Quantum Field Theory I - Lecture 13

welcome back this is uh lecture 13 in our series and and today we're continuing in our quest to put uh interactions back in to put interactions in our our theories here so with a few lectures ago we defined scatterings right and and define observables in terms of those scatterings and of course we know that unless these excitations these particles that are the excitations of these fields cannot interact between themselves there'll be no scattering and i told you that that interaction will come from the non-gaussianities of the the lagrangian or the hamiltonian right we're looking at terms that are not just the field squared which i which i have been calling the free uh uh part of the hamiltonian in which we have shown right we have seen that uh the the um the hubert space of the gaussian part of the hamiltonian is just a set of a lot of oscillators which are all independent right so there are no no interactions and now we want to bring those interactions in right last lecture i define the interaction picture which allow me allowed me to prove this theorem right this is fyman's theorem and it is a nice um very useful relation because it writes uh the endpoint correlator of a number of operators in my in my fully interacting theory and and now i'm talking about operators that can even be to be powers of the fields or it could be the fields themselves right or composite operators of the fields right in the heisenberg picture uh between uh vacuums the vacuole of the interacting theory and it expressed that in terms of a limit of the vacuum of the no the free theory right and as long as i change these operators and this is wrong to the interaction picture so now on this side the operators will be in the interaction picture which separates uh the the free hamiltonian from the interacting hamiltonian in the evolution so now operators evolve with the full uh hamiltonian and and uh with the interaction hamiltonian and and um states of over the free one and this is a nice expression because now the whole interaction part of the hamiltonian which is what is this h1i is this h1 means the interactions and i means i am in the interaction picture right is in this exponential which i remind you is the time order exponential so i this exponential is defined as a series in these operators remember everything here is operators i'm not putting the head notation but they are they're all operators but these operators this product of operators appears here will be applied in a time-ordered way including the ones multiplying the exponential so just have a number of operators multiplied here and they are always applied in a time order manner so the point of today's lecture and this is what the title is about this vic's theorem is how do i calculate this uh product of many operators right you agree with me in the end the moment i try to calculate these these are the important objects to material i already told you if i have green functions i can look at very singular points in the green functions and that will give me amongst other things that will give me s matrix elements which i can use to calculate cross sections right so um when i'm trying to calculate this it will boil down to calculating this numerator here at least right i have to deal with the denominator but you see that's simpler we will talk about the denominator later but the numerator will actually boil down to to calculating the following thing right i'll have a time order product i'll have these operators right here all these operators will be here but this potential will never really calculate exponential right i i'm assuming at some point i can truncate these at some order and calculate some finite number of of powers of h i t and that will look like these right so i have this integral let me bring it a little bit to the right h 1 i t to some power n m let's put m because i already use n for the n positions here right so to sum for m and this is the kind of object that i have to calculate right assuming i can truncate the exponential so the first term i will have to calculate this m equals 0 then if i want a better approximation i'll go to m equal 1 m equal 2 3 whatever right i'm assuming i can truncate the series and still have a um a good approximation to the interacting theory right we'll deal with um how well this series converge later and what are the exact conditions for now it's either death or or nothing right either i can truncate the exponential or i won't be able to do anything and of course i remove all numeric factors so there's a magnus i here that i don't care is out out here vic's theorem helps me with this product right this is the main object i i want to attack for now and notice that what i want to calculate is even simpler than that because in the end these uh the whole system only has one degree of freedom let's assume that we're talking for enough a theory with just one field phi right so there's only these phi in the interaction picture so the interaction hamiltonian will also be phi to some power n which then to some power i don't know k right i don't care maybe the interaction is a fight to the cube maybe interactions the five to the four maybe both but in any case the hamiltonian is just some fight is fight with some power which will then be also taken to the power m so we'll have to phi 2 km right but in the end it's just a power of phi same goes for these operators doesn't matter how complicated these operators will be they'll be defined the most complicated cases power series in terms of these files right this this operator's phi so in the end calculating this right amounts to really just calculating some product of many five eyes at different points right the points themselves depend uh i can have many different uh combinations right suppose for instance this is just time because this is the full hamiltonian if i go to the density then there's also an integral in in position here let's say i call the position in which i'm calculating this hamiltonian z so i could have many phi's to the z depending on this power depending on the power of phi that is inside here this operator could be phi square so i have x one twice and then x two twice and then x three twice right but in the end it's just a product of a lot of files a time-ordered product of many uh operators field operators and this is what i want to calculate our mission for today is very very straightforward i want to calculate a time order product of many fields so let's see how can we go about that and the first thing i'll do i'll separate the field in two parts so i'll take phi in the interaction picture and write as phi i plus be careful because because my plus is very similar to my dagger so uh pay attention to the contest context and and and so i'll divide my field into this part of course everything here is a function of space-time position where phi i plus x is just the integral of d three p over two pi cubed one over square root of 2 e p a b exponential of minus i b x remembering that x0 is being called t minus t0 in the interaction picture remember i have some reference time and then the time i'm calculating this and p0 is equal to ep okay so that's the guy i call five plus this is to be consistent with the notation most books use especially basking right but keep in mind very misleading because five plus has the annihilation operator and the negative spanish right so not very good plus but that's the definition and phi minus of x is just the other part right remember i'm just dividing the definition we had uh for the the field in terms of a a dagger and separating the creation part from the annihilation part and phi minus has the creation part has a dagger of p exponential of i p x and again same conditions here x naught is just t minus t 0 and p0 is ep and pay attention to what i did here this is interaction picture operator so i started from the very definition of how they evolve remember operators evolve with the free hamiltonian here from some initial point in time and this initial point in time is the same as the uh let me put this in a different color is the is the same as the schredinger picture operator right this this s of x that does not depend on time at all it's just d three p two pi cubed 1 over square root of ap same thing as if you find here right and let's focus on the a part i have also the a dagger part but it is the same so there's this plus the dagger part and here's just the three vectors just the space part of the operator then these two explanations come together interaction picture i have to apply these two right they come from the left and to from the right and act on this uh a operator so you get something like this a of b exponential h naught t minus t 0 which just writing writing these exponentials as series and using the commutation relations with a with h naught you can show that that's the same as a b exponential of minus i e p t minus t 0. and this is the time part that went into here that's why x naught is t minus t 0 and p 0 is ep right so i included the time evolution here if you don't remember how this commutation part uh works i i have proven that we have used this before it's on page 60 of uh of the lecture notes uh you can go there and take a look but they're important points here right many so for one i'm not changing even if now i'm going to interacting theory i'm not changing the commutation relations between the operators that means the commutation relation between a and a dagger are preserved and and that will mean also that the creation and annihilation interpretation of it will be kept in the interacting theory the second thing is since operators evolve just with the free hamiltonian the time evolution here is the same as in as in the free theory right that's why i ended up with the time part these a's here have no time evolution it's all included here and it and just h naught is contributing so you have no new parts here you would expect energy to be different but the energies that show up here is just energies for uh uh the fritherian which we we know by the way that this ep satisfy the relativistic dispersion relation so that is uh convenient right another convenience of the interaction picture but the point here is since these guys are also acting on the vacuum of the free theory then i know a lot about how they act in particular and very useful we know that phi i plus acting on the vacuum is zero