs/0909.1327" . Like the MN E6 theory, the Tn theories can also be coupled to other fields, becoming sectors in a larger theory.

I need to think for a while about the various limits in AdS/CFT before I can answer the second set of questions.

 

[suprised]

Fully agreed to mitchell. Some extra comment:

In string dualities one can have the phenomenon that quantum corrections and classical geometrical quantities can be exchanged between different formulations. That is, in one duality frame you may encounter quantum corrections to certain quantities at arbitrary loop order (genus of world-sheet), and in another frame these same expressions arise from the classical geometry of the compactification manifold. So there is no absolute notion as to what is quantum and what is classical.

 

[tom.stoer]

I still do not understand what "always strongly coupled" or "semiclassical limit" and "no Lagrangian formulation" have to do with each other. Perhaps this has something to do with "quantization of a Lagrangian" or "perturbative treatment" .- which would be missleading.

In QCD you can write down a path integral based on quarks and gluons which is valid below Lambda-QCD and which does not require any (high energy) asymptotic freedom.

 

[Finbar]

Its basically as mitchell said. If the quantum theory has a classical regime it means that the integrand of the path integral is very strongly peaked around the minimum of the action. So the integrand is

$$e^{-\frac{S[\varphi]}{\hbar}$$


The action will be proportional to the inverse of the coupling
$$S \sim \frac{1}{g}$$
So for a classical regime to exist we need
$$g \to 0$$ i.e. weak coupling. You can see that this is equivalent to taking hbar to zero as well. Then

$$Z =\int D \varphi e^{-\frac{S[\varphi]}{\hbar}} \approx e^{-\frac{S[\varphi_{cl}]}{\hbar}$$

 

[mitchell porter]

Regarding atyy's second set of questions:

For the original example of AdS/CFT (stack of D3-branes), the string theory and the gauge theory are perturbatively calculable in different parameter ranges. http://www.springerlink.com/content/9p632240j7314480/" (section 3.5):

the string theory on AdS₅ x S⁵ is currently only really calculable in the classical supergravity limit where g_s << 1 (so no string loops) and l_s >> R (so no alpha' corrections). In terms of YM parameters this means that N >> lambda >> 1, which is the planar ’t Hooft limit, but at strong ’t Hooft coupling. On the other hand, the YM theory is only under perturbative control at small lambda and finite N. A great deal of the power of Maldacena’s conjecture comes not just from the fact that it is an explicit realization of the AdS/CFT conjecture, but also that weak coupling on one side of the equivalence is strong coupling on the other.

Because one is varying several parameters here - rank of the gauge group (N), coupling constant (g, where g_string = g_Yang-Mills²) - or even their product - lambda, the 't Hooft coupling, is g_stringN - it can be hard to keep track of the relations between these limits. But maybe the important conceptual question, for the present discussion, is whether the existence of a calculable framework on one side of the duality implies the existence of a "classical limit" on the other side of the duality. I suppose the answer is "yes", but to visualize or comprehend this limit, you have to use the variables on the other side of the duality.

So maybe the best initial answer to atyy's challenge - how can these SCFTs not have classical limits when they are dual to classical supergravity in the bulk - is that, yes, these SCFTs do have classical limits where they "shouldn't", but the only classical characterization of those limits is precisely in terms of the dual, bulk variables (which can all be defined by the right combinations of operators from the boundary theory). You wouldn't be able to see it if you were just looking at the "original" variables.

Now, returning to John Schwarz's talk, the three primordial examples of AdS/CFT are for D3-branes, M2-branes, and M5-branes. In every case, you have a stack of coincident branes with a worldvolume theory that is decoupled from space-time far from the branes (think of the causally disconnected regions that can show up in Penrose diagrams), and the worldvolume theory is equivalent to string theory in an AdS space. For strong gravitational back-reaction, such that the branes form an event horizon, AdS is the actual near-horizon geometry. For weak gravitational back-reaction, such that the branes are existing in flat space, the AdS space seems to exist as a manifestation of energy scale in the worldvolume theory, akin to Guifre Vidal's MERA construction (but this is one of the conceptual issues that is still being worked out). So whether you're at weak or strong 't Hooft coupling, AdS is there.

