 So, since we have said a 0 already equal to 0, note that divergence E or let us write d i of f i 0 is equal to d i of d 0 of a i right because the others are all 0 the a 0 is 0. So, there is probably with a minus sign right and let us put in the gauge index So, the electric field is d a by E equal to which is what I started writing d a by d t minus grad phi, but the phi a 0 we have set equal to 0. So, basically it is this and therefore, if you this is the gauge condition this is c. So, varying c with respect to the gauge condition means that you have to calculate this. So, now you put the that is same as d i of minus d 0 of delta a i a by delta lambda b. So, that is equal to d i of minus d 0 of now we are varying this with respect to this. So, there will be d and mu somewhere right. So, after all these are. So, a i is what I have to write a i a. So, there will be d i of right and then delta a b and a delta 3 of or delta 4 of the arguments of f and arguments of this. So, we have to be now careful about this x and say y. So, this matrix this is a gigantic matrix it is not just these labels, but also their coordinate labels. So, it is going to be delta 4 of x minus y right. The delta a b can of course, come out, but this will be delta i of x will act on this delta function and then plus f a b c a mu b and delta c b and again delta 4 of x minus y. So, that is the formal expression for what this d c by d lambda is. So, I went through this in great detail because that is that crystallizes what all these fancy abstract reasoning. So, in practice of course, any good physicist would have done all suffered all this first and then said I have a clean picture you know then you say I am Dirac constraints and there are first class constraints and second class constraints and there is canonical structure on the space of constraints etcetera, but this is what the Granje calculation boils down to. So, this is the determinant you will need to insert in the path integral to make sense of it. Now, this was observed by Feynman in his own way without suffering all this, but suffering other things by in practice looking at what the diagrams needed etcetera are and that by in trying to impose gauge condition you had to remove some of the diagrams. So, we now claim w j yes. So, right and these are all star if you like only the real ones remain now sorry I am very sorry unrestricted unrestricted here. So, these are the full set because we have put in the constraints times e raise to i times action and plus i times integral j dot j dot a. So, now we are stuck with this delta function and there is another series of tricks which then gets you to Fadya Popov answer. One of the tricks is that simpler trick is to rewrite. So, simpler trick is to rewrite there is I am remembering here that there is exercise please know somebody note alternative argument for the above in Ramon's book which is actually quite popular with lot of people it introduces some formal thing called delta C inverse which is equal to integral d over gauge field space of delta of yeah C a. So, one should go through this to see there is another formal elegant argument which also leads to the same answer it is the same in spirit, but done a bit differently. So, we deal with this delta function of C in a slightly trivial way and it is simpler of the two problems. One is that think of so, so reinterpreting or rewriting delta of C. So, here we claim that think of this as C a of the canonical variables minus some constants which do not depend on the in Dirac's language when you convert the C gamma language to the canonical language the second class constraints which can be explicitly eliminated will not appear and this will remain only on the first class constraints. So, the a 0 and pi 0 are not required. So, we can also say that. So, all this is only dealing with the first class constraints once that cannot be trivially said to 0 or solved for. In fact, if you will read the slightly lengthy discussion in Ramon's book he does. So, for the Abelian case you can get by without doing these technical things because you actually just invert the gauge condition dive a equal to 0 in the Coulomb gauge you just get an awkward looking long expression which is end of the story no determinant to be solved for you explicitly solve and stuff it in. So, then you only have two variables to integrate over. So, here the more superfluous one the second class one does not even appear anymore and it is the second one that is one is trying to do. And by the way this argument which is the alternative one basically says that we want to get rid of superfluous integrations over gauge copies of a. So, effectively you have to factor out the volume of the gauge group except that it is function of x. So, the gauge group functional or gauge your gauge valued u valued u of x gauge transformation. So, it is that which is being removed for that you do not the superfluous one can be assumed to be already gone and then you remove the gauge volume for the remaining gauge condition. So, to deal with this one then says yeah just let me finish here. What we note is that look at this formal integration this G is some function of x integral dG of delta function of C A minus G A times this Gaussian stuck in is just equal to replacing wherever I see G by C's. So, it becomes this, but we note that this whole expression which is a Gaussian integral is at best. So, is at best giving you some answer which is not dependent on canonical variables. So, this right. So, if you look at this answer is some Gaussian right it is determinant of let us say I mean multiple of square root of alpha at infinite number of points in space and time. So, it is some constant. So, we take this expression which is a constant and multiply it to this where we shift the C A by this constant without affecting the Jacobian of the this integral ok. So, the big functional integral is going to maybe accumulate some Jacobian again which is not going to depend on the canonical variables. So, I went in reverse. So, think of this as firstly you do this transformation. So, this is this is 1, this is 2 and 1 this and. So, if you believe these two that this is nothing, but some constant similarly shifting C A actually does not cost you anything in this functional integral. Then you first shift the C A do the number two steps first then multiply on the left by this ok. Now, you do G integral. So, this is step 3 you do it in this order 1, 2 and 3 in reverse actually ok. So, note that shifting this by some constant some C number function not a canonical variable did not cost anything also we multiply by an overall Gaussian which is not going to change canonical