 How do we give a general outline of how to do the calculation of the annulus partition function. So, today we will give more details, but for that I have to give you the second dose of the bitter pill. So some introduction to open bosonic string field theory. And I will assume that you are at least somewhat familiar with the wall sheet formulation of string theory. And we will be talking about the open wall sheet, so it is simpler than the closed wall sheet. So we will denote by h the vector space with a vector space of open string states, but this will include matter and host sector. And now let me introduce a notion of an off shell, off shell classical string field classical open string field ok. This classical I will explain in a few minutes, psi this by definition is a state in H with ghost number 1 ok. And let me remind you that the ghost number definition that we have given is that C carries ghost number 1, B carries ghost number minus 1 and the matter fields of course do not have any ghost number. So we will you know introduce a basis state. So let us call phi n r basis state of ghost number n in H. So then we can expand psi as a linear combination of the basis states. So let us call this psi r phi 1 r and when I call these the string field that basically means that these coefficients of expansion are the dynamical variables dynamical variables of the theory. So path integral is integral over the this psi r's ok that is the meaning of the string field being a state of ghost number 1 will give the action in terms of the psi r's in a few minutes. And this is a classical string field in the same sense that the usual Maxwell field ok, Mu is a classical field, but in the full quantum theory of course you will also have the ghost right. So when I say classical string field that only means that you have not included the ghost yet ok, the ghost that comes from the further up of gauge fixing that we have to do separately. So now I have to give the action right because this path integral you have to have a weight factor e to the minus s in the Euclidean theory. So the action should be some number or some function of the psi r's and here is the form of the action where this QB is what is called the BRST charge ok and I will write down the explicit expression for this. I will write in the language of conformal field theory as contour integral of C of z, Tm of z ok. And this is of course a properly normal order this Tm is the matter energy momentum tensor ok, which has c equal to 26. When I write t without a subscript that corresponds to the full energy momentum tensor. If I am using other specifically matter or ghost I will write a subscript. So this QB is some operator in the Hopon-String Hilbert space and then this is the inner product. So given any set of any set of variables values for psi r this generates a number right and that is what the action is supposed to do. It is supposed to generate a number for a given string field configuration. Is this notation clear? Ok, so this QB has some nice property QB squared is 0 and you can also see from the explicit expression that it has ghost number 1. Oh I should have said that this is when I wrote down the classical string field theory action this is only the quadratic part and then there are interaction terms which at least for today's lecture we will not need, which at QB can hire already in coupling in the psi. So because of this property the S is invariant under a gauge transformation of the form delta psi equal to QB on lambda by lambda is a state of ghost number 1, a ghost number 0, state in H ghost number 0. So in particular we can expand lambda as sum over r lambda r phi 0 r and these numbers are what we will usually call the gauge transformation parameters and the action is invariant simply because of this fact that the square of the BRST charges 0 so you know transform psi by QB times lambda ok QB squared just gives you 0. So if you have to gauge fix now then I may of course define possible gauge choices but the one that is commonly used and which is what makes connection of the world sheet theory directly is what is called the Siegel gauge put the condition that B 0 and psi equal to 0 and in this gauge the action takes a simple form and that is half psi C 0 L 0 psi this basically follows from the fact that when you look at the expression for QB the only term that involves C 0 is of the form C 0 L 0 and then there are terms without C 0. So when you take the matrix element between two size which are all both annihilated by B 0 right which is this gauge condition then only the term that has a C 0 contributes and you get this as the gauge fixed action. This is the L 0 matter right. No no not L 0 matter the this is the full L 0 right. So the point is here the L 0 matter comes from this term but this term has an L 0 host ok which is hidden here. So when you actually look at the expanded in the modes and look at the coefficient of C 0 that it is exactly total L 0. Total ok. Once you have gauge fixed of course you have to introduce further up of a host so need further up of a host. I am going to state a result without giving a proof that the full action to S total ok which basically means that S matter plus a S gauge fixed plus S host has a form half psi tilde C 0 L 0 psi tilde where psi tilde is an arbitrary state in H ok some particular cases of this you will see later. So the only difference between these and this is that here there are more variables right because psi tilde is an arbitrary state in H. So this of course contains this but this contains many more ok all the other host number states and the interpretation of those other host number states host number states is that they are the further up of hosts that count of the gauge fixed ok. So one when one does a further procedure for finding out what the kinetic term is you find that it is basically this with the conditional