remember plus here means annihilation right there's a here and not a dagger and also means that the opposite is true right if i act with phi minus interaction to the left that's also zero because i'm putting an annihilation operator on the vacuum now i want to use these to notice a very uh not very intuitive relation between the time ordering and the normal ordering for the rest of the course almost i'll substitute my notation for the normal ordering we have been using we didn't use it a lot but it few times it showed up that's how we were indicating normal ordering now we'll be using normal ordering so much that i i prefer to use a more explicit visible notation will be this one so this is the normal ordering now from now on right and consider now a particular time ordering right let's start with just two operators take the time ordering of phi i x phi i y right just two operators at different points in space time for sake of definiteness let's assume x 0 is bigger than y 0 then we know how this organizes and let's expand these two guys in terms of 5 plus and 5 minus right so i have 5 plus of x plus 5 minus of x times 5 plus of y 5 minus of i if i time order that i just have 5 i plus x i plus y plus phi i minus x phi i minus y and then the cross terms which are phi plus i x phi minus i y plus 5 i minus x phi i plus y right nothing here just wrote very explicitly the time ordering all y's are to the left of all x's which they were they start anyway and and that's fine but now notice that i want to find a relation between time ordering and normal ordering and normal ordering in this language means putting all plus whatever the position is you're calculating to the right of all minuses right that's what our normal ordering looks like you just have 5 plus 5 plus 5 plus until you have the first 5 minus and then phi minus from that point on because then i'm acting with annihilation operators on the vacuum before i act with the creation right that's normal ordering when i look here this is almost already normal ordered because this is just plus so normal ordering is by definition already done this is just minus this i in this term i have the plus to the right of the minus so this is normal order but this guy is not that's the only term which is actually not normal order right so let's let's do it let's normal order this guy i can write it as the commutator of phi i plus x with phi i minus y plus the opposite uh uh ordering right because this is will phi i minus y phi i plus x right if i write this commutator explicitly this term will conceal with the negative term that comes from in here and you get the one above so i can rewrite this guy in this way and then everything but this commutator is normal order so i can rewrite the time ordering of two operators for that case right is the normal ordering now i'm collecting all these terms these these these and this one right this is the normal ordering of phi i x phi i y and the difference between the time ordering and the normal ordering is this commutator let me copy it right here and color it accordingly all right so essentially i rewrote the norm time ordering in terms of the normal ordering plus the term that relates both right this is true for x 0 bigger than y 0 but i could have done the opposite x 0 smaller than y 0. that means every x would be exchanged by every y here but nothing else changes right i can just replace x by y in this expression right here right so in the case where y 0 is smaller than x 0. let me put this explicitly as to not create confusion so this is only true for x0 bigger than y0 but now i have another expression which i on the right side here i just replace y by x y by x and this is true in the other ordering right for the time in here inside the normal ordering x and y have no meaning because in the end i have to expand these and apply it in a normal way so in any case i can just commute these two in here because the real ordering that i'll get applying these operators is the normal one but in here of course i cannot right if i i try to exchange x by y here i have to to to be careful right this is it's not true that i can just change all right so these two expressions are not the same although i can uh i can change x by y inside the normal order so what i want to do next you see the difference between those two is really in this commutator which is different here so i want to define an object which is called a contraction let me define a contraction which will indicate like this so the contraction of phi i of x with phi i of y and the contraction is indicated by a symbol like that this is defined as this commutator when x naught is bigger than y naught and this one when the opposite is true okay and and notice that since these are commutators and commutators are functions they are not operators anymore right the commutator of two operators will be proportional to some dell direct delta and or green function whatever right i can actually put these contractions these commutators inside brackets right so i can take for instance this guy and just put it inside the bracket this is the same as the operator because i cannot same as the commutator because i i can always take the commutator multiply by one this is just 1 and then put it inside and this is convenient because then i can calculate this commutator let's write it explicitly right this is phi i plus of x phi i minus of y minus i won't even write x and y but this part is zero because phi plus acting on the vacuum is just zero all right it's annihilation acting on the vacuum so that means this guy is just 0 5 i plus 5 i minus acting on zero remember this is y this is x and we know we have seen this object before right this is just the propagator this is just the propagator go back and take a look at the definition of the propagator and you see that's exactly it i'm creating a particular y and inhibiting it at x this is a two-point function and the same is true for the other other option right if i take this operator now this is this is what the contraction looks like when x naught is bigger than y naught but if i take this one and put inside inside the bracket in the vacuum i'll i'll get to the conclusion that this is just the propagator from x to y because again i can put zeros here and then use the same kind of logic to disappear with the other commuting uh the one that actually puts a an inhalation on the vacuum so what i'm i'm getting to here is that this contraction that i define here right this is definition but the the consequence of that definition is that the contraction remember this is the propagator from y to x that that propagator that was not causal right it's just the two point function right it is the the transition from i state with well-defined position y to a state of well-defined position x this is the one we used to show there was not causal right but in one combination x naught is bigger than y zero and the other one is the other option we have seen this before take a look at page 68 of the lecture notes this was the final propagator it's the propagator that is this guy when you have one time ordering and this other guy when you have the other time ordering and that's a important thing right so now we were just innocently let's say looking for the relation between time ordering and uh normal ordering and we have suddenly uh propagator being essentially the difference between the two i see here is that the time ordering is the normal ordering plus a five month propagator because now i can put these two equations together right because the final propagator has in its definition the two options right the two time orders which actually are wrong here this is y not bigger than x zero right and and and that's the deal and again since this is not an operator anymore i can include it inside the normal order you know time ordering because there's no operators here anymore this is just a function right and finally i write a very important relation which is the time order product of field operators is equal to the normal order of the same product plus the contraction which is fireman's propagator and this is the particular case of the fixed theorem this is the victorium for two fields let's see how the generalized version looks like right the generalized version which you prove as an exercise and it's it's just proved by you looking at three field looking at four until you convince yourself that that's what happens right it's just the following right the time order product of a product of n fields and fields is just the normal order product of the same fields plus all possible contractions and i'll define pretty soon what all possible contractions mean means but when i say contractions i'm talking about uh things like that right and so let's state the theorem it's proven just like i did you can just put a third field here see what you have to do to go from the time ordering into the normal ordering you you have a bunch of commutators appearing every time you try to change the position of two guys you get a commutator these commutators are these firemen propagators right these contractions and you do that until you have these three fields ordered normally on the right side and you see what kind of commutators appear and that's what will make all possible contractions means all possible commutators like that that will appear and this is called vixx theorem it's a very important result very useful for particle physics and to get uh observables out of out of this theory just to make clear what all possible contractions mean let me show an example say we had a time order product of four fields now i'll really simplify my notation because i'm carrying this interaction picture index here and carrying this to say x1 i'll just call this phi one because i'm lazy fi one right one means x one and i'll ditch the interaction picture completely because we'll be working the interaction picture for the rest of the course all right so and now i'm talking about this 5 1 5 2 5 3 5 4. it does not mean i have four fields it's just five calculate the position x1 until position x4 so according to vic's theorem i'll have the normal ordering of the same thing plus all possible contraction what's a possible contraction a possible contraction here would be to contract phi one with i2 and leave these two uncontracted that means this is the product of a final operator times two final propagator times two operators which