For D3-branes, I quoted Argyres on how perturbative limits exist in two distinct parameter ranges that lie on opposite sides of the duality. As Schwarz says, for some time it was believed that the worldvolume theories for M2- and M5-branes were non-Lagrangian, meaning that there was no perturbative formulation. Following the discussion in previous comments, this would have meant that the only classical limit for this quantum theory was in terms of bulk variables, like supergravity. But ABJM showed that the M2-brane worldvolume theory was a super-Chern-Simons theory, and the Chern-Simons level "k" was able to play the role of Yang-Mills coupling g_YM, so there's a 't Hooft coupling for these theories, kN.

The remaining question is whether the M5-brane worldvolume theory also has a perturbative formulation, or whether it really is non-Lagrangian. ("Always strongly coupled" must mean that the parameter, which you might have wanted to use for a perturbation expansion, remains large at every energy scale - there's nothing like asymptotic freedom.) And this is the subject of ongoing research.

 

[suprised]

Perhaps this is a stupid question: what does ít exactly mean to have a theory w/o Lagrangian formulation? What is the definition" of such a theory?

Some theories are just defined in terms of their scattering amplitudes, or S-matrix, without any lagrangian formulation. This is pretty familiar from d=2. For example, it has been claimed way back that the SU(2) WZW model with E7 modular invariant does not have any lagrangrian description. I am not sure whether this statement is still considered true or not, I didn't follow the lit.

The many dualities that have been discovered may in fact teach to think in this direction more seriously again. They imply that lagrangian formulations can be ambiguous and may blur the view to the essence of a theory, which is its scattering amplitudes.

A canonical example is given again by the Ising model: it can be realized in terms of free fermions, psi, or bosons, in terms of which psi= exp(i phi). The bosons are periodic, thus can be viewed as compactified dimension. But it can be misleading to give a deeper significance to this "extra dimension". If one would just study the S-Matrix, one would not fall into the trap of attributing a higher-than-deserved significance to a particular lagrangrian representation of the theory.

 

[tom.stoer]

Its basically as mitchell said. If the quantum theory has a classical regime it means that the integrand of the path integral is very strongly peaked around the minimum of the action. So the integrand is

$$e^{-\frac{S[\varphi]}{\hbar}$$


The action will be proportional to the inverse of the coupling
$$S \sim \frac{1}{g}$$
So for a classical regime to exist we need
$$g \to 0$$ i.e. weak coupling. You can see that this is equivalent to taking hbar to zero as well. Then

$$Z =\int D \varphi e^{-\frac{S[\varphi]}{\hbar}} \approx e^{-\frac{S[\varphi_{cl}]}{\hbar}$$

This seems to be wrong! Look at gravity. Perhaps GR is indeed the classical limit of some (yet to be identified) theory of quantum gravity, i.e. h=0. But of course this is NOT equivalent with G=0; GR does exist at non-zero G and we are all happy with that.

 

[atyy]

This seems to be wrong! Look at gravity. Perhaps GR is indeed the classical limit of some (yet to be identified) theory of quantum gravity, i.e. h=0. But of course this is NOT equivalent with G=0; GR does exist at non-zero G and we are all happy with that.

Perhaps he is thinking of GR as a field on flat spacetime?

 

[tom.stoer]

Hopefully not!

GR is much more than that. There are indications that QG (quantized GR) could be consistent for G>0 but inconsistent for G=0. G=0 could be interesting for toy models but totally irrelevant for nature.

 

[atyy]

Hopefully not!

GR is much more than that. There are indications that QG (quantized GR) could be consistent for G>0 but inconsistent for G=0. G=0 could be interesting for toy models but totally irrelevant for nature.

Is GR really much more than that in physically relevant cases?

http://relativity.livingreviews.org/Articles/lrr-2006-3/ , Eq 62 gives GR as a field on flat spacetime provided harmonic coordinates can be used.

