## Connes & Marcolli paper on renormalization

> Chris Oakley wrote:
>
>
>>>No, I'm saying that you have no right to expect that the theory works to
>>>infinite energy when you've only tested it to ~1 TeV.

>>
>>I am not sure that I am following you here. Before renormalization QFT as
>>normally practised leads to infinite cross sections and decay rates. I have
>>seen it claimed that the infiniteness is because there is something going on
>>at extremely high energies that we do not fully understand that rescues the
>>situation at low-energy - the "ultraviolet completion".

>
>
> This is nonsense. Infinities arise because integrals taken over
> unbounded momenta don't exist; so doing it leads to nonsense.
> Instead, proper QFT takes regularized integrals, for example by
> adding an explicit cutoff $\Lambda$. This simply means that they
> calculate with an action that depends on $\Lambda$ as an additional
> parameter. Once this is done, everything is finite, but
> $\Lambda-dependent$.
>
> The only problem with that is that the cutoff destroys Lorentz
> covariance - apart from that it would be a completely respectable
> field theory in itself. Now Lorentz invariance is violated only
> at momentum $> O(\Lambda);$ hence to have the theory conform to
> physics that can be checked it suffices to take $\Lambda$ large.
> But for aesthetic reasons or since we believe that symmetries are
> fundamental, we want to have fully invariant theories. This requires
> that we let $\Lambda go$ to infinity.
>
> But in order that the results have a finite limit we must at the
> same time make the coupling constants g dependent on $\Lambda$.
> If this is done in a correct way (and the textbooks on QFT teach
> one or more of the known correct ways), one encounters no infinities
> at all in the whole process.
>
>
>
>>I see the Feynman-Dyson approach to QFT as failing to give sensible
>>or even finite answers to physical quantities, but I do not see this
>>phoenix-like rebirth after all the infinities have been conveniently wiped
>>out as anything more than just wishful thinking.

>
>
> The approach that gives infinities is the pre Feynman-Dyson approach,
> while what you call 'wishful thinking' is the advance that earned
> Feynaman, Schwinger and Tomonaga the Nobel prize. They showed how to
> _avoid_ the infinities completely, on the perturbative level, and the
> renormalization group explains why the result is still meaningful as
> a 'perturbative' approach, although at first sight, very large corrections
> seems to be involved.
>
>
> Arnold Neumaier

Your setting $\Lambda$ to infinity is fine. That's what Feynman et al. did,
and how they obtained a finite and accurate S-matrix. The problem with
this approach is that masses and coupling constants are infinite. In
other words, the Hamiltonian of the theory is infinite. In other words,
you cannot do time evolution. You may say that nobody needs time
evolution in high energy physics. And you will be partly correct: almost
all existing experiments are well described by the S-matrix, i.e., the
relationship between remote past and remote future.

However, I presume that every physicist believes that there is
some non-trivial time evolution of interacting physical systems
between the two time extremes. Such time evolution is evident
in macroscopic low-energy systems, for example.
QFT ignores that. That's the main
problem, why renormalization approach of Feynman, Tomonaga, and
Schwinger is not the final word in QFT.

Eugene
www.geocities.com/meopemuk

>



Chris Oakley wrote: > Let's just remember how we got to where we are: > > (i) We solved free field theory, including the expansion in terms of > annihilation and creation operators. > (ii) We introduced an interaction term in the Lagrangian density. > (iii) We introduced a unitary transformation that turns free fields into > interacting fields using the Interaction Picture. > (iv) We wrote out the Interaction Picture time evolution operator as > time-ordered products of the interaction. > (v) We reduced the time-ordered products to normal-ordered products. > (vi) We obtained thereby the S-matrix for scattering processes involving a > finite number of incoming particles with definite four-momentum scattering > to a finite number of outgoing particles with definite four-momenta as a sum > of Feynman diagrams. > > Many of these Feynman diagrams are pathologically divergent integrals. > > My response: This is because the theory is fundamentally flawed. Go back to > (ii) and try again. > I would like to suggest a different path (similar to what Weinberg is writing in his book): (i) We constructed the Fock space as direct sums of tensor products of 1-particle spaces (ii) We defined a non-interacting representation of the Poincare group there, including non-interacting Hamiltonian $H_0$ and boost operator $K_0$ (iii) We defined interaction terms in the Hamiltonian and boost operator $H= H_0+V, K = K_0+W$ (all operators are expressed through particle creation and annihilation operators, no fields) (iv) We found that S-matrix is ultraviolet divergent (v) We added renormalization (infinite) counterterms to the Hamiltonian $H' = H+R ,$ so as to make S-matrix finite and in agreement with experiment. This is where QED stands now: finite and perfectly accurate S-matrix obtained from infinite (wrong) Hamiltonian H'. My suggestion is (vi) Apply unitary dressing transformation U to the Hamiltonian H', so that 1) the new Hamiltonian H'' $= U H' U^{-1}$ is finite 2) The S-matrices obtained with H' and H'' are exactly the same. Now you can forget the nightmares of QFT, canonical quantization, Haags theorem, etc. You can use H'' in all usual quantum mechanical formulas (obtain time evolution operator, S-matrix, bound states via diagonalization, etc) without any need of regularization, renormalization, and other tricks. Eugene www.geocities.com/meopemuk