will have to be normal order right only the contractions ignore or normal ordering because they're already functioned that there's also another possibility which is phi 1 contracted with i3 and then fact 2 with five four five three five four but also double contractions so phi one we fight two contracted we fight three five four that's another possibility goes on right another one would be phi one with five four two with three you know everything every possible contraction two by two you'll be combining these fields two by two and whoever is left uncontracted fine it's still inside the normal ordering and you can see that the way you prove it's theorem in the the canonical quantization you see this is much easier to prove in the path integral quantization it'll show there but we'll see that later but for now you can see that the reason you have all possible contractions here is because after you write it in the time ordered way they'll be every possible commutation between any two fields will be necessary to bring that into the normal order you have at least one instance of a plus to the left of a minus say for instance between phi one and phi two and you have at least one five plus to the left of i minus between phi one and phi three so you have to use all commutation relations possible to get to the normal order and that's why you get all possible contractions right you can see that very clearly if you do three if you're really brave i mean or or really stubborn and didn't get confused by the tree the share amount of work you need to do four will convince you right i'm sure right the moment you try to do four at the halfway through it you say yeah yeah that's what happens and i'm convinced right but the the bottom line is that i get objects like feminine propagator from x1 for instance here x2 times the normal order of 5354 because by the way the normal order of this sum is the sum of normal orderings right so then i can take these propagators out of any particular normal ordering but you also have terms like this one say in here is x1 x2 x3 x4 which are objects that should be familiar for people that did the exercises a few lectures ago when i asked for the four point function of of harmonic oscillator we had products of propagators like that it was like delta delta between times right four different times very similar thing to what we're getting here and that's not an accident again i'm calculating the four point function here so now now let's see how useful the vix theorem is to calculate what we want to calculate remember going back our target is to calculate um sorry here products of time ordered uh time-order products of the field operators many of them right let's see how that that helps us the big actually enormous advantage of this is that is it comes when you remember that since we're calculating those time order products in the vacuum right so in my new abbreviated notation i'm calculating things like that right and filled in the vacuum then these normal orderings just go away they just go away because if i apply if i put an operator in a normal ordered way say doesn't matter how many fields i have here acting on the vacuum the first thing that acts on the vacuum here is annihilation operator so this is zero even for terms like this one what i'll have is like zero d f x 1 minus x 2 n 5 3 5 4 vacuum of course this guy goes outside but this normal order product in the vacuum is zero again so really when you're calculating time ordered products of fields between vacuum states the only terms of this whole sum that survive are terms where no operator is left uncontracted because if any of them is left uncontracted it will be normal order and will be zero in the vacuum that means that this guy calculated in the vacuum between vacuum states it's very simple i just have to think of all possible full contractions i cannot leave anyone uncontracted and that i can do by just looking at the form right is x1 minus x2 df x 3 minus x4 which is this contraction plus let's do it in colors in the name of didetics this is x1 minus x3 df x2 minus x4 and finally there is this one which is just x one minus x four x two minus x three and remember five and propagators are also symmetric by exchange between x1 and x2 so it doesn't matter which order i put inside the arguments of the propagators again this is very familiar to that same exercise i just mentioned right in which you get every connection possible between these four points x1 x2 x3 and x4 which i hope is something you notice when doing the exercise for the four point function of the harmonic oscillator so you see vic's theorem really simplifies calculating things like that it will be in the end just a product of final propagators which is pretty uh interesting so the next step now uh is uh to show a as a specific interaction this is as far as i can go without specifying an interaction so let's take a look at a specific one it's actually a very popular one not because it applies to many physical states although it applies to some but mostly because it's a very simple one so it's very good to to use for for understanding how interactions work in field theory before we go to models which are more useful for particle physics or or other physical systems let's then talk about our first interacting theory which will be lambda five form the favorite toy model of quantum field theory books everywhere is not i mean i can tell you that there's no elementary particle that has only a 5-4 interaction but for instance the higgs particle has are supposed to have because we didn't measure directly that one yet it's supposed to have a 5 4 interaction but it has also a 5 3 fi 3 you call interactions more complicated interactions so this is a very simple interaction that is usually used only to model the the simplest behavior of instance right but it is useful for us because of the simplicity right so what what does it mean phi 4 that means a particular assumption for the interacting part of the hamiltonian right which of course we i'll write as a hamiltonian density because we're talking about local theories so this density depends only on one position position right and the form for these uh hamiltonian density the interaction part is still summed up with the gaussian part right is just uh lambda which is just a number over four factorial you understand why i want this 4 factorial later of course i could absorb that into lambda it's just a number anyway but i prefer to keep it out here 5 to the 4th of x so this is just a product when i turn phi into operators of four times the operator all of them calculated at the same point right because this is a density that depends only on one point there's a local theory all right so we will carry that forward and see what kind of theory that gives us we'll stick with lambda five five four for a while so by the end of this course i hope you can all say just at least i understand lambda 5 4. so let's see but first some general considerations right remember that in five months diagram i had on the right hand side of five months uh theorem i had an exponential of this guy right so the left-hand side looks like that phi of x1 until phi of x and omega these are heisenberg picture right and on the right side i had this limit of t going to infinity 1 minus i epsilon there was some normalization which i'm ignoring for now right is that that vacuum vacuum transition without these operators but then you had what matters for us now is the vacuum of the free theory and this product now everything in the interaction picture times this potential minus i minus t to t d t h one i d vacuum right that's what you had and now i want just to make some general considerations on when i can to truncate this exponential because these guys are operators it's important not never forget right the the hamiltonians here are written in terms of operators and that means that in general i can write h1 s lambda lambda now general it doesn't need to be 5 4 of course we are always thinking of i4 now but it could be other different interactions could be 5 3 5 5 more complicated stuff right but what i have to assume here is that lambda is small right let's say much smaller than some number much smaller than one if you want not necessarily one but lambda is small let's say it's even close to zero right that means that if i get the expectation because operators cannot be small or big right operators act on states but expectation values of operators can and i have zero here on both sides right but in general for any state not necessarily the vacuum i can write h1 as lambda and the expectation value of that operator that characterizes the interaction now it's and just indicating this to be any state and that also means that expectation value of h1 times h1 right h1 square will be proportional to lambda squared of phi square right this is still keeping general right in this case in the case of lambda phi 4 you have a 5 4 here 5 to the 8 here but the lambda 1 is good outside right that's why it's not necessarily smaller than 1 because now i have a 4 factorial here that helps right right so that means that if if lambda is more enough right maybe even if this expectation value is not much smaller than this one maybe it's even bigger right if lambda is small enough then the suppression created by multiplying by powers of lambda can win over these which i don't know maybe this guy is bigger than this one right but if lumber is small enough that can win and that's the condition i want i want this right the operators these much smaller than these which is just the normalization of the state and so on and so forth not only i want that i also need because you see these operators here i also need phi 1 phi n h 1 square it's smaller than phi 1 5 n h 1 and the product without the interaction i need that to be true for any number of operators and it's not trivial to show in general right for for for any theory that this will be true but that's what i am assuming so it's important to have that in mind right that that's what we are assuming here it's it's kind of intuitive that will be true for lambda as small as i want but nature is not that uh kind with us right most of the times lambda is a number and we know how much it is it's not zero right uh sometimes it's not even that small so have to be careful with these compositions maybe test them numerically there's all kinds of approach but that's the assumption