host number relaxed is this statement here. Just one part of this I will illustrate in a few minutes. Let us now introduce a new basis states phi n r as the basis states in H of host number n satisfying the Siegel gauge condition because psi tilde will be expanded in this basis. But using this we can also find a basis states of the classical phase. So the classical can be expanded some over r chi r phi 1 r ok this is similar to what we had earlier except that now this includes only those basis states which are added by v 0 and then the out of Siegel gauge modes right which are there before you gauge fixed those can be written as sum over r phi r see what I have called it phi r c 0 on phi 0 r. In other words the basis states of the in ghost number 1 the total set of basis states in ghost number 1 can be thought of as a set of basis states in ghost number 1 that are added by v 0 and those that are not added by v 0 right they are added by c 0. So those can be written as c 0 acting on basis states of ghost number 0 these are added by v 0 ok. So this together forms the total set of basis states of ghost number 1 ok this is just a convenient way to write the classical strength field ok for reason that it will become will become clear soon. So now we see that if we take lambda if lambda we let us take some particular class of lambdas for which because lambda has ghost number 1. So this is a linear combination of states of ghost number 1 I am not including the ones which have also c 0 right I have just chosen a particular class of gauge transformation parameters. Then delta psi which is q b lambda q b recall that it has a c 0 l 0. Just a question yes this lambda is different from the lambda the previous one this ok you can say it is the same except that the sum is over a subset right because this is running over only the basis states of ghost number 0 with an added by v 0 that lambda was multiplying the full basis states I have just chosen a particular subset of gauge transformations ok which have this property thank you. So in q b lambda has a term which is c 0 l 0 acting on sum over r lambda r phi 0 r and then there are terms without c 0 because q b the c 0 part is this. So now if you compare this right see this is the expansion of psi the classical strength field if you compare this with this we will find the delta phi r delta of sum over r phi r. So delta of this is c 0 sum over r lambda r l 0 phi 0 r. So this basically tells us that in this basis the operator that relates the gauge transformation parameter to the fields involve this l 0. If you write in components I can write this as delta phi r as h r times lambda r where h r is the l 0 eigenvalue of phi 0 r is this clear because this one is will be c 0 sum over r lambda r h r phi 0 r ok. So if we just compare this with this you will see the delta phi r is lambda r h r right which is what I have written here. So this says that in this basis the relation between the fields and the gauge transformation parameters is given by this h r which is the l 0 eigenvalue. And that is what is supposed to enter the Gauss determinant if you recall when you have gauge fix if in the Siegel gauge you gauge fix these objects right phi r's are out of Siegel gauge. So the Siegel gauge corresponds to getting phi r equal to 0 and it is a relationship between the fields that you have gauge fixed and the gauge transformation parameters involve the l 0 eigenvalue h r. And that is the reason why in the Gauss kinetic term also you get this easy l 0 is this point clear right. So I am just giving a argument as to why l 0 appears in the Gauss kinetic term right. Of course the fact that l 0 appears in the original action is a consequence of the fact that there is a QB. But it also acts in the Gauss kinetic term because of this observation that the part of QB that is relevant in the Siegel gauge is the C 0 l 0 part. And then l 0 eigenvalue will be the will be will enter the Gauss determinant ok. So let us start with that gauge the full action S total and half Si tilde C 0 l 0 Si tilde and now we want to normalize the basis states in such a way that this has the form that we encountered earlier. Last time you recall that you have identified certain quadratic term as the open string fields for reaction ok. So today we are going to show that this has that form ok. But for that you have to appropriately normalize the basis states because the coefficient of string fields are the coefficients of expansion in the basis state right. So you can write Si tilde in general sum over write anything of Si tilde r ok. But these include the matter fields ghost fields the classical field ghost fields and now we compare this with half sum over B HB Si B square plus sum over F HF UF VF. This is the form we had in the last lecture right when you try to express the annulus partition function in this form. So you see from here that we should pick up the coefficient of Si B square should be HB and this can be achieved if can be achieved in the ghost number 1 sector right because then this will be of the form HR Si R tilde square if we normalize the basis states this way ok. Let us substitute this expression here and use this normalization ok l 0 of course will just pick up the l 0 Eigen value and we will get this form. So this is for the bosonic field this is a classical field. But in order to get this we need to organize so let us look at the ghost number 2 ghost number 0 sector phi 2 R C 0 phi 0 S should be delta R S. So here I have used the fact that the total ghost number must add up to 3 in order to get a non-zero for non-zero matrix element. So let me then summarize if you have not followed everything ok. So this is the gate fix form of the string field reaction ok which has matter ghost everything. In order to make connection with what we had earlier ok because earlier we had a proper integration measure ok. But we just declare that