Haag's theorem does not apply since the theory with cutoff is not > Lorentz invariant. One cannot be so glib. In the only case of physical interest $(\Lambda =$ infinity) the theory $*must be*$ Lorentz invariant.



Chris Oakley wrote: >>Well, you can throw out the theory, but that seems a bit drastic. > > > I believe that one has no choice. One can in any case find holes in the > cutoff procedure. Supposing, for example, we were to choose $\Lambda$ as a > function rather than a constant. Furthermore, let us make it a function of > variables that never even entered the original equations. There is no reason > in principle why we should not be allowed to do this. We then very rapidly > get nonsense. You will then say that it was my own fault for not making it a > constant, but all that means is that you can find a path through the > minefield. It does not mean that the minefield has ceased to exist. > > There is no such thing as "reasonable" in theoretical physics, there is only > mathematical logic. The pillars of our current understanding of physics, > namely quantum mechanics and special relativity, both defy common sense. One > becomes aware of this very quickly when one tries to explain them to an > intelligent non-physicist. You and Dr. Neumaier say that it is > "unreasonable" to have four-momentum integrals extending to infinity, > leading you to want to cut them off. I say that if the mathematical logic > leads one there, then one has no choice other that to accept it, and if this > leads to nonsense then one has to accept that that something is > fundamentally wrong with the approach. > > Let me be clear on this - I could not do any better. All my attempts to find > a rigorous way of doing Feynman-Dyson perturbation theory failed. This puts > me in good company with Dirac, Pauli, Feynman, the axiomatic field theorists > and many others. But in the end, I just decided that it was not worth > rescuing, tore everything up and started again. And - intermittently, at > least $- I am$ still working on that. > > Hi Chris, I share your frustration with renormalized QFT. I think I know the way to fix the problem of infinities. The idea is very simple: QED simply uses a wrong Hamiltonian. There is a rigorous way how to find a new Hamiltonian of QED (while preserving the relativistic invariance and cluster separability) which is 1. finite 2. yields the same accurate S-matrix as renormalized QED. (so it has the same predictions as far as experiment is concerned) The details are given in E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys. 292 (2001), 139 and in online book available at www.meopemuk.com I would appreciate any comments and criticism of this approach to QED. If you don't think it completely solves the problem of ultraviolet infinities, I would like to know why. More papers can be found at www.geocities.com/meopemuk Regards. Eugene.



In all the references I have the arguments rely on the interaction being a > > local product of fields. I do not know what one can do to make the theory > > work when this is not the case. > > One simply gets more complicated Feynman rules. It is not very difficult > to generalize them to arbitrary nonlocal interactions. In momentum space, > the formulas become the standard formulas, but with explicit cutoff > included. I have never, *ever* seen this. Can you give me a reference? I have only seen Feynman rules for *local* field equations being used and then the integrals being cut off *after* we get the expressions for the loops and have not liked what we saw. > Thus to do the suggested exercise, you should always work in the momentum > representation, and just apply the arguments of the textbooks in a sensible > way to the regularized situation. I am familiar with the momentum representation (which I use more or less exclusively in my own work), but cannot see how to adapt the arguments of Feynman-Dyson perturbation theory using this to achieve the required result. Please enlighten me.



