 Thank you for the invitation and we're really happy to talk at this conference in the honor of Dirk Who I met actually about 30 years ago When he was still a PhD student of Jürgen Körner and Mainz At the time we had a joint interest in the gamma-5 problem of dimensional equalization Oh I was visiting Jürgen Körner. It's a time Yeah Yeah, I met Dirk 30 years ago and At those days in fact From what I remember Nobody thought that Feynman diagrams were something very interesting mathematically it was something very tedious and It had to be done, but nobody thought that there was any interesting mathematics in it now 30 years later Feynman diagrams have turned into something like a gold mine for pure mathematicians and I think everybody here agrees that Dirk has a lot of credit in this So today I'm going to talk about Not really about a lot of algebraic methods, but about my own specialty, which is world-line integration So for those who don't Anyway, a few words of history for those who Don't know about the world-line formalism The world-line formalism is an alternative to Feynman diagrams and it's as old as Feynman diagrams themselves Because Feynman invented some Feynman invented work on passing the girls at the same time When he invented all the Feynman diagrams in fact that seems that he used them as a guiding principle for finding Feynman diagrams and Seven year Feynman diagrams. He was so happy with some city basically forgot about all those Relativistic passing the girls So let's start with scalar QED and the sequence function and X space was a client-garden operator We have to work in Minkowski in the completely in space and Many people know probably is that You can Exponentiates a proper gator and then convert it into a quantum mechanical pass integral Which is Maybe just you could try if you go to the menu bar, maybe in the view you can change to a single page View if you go With your mouse to the I'm not familiar with tech shop if you just open it with You can use preview display format In the tech shop menu Go to preview you Display format Okay And then full screen No, let me go full screen like this. Is it okay? Yeah, I think that's better actually that way then Yeah, I think that's that Let's go go back to a finance work in 1950 in modern notation he constructs a Pass integral represent representing the proper gator of a scalar particle in Euclidean X space going from X prime to X and overtime T and one has to integrate over all Pass connecting experiment X in Euclidean space time and That's what nowadays called Photon rest over Gator and it's a diamond diagram that is down here But it will be important for the following set the photons are not ordered here Photons have a fixed momentum a fixed number But it is not fixed in which order say up to the arrive emitted Yeah, but here we This could either be the propagator and an external field or you can convert it into the amplitude This and photons specializing the field to plane waves If you do the same thing for the one loop effective action you start with a trace lock of the Klein Gordon operator and Then instead of pass that go from X prime to X and space time You have to integrate over all closed pass and space time and again if you leave the external field General then it's effective action You expand the field in plane waves, so then you get see one loop and photon amplitudes and Now the structure of QED is so simple said it's quite obvious that just from these two elements Connecting photons in pairs you will be able to construct the complete QED as matrix and since QED doesn't really have non-trivial symmetry factors There's also no problem this this overcounting which would actually be an issue in five five to four theory and Feynman himself one year later generalize this to a spinner QED So he found what is called a Feynman spin factor So you start with scalar QED and then for a fixed pass you insert Along the pass you insert that spin factor here you calculate its trace and the d to the contribution of spin and after the invention of grass man pass integrals Fratkin 25 years later found the modern way of Presenting the spin factor in terms of a grassman pass Why do we prefer this for two reasons first? You don't have to pass order. Yeah, that would be very important You don't have to fix an ordering of selects along the loop and Moreover, it makes a super symmetry money test between the orbital and spin degrees of freedom which Quantum mechanics we have already there's a Pauli equation But in the standard drug approach, I can look at some baton and quantum field theory now a These buzz integrals we are not much used for calculations until the late 80s Then people had learned in string theory how to some methods of calculating buzz integrals grassman integrals usual G2 symmetry Etc. So Oh Yeah, of course overstress It's a box that eventually led to the idea that one should calculate such pass integrals like in a one-dimensional innovation theory that means Orbital pass integral should be calculated using a greens function said is for the ordinary second derivative, but adapted to the proper time circumference that fixes the periodicity and The psi bars in the third percent of spin factor should be calculated using a greens function such just the signal function and Effectively contains all the manipulations that normally you do this the drug matrices so in a way so psi Functions are like the drug