for for us right now from a theoretical point of view uh lambda is what help us control these uh series right if that is true if all these uh relations uh uh all these yellow keys in size are true then i can truncate exponential and actually just take as many terms as i want to get the precision right because i have a first approximation which is just not taking the interactions the free theory then the first interaction of lambda will give me a little interactions the second one will correct that one and i go as far as far as i want depending on how precise is the measurement i'm trying to to compare my theory with right the how far i'll go into this series depends on on what kind of approximation i need with that in mind let me do this expansion for these n point function that i'm trying to calculate here so i'll take this object actually all the way until this one over n here and do a power series in lambda right remember there's also a exponential of the interaction down here which also would bring powers of lambda so this expansion might seem a little bit strange but bear with me let me do the expansion for the numerator first and we'll see what happens with this denominator later right so the first term in this expansion is pretty obvious right it's just the identity part of the exponential right just copy this thing down here and throw the exponential away so that would be the first term in my perturbative expansion on the interaction the next one is given by this and now i have the first term in the exponential expansion which is just minus i minus dt dt h i in h1 interaction t and the the third term is just that square over uh with a factor half right so i just put it let me put it a little bit down that would be just one half of these square right which is this series and it goes on right just to define some notation right we can say that this is really the endpoint function it depends on x1 2 xn right but these other ones we usually indicate them like this which means it's the same endpoint function calculated to zero order in the perturbative expansion right it's just order zero so this one right here would be just g n one depending on the same variables this is g and two right where n stands for the number of insertions of the operators that means how many different positions this green function depends on right and the number on top here is the order of the perturbative expansion that i'm calculating that object right so and the full object of course i have to uh resum the whole series and get exponential that was up there so now um let's uh take a look at some examples right of of these this calculation and let's start with the one we saw already right which is let's let's start with zero order in perturbation theory so i'm taking order lambda to the zero right and let's take a look at the four point function for that so g for zero of x1 x2 x3 x4 this is just this object and again remember i'm using a simplified notation phi 1 means phi of x 1 in the interaction picture phi 2 5 3 5 4. so this depends on four different positions well we already saw that one once we use vicky's vic's theorem here i'll get all possible contractions and that gives me a product of firemen propagators let me write them real quickly because what i want to show here now is that there is a graphical way of getting the same expression which serves us well especially when you have many uh and many uh operators to contract like four operators is easy to look at all possible contractions but even intuitively right if i keep increasing the number of uh fields in there it's not difficult to imagine that at some point trying to get all possible combinations two by two can be a lot of work all right so there's there's a shortcut for that right and and that shortcut is what we call a very important concept finding where's my text um here phenomenon diagrams right these five on diagrams are a graphical way to represent the same mathematical expression so we'll have one-to-one relation between these drawings i make and some mathematical expression and to understand how that work right in this case again it looks like i'm just creating these drawings to make my life harder but the bigger and bigger the the calculation you want to make the more and more advantageous is to use the diagrams instead of actually just using this theorem right so let's see uh the idea here is i i'm calculating a four point function so it depends on four different space-time coordinates and what vic's theorem did was connect these two by two in all possible ways right take one and two three and four etc right and and that means uh i can graphically represent that i had three options i can graphically represent these three different uh possible contractions like that this one represents the first one x1 x2 3 with 4 right i can put a 1 that x 1 connects to 3 and 2 connects to 4 which is this one and i can put another one that one connects to four and two connects with three to avoid confusion i'm leaving a gap here to say goes under right there they're not really really touching each other because eventually i'll have some interpretation from for when they actually touch each other so they're going under i could even draw x2 here and x4 there but i want to make my life easier and always keep these points fixed because then it's clear that this diagram is different from this one all right and and that represents that uh same sum there right i'm just summing up so now i'm writing g for zero as this sum and then of course if this is true that also sets a rule for these red lines that i'm drawing there right so what i'm saying is that i have a feminine rule which are the rules that assign mathematical uh expressions to definement diagrams right and the final rule here is when i have a line like that connecting two space time points x1 and x2 that by definition is the fireman propagator between one and two again the final propagator for the scalar field is symmetrical under exchange of x1 and x2 so i don't have to worry really if it's one to two or two to one is is symmetric right so that is the final rule for the propagator and of course once you get into the business of using uh five month diagrams a lot you start calling these the propagates so draw the propagator all right because really the the diagrams are so useful that sometimes it seems like they are the theory and actually uh uh for a little while people really believe that the most fundamental thing here was the final rules but now remember too that we can interpret these at least one of the directions right the female diagrams has two components for two different time orderings right and and you could interpret these as a creation of a particle at one of these points and annihilation in the other right so it's common to to read these as uh oh so these means that created the particle in one and and annihilated into it propagated from one to two all right and uh of course that's how we read it but keep in mind there's two components there's another component that goes from two to one and you have to be careful not to over interpret these right as propagation because again this guy is not even causal right so it's more a correlation than a propagation but the common parlance is really saying oh the particle is created here goes to there and and we'll see how far can we we take that seriously right so now uh let's go to a simpler uh again this is just a free theory right because i'm taking lambda at order zero this is just the free theory uh let's go to a simpler uh endpoint function let's take the two point function and put interactions in right really look at the two point function long enough to be able to put interactions so the two point function is this right the full complete two point function is something that depends on two points and is defined as these right phi one phi two omega and then i use the final theorem and big theorem to get the uh order by order expansion so the order lump the zero is pretty easy right that's really just uh the contraction so g2 at order 0 is just 0 d phi 1 phi 2 zero that has only really one possible contraction which is this propagate right this one so the two point function is just a propagator and you can see now that the four point function at zero order in perturbation theory is just products of two point functions right the four one is written as products of of the two point function now let's see finally introduce an interaction so now let's look at the two point function at order lambda to the first power now uh remember my interaction just i wrote upstairs the interaction we're dealing with now is the lambda phi 4 interaction which means that my hamiltonian looks like this right this is the full hamiltonian this is the hamiltonian density right here right and that means that when i calculate these objects g two one x one x two the two point function continues depending only two space-time positions but in here i'll have phi one i'll have phi two but then i have this integral of minus i d t and i have to choose some space-time coordinate to that t because it's an integrated variable right integration variable zero if i substitute that in here you see that i don't get i get one field two field but i get another four uh operators right by just one insertion of the interaction i get four operators right and this dt goes together with this space component i will use z for the variable so i'm using z for the this integral right here d3 z has to not confuse with the x1 and x2 that is here and this will be dt z right the time uh for that coordinate that means that um these will become just i can i can put together this integral in time with this integral in space and just have a d4z phi over 4 factorial phi z to the fourth so have four insertions of this phi z to the fourth in a way that i end up calculating and now figuring out all the factors right so this minus i comes out here lambda is just a number four factorial right even the integral goes outside right i have an integral of verb d4 z right off this object now phi 1 phi 2 and i will abbreviate phi of x as phi x 5 z phi of z as phi z so f i z four times and i have to apply vic's theorem now to get all possible contractions between six fields you see how how quickly if this is the simplest function you can imagine it's just a two-point function of the simplest theory possible and already i have to find all possible