Smearing an interaction in time just amounts to using a different > time-dependent potential, constructed at time t from the original > time-dependent potential (in the interaction representation) at all other times. > This is best seen by using the representation of the S-matrix > as time-ordered exponential. This holds for arbitrary time-dependent > nonlocal potentials, and one can go back from any nonlocal 4D interaction > to a time-dependent potential formulation. Thus smearing in time is not a > real problem: One starts with the smeared interaction defined by the > cutoff, uses the time-ordered exponential to work out the corresponding > Hamiltonian interaction, and proceeds from there. This may be obvious to you, but it is not obvious to me, I am afraid, so please spell it out. > > In all the references I have the arguments rely on the interaction being a > > local product of fields. > > This is not even the case in nonrelativistic field theories: > Coulomb interactions are not describable by local products of fields. I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1, chapter 6. The arguments assume that (e.g.$) H(x_1),$ etc. in equation 6.1.1 is a local product of fields defined at the particular spacetime point. He is already considering this interaction Hamiltonian density H in the Interaction Picture. My point is that one can only transform a product of fields to the I.P. in the naive way if they are all for the same time, so one cannot get to this equation if H is a smearing that is non-local in time.



Eugene Stefanovich wrote: > Chris Oakley wrote: > >>Let's just remember how we got to where we are: >> >>(i) We solved free field theory, including the expansion in terms of >>annihilation and creation operators. >>(ii) We introduced an interaction term in the Lagrangian density. >>(iii) We introduced a unitary transformation that turns free fields into >>interacting fields using the Interaction Picture. >>(iv) We wrote out the Interaction Picture time evolution operator as >>time-ordered products of the interaction. >>(v) We reduced the time-ordered products to normal-ordered products. >>(vi) We obtained thereby the S-matrix for scattering processes involving a >>finite number of incoming particles with definite four-momentum scattering >>to a finite number of outgoing particles with definite four-momenta as a sum >>of Feynman diagrams. >> >>Many of these Feynman diagrams are pathologically divergent integrals. >> >>My response: This is because the theory is fundamentally flawed. Go back to >>(ii) and try again. >> > > > I would like to suggest a different path (similar to what Weinberg is > writing in his book): > > (i) We constructed the Fock space as direct sums of tensor products of > 1-particle spaces > (ii) We defined a non-interacting representation of the Poincare group > there, including non-interacting Hamiltonian $H_0$ and > boost operator $K_0$ > (iii) We defined interaction terms in the Hamiltonian and boost > operator $H= H_0+V, K = K_0+W$ (all operators are expressed through > particle creation and annihilation operators, no fields) > (iv) We found that S-matrix is ultraviolet divergent > (v) We added renormalization (infinite) counterterms to the > Hamiltonian $H' = H+R ,$ so as to > make S-matrix finite and in agreement with experiment. > > This is where QED stands now: finite and perfectly accurate S-matrix > obtained from infinite (wrong) Hamiltonian H'. My suggestion is > > (vi) Apply unitary dressing transformation U to the Hamiltonian H', > so that > > 1) the new Hamiltonian > > H'' $= U H' U^{-1}$ > > is finite > > 2) The S-matrices obtained with H' and H'' are exactly the same. > > Now you can forget the nightmares of QFT, canonical quantization, > Haags theorem, etc. You can use H'' in all usual quantum mechanical > formulas (obtain time evolution operator, S-matrix, bound states via > diagonalization, etc) without any need of regularization, > renormalization, and other tricks. Well, this is a research program - the question is whether you can actually carry it out. How do you construct U? It is there that all the renormalization issues will recur. Since H' is an ill-defined object one must regularize it, then construct U for the regularized theory, and then go to the limit and show that it really exists. This poses the same problems as the original renormalization program. Arnold Neumaier



I share your frustration with renormalized QFT. I think I know the way > to fix the problem of infinities. The idea is very simple: QED simply > uses a wrong Hamiltonian. There is a rigorous way how to find a new > Hamiltonian of QED (while preserving the relativistic invariance and > cluster separability) which is > > 1. finite > 2. yields the same accurate S-matrix as renormalized QED. > (so it has the same predictions as far as experiment is concerned) > > The details are given in > > E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys. > 292 (2001), 139 > > and in online book available at www.meopemuk.com > > I would appreciate any comments and criticism of this approach to QED. > If you don't think it completely solves the problem of ultraviolet > infinities, I would like to know why. > > More papers can be found at www.geocities.com/meopemuk Eugene, As you may remember we met and talked in California three years ago. It is nice (and rare) to see someone who has bothered to figure everything out for themselves. I don't claim to understand every detail of your work, but I do take Haag's theorem seriously and therefore don't believe that the U, W and F operators in section 2 of your 2001 Ann. Phys. paper exist. Not in a fully relativistic theory, anyway. Also I notice that the regularization scheme is still exactly that ... i.e. you get the infinities and then devise a scheme (in your case a "clothing" operator) for dealing with them. This is still subtracting infinity from infinity, and I really don't think that people ought to do that. Chris