matrices, but they are kind of flexible and not bound to appear in any fixed order We look at it nowadays now in Going back to skill at you a dean if you want to do this Explicitly you go back to the one loop effective action You put the field in terms of playing waves you expand You take out the terms that contain every of the improvisations and momentum Once the door would be called the total mix term That was the logic way of doing through the transformation And then you get the one loop and photon amplitude in a representation where each photon is represented by a vertex operator and Is integrated along the trajectory And it's very important that at one loop you have Here a problem said before doing your passive the curl you must actually fix the zero mode Which is the zero mode is Tells you that you could actually Change a loop just by translating it without changing the work on action So that would be a flat direction for your Gaussian passing the curl. So you have to fix it and said Eventually, they just produce energy momentum conservation and then to Actually, well at this stage your passing the grill is already Gaussian. Yeah, of course the orbital X of tau coordinate appears only linearly and exponentially Together the closed formula you formula you do this a little exponentiation tricks. It's also well known from string theory And then you do the Gaussian integration formally In flat space, you can do that formally you never have the problems and pop space. It's a bit more tricky You'll get what is called a von Kosovo master formula So you have seven loop and Photon amplitude and scalar QED written in a very very you have Here's a global proper time of the particle the scalar particle in the loop You have still one integration for each photon also one of them is redundant and Then you have this exponential Which is kind of fake because eventually these guys will have to be expanded out and we have to take the terms linear And also in each polarization But the important thing is that it contains only these Worldline dreams as well and dreams function G and it's the relatives and And yeah, those who haven't seen this before should keep it should take notes that Yes, the data function appearing here. Now, this is a very nice formula because it's a One-line formula having the full information on the one look at photon amplitude It's valid off-shell that means you can use it for going to higher loop orders by sewing Very important to write it down. We did not have to fix the ordering of the photons along the loop But in Kosovo found also sets this formula as information on the cases where the scalar Becomes a spinner or even a glue on sees as a famous one Kosovo rules and And Strassler actually investigated more the case of QED itself and Study what happens when you actually remove all the cortic vertices. Yes Yes, these data functions that create seagull vertices, which Obviously we have in scalar QED and In a spinner QED normally we don't have some but here we have some because effectively we are using a second-dollar formula soon but in any case you can remove them and Set after sets the integrant becomes actually much nicer Which is also kind of familiar nowadays from things like K K LT or double copy many many nice things nice nice algorithms start working only When you remove all quarters cortic or higher vertices and everything becomes three volume Let's look at the four-fold room case. He's a scalar spinner QED Make very little difference here. So normally we would Like it's been a QED The sum of six promoted fine-man diagrams and scalar QD some more of this the seagull vertices In In the western formalism if we Just use the master formula in a straightforward way We would in fact be doing more less the same as doing on ordinary fine-man diagram Calculation using Schringer parameters It would be the same amount of algebraic work. It's the same type of integral the same time to reduction So that wouldn't be a very interesting thing to do The formalism becomes interesting only when you use the freedom To actually do something with these integrals and what you can do is You can massage them by a large number of integration by parts and Then they actually reduce to five Tensors structures which Actually are already gauge invariant. So here we introduce a field strength tensor for each of the four photons and Here you have the traces of products of the strength tensors and then here you have some chains of the strength tensors and It is very curious that exactly this basis of five tensors was found in 1971 by Costandini et al Using security water temperature. So they started writing down the 130 a tensors which you can write down at four point and then systematically reduced using the QD water and the Promotation of variance. It is very curious that We just by integration by parts generated exactly the same basis just trying to Just trying to Why it's the integrant as compact as possible. So in this space is The Coefficient function for for spinner QED are these five here You see the involve In the exponent you still have the green's function G and the pre-factor you have only the second to the first derivative you don't have the second derivative anymore and More over the parts that comes from the spin factor Always appears in sees a combination. Yeah, you're always close cycles of indices and then a difference of Orbital term and the spin term and that is