two by two contractions of six fields right then the number of fields here goes up really quickly as you go to more and more uh complicated objects so i could just do this brute force and and and and and just do all possible contractions and write a immense uh expression but let's try to do it more systematically right so that we can classify the kind of contraction that can show up there's only really two classes of contractions here there are many contractions but they fall into two very distinct categories right the first option is to contract phi 1 and with phi 2. there are many contractions that start with contracting phi 1 and phi 2 and then you can do all kinds of of contractions between these many phases right so what i have in mind is something like this where i'm considering all contractions that have these in them right and and then it's easy to see there's only three others right because it's just this contraction between these four operators there's this option there is these one right this guy with this one and and these two and there's also uh a third one which is these right but this three different kinds of contraction will give me the same answer right this is just a propagator a final propagator from z to z i know it sounds strange to write it like this this is really the propagator of zero right but bear bear with me i'll keep it like that because we want to see the logic behind these uh these contractions right so i have basically just these right for the three a possible contractions will give me the same result so i just i have i'm adding over all possible contractions so i just get a factor three right which i'll put explicitly i want to keep track of this factors so i'll put a factor 3 here and the rest is minus i learned that over 4 factorial i have this integral d 4 z and i have a d phi'man from x1 to x2 which by the way can go outside this integral because it does not depend on x on z sorry on z and then the two contractions of z i'll find uh the two final propagators from z to z right so that's the expression this is all possible contractions in which phi 1 is contracted with i2 the second class of contractions is given by by those contractions in which phi 1 does not contract with i2 now phi 1 will have to contract with one of the z's let me give me a little bit more space to draw the contractions so now i'm thinking that phi one is contracting with one of the z's and there are four possible contractions there right there's phi one with the first d five one with the second z five one with the third and five one with the forces there's four options but it's always in in all of these options is always i'm always getting the final propagator between x1 and z so they're all equal there's a factor four there that i want to keep and once i contract phi with one of the z's i still have phi two uh contracting with one of the other z's and that's three options for that right take the blue one here inside the blue contraction i still can have fight two doing these these or these these are three options multiplying that four and that's true for every contraction of phi one right once i contract phi one phi two has three other options to contract and then of course once i contract phi 2 say i take this option this phi z is contracted this phase is contracted the only option for the two phases that are left is to contract between themselves so the blue contractions here is one of the many possible diagram contractions that i have here in fact many means 12 in this case there's 12 options right and that's what i counted here that's the numeric factor in front right but the rest is just writing one of these contractions because they are all the same now nothing drops outside the integral because i don't have a propagator between x1 and x2 i have a propagator between x1 and z sorry z another propagator between x2 and z and another one from z to z and that's that's the expression i have now between for these contractions so now i found all contractions but i saw there's only two different expressions i could make but this one appears 12 times these one a pre appears three times for a total of 15 possible contractions there but in a more systematic way right and these numbers in red will become very important pretty soon and you'll see why so let me write my expression here what i'm saying is that this two-point function at first ordering perturbation theory so now i'm i i'm that's a result that's a green function for the interacting theory right in the perturbative approximation plus this one right so that's that's the first result of the course for a interacting theory now let's see how i would again i'm i'm telling you that we want to get rid of doing these combinatorics as soon as we can right because you can already see i put one more insertion of the interaction and i have an extra four fuels and and suddenly i'm doing combinatories for 10 fields all possible contractions of 10 different fields i don't want to do that every time i need to do a calculation right i'll do it a few times here just to get the rules straight but i don't want to be doing that forever so what i want to do is it's it's see how the define rules work here for the interacting theory so let's think what's happening now i have x1 x2 which are the two variables this function depends on but i also have z this is integrated over so is integrated over all space so it's it's a space-time variable that is does not appear in the the answer but it appears in the middle of the calculation so let me put those three points here i have x1 x2 and z and it's very important to know the distinction between these three points these are external points there are variables which are fixed right and i i tell you how what x1 and x2 are and and you calculate the green function but z is not z is an internal point let's say and now let's see how i'm connecting these points for for the first class of of contractions this one what i did i put here option one which was these uh first class of contraction what i did was connecting x1 to x2 with a propagator right so i have the propagator from x1 to x2 and i have these two propagators that go from z back to z which i can draw like that again these lines are propagators but they go back to the same point that's why i'm indicating like that if i put 0 here i don't know which point is that that they are going back to right but it is z and the second class of diagrams let me erase this to that i'm adding a second class of diagrams that is actually a little bit different and now z is in the path from x1 to x2 because i have this propagator that goes from z to x one i have this propagator that goes from z to x 2 and i have this one that goes from z to z and that's how the diagram looks now and then this is this is interesting right because this this these diagrams they have many components but if i told you right you have these three points connect them all the way all the way as you can with the following rule i have four lines coming out of z you see there's always four lines coming out coming out of point z right and and you can understand why this is so right we have four fields in point z that's because the interaction is a phi to the fourth right so i'll always have four fields calculated at that point so when i do contractions i always have four lines coming out of the z so if i told you i have this point z i have i want four lines coming out of z i have just one line coming out of x i would just want one line coming out of x2 because there's only one field acting there connect them in always possible you cannot leave any of these lines hanging without a point right they have to connect these points where the fields were you get these two drawings there's no other drawings you could i mean you could draw them differently but they they are the same expression right the same mathematical expression the tricky part of these and then you you can rationalize that you can guess that this point that is where the interaction is will be integrated over this integral you can guess too because you know that the final answer does not depend on z but these factors like these three times everything here especially the 3 and the 12. those do not look obvious just by staring at the diagrams at all right if you of course you can get them by doing the combinatoric analysis that we did above but that's what we want to avoid mostly we want to a quicker way of finding these uh factors these numeric factors right and that's that's the tricky part right uh that's uh by the way this was option two uh that's the tricky part how do we how do i guess right i know that there are three diagrams three contributions equivalent to this diagram there are 12 equivalent to this one but how do i i guess that by looking at the diagram right so let's work on that and for that i'll get amazingly that's the kind of thing that you see by looking at a much more complicated case because then you have every effect possible and and let's look at that let's look at a really complicated case so this is what we will attack now right we're skipping one ordering the perturbation theory and going straight to level that to the cube right uh you know because we like it rough we we like difficult challenges and now what we will have right since it's the two point function it only depends on these two external points x1 and x2 but now since we're taking three insertions of the interaction right we're going to the fourth order in that expansion of the exponential will have three internal points which i can name whatever however i want because they will be integrated over right these internal points where the interactions uh that are in introduced by the interactions uh by the way they are called vertexes right so these are the the vertex of of the theory and and and we have three of those so since we are taking lambda to the cube and we know that for each power of lambda you have together a four factorial and a minus i that is uh in front of the hamiltonian you have minus i lambda 4 factorial to the cube that's what you expect since you're expanding exponential until the cubic term you have 1 over 3 factorial in front here right that's uh that's a given and now the three vertexes which i'll name z so at z you have phi to the fourth so four insertions of the field the second vertex i'll call w and here it is just four fields acting on the same point this is phi w phi