Chris Oakley wrote: >>Haag's theorem does not apply since the theory with cutoff is not >>Lorentz invariant. > > > One cannot be so glib. In the only case of physical interest $(\Lambda =$ > infinity) the theory $*must be*$ Lorentz invariant. We only know that the universe is governed by Lorentz invariance to a certain accuracy, which means that the theories with finite but large cutoff (significantly larger that the largest energy we can create or measure) already are in full agreement with nature. Thus $\Lambda=inf$ is not needed for results of physical interest. Arnold Neumaier



Chris Oakley wrote: > I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1, > chapter 6. The arguments assume that (e.g.$) H(x_1),$ etc. in equation 6.1.1 > is a local product of fields defined at the particular spacetime point. He > is already considering this interaction Hamiltonian density H in the > Interaction Picture. My point is that one can only transform a product of > fields to the I.P. in the naive way if they are all for the same time, so > one cannot get to this equation if H is a smearing that is non-local in > time. Rewrite in each of Weinberg's formulas the fields in x as integral over fields in p. This gives you completely equivalent formulations, now only involving momenta. Once you have done that, it is very easy to take care of a cutoff. Arnold Neumaier



Chris Oakley wrote: >>>In all the references I have the arguments rely on the interaction being a >>>local product of fields. I do not know what one can do to make the theory >>>work when this is not the case. >> >>One simply gets more complicated Feynman rules. It is not very difficult >>to generalize them to arbitrary nonlocal interactions. In momentum space, >>the formulas become the standard formulas, but with explicit cutoff >>included. > > I have never, *ever* seen this. Can you give me a reference? No; I don't know of any. I had to work it out for myself; it was a hard road to gain this understanding. I gave you instead an exercise. Physicists are too lazy to do things as carefully as you wish, so you have to do it yourself. I gave you the outline, so that you can fill in the details. > I am familiar with the momentum representation (which I use more or less > exclusively in my own work), but cannot see how to adapt the arguments of > Feynman-Dyson perturbation theory using this to achieve the required result. > Please enlighten me. Please start doing it and you'll see that everything can be done. Begin by writing the interaction in the momentum representation as a multiple integral, add the cutoff, and then start copying in spirit each line of Peskin & Schroeder or your favorite textbook, adapting it to the new situation at hand. You can hardly go wrong, but of course it takes some effort to get familiar with the nonlocal case. It is very hard writing out pages of details in ascii, and I have no intention to do so. Should you ever want to visit Vienna, I'd show you on the blackboard what to do. Arnold Neumaier



Eugene Stefanovich wrote: > Arnold Neumaier wrote: >>But in order that the results have a finite limit we must at the >>same time make the coupling constants g dependent on $\Lambda$. >>If this is done in a correct way (and the textbooks on QFT teach >>one or more of the known correct ways), one encounters no infinities >>at all in the whole process. >> > Your setting $\Lambda$ to infinity is fine. That's what Feynman et al. did, > and how they obtained a finite and accurate S-matrix. They don't set it to infinity; they keep it finite and take the limit at the end. > The problem with > this approach is that masses and coupling constants are infinite. The bare masses and bare coupling constants are finite but depend on $\Lambda$. In the limit they would diverge. But one never needs to take limits on bare objects. The renormalized masses and coupling constants are finite. > However, I presume that every physicist believes that there is > some non-trivial time evolution of interacting physical systems > between the two time extremes. Such time evolution is evident > in macroscopic low-energy systems, for example. > QFT ignores that. That's the main > problem, why renormalization approach of Feynman, Tomonaga, and > Schwinger is not the final word in QFT. One, not necessarily the main one. The fact that the S-matrix exists only as a formal series instead of as a unitary operator on some Hilbert space is another important reason. Arnold