actually related to the water and super symmetry and Was first derived the Brown and Cosser one and This Combined the cases of spinner and scalar If you want to go to scalar QED, you just Delete all the terms that are here So this is actually a recent result previously we did not have such a Such a compact version and we also didn't have said reduction of the tensor basis This is what we have in progress We use this for the act for the first calculation of support photon amplitudes Totally off shell, which was never done. In fact said work by Costardini at all There's a calculate this two legs off shell and two on shell So we do certain general kinematics as well as there's one or two photons taken in the low energy limit and Also Not only is integration, but also the tensor reduction is actually Is a kind of Really adjusted to the world and formalism Our guiding principle is we never want to really split the amplitude in the order sectors so it We do we cannot use tensor reduction algorithms because all of them work only once you have fixed the ordering of the external legs So we have been better than that That actually works without fixing them So why what is the motivation for this? Well, this is not really the workshop to talk about this, but for example In G-2 calculations, there's no going to six and seven loops with the right by light sub diagrams Figure very prominently. They are important here and having Nice simple formulas for them, which do not Make it necessary to fix the ordering of selects It could be quite important Moreover, you can also use one of the photon amplitudes of course to go to a higher orders in the photon propagator like Many years ago already be constructed This Michael Schmidt be constructed to loop diagrams What normally would be these three diagrams and the water and formalism and Managed to calculate the two loop either function without encountering any non-trivial integrals and actually also without Having to split them to see two different apologies Also, we never managed to carry this on to the three-year-old case. We still would like to do that In fact, then We all met it's a multi-loop workshop in Espen 1995 that was there with protest to Turkey and under the video Jeff In fact is a three loop and higher loop photon propagator and its beta function was Very big topic. It's a time What had been calculated is the follow beta function and spin off you a D and David just had calculated the three loop either function scalar QED and In both scalar and spin off QED it's a weapon set at three loop See the three contributions that cancel out and At four loop in the spin off QD case There also see the fives that cancelled out So it's a time everybody was convinced that the QED beta function it's a quench level was rational and David the game of the logo even constructed kind of a roof but Years later actually circles rule group calculated the five-loop coefficient and it there was some see the trees that just refused to go away and So we know the cancellations are not complete but that's still I Think this is a puzzle why it's they happen in the first place is still Is still Unresolved and We're thinking about In fact just this week I was in contact with Jonathan Rosner Who had actually calculated the three-loop QED beta function and he told me that he was still very much like to know Why see the three cancels out Now why is it why is it difficult to apply is about one formalism. This is kind of problem That is because the advantage of having all the diagrams in one big integral is unfortunately Some but Let's say formal As long as you don't know how to calculate integrals that have absolute values in the exponent Yeah, don't try to give mathematical integrals which have absolute values in in the exponent for example Computers are not good at this kind of the sink. Yeah, and it's a pre fact. Yeah, so to begin the fundamental integral that we have to calculate is in at least in a billion theories in the modern formalism is An endpoint integral where all end end points are running over the full circle You have this is what we call the universal exponential linear and the greens functions between the lines points, which is lumped of square is usually At one loop it's P i dot P j it higher loops it can become something more complicated the pre factor will be a Pre factor will be a polynomial in the G dots cost derivative of the greens function, which has a signals function So the basic challenge is how do you calculate integrals like this Without actually splitting them into order sectors or making case distinctions for the remaining points usually you would just Start integrating over one variable Fixing the others, but even send or evaluate Evaluate the signal functions. You would have to fix another thing. So We know that once once you start doing that you lose all the advantage so we don't want to do that so What would a master magician do he would start with simple classes of in integrals start with polynomials and then go to Exponentials, etc the symbols non-trivial integrals are what we call a chain integrals and here you see if you integrate Product of G dots. We are seeing that indices for a chain. Then you Then you get a Bernoulli polynomial to the same as a GF's you get an Euler polynomial They more about this in a moment Yeah, you have you can have a totally an integral that is actually totally trivial in a in a given sector like