w etc and the third one i'll call u right so i have points x w and u these are just same with space-time points like x1 and x2 but they are internal points they are the vertices of these uh of these theory now i have two ten fourteen fields and have all find all two by two contractions i'm not doing that right what i'm doing is again we have these many classes of different contractions each of these classes will be represented by one diagram like it we did before you saw i have two classes i have just one drawing for each class the diagram represents the whole class but i have to find the multiplicity of the class i have to find how many times that exact uh mathematical expressions appears again and again right i have to find that multiplicity in order to do that i will take what contraction one specific contraction and try to see how many times that one would be repeated so just for sake of the argument this time instead of calling this x1 and x2 by the way i prefer to call it x and y just to keep the notation in line with my lecture notes let's do x and y and let's take a particular contraction so let me see i'll use the exact same contraction i used in the nodes uh let me write it here so this is a very specific contraction i'm contracting x with one of the z's y will be contracted with a w let's say two z's are contracted together z is contracted with a w that these two w's i contracted with u's and two of the u's are contracted amongst itself so you see i left nothing uncontracted so this is one of the options that survives the normal ordering right and it's just one of the many contractions possible it just took a particular one that allows me to show all the kind of complications that you could have here for that contraction i know how to write the mathematical expression now it's pretty easy right so i just bring those factors out uh all the integrals d for z d for w d for u and then there's a bunch of final propagators so there's one from x to z there's one from y to w there's one from z to z there's one from z to w this one there's actually two going from w to u right so i get the f w u square because i have this guy multiplied twice here right there's a pump propagate from w2u another one and finally there's one from u to you so the question i want to attack here is how many times this exact expression shows up another contraction right the same expression and and that's doable we can without actually doing all the contractions we can find how many they are so the first thing for that we have to think a bit about the expression itself and how to get to this expression the first point here is the integration variables or what you could call the vertex naming i named this vertex z this one w this one u but those are just integration variables i could change them in particular i can call z w i can call w u i can i mean i can change their names so if i get other contractions that are basically the same form as this one but just different in in the exchange of a zip or a w or a zipper u or even a three-way change here it will be the same expression because i can just change variables in the integral and uh that would be that okay uh one way of thinking about it is that i i of course i can do different contractions maintaining the position of the integrals here but i could also do this right that's a change in variables names that will give me exactly the same contraction the same expression just exchanging w by z and there's obviously obviously three factorial orders i could choose for these three integrals here right these three blocks i could reorder them in three factorial ways so that means that just fro from the fact that you can rename z w and u i'll have three factorial expressions which are exactly the same i get exactly the same expression if i just rename these guys right so that's already a multiplicity which is quite big already okay uh another one the second one comes from the contractions at vertex z at this vertex right and and that's easy to see right so z is basically there's one contraction coming from x so this line coming from x could end up here here here or here and you get the same expression there's another line coming from w that could connect anywhere either and of course the third the third line you have no option once you fix where the x line is connecting and where the w1 is connecting the two others have to connect between themselves to get the same expression so the line coming from x has four options where you could connect right the line coming so if you think about it you have phi z phi z phi z i have x comes around and connects uh with one of them so i remove already the one that got x and then the three left over have you have the line coming from w that can connect in three different places here say here or or here so once you took one of the fields to contracted the x you have three options for the one that connect with the w and then the two orders for the blue option is he these for the red option is these and for the white option is this the two other have to connect between the sub so there's no more multiplicity so essentially what you get here is 12 options that multiply these three factorial right it's multiplicity on top of multiplicity then you have even more multiplicity for let me put this to this side because i don't want to lose my expression vertically here i would need a bigger blackboard to do this right so let's think about the contractions at w right it's it's a similar game right i have a line coming from z that gives me uh 4 factor 4 because that line coming from z can connect anywhere so let me schematically put this here so there's a line coming from z then there's a line coming from y which now can connect to three different places and now there's two lines coming from u right u is throwing two lines in this direction they can connect basically in two ways right they can connect let's call this u u this two line can connect like this right or they can connect like this so there's two options right i can put like these or like this so that's another factor too here which gives me even bigger multiplicity but it never ends right there's also the contractions at you now i have uh two lines coming from the w right so one of them can connect to four different places oops one of them can connect to four different places the other line now has three options right these two lines and then the leftover u's have to contract among themselves so this is all but now i have to be careful right because these two lines coming from w to you make it so that the analysis i did here for w independently of you to to be a little bit wrong right i did a double counting here that is possibly not obvious right what what happens is let's forget about z and y that i did here i did these right and then for these lines going to u right i said well there's two ways i can connect these two lines here and again i said well these lines can connect anywhere with you on you here but if you take any pair of fields calculated point u right you see that if i put a factor two here saying it could go like this or uh these two options right let me let me do it like this could be like this or crossed option right and again i think this also here has two options right here or crossed i would come to the conclusion there are four ways i can connect these two field to those two fields but there's not right there's really really only two possibilities this one and this one there's no other so in the way i did i actually counted twice every time i have two lines going there so i put a factor one half to take care of that is easier to do like that it's easier to first look at the multiplicities independently and then remove the double counting only here i have this problem right you see i never never have two lines going in the same way this is connected with this this df square that is here right okay so i can solve that by just taking a factor half and multiplying all these factors by half in there so let's put all together right i'm my conclusion here is that that exact expression appears 3 factorial times 4 times 3 times 4 times 3 times two that comes from here times four you'll be clear why i'm writing it in this stupid way pretty soon right right here let me organize these so let let me put an extra factor 2 here to make these a 4 factorial i also make this one a 4 factorial so i need extra factor of 4 in the numerator that means that i can write this as 3 factorial 4 factorial to the cube right i make these 2 a factorial 2 but then i get a factor 4 in the denominator right the reason i'm writing it like this is because this number is very similar to this one and this is not an accident you see pretty soon but this number is actually the number 10 368. so by now i should be glad that i didn't really go looking for every contraction and this is just from this class this is 10 368 possible contractions that will give me exact same expression right if i have a 500 diagram that allows me to get these and this numeric factor without going through the contractions i say we get a good two right instead of doing all the 10 368 uh options right what's the chance of getting it right so now let's see how we can get these factors from the diagrams the first step is to actually understand a very particular cancellation here so now i said i have this 10 000 something ways of obtaining the same expression that means that the contribution of this class of of contractions to the final uh two point function will be just ten times ten thousand whatever times these right but i use it in this form and i see that this 3 factorial goes away with this one this 4 factorial cube goes away with this one and i get just minus i lambda over 8. and this is kind of amazing right after all these huge numeric factors i just get i one i just dividing by eight in the end and everything else is just a copy of this sorry i missed a cube here so this is minus i love that to the cube which is actually just minus i lambda cube over a or plus this huge constellation of pneumatic factors of course it's not accidental it is part natural and part uh design so let's see uh the first part is the constellation of the three factorial right these three factorial is coming from the renaming of these three integrals right the three internal points of uh of um the diagram right the three internal the three vertexes so for and we know that if you