the exponent whatever appears in the exponent is the string field for reaction. In order that that form matches this we just have to choose the basis states satisfying this conditions right and then there are other conditions for the ghost number which I will not read. Sorry I lost the point that when we expand phi tilde as phi tilde R n phi sorry psi tilde that one. Psi tilde was any state of the Hilbert space right. Subject to b 0 and psi tilde I call the 0. We are gate fix right. So psi tilde is any state in the Hilbert space but all satisfying the single gate. Didn't like when we added back the string field theory ghosts didn't we get back the the whole Hilbert space. No no maybe I if I didn't mention it let me say it now that psi tilde arbitrary state in H ok satisfying b 0 psi tilde equal to 0. Thanks. Yeah. Because if you had gotten arbitrary state then even in the classical sector you would have gotten back the ones that we have gate fix right. So you need to this this is a statement that psi tilde is an arbitrary state satisfying b 0 psi tilde equal to 0 and that is how you can expand psi tilde in this basis. So compared to what we had yesterday right let us just recall what we have achieved. We have just adjusted so the action string field theory action take the same form as what we had yesterday right. So that is not what is a new achievement. But the new achievement is that now we know that when we expand the string field theory in a basis how the basis states should be normalized. And that will be important because this basis states that multiply these those will be the vortex operators of the states that we have to use right and it fixes the normalization of the vortex operators. Because now when you calculate various correlation functions we have to use the vortex operator with this specific normalization ok and we will see that that will be important in how we normalize the vortex operator. So now let me be more specific and try to expand the classical string field and here I will just 2 in the L 0 equal to 0 sector because that is why the problem came for L 0 less than equal to 0. That is why the problem problematic states came right. We saw that for L 0 larger than 0 the integrals are perfectly convergent and so there is no issue. So far this was quite general right now we go back to again 2D Boson string ok. So the classical string field expansion which is before gauge fixing. So this is before gauge fixing is let me write the all the terms and then I will explain various ambiguities etc which might drop off. So the gauge parameter again we are looking at the L 0 less than equal to 0 sector. So this is i theta and then the in the gauge fix theory we put psi tilde as so this alpha minus 1 is the oscillator of the x field of x satisfying alpha minus 1 alpha 1 sorry alpha 1 alpha minus 1 ok that is how we normalize it. And using that we can check that this basis state does satisfy this condition ok the norm is s 1 because c gives you norm 1 c minus 1 c 0 c 1 is 1. This one I will the normalization of this I am going to come back in a minute because this is not fixed by that right this is only for the basis states with which are in the signal gauge. So this I have fixed to be just the vacuum is the only one the normalization of this in fact is not very important ok how you normalize ok and the for reasons that I will explain ok. All we need is that this normalization that whatever you normalize this state this state should be c 0 on that ok that is this feature that if you have taken a basis states normalization of the basis states in some way ok then the out of signal gauge states will be c 0 on those basis states that is the statement. So if I have taken this to be the basis state to be i times the vacuum so this should also be i c 0 acting on i on i that is the vacuum. Because you will see that if you normalize if you change the normalization of this and normalization of this then both phi and theta will be normalizing the same way and that in fact does not affect the final result because if you recall in the final expression we had an integral over phi in the numerator and an integral over theta in the denominator because if you both normalize both phi and theta it makes no difference. This i I have just put in because with this i it turns out that phi and theta are real without this i phi and theta will be imaginary variables but again at the end it does not matter because it will just cancel out because you have integral d phi and integral d theta ratio here ok. So these of course we have already fixed this is you can recognize this as the L 0 equal to minus 1 state right that is the one which gave the h b equal to minus 1. So these two states ok this is Gauss number 0 this is Gauss number 2 I have to make sure that the inner product between this and this with c 0 in between is 1 right that is this statement the Gauss number 2 Gauss number 0 states should be normalize this way but that does not fix individually the normalization of 0 or this right you can scale this down scale this up but the inner product should be fixed which means that if you scale this down and scale this up then u will be scaled down and v will be scaled up but that again does not affect the final result because we had integral du dv ok. So du dv is unchanged under this risk area ok. So while there are ambiguities in this basis expansion as long as you satisfy these constraints all those ambiguities go away in the final expression. So now let me explain how we find the relationship between psi b 0. So relation between psi b 0 and y because all this was done to do that right relation between psi b 0 and y and the relation between theta and alpha. So from here we see that a psi 0 is multiplied so psi b 0 appears in this combination psi b 0 c 1 alpha minus 1 on vacuum and this I can write in the vortex operator language as psi b 0 i root 