This is not such a bad restriction, since harmonic coordinates can penetrate the event horizon. http://relativity.livingreviews.org/Articles/lrr-2000-5/ , section 3.3.2 and ref. 42

Other claims I've seen that GR is equivalent to a field on flat spacetime are in
http://arxiv.org/abs/gr-qc/0411023
http://arxiv.org/abs/gr-qc/9512024 (Eq 12)

I don't know if there is an implicit assumption of harmonic coordinates in the latter 2 references. Deser's paper does explicitly claim to achieve the full theory, with no special gauge involved.

 

[Finbar]

This seems to be wrong! Look at gravity. Perhaps GR is indeed the classical limit of some (yet to be identified) theory of quantum gravity, i.e. h=0. But of course this is NOT equivalent with G=0; GR does exist at non-zero G and we are all happy with that.

It's right and works just fine for gravity! The limit is not G to 0 but g=Gp^2 to zero the dimensionless coupling. p^2 can be thought of as the momentum of an individual graviton. In other words its the limit p^2<< Mpl^2 is weak coupling.

The limit G to zero is actually the strongly coupled UV limit of the theory at a fixed point.

 

[tom.stoer]

If the quantum theory has a classical regime it means that the integrand of the path integral is very strongly peaked around the minimum of the action.

OK, a quantum theory w/o classical regime means that there is no strongly peaked integrand of the path integral. That's fine.

But still there is a path integral.

I still do not understand why the absence of a certain limit means that the theory is not defined at all; it simply means that certain approximations do not work. In QCD the limit g=0 isn't a reasonable limit in the IR; but QCD exists in the IR, it has a well-defined Lagrangian or path integral description, it can be formulated in terms of elementary quarks and gluons, even if g=0 is nonsense.

 

[fzero]

OK, a quantum theory w/o classical regime means that there is no strongly peaked integrand of the path integral. That's fine.

But still there is a path integral.

I still do not understand why the absence of a certain limit means that the theory is not defined at all; it simply means that certain approximations do not work. In QCD the limit g=0 isn't a reasonable limit in the IR; but QCD exists in the IR, it has a well-defined Lagrangian or path integral description, it can be formulated in terms of elementary quarks and gluons, even if g=0 is nonsense.

If there is no RG flow between $$g=0$$ and $$g=g_*$$ (in either direction), one cannot say that these points are limits of the same theory. For example, the 6d (0,2) theory is related to 5d super Yang-Mills theory as a strong coupling limit. However there is apparently no weakly coupled 6d theory that is connected to it.

 

[Finbar]

OK, a quantum theory w/o classical regime means that there is no strongly peaked integrand of the path integral. That's fine.

But still there is a path integral.

I still do not understand why the absence of a certain limit means that the theory is not defined at all; it simply means that certain approximations do not work. In QCD the limit g=0 isn't a reasonable limit in the IR; but QCD exists in the IR, it has a well-defined Lagrangian or path integral description, it can be formulated in terms of elementary quarks and gluons, even if g=0 is nonsense.

Well in the IR there are no propagating quarks and gluons so its far from straight forward to define the theory in the IR its one of the Millennium Prizes

http://en.wikipedia.org/wiki/Yang–Mills_existence_and_mass_gap

In practice people use chiral perturbation theory in the IR instead not in terms of elementary quarks and gluons.

So its not true, to the best of our knowledge, that there is some nice Lagrangian description of QCD in the IR in terms of yang mills fields and fundamental fermions.

The classical limit of yang-mills is in the UV where the theory is weakly coupled and well described by the classical yang-mills Lagrangian.

 

[tom.stoer]

Well in the IR there are no propagating quarks and gluons so its far from straight forward to define the theory in the IR its one of the Millennium Prizes

...

In practice people use chiral perturbation theory in the IR instead not in terms of elementary quarks and gluons.

So its not true, to the best of our knowledge, that there is some nice Lagrangian description of QCD in the IR in terms of yang mills fields and fundamental fermions.

I am sorry, but QCD in the IR can be defined via lattice gauge theory using fundamental quarks and gluons, and this approach (path integral based on a Lagrangian) is able to predict various nucleon quantities like masses within a few percent.