this one But no computer will actually give you this nice formula Because he will split it in the bias sectors and he will never be able to To recombine the result synthesis For what's in sense, there's also issues that's from at three point already Is the bias G dots are not algebraically independent anymore? so generally It is also possible to write down polynomials and G dots and G That are actually zero, but in each other sector for a different reason Over Miller has studied the algebra of these polynomials now some years ago I Solved the general polynomial problem in in a recursive way by Foundings by deriving this formula here And also then when you just want to integrate out one Viable over the circle So you calculate the integral u with any ordering of the remaining u1 to u1 and You have G dots to arbitrary powers Then this formula tells you how to remove set how to do this one integral and write the result again in terms of The G dot functions of the remaining points as a polynomial So applying this record recursively Obviously, you can calculate you can calculate any integral you want if it's just polynomial Now why do we see Balooly Bernoulli numbers and Bernoulli polynomials here? Well, that is because Remember we started with passing decals said go over periodic functions and Then we had to fix a zero mode and that takes out the constant functions So we are naturally now in the Hilbert space of periodic functions or soccer mode as a constant functions Actually being also gone out as a constant function makes a big difference because it makes the ordinary and derivative invertible and See ends Yeah, see the inverse of the ends derivative in the space of periodic functions without the zero mode is essentially a Bernoulli polynomial except we have some signum in the odd case and The zeros power is special because it's not the delta function But delta minus one because of the zero mode subtraction. That's what represents the unit operator now It's curious that this formula as it stands. In fact You can never find in any book on functional analysis or integral operators many books you can find The formula for the for the Fourier series of the Bernoulli polynomials Which is actually equivalent to this formula If you assume that x is positive here But it misses all the fun stuff that x equals zero the signal and the delta that actually give you a nice closed Algebra of integral operators on the circle. Yeah for that reason in this formalism Bernoulli numbers Bernoulli polynomials appear all over the place for example in years ago Our previous speaker Gerald we were calculating the two loop self dual or Heisenberg Lagrangian and from said we got We got Yeah from said we got corresponding photo amplitudes in the Low energy limit with all equal helicities You see is there are some Yeah, at one loop this amplitude would be would essentially be a Bernoulli number There's some stuff here which is it which takes care of the Momenta and Polizations, it's just spin electricity stuff But the coefficients are just the newly numbers and the two loop you start seeing folded sums of the newly numbers And that's a story I want to tell because it's somewhat involved because I was just visiting the eye HES when Then Yeah, we were doing this and actually We had actually two formulas Power enforces to look effective Lagrangian Which both involves Bernoulli numbers, but they actually look very different so I asked Who can help and See whether and tell us whether this is already known. This is a known identity Which allows us to identify as these two formulas And he said that only Richard Stanley could help with this So we wrote a letter to Richard Stanley and after a couple of weeks he indeed told us said this is equivalence between These two versions of the Expansion coefficients of the two localized by Lagrangians can be actually are equivalent to Oilers identity to combining oilers identity for the Bernoulli number, which is very well known This this most with Mickey's identity, which at that time was considered the most non-trivial Identity known known for Bernoulli numbers Yeah, so we learned from Richard Stanley said actually The Bernoulli number identities that involve only bn over n factorial are usually easy Because most likely you can get them from the generating function But identities that involve bn over n are considered difficult Mickey's identity involves actually both and Was considered very mysterious at the time And actually it turned out at about the same time Farber and Pantary Pante published a similar formula Which they found in a string theory calculation And couldn't prove actually the guests that they couldn't prove it since they asked Zagia And Zagia really managed to prove it and send this after getting this information by Richard Stanley and Yeah, we send this Gerald we send Figured out how to use a word and formalism as a guiding principle to not only give a unified proof of both identities Mickey and Papa under the bandit Zagia and Also, we generalize them in various ways. It's a quadratic level. We generalize them to the cubic level and We in fact pointed out a systematic construction of such Identities to arbitrary orders only said as far as I know So far nobody has really done this That's not something we can our students can ask our students to do Unfortunately Anyway, coming back to my real topic after settling's Problem of integrating an arbitrary