variable answer. Then we basically note that after doing that manipulation I can carry out the G integral and replace my delta function of C by this Gaussian version Gaussian version of it. Why is that good? Because instead of having to interpret some formal delta function over some condition in the phase space we have replaced the phase space expressions in the same line as the action functional and if the C A are typically linear in the variables then we just have added some quadratic terms in the effective action ok. So, by ok. So, that is not so bad. So, it just makes it life more convenient. So, let me remind you of the other main philosophy why all these very slight of hand works as I was saying in the very beginning all of these are fortunately not any numbers you are calculating. What I tried to emphasize was the path integral or the functional approach is never a very useful tool for doing quantum mechanics or chemistry. It is primarily a tool for deriving identities and relationships between greens functions. And so, it is its symbolic dependence on the canonical variables that is all that matters. It is not meant to actually give you any numbers. So, if you have infinite factors multiplying it it does not really matter. Eventually you will vary and you are also going to divide by you know when you take expectation value of something it will be equal to insertion of this, but divided by the thing without any insertions. So, all the infinite volumes are going to cancel out. So, if it overall changes any numerically the numbers we do not care about it. It will in this stack of things I was drawing instead of being this surface it will become some another surface that is all it will do. The identities that you will derive by say you are calculating some operators by doing integral O O divided by integral these things would not change those relationships would not change. It is actually meant to derive relationships between operators or observables in or I mean identities between eventual observables even greens functions are not directly used as observables you convert it to S matrix in the end, but those things are not going to be affected by these manipulations. So, the path integral itself the functional integral itself does not actually calculate anything, but it represents the it shows that the dependence on the canonical variables is essentially Gaussian exponential or exponential of whatever the expression is, but that is all it really captures. So, these are in the these are as it is called generating functional. It is basically a generating functional of various identities and not itself any value of any numerical value to be calculated. So, this will be replaced by this which is nice we got rid of one delta function. The other one has a really bizarre story and which is where the Russians come in and we will only start it today and then complete it next time. So, rewriting delta of c by d lambda. So, here clever people observed that I mean the only good thing we can do here is Gaussian integrals. This is what however illegal we have all accepted as friends that we can do Gaussian integrals. So, what Gaussian integral can give you a determinant. So, of course, you are very clever. So, you know that determinant can be got by doing. So, if I had a multivariate Gaussian and then d x i this Gaussian is 1 over square root of determinant a. So, we got a determinant, but heck it is a square root and in that denominator. So, we say we can do a little better that is effectively secretly bringing in two variables. So, it actually becomes determinant of a, but now this silly thing is in the denominator, but we want it in the numerator. So, there was this man Berezin. So, grass min variables are the things that are anticommuting whatever the thetas are they are not numbers. So, and there is some multiplication among them. So, there is some kind of an algebra some kind of variables called theta some number of them i equal to 1 to n and some multiplication defined over them a binary operation. We make up the we invent a calculus for this. So, we are reaching 7 o clock. So, let me first tell you the answer then a then a calculus can be invented such that integral d theta i d theta i star e raise to a i j theta i theta j star is equal to determinant of a in the numerator. So, this is what we will check next time, but let me start it off. So, but anyway just to also jump to the answer if we do this then using this we can rewrite this as I am sorry to say I have used the same symbol c for eta good. So, we I do not need to. So, continuation of that then we can d lambda b be replaced by as we worked out in that example it is a complicated expression usually, but it can be found. So, it will become a function of x and y and will have some delta functions in a and b. So, this is what happens. So, you replace in then that determinant also by something which again as you may imagine if the c's are essentially linear in the canonical variables then this variation will at best produce because the covariant derivative is after all sort of quadratic in the a's it will with some residual a this m a b will be again at best quadratic in a it will be something functional of a, but new variables eta and eta b which are anticommuting complex complex anticommuting variables over which you have to do the integral. The advantage is yeah. So, we will see the mathematics of that next next time, but the advantage of this is that we then interpret the eta's as if they are anticommuting complex scalars well complex is not so crucial. So, fermionic Lorentz scalars which basically means that their loops have overall minus sign and their kinetic term. So, to speak gets decided from here because recall that a b times some derivatives of delta 4 of x minus y. So, which produces the kinetic terms usually looking exactly like it is a Dirac spinner you know eta bar d i box box right. So, it becomes box it is like a scalar of course yeah. So, not fermionic. So, it is. So, it will give some scalar like kinetic term to them, but they have to be treated as fermions because they are anticommuting. And the good thing is that your genuine fermions electrons etcetera have to be dealt with also by Berezin integral only. So, and if you do it then you actually recover QED Feynman rules of course, nobody actually bothers to do QED by that method because it is easy to quantize more directly, but if you did then it will give you the correct answer. The eta are called Gauss well appear in the calculation and have the wrong statistics are called Gauss. Let me write Gauss properly otherwise Gauss may not be happy with me. So, I think we can stop here this time.