Please don't confuse the problem of string theory, namely to identify the fundamenatal description with the problem in QCD where the fundamental degrees of freedom are well-known and only for some calculations different descriptions are suitable (and even this is not true if you take lattice gauge theory into account).

The relation between chiral perturbation theory and fundamental quarks and gluons seems to be somehow similar to dualities in string theories (perhaps I used this as an example), but it can be misleading: in QCD everybody would agree that in principle all regimes including the IR sector are accessable using the fundamental d.o.f. Especially the dynamical phenomenon of color confinement cannot be explained using effective degrees of freedom.

QCD serves as a good example what should work for strings but where this program seems to fall short: in QCD w/o lattice gauge theory the conclusion would be that we have different regimes related via "dualities" which haven't been constructed rigorously, but which are expected to cover the full theory space. But b/c we have lattice gauge theory, and b/c we know about the fundamental description, and b/c we can do the IR calculation using the same d.o.f. as for the UV regime, we conclude that the low-energy effective theories are just that: effective theories valid in certain regimes as approximations to the fundametal theory. Denying the existence of a fundamental description of string theory and being satisfied with all these beautiful dualities actually means surrender.

 

[Finbar]

Yes QCD has a weak coupling regime in the UV so it does have a classical regime where the classical YM Lagrangian gives a good description of the physics.

A Lagrangian is really a classical concept you integrate it over space-time with appropriate boundary conditions to obtain the action which is then minimised to find the equations of motion. If the quantum theory has a regime for which the procedure gives a good approximation to the quantum theory then this means that we can quantise this Lagrangian to obtain the quantum theory i.e. the path integral.

However we may have a theory that is a perfectly well defined quantum theory with no classical regime.

QCD in the IR, while it is the quantisation of a classical theory in the UV, doesn't have a Lagrangian description in terms of fundamental fields in the sense that minimising some quantity will not give you a useful description of the physics. Thats not to say that it doesn't have a path integral.

 

[tom.stoer]

Yes QCD has a weak coupling regime in the UV so it does have a classical regime where the classical YM Lagrangian gives a good description of the physics.

True, but irrelevant for strong coupling.

A Lagrangian is really a classical concept you integrate it over space-time with appropriate boundary conditions to obtain the action which is then minimised to find the equations of motion. If the quantum theory has a regime for which the procedure gives a good approximation to the quantum theory then this means that we can quantise this Lagrangian to obtain the quantum theory i.e. the path integral.

No. If you are able to write down a mathematically well defined PI you don't care if there is a classical / weak coupling / perturbative regime. This is irrelevant.

QCD in the IR, while it is the quantisation of a classical theory in the UV, doesn't have a Lagrangian description in terms of fundamental fields in the sense that minimising some quantity will not give you a useful description of the physics.

Not true. You have a path integral (lattice action) in terms of fundamental fields (quarks, gluons) and you can calculate physical quantities. Done!

----------

There is no single step where any classical / weak-coupling / perturbative regime is required. The QCD lattice PI would work even w/o asymptotic freedom, chiral effective theories or whatever. It is a complete decsription of a QFT w/o any restrictions (except for the fact that it's hard to do certain calculations :-). So to say that a Lagrangian is a classical concept is missleading. It is used in classical physics, it is used for quantization. But if you are able to guess a Lagrangian plus a PI measure plus observables this completely defines a quantum theory. The problem is different: w/o having a classical theory or a weak-coupling regime the calculations may be more difficult or nearly intractable, but that doesn't mean that the theory doesn't exist.

Assume I go back in time and hand over the QCD lattice PI to Newton. Would he accept it as a physical theory? What about Einstein? Heisenberg? What if this lattice PI would exist w/o the nice perturbative calculations and w/o any LHC / Tevatron / HERA / ... experiment? It would still describe QCD.

I think you have something in mind regarding string theory and you want to conclude that certain things in string theory are different b/c certain well-known formalisms are no longer applicable. What are your ideas? What is missing? What does it mean that "a theory lacks a Lagrangian description"? Or that it "has no classical or weak coupling limit?" I doubt that you are able to proof that there is no PI available that fully describes the theory.