polynomials now we would like to attack the problem of including the exponential factors, which is much more difficult and Here it makes sense to go back to the symbols case so the scalar case so we have the pure exponential factors no pre-factors This formula was apparently written down for the first time in Poliakoff's book and Is that means we are now talking about the same off-shell scalar functions, which for example appeared in Andrea the video Jeff's talk Now remembers that on shell that stuff is easy in our days, but off-shell still not and In fact, not many people even want to work on such things like And really Egypt or dream I'm only the rest of actually all have worked on On the three and four point functions and see all agrees that in the three point case You need to have a geometric functions to f1 and the appeal function f1 in the four point case You need traditionally Allow read what is called a lauricella Saran function So at least one knows in principle what one is looking for. Yes, this these forms This looks deceptively simple Is it the same formula can also be written down for the effective action? in that case in In fact, it becomes more general because for the effective actions There's actually no reason to assume that it's five cube. It could be as well any self in the action so here's a close formula for the effective action and Of course here normally people don't talk about the average functions. What they do is they just expand and powers of these derivatives and Then they collect terms of equal mass dimension and That gives you see it from the expansion Still under the T integral that you can do doing the T integral does make any sense because it just creates Inverse powers of the masses and more over would start depending on the space time dimension But these operators are not so if you Would want to do a very brute force approach to calculating the effective action in five cube Theory you would now just expand all the exponentials and You would have this integral here This all possible powers of all possible greens function And again, this is an integral that is any particular such integral is totally trivial, of course But to obtain a closed form expression for arbitrary n and arbitrary exponents is a quite formidable challenge It's only reason these that we have made progress in fact towards Doing this namely See the essential idea is actually to take each exponential factor And not directly expanded in powers of the greens function, but the rasa in inverse powers of derivatives So if you take an X this is exponential and you ask yourself How would the what coefficients do I get if I actually expanded In the matrix elements of the of the inverse derivatives on the circle You you get that there is the following formula which involves The two ends Bernoulli polynomial minus its coincidence limit with its which is the corresponding Bernoulli number and then it involves an odd Hermite polynomial and Yeah, you can integrate this formula from zero to one and you get for a formula for the ever function which actually expands it in terms of Bernoulli numbers and Hermit polynomials and Yeah, if you see this you Would assume that it must be known to mathematicians, but so far we have not been able to find it in the literature So if anybody in the audience has seen such formula, I would be happy to know about it So now let's see how to use this formula in the three-point case So we in the three-point case we have three exponential factors on each your place is expansion and now it We have to remember that the space we are working on is such that You're always orthogonal to constants. Yes, that means also that These matrix elements that involves Bernoulli numbers actually Become zero when integrated on either side that means we Can say is that if I want to use this factor here I must also use this one and use this one and form a closed cycle So as not to have any That's a loose end that would Indicate a zero So if we look at the most non-trivial case where that we take the product of C3 then you get this integral and then you use a Completed relation, yeah to convert this into a trace. That's a trace of Inverse derivatives in the space is against is against is again a Bernoulli number so without Any real work what you get is a closed formula for what before we called I3 this arbitrary Exponents ABC Where you now have C's finite sums up here You have coefficients H that come from the Hermit polynomials and Here you have the newly numbers And Yes, yeah, we can essentially see already the general structure that you're going to get Only is that it's starting from four points We will not Immediately be able to do all the integrals using completeness Because at four point you have will have integrals There is a variable which you want to integrate actually appears in three But in three factors, and that's what we call a cubic workland vertex But now it actually paid off said many years ago actually this uberbüller the constructed Precisely what this is like this when we actually try to build Yeah, we once built a Kind of a toy world and quantum field theory Where an arbitrary multiple setter value can be encoded in an assignment diagram and an arbitrary Identity for between setter multiple setter values Can be derived by performing integration by parts on those Feynman diagrams because It's exactly the same procedure that we need here Remember that