go to order n in the perturbation theory you have n vertices and internal points and that means there will always be n factorial way of reorganizing renaming them right this so this is three factorial but if i had an extra would be a four factorial and so on right this is always true but it is also always true that this power of the vertex is also is coming from the expansion of exponential and this potential will have 1 over n factorial here right so these n factorial ways of renaming the vertices will always conceal with the one over n factorial of the exponential that's the natural part there's no way around it and and i don't have to worry about this three factorial at all right it will always go away with this one so that part i don't have to worry about the second cancellation comes from the fact that think of any vertex any any vertex take z for instance there's four lines going out of z so innocently i would say if this all these lines are coming from different places the first one will connect to one of the four z's there's four ways to do that the second one will connect to the three others three ways and the last one will connect to two orders there's two ways and then the fourth line coming has to connect to whatever is left there's four factorial ways of connecting lines to these five to the fourth right four options for the first line three options for the second two for the third and and so that's four factorial ways right and that four factorial for each vertex is always concealed by the four factorial that i put here when i define lambda phi four i could have defined it like that but i've put a four factorial there and i said well eventually you understand why i want this number here i could include it in the definition of lambda but i don't because that for factorial does away with the four factorial from the combinatorics at that point right this second rule while designed the four factorial here to conceal with the four factorial in the numerator it is kind of tricky because it there's some situations where this is exactly not exactly true the first one you can see at z right i have a line coming from x four ways a line coming from w again four ways but then i'm contracting two z's together and that makes me miss a factor two if the other lines that are connecting from zero or coming from different places then i would have a four factorial here but instead i lose the a factor two because i'm contracting to this together so that's one of the factor twos that i i missed and since this goes in the numerator right this factor goes in the numerator of the final expression and i have a 4 factorial in the denominator in the end the 2 that is missing here will be left over into the denominator of the final sum overall options another way that these four in w everything is fine you see there are lines coming from everywhere right but between w and u there was a problem remember there was a double counting because the two lines are coming from the same place and i have shown you here that if you do it innocently you put a four factorial here and a four factorial there you get a double counting so i divided by two again that's another place where i get an extra factor two in the denominator when i do double countings like that and if you look at you right then you have both right you have a factor two missing here so i don't have four factorial and there's also the double counting that i put at a factor half a fatter half this factor half does not appear twice at w and u is it's a effect that appears between them so you have a factor two that is coming from this in the denominator another one that is coming from these in the denominator and another one here you have three factors two that spoil this perfect constellation that i was aiming for and three factors two is two times two times two is this eight that i have here so this eight that appeared is actually deceptions to the perfect cancellation i was aiming for when i put this four factorial in the definition of the interaction right and the three factorial which is just a fact of being the interaction being exponentiated that one goes away always right there's no way i can actually spoil that one so in the end this factor eight is totally understandable just by these three happenings that happen that that that you have here right that spoil the idea that i have just four fields and they connect from to other places if these places are the same or the connection is like that you can spoil your your naive consolation so now we have the now we have to to ask can i see that just looking at the diagram right this by the way is called the symmetry factor of the diagram why symmetry well let's look at the diagram and see so now this is the expression right and then remember i have a diagram to represent all 10 368 contractions that give me the same expression one time and diagram will represent that let's draw it right again i'm calculating the two point function that depends only on x and y right but now i have all this junk in the middle so let's see from x i go to z all right let's try to rebuild the story here from z i have two options right i have a line that goes from z to z that one and i know i have another one that goes to w in w what happens i have this line that goes to y i actually put y too far away i didn't need that much space and i have two lines going out of w and going to u and finally i have a line that goes from u to you right half the fun of being a particle physicist is naming these diagrams right you find penguin diagrams you find tadpole diagrams i'll call it the desert cactus diagram so now we want to see where that factor 8 is coming from and why we call that a symmetry factor the point is that these two lines that go out here the only point that is are really fixed are x and y right z is now fixed by its connection to x because this is integrated over right so it's this is kind of a topological analysis i'm doing here the exact place where these lines are i could have drawn it in very different ways the important is this these connections right i could distort these lines and the fact is that i can flip this line in this direction and say this is a factor 2 symmetry because i can flip the line remember this is just a way of actually seeing the fact that this connection breaks the perfect cancellation that we saw right there's a factor half missing here on the denominator so that's the symmetry factor same between these two lines i could just exchange this line by this one that that was the double counting we did right i can exchange this two line because both are going to w and u and get the same diagram so that's another factor too and there's the same thing here with these two lines going to you that i can just exchange these two lines right you can also think it some books i'll actually say it like that i have you and this represents two places you could connect a line and you can put it like that or flip it and put it again i don't know some places indicated like that right and i have two options on how to connect it there so that's a symmetric symmetry factor of two right but remember they are coming from here factor 2 here factor 2 here and the double counting here so the total symmetry factor of this diagram is 8 right is the product of the individual symmetry factors you see that can get more complicated right if i had say three lines connecting w with u the double counting could be much worse all right so you see that symmetry factors are not always two right this is just amazingly one of the simplest cases so that's the symmetry factor of of this diagram now just to narrow it down let's go back to the simplest simpler case we already did right but do it with diagrams now so take two point function at first order in perturbation right from x1 x2 that we did uh previously let's try to just draw the diagrams right so we had the diagrams before so we know which one they they they were so i had one that x1 went to x2 and then the z fields were all connected to the same point z so they're contracted amongst themselves and there was this second option in which x1 and x2 were connected to z x1 and x2 and z and then there was one propagator from z to z and let's try to figure out the expressions just by looking at the diagram so let's start with the symmetry factor since we're on it on this one that's easy there's only one case here that i can really invert and get a factor 2. i cannot exchange x2 to x1 because these points are labeled right and and so changing x1 2 by x1 would actually mean something they are variables that i know the value right so i can distinguish these inversions so the symmetry factor here is just two this case is trickier now because i have this inversion i have this one but i could also flip it like take this whole loop and bring down or take this whole loop up right so there's another factor two here that comes from that symmetry of of these this drawing right if you want to convince yourself just take 5z to the fourth and see how many ways you can connect it together and see how different that is from four factorial and you see there's a factory eight there right so the symmetry factor in this case is eight that allows me to write the expressions for these diagrams right so this is minus i lambda over eight and pay attention right the symmetry factors go into the denominator because it's usually it's usually no it's always the fact that the cancellation didn't happen so i have leftover numbers in the denominator and then i just have to write the propagators right so this is a propagator from four one from one to two x one minus x two right i know that the internal points are integrated over so that's part of the rules for this diagram so any point that is not an external point needs to be integrated over the f z minus z square because i have two propagators from z to z and that's the factor a again here i have minus i over lambda over two which is the symmetry factor and the integral in the internal point that's another rule and then the propagators right i have a propagator from x1 to z i have one propagator from x2 to z and another one from z to z by the way the order of the propagators here also does not matter because these are just functions they commute right they are symmetric and they commute they're very easy to work with factors so we understand the symmetry factors we understand the propagators we