On Thu, 21 Oct 2004 07:47:32 $+0000,$ Chris Oakley wrote: > >> > In all the references I have the arguments rely on the interaction >> > being > a >> > local product of fields. I do not know what one can do to make the > theory >> > work when this is not the case. >> >> One simply gets more complicated Feynman rules. It is not very difficult >> to generalize them to arbitrary nonlocal interactions. In momentum >> space, the formulas become the standard formulas, but with explicit >> cutoff included. > > I have never, *ever* seen this. Can you give me a reference? I have only > seen Feynman rules for *local* field equations being used and then the > integrals being cut off *after* we get the expressions for the loops and > have not liked what we saw. One does not have to go far to see this. As Arnold mentioned in a previous post, the coulomb interaction in non-relativistic many body systems is a standard example. When drawing diagrams in momentum space instead of a point interaction vertex we an internal line which represents the interaction. However, the internal line does not represent a photon as we would expect in full QED, but simply the Coulomb potential. The electron vertices are allowed to have distinct positions and the Coulomb potential depends on their difference. Since the Coulomb interaction is translation invariant, there is still a nice picture of the interaction in momentum space: the internal interaction line represents a momentum transfer. All the necessary Feynman rules are worked out for both position and momentum space in _Methods of QFT in Statistical Physics_ by Abrikosov et al. Even the cases when the interaction is not translation invariant are treated. Igor.



However, I presume that every physicist believes that there is > some non-trivial time evolution of interacting physical systems > between the two time extremes. Such time evolution is evident > in macroscopic low-energy systems, for example. > QFT ignores that. That's the main > problem, why renormalization approach of Feynman, Tomonaga, and > Schwinger is not the final word in QFT. It is interesting that undergraduate quantum mechanics does better than QFT for processes that involve a (finite) time variable. QFT ought to be an extension of this, not an alternative.



>>>In all the references I have the arguments rely on the interaction being a > >>>local product of fields. I do not know what one can do to make the theory > >>>work when this is not the case. > >> > >>One simply gets more complicated Feynman rules. It is not very difficult > >>to generalize them to arbitrary nonlocal interactions. In momentum space, > >>the formulas become the standard formulas, but with explicit cutoff > >>included. > > > > I have never, *ever* seen this. Can you give me a reference? > > No; I don't know of any. I had to work it out for myself; it was a hard > road to gain this understanding. > > I gave you instead an exercise. Physicists are too lazy to do things as > carefully as you wish, so you have to do it yourself. I gave you the > outline, so that you can fill in the details. I have not done it because I do not believe that it can be done. Products of field operators in the interaction Hamiltonian density must refer to the same time if we are to be able to transform to the Interaction Picture. I might just add that I am not the first person to be bothered by the sloppy mathematics in QFT. People have been working on this, on and off, since 1928. If there was a way of making the procedure rigorous, it would have found its way into the text books by now. However the best QFT text book *I* know of - Weinberg - makes no mention of the possibility of defeating infinite subtractions via non-local field equations. Neither does the supposedly more rigorous Scharf. > Should you ever want to visit Vienna, I'd show you > on the blackboard what to do. A kind offer, but be careful - I might take you up on it.



I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1, > > chapter 6. The arguments assume that (e.g.$) H(x_1),$ etc. in equation 6.1.1 > > is a local product of fields defined at the particular spacetime point. He > > is already considering this interaction Hamiltonian density H in the > > Interaction Picture. My point is that one can only transform a product of > > fields to the I.P. in the naive way if they are all for the same time, so > > one cannot get to this equation if H is a smearing that is non-local in > > time. > > Rewrite in each of Weinberg's formulas the fields in x as integral > over fields in p. This gives you completely equivalent formulations, > now only involving momenta. Once you have done that, it is very easy > to take care of a cutoff. You are missing my point. Limiting the integral in $p^0$ space results in a smearing in the time integral. This is just basic Fourier analysis. It is the latter which prevents a transformation between the Heisenberg and Interaction pictures for the field operators.



Chris Oakley wrote: >>However, I presume that every physicist believes that there is >>some non-trivial time evolution of interacting physical systems >>between the two time extremes. Such time evolution is evident >>in macroscopic low-energy systems, for example. >>QFT ignores that. That's the main >>problem, why renormalization approach of Feynman, Tomonaga, and >>Schwinger is not the final word in QFT. > > > It is interesting that undergraduate quantum mechanics does better than QFT > for processes that involve a (finite) time variable. QFT ought to be an > extension of this, not an alternative. > That's exactly right, and my approach (with finite well-defined Hamiltonian) makes time evolution in QFT not more difficult that in non-relativistic undergraduate QM. The only difference is that 1) relativity is respected; 2) creation and annihilation of particles is allowed. Eugene www.geocities.com/meopemuk