Delta to the power zero gives a delta function minus a constant So what we do is we do integration by parts lowering One indices and raising the others until one index becomes zero that happens here When that index becomes zero Then you have a better function that kills the integral or you have Minus one that doesn't kill the integral, but at least it removes that factor and sends the remaining factors Already have involves integration variable Only twice so then you can actually do the integration like you did in the two-point case So it becomes Very simple combinatorial problem so we can actually foresee that We should be able to get closed formulas for these coefficients in Not only for three or four, but probably for five or six point And then of course There is the question how do we get back to the standard description in terms of hydrogemetric functions Like these coefficients must must must be summed over and they must get known by the geometric functions But there are not no formulas known for hyper geometric functions that involve Bernoulli numbers and so coefficients as far as I know Some unknown identities to do that here and also previous experience this However with Gerald and others has shown that having Bernoulli numbers It's always nice also when you want asymptotic estimates because the Bernoulli numbers are Related to C of 2n which rapidly converges to one so you have a very rapid convergence of the Bernoulli numbers To this simple factor here so it might be interesting to getting the Large and asymptotic of C1 to n point functions. Yeah, and then of course you would like to Go back to gauge theory puts a pre-factor polynomial polynomial And eventually back to multi-loop All right, that's all Happy birthday Dirk Thank you, Christian If you have questions, please feel free to unmute yourselves and raise your hand I was just wondering that the expansion show at the very end with this mid-polynomials and you said that you did this in the phi cubed scalar case Yes, these coefficients Appear when you take standard one loop end point There come. Yeah Yeah So or Equivalently see the active action But that was in the scalar theory, right? In the scalar theory and any gauge theory you will have some pre-factors involving G dots But this idea of this kind of expansion. Is it is it useful in any way also in the QED case or does it just not apply? Well, I'm Mostly interested in the QED case. Yeah but I Yeah, but we have to go step by step In fact In 90 in 1990s with Michael Schmidt and Danny Flegner We wasted a lot of time doing multi-loop calculations that never worked out Because He didn't understand that To calculate these integrals you have to develop your own message. Yeah You shouldn't try to use too much of the known stuff Yeah, but seems like you're making quite some progress, so I'm excited to see What's coming up? We have a question from David Well, it's a comment really it's to thank you for Reminding us of the work that dick did with Bob Dahl, Borgo and me in 1995 now They in the chat room. We had a wonderful tribute from Bob to dick for a moving. I think this is an opportunity for us To me to give our appreciation to Bob and you put your finger exactly on the place where Bob Dahl, Borgo's wisdom showed when you said that we also not only did three and four of calculations and Spinner QED, but in scalar QED Where you would think there will be extra diagrams from seagulls But Bob told us no Well, I was all set to draw 21 extra diagrams and he said no You can represent your scalar by a spinner Using the Duffin gamma Petyar formula on all I had to do was to take my code and just change And just change the propagator of the spinner. So that's a wonderful example. And that was How you really benefit when you have a senior calamarita Actually, I was expecting you would talk about this tomorrow in your Tasmanian adventures I've not chosen to which is why I've Mentioned it now, but it was wonderful to have a very wise Older collaborator who was still continued to be productive and focused and clear-minded Well above my modest age of 73. So dick has a longer period of intellectual activity to look forward to if he can manage as well as Bob has done Thank you for sharing this May I have a question? Sure Hi, Christian. Thanks for a good presentation. So I Actually, it's not a question but a comment. So I think I have the Formula you were looking for for these error function in terms of Hermit polynomials Yes Yes, and if you allow me to share my screen, so I will show you the paper There's a formula in terms of Hermit polynomials, but I haven't seen one involving Bernoulli Yes, it's actually in journal of chemical physics Oh 1999 so I can send you this paper, but I can show also to everyone You can also put a link in the chat Oh, no, no, it's well actually and I downloaded it in complicated ways. So it's Well, maybe then you shouldn't show it on a very recorded Right, but I Christian I will send it to you can do that privately, right, right Has it been used for something in the paper? Well, it's just called Hermit polynomial expansions of the error function and related of zero of your integral Journal of chemical physics Thank you that Andre this sounds like exactly the kind of interesting connection observation that only happens at conferences when people listen to each other Let's thank every speaker of the this afternoon session and the evening session again