 

[atyy]

If one would just study the S-Matrix, one would not fall into the trap of attributing a higher-than-deserved significance to a particular lagrangrian representation of the theory.

How is time evolution described when there is neither Hamiltonian nor Lagrangian?

 

[tom.stoer]

Afaik the S-matrix approach has failed and I do not see how it could be raised from the dead. And I do not see why it should be easier to extract bound states physics from scattering states - even if there may be no scattering states in a certain regime at all.

Example: how would you extract the well-known QCD form factors or structure functions from the QCD S-matrix? analytically, not experimentally?

 

[suprised]

How is time evolution described when there is neither Hamiltonian nor Lagrangian?

There is of course an abstract Hamiltionian, one just cannot write it down explicitly.

 

[suprised]

Afaik the S-matrix approach has failed and I do not see how it could be raised from the dead. And I do not see why it should be easier to extract bound states physics from scattering states - even if there may be no scattering states in a certain regime at all.

Example: how would you extract the well-known QCD form factors or structure functions from the QCD S-matrix? analytically, not experimentally?

In fact what happens these days IS a resurrection of scattering matrix/amplitudes techniques, and this expressly goes against lagrangian formalism. Pages over pages of complicated feynman diagram calculations can be replaced by a few lines when employing the new twistor-based techniques. Just listen to recent talks of NAH, where he very strongly (perhaps a bit too strong) spells out how the old traditional QFT methods based on Feynman diagrams should be superseded by the new techniques.

 

[tom.stoer]

In fact what happens these days IS a resurrection of scattering matrix/amplitudes techniques, and this expressly goes against lagrangian formalism. Pages over pages of complicated feynman diagram calculations can be replaced by a few lines when employing the new twistor-based techniques. Just listen to recent talks of NAH, where he very strongly (perhaps a bit too strong) spells out how the old traditional QFT methods based on Feynman diagrams should be superseded by the new techniques.

Again it seems that you confuse QFT and the Lagrangian formalism with perturbation theory and Feynman diagrams. Neither Feynman diagrams nor perturbation theory are fundamental. Old traditional approaches based on Feynman diagrams are partially outdated, but not due to twistor strings or something like that, but due to non-perturbative methods developed (again) for QCD - based on Lagrangian or Hamiltonian techniques.

Where does this impression come from that writing down a Lagrangian automatically implies that it has to be treated perturbatively? Or that perturbation theory itself IS QFT?

Comparing QFT with Feynman diagrams is like comparing calculus with Tayor series.

I think we should stop this discussion.

So let me ask again: What does it mean that "a theory lacks a Lagrangian description"? Or that it "has no classical or weak coupling limit?" I doubt that you are able to proof that there is no PI available that fully describes the theory. What does it mean that "there is an abstract Hamiltonan which cannot be written down explicitly?"

 

[Finbar]

True, but irrelevant for strong coupling.


So to say that a Lagrangian is a classical concept is missleading. It is used in classical physics, it is used for quantization. But if you are able to guess a Lagrangian plus a PI measure plus observables this completely defines a quantum theory.

What if I hand you the Hilbert space of a theory and the observables. You can have a well defined quantum field theory without ever having to write a Lagrangian or a path integral.

As a matter of definition a Lagrangian is a classical concept. Yes, when you quantise a theory you use a quantity which has the same structure as the Lagrangian in the PI. But in a theory which has not been obtained by quantising a classical theory there is no Lagrangian any where in the theory.

 

[suprised]

Again it seems that you confuse QFT and the Lagrangian formalism with perturbation theory and Feynman diagrams.

Emphatically not!

Neither Feynman diagrams nor perturbation theory are fundamental.

This is exactly what I wrote over and over again.

Old traditional approaches based on Feynman diagrams are partially outdated, but not due to twistor strings or something like that, but due to non-perturbative methods developed (again) for QCD - based on Lagrangian or Hamiltonian techniques.

_Also_ due to twistor strings, and precisely this is my point. By now scattering processes are being computed in completely different way as before, which goes in the direction of analytical S-matrix.

Or that perturbation theory itself IS QFT?

Who ever wanted to claim this?

I think we should stop this discussion.

You bet.