2 is normalized as this is standard normalization that one uses for the versus scalars plus non singular. So because of this 1 over 2 here I need a square root of 2 in the definition of alpha minus 1 and because of this minus sign I need an i. So now the way we proceed is the following that we know that y dependence y dependence of an amplitude of any amplitude should be of the form e to the i omega y where omega is the total energy yes we can write as 1 plus i omega y plus higher order trunks as a Taylor series expansion. So this basically says that the y insertion should produce a factor of i omega we compare this with psi b insertion from the state that multiplies it we can conclude the psi b insertion what it does is that it inserts integral i root 2 del x of z dz on the boundary okay this is by the standard state operator correspondence and then you have to do this from unintegrated to integrated vortex operator but just from the structure of the vortex operator we know that this corresponds to inserting this on the boundary of the disc or what are the boundary of the wall sheet. So you can take a simple wall sheet like a disc and put some closed in vortex operators v1 v2 up to vn with the energy say omega 1 omega 2 up to omega n sorry with a momentum yeah so we have to which is also the energy in this case because it is the momentum conjugate to x and then we can use the operator for expansion OP is the del x of z is our vi of w is i omega i the minus 2 z minus w let me call this w i okay so this OP basically follows from this okay because vi has a factor of e to the i k i omega i x w i okay using that you can read out this OP. So I should have said something else okay so when I say it is this there is a another factor here g o which is the open string coupling constant okay this is related to gs but we will see that we don't we don't need it this comes because anytime you insert an open string you should accompany it by a vortex by a coupling constant right anytime you insert a closed string it carries a factor of gs anytime you open you insert an open string it will carry a factor of g o. So what you have is that you have this operator i root 2 del x inserted at the boundary we are integrating this and we know that the z with w i OP has this structure so you can do this by Cauchy's theorem right if you know the acd at every fold we can use Cauchy's theorem to evaluate this and let me just get the final expression so final expression is that it is i i pi squared up to sum over k omega k times the original amplitude amplitude without psi b. So now we see that we can compare this with what we had we would have expected i omega y. So this tells us that i omega y should be i pi root 2 g o sum over k omega k psi b sorry this is of course just omega what we call omega here this cancels to time psi b okay or another psi b is y over oh here there is omega but omega cancels y over pi root 2 okay so this is what we called k1. So this is what we had called k1 y well this is the way we calculate k1 again the details of the calculation you can work out but the philosophy is simply that you have to first fix the normalization of the basic states in that in which you do the string field expansion okay and that normalization is fixed because we want to reproduce the answer that the annulus amplitude get right and that required that we normalize we have certain form of the action okay and so for to get that form of the action we have to normalize the basic states in a certain way okay and once you normalize the basic states that way we can compare the coupling of y to coupling of psi b and get the result are there questions. So let me briefly say about how to determine the relation between theta and alpha so the strategy is more or less the same to take a state xi any open string state with one end on that instant on the instant on okay now of course open strings have two ends right so if one end is on the instant on the other end has to be somewhere and the point is it does not matter what that something is but you need to add some spectator brain it could be another d instant or also spectator d brain and any state of this open string I am calling xi and the standard u1 gauge transformation is xi goes to e to the i alpha times xi right to which sorry xi goes to the i alpha times xi which you can write as 1 plus i alpha xi plus higher order terms so delta xi is i alpha right this is the standard infinitesimal gauge transformation the with standard normalization of alpha and this has to be compared with the string field theory gauge transformation of xi and I should say here that this part of the gauge transformation is not what I have written down so far okay because here you need the non-linear part the linear part is like delta i equal to the derivative of lambda derivative of theta right that what we are written down but there is a you need to because a non-linear part because this is really non-linear gauge transformation right it has a product of the transformation parameter and a field but this is known of course and the general form that it takes is delta xi equal to some constant times theta and this constant is given in terms of 3-point function theta xi 3-point function theta xi and its conjugate state which are denoted by xi c so these have their own vertex of quarters which have to be normalized okay but the point is that the theta normalization has been fixed right you see that this is i times the vacuum we have already fixed that is the vertex of quarter for theta so the vertex of quarter for this is i times identity okay since theta multiplies theta multiplies i times on the vacuum and because it is proportional to identity okay this calculation simplifies because now you have a 2-point function of xi and its conjugate state but that is basically one because xi is normalized right because once one is identity that this has 