understand this integral maybe it's these guys that still need some explanation right but it's easy you see that every time you have a vertex right this is part of the vertex every time you have a vertex that means you inserted the interaction in and in your exponential you had right minus i times the interaction minus i times this interaction and in here you had lambda the 4 factorial goes away we already discussed that so every time you take one power of this you get a minus i lambda so for each vertex after all the other constellations happen you still have that for each vertex you have minus i lambda now we did it with the diagrams let's check right so if there's a factor eight here and a factor two here we actually did these contractions before right they are up here right those are the two so in here you can see these three will conceal the three in the factorial here and you have a factor eight is just four times two and the same here the 12 will conceal the 12 down here and you're left only with a factor two here and a factor eight up there which are the two and the eight that we got down here straight from symmetry factors so i can get just by looking at the diagram get the expression for every contraction that will give me the same diagram the diagram takes over the whole class right i and by drawing diagrams i can quickly see that there's no other really different diagram that you can draw just try with these rules x the vertex always throws out four lines no line can be left without ending at any point right and x1 and x2 have only one line getting to them that's basically all you can draw of course you can draw these these these sideways but the symmetry factor will be the same if you draw it sideways you know uh it's it's this is all this is all and that allows us to deduce the final rules for this interaction right so the first thing we saw here it's pretty easy is the rule for the vertex so now i have these uh definition of the vertex inst in terms of uh fine rule so every time i see a vertex at a point for this particular theory the point does not even matter for some theories it will be more complicated but for this theory it's just a complex number it's just minus i lambda so if you have many vertexes you just make a product of these and that's what happened up here right here i had three vertex three vertices right so i got minus i learned that to the cube right one minus i lambda for each vertex and this is the final rule for the vertex with a lot of uh implied things right you you have to later integrate over the the vertex you have to to put a symmetry factor right so let's make a summary table of that and here is the final version of it right so these are the five rules for limit of five four four green functions and in position space these two things in parenthesis are actually very important because it's very common for to hear students saying like well but i looked the final rules in these or that book and there are other five numbers so there's two there's a lot of caveats there one is all the conventions like metric uh the way you include uh four factorial in the interaction or not some books don't right but that's that's expected the other thing which is less obvious is that you can have right final rules for many objects in your theory in here i'm doing it for the green functions in position space of course eventually we'll have to worry and that's the actual actual thing we want is s matrix elements and we know that we have to go into momentum space and then do the l s z reduction formula which will eliminate some propagators that are here so keep in mind these are green these are the final rules to obtain the green functions not s matrix yet so you'll find books that do final rules at this point and sums that that goes straight to the s matrix and only write final rules over there that said the rules are pretty simple for each pro you draw the diagrams at some order you decide which order you're working at you draw the diagrams and then for each propagator you put a final propagator for each vertex there's a factor minus i lambda and an integral over the variable for the position of that vertex for each external point these are the points that the endpoint function actually depends on there's nothing this seems like why you're writing that but you see that for more complicated theories there's actually uh functions and numbers associated with these external points is just that lambda phi 4 and especially since this is a scalar field there's nothing to this external point so for now this is just a statement of a trivial thing and finally you have to define divide by the symmetry factor of the diagram so once you know the rules the application of the three first rules is easy right what deserves some training and needs some practice is actually finding out the symmetry factors so let's look at a few more examples before we finish so take this diagram now i'm just drawing uh diagrams out of the blue i'm not even worried about what kind of green function i'm trying to calculate with them right but it's all in lambda phi 4 because that's the one theory we know right so the external points are x1 and x2 so this is a two-point function and i'm doing it that order lambda square because i have two vertices right so the only thing to to a lot of things are obvious here so let's apply the rules for each propagator i have a fireman propagator for each line so there's one from x1 to z1 there's one from z2 to x2 remember the order here does not matter and there's three of them so cubed from z1 to z2 that part is easy i have two vertices so minus i lambda square and i have to integrate d4 z1 d4 z2 now symmetry factor the only part that actually needs you demands that you think a little bit so now the symmetry factor obviously come from these three lines because i cannot exchange z1 by z2 because despite z1 and z2 are integrated over so their labels are not significant but z1 is connected to x1 which is not integrated over and z2 is connected to x2 so i cannot flip this over what i can do is exchange this line with this one or these with this one or these with this one so there are three lines that i can exchange in all possible ways that's a three factorial symmetry factor that goes in here so that's the final expression there are three factorial ways if i label these three lines there are three factorial arrangements i can do here another example now is x1 oops wrong color x1 wrong color again stuff z1 x2 then crazy stuff throwing two lines out of z1 and i put two more vertexes there called z2 and z3 you see i'm respecting all rules of lambda phi four so each interaction has four legs coming out of it and the external points have just one leg so this is a valid diagram at order lambda cubed right three interactions let's write it right so propagators let's already put the interactions first so minus i lambda cubed i have to integrate over z1 z2 and z3 and write a bunch of propagators so there's one like this same thing for x2 because z1 is connected to x1 and x2 right and there's two propagators from z1 actually one from z1 to z2 and another one from z1 to z3 and now three of them cube from z2 to z3 right now symmetry factor right one of them is obvious because it's the same that is up here right these three lines again have a three factorial right once you know the one above you you know this one but there's one that is not obvious at all and it comes from the fact that z2 and z3 are integrated variables so these points are not different from each other in any way i could remove the label right and they are both connected to the same point so that means if i bring z3 over here and z2 over there i have the same diagram in other words i could exchange this line could connect to the leftmost socket on z1 and this one could connect to the rightmost socket or the other way around on z1 so there's two ways i can do it and that's another factor too so the factor the symmetry factor here is actually 12 and not just six like this one so this those are two examples of how you you do this right and the last comment and and there'll be some exercises on symmetry factors for you guys but the last comment i want to make is that of course if you're doing that expansion right say for the two point function phi one phi two it's the limit and whatever i'm not writing everything again one over n you have this expansion right so the first term for the two point function we calculate x1 x2 this is order lambda 0 right when you go to the the next order we saw two diagrams right you have to include both so the game here is that at any order in perturbation theory you have to include all possible diagrams of that order otherwise i mean this is the same order in lambda as this one this is order lambda 1 right this is of the same order as this one so you cannot justify saying this is smaller than this and throwing one of them out right in fact many symmetries of the theories we'll see that are only satisfied order by order in this expansion but in order to get the symmetry you have to include all diagrams at any particular order so if you go to order uh lambda square you'll get a lot of diagrams right so this is one of them again physicists go nuts with naming this this is called the sunset diagram and but we also have uh this is order lambda square right you just have to count vertexes there's this there's one that looks a lot like my cactus right it goes on so and these are the all the classes right for for for that particular uh order in the perturbative expansion that goes on there's there's a lot of others still at other lambda squares so keep that in mind that at any order you must strive to find all possible diagrams of that order which is not trivial too right it's interesting computational problems try to find all the diagrams especially higher orders how many diagrams there are and a lot of smart computing people actually attacked these in the past and now we have softwares that do that right fairly consistently okay so that's it for today uh i'll give some exercise for you to practice your feynman diagrams and your final rules other than that see you next video

Channel: Instituto de Física Teórica, IFT-UNESP

Views: 723

Keywords: ift, unesp, field theory, quantum field theory, Wick Theorem, lambda phi 4, Feynman Rules, Real Scalar Field

Id: DwlODczVEhE

Length: 135min 33sec (8133 seconds)

Published: Thu Sep 17 2020