2-point function is just one so basically what you find is a delta xi in this language becomes i times g open theta okay any 3-point function again you have to insert a factor of g open so now compare this with this and that gives you theta equal to alpha over g open okay and this is I think what we had called k2 are there questions so let me just put things together so exponential of the annulus partition function will be given by i okay this came from the integral d chi over square root of 2pi e to the chi square by 2 that is the tachyonic then you have a root pi this came from integral d phi e to the minus s okay and s turns out to be phi square okay this in fact we saw in the gauge theory language yesterday but from the string field theory also if we just evaluate s in the background phi phi is the state that multiplies c0 on c0 on vacuum right evaluate the action for this step you will find exactly pi square so this gives you root pi then the psi b we had a 1 over square root of 2pi okay in fact I think yesterday when I wrote down the expression I had missed out this term but this is of course there because the integration measure is g psi b over square root of 2pi then this k2 right the relation between psi and pi so that is 1 over pi root 2 you know and then 1 over integral d theta 1 over so this whole thing comes from psi b and the theta integral gives you 1 over g 1 over g open times 2pi okay because alpha integral gives you 2 pi so integral d theta is 1 over g open integral t alpha which is 2 pi okay so g open cancels in this case okay but it does not always happen sometimes there are factors of g open left over and I think if you put everything together you get i over 4 pi square and very important the y integral is left till the end okay as we described and so you will at the end we will have a 2 pi delta e okay this comes from the y integral integral d y to the i e y and this result in fact agrees with the dual description right we already had known the sense result from a dual matrix model calculation and it indeed agrees with the matrix model results are there questions so let me then finish by writing down the list of examples where this has been done right I described just once example c equal to 1 2d bosonic string theory but let me list all the examples and I will not give the references but you can find the references in the notes that part of which has already been uploaded probably cases where this has been done the 2d string theory I already showed there is a class of theories which are called c less than one string theory okay these are similar to 2d except they do not have any x and instead you have a CFP with central such as less than one then the supersymmetric cases various supersymmetric cases 2b in d equal to 10 2a or 2b on karabi or 3 fold so this means three dimensional karabi or six dimensional karabi or space six real dimensional karabi or spaces this is relevant for string phenomenology then the orientive folds of these then some formulaic kind of string theory type 0b okay and something that is still not published but come out soon is the sine level theory these are more complicated version of the 2d string theory that I described where the scalar the x and pi the x and the sine level field instead of being independent fields they are coupled okay so that is the sine level theory and what one finds is that wherever the answer is known from a dual description okay this gives results in agreement when the answer is not known from a dual description then this just gives new results okay for the d instant on aptitudes okay so this is as much as I will say about the normalization okay so tomorrow I will describe the higher order calculations okay which includes for example the d 2 point function and annulus 1 point function okay and we will see that it requires a little slightly different kind of technology okay other than what we have studied so far okay so you have to learn a little more about string field theory and then we will see how we handle those is this clear so far so in the remaining time I will try to give a introduction to the last part of string field theory what we need sorry I have a question yes so you mentioned the check with the dualities when you have a dual description yes so in those cases do you see that these instant on corrections are all the instant on correction or do you see signs of or of other corrections that are beyond this yeah I think that in the dual description the only ones that are known that have I mean that follow from supersymmetry etc are the d instant on corrections I think in these examples in type 2 or type 2 and Calabria 3 4 there are some signs of the ns fiber instantons yeah but those are I mean even in the dual description they are not very well studied but certainly there is signal that those are there but these are the dominant in sent on because ns fiber instantons are certainly sub dominant there is a question so from this the gauge fixed action let us go back to the gauge fixed string field theory action it had this structure of half psi tilde c 0 l 0 so this tells us that the propagator is proportional to 1 over l 0 okay this is for the open string okay so let let us just keep this in mind for a while and then describe how exactly string field theory works because I said that it is designed so that it reproduces the usual worksheet amplitudes that is formally right but let me just explain how it does it so string field theory because it is an I can ordinary quantum field theory you can calculate the amplitudes using Feynman diagrams in SFT we can compute amplitudes using Feynman diagrams okay so for example if you are calculating a 4-point function 3-point function typically is just this but a 4-point function at tree level will include terms like this and then there is also contact term so it turns out that each of these Feynman diagrams so each of these Feynman diagrams gives integral over a subspace of the Riemann surface motorized space correct integral okay now let me explain what I mean by correct integral so I said that the worksheet expressions the worksheet expressions involve integrals over motorized space of Riemann surfaces right for example for the that is 2-point function one point you can fix at the origin okay because of the SL2 are invariant but other one you have to integrate along a line so that is the integral kind of integral you are talking about right so each string amplitude involves certain integrals and the string field theory is designed so that each Feynman diagram gives part of this integral okay but when I say part of this integral okay integral is exactly the same as that we get in the worksheet okay is the integral runs over a subspace of what it is supposed to run over okay so for example here if the integral was supposed to run from 0 to 1 say from the origin to the boundary okay one of the diagrams will give you say from origin to somewhere some place and the other diagram may give from that place to the boundary okay this is the way the string field theory Feynman diagrams give the worksheet expressions is this statement clear now what is a little surprising here is that normally when you talk about Feynman diagrams like this okay you do not have any integrals okay sometimes in the loop moment in the loop you have momentum integrals but otherwise the Feynman diagrams like this do not have any integrals right so you can ask where does the integrals that come in the worksheet come from and the short answer is that the integrals come come from swinger parameter representation so this more specifically this 1 over L0 in the term in the propagator the propagator may also have some numerators that are not important but now the 1 over L0 okay in swinger parameterization it replaced by dt e to the minus t L0 okay I do not know whether you are familiar with this typically when you have 1 over k square plus m square in the propagator right you replace it in the swinger parameterization by this okay that is the swinger parameterization and it is this integrals over t that become integration over the model line on the worksheet okay after appropriate change of variables you will find that the integrals over t become the integrals over the model line on the worksheet is this statement clear so now that you have understood how the integrals in the worksheet integrals emerge out of strain field theory we can ask how do the divergences that you see in the worksheet description emerge in strain field theory right what is the origin of the divergence so for this let us examine this formula a little more okay let me write this again 1 over L0 equal to integral 0 to infinity dt e to the minus t L0. Now if you look at this formula you will find that this is of course an identity for L0 again for the positive given any L0 again for positive L0 again for this integral can be done and it is clearly equal to the left hand side but for L0 less than 0 the right hand side is divergent left hand side is perfectly finite so this is one of the ways the divergence in the worksheet is skewed by strain field theory that if we have in the worksheet right worksheet doesn't know that we are this is a representation like this right the worksheet just uses this and if you use this you will get a divergence from t goes to infinity and for negative L0 but if you use this then there is no divergence so this is one way strain field theory queues the divergent integrals in the worksheet the more subtle ones are L0 equal to 0 states for L0 equal to 0 both sides diverge that so the strain field theory by itself doesn't help nevertheless you see that on strain field theory side we have a physical understanding of where the divergence is coming from right so in SFD the divergence comes from an internal propagator going on shell because you are basically at the pole of the propagator right one over L0 is the propagator so you are basically sitting on the pole and in quantum field theory you know that whenever an internal propagator goes on shell there is a procedural defect that is there is something wrong you are doing with that procedure otherwise you should never encounter a situation where an internal propagator goes on shell right we are just trying to use for example Feynman diagrams if we try to use Feynman diagrams with zero modes right then you will encounter such things but you know that the zero modes are not to be treated as using Feynman diagrams so while strain field theory doesn't directly cure these divergences it tells us how to that it gives us a physical interpretation in the language of quantum field theory and once you have that physical interpretation then you can use insights from quantum field theory to cure these divergences okay and that will be the basic philosophy how we remove these divergent in divergences from that he encountered in the worksheet integral and finally I should just say that all the divergences that he ever encountered in the worksheet calculation are of this kind right once you translate it to the language of strain field theory you find that all the divergences are coming essentially because of the wrong use of swing environment okay so once you understand so this is the reason why strain field theory worksheet theory is very useful because it gives you it gives a very compact way of writing the amplitudes okay but when worksheet theory fails in the form of divergences then you have to go back to strain field theory and try to interpret what the divergences are due to and then resolve it using insights from quantum field theory so this is what we'll try to do next time okay let's thank Ashok for the next lecture if there are no urgent questions we can postpone the question